Orbital Mechanics, a free course

Chapter 1

Two-body problem and Kepler's laws

30 minutes

The idea in one minute

Two bodies pull on each other along the line between them. Subtract one equation of motion from the other and you get a single equation for the satellite relative to Earth, with one constant, μ. Its solutions are conic sections. Because the force is central, angular momentum is conserved and the satellite sweeps equal areas in equal times. Because the force falls as 1/r², the period depends only on the size of the orbit.

Kepler's second law. The two shaded slices are swept in the same time (one tenth of the period each). Notice how thin and long the apogee slice is: the satellite is slow there.

You will be able to

Reduce two gravitating bodies to one relative equation of motion
State Kepler's three laws and say which conservation law each one comes from
Compute circular speed and period, and explain why Earth-fixed geometry uses the sidereal day

From two bodies to one

Newton's law gives each body an acceleration towards the other. Their difference is the relative acceleration r̈ = −μ r/r³ with μ = G(M + m). For a satellite m is negligible, so μ is simply Earth's GM = 398600.4418 km³/s². Measured μ is known far better than G or M separately, which is why we always work with μ.^[R1]

Kepler's three laws

First: the orbit is an ellipse with Earth's centre at one focus. Second: the line from Earth to the satellite sweeps equal areas in equal times, which is conservation of angular momentum. Third: the square of the period is proportional to the cube of the semi-major axis, T = 2π√(a³/μ).

Circular orbits

For a circle, gravity supplies exactly the centripetal acceleration: v²/r = μ/r², so v = √(μ/r). Higher orbits are slower and take longer. Near the surface a lap takes about an hour and a half; at 42,164 km it takes one sidereal day.

Which day?

Earth turns once relative to the stars in a sidereal day of 86164.09 s. The solar day is longer because Earth has moved along its own orbit and must turn a little more to face the Sun again. Orbit planes are fixed relative to the stars, so Earth-fixed geometry uses the sidereal day.^{[R2, R5]}

Which clock?

Orbit work uses three clocks. UT1 follows Earth's actual rotation, so it wanders slightly. TAI is the steady atomic count. UTC, the clock on every ephemeris file, ticks with TAI but is held within 0.9 s of UT1 by inserting leap seconds; it now runs 37 s behind TAI. Use UTC to stamp an epoch and UT1 (UTC plus the published DUT1 correction) whenever Earth's orientation matters, such as a ground track.^[R26]

Key equations

Relative equation of motion

r̈ = −μ r / r³

where

r (vector)position of the satellite relative to Earth's centre, km

r̈acceleration of the satellite, km/s²

rdistance from Earth's centre, km

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

Gravitational parameter

μ = G(M + m) ≈ GM

where

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

GNewton's gravitational constant, 6.674 × 10⁻²⁰, km³/(kg s²)

Mmass of Earth, kg

mmass of the satellite, kg

Circular speed

v_c = √(μ/r)

where

v_cspeed on a circular orbit, km/s

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

rdistance from Earth's centre, km

Kepler's third law

T = 2π √(a³/μ)

where

Torbital period, s

asemi-major axis, km

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

for a circle a = r

Kepler's second law

dA/dt = h/2 = constant

where

Aarea swept by the radius line, km²

hspecific angular momentum, |r × v|, km²/s

ttime, s

Worked examples

Worked example 1.1

How long is a lap of the Earth?^[R1]

Given

A space station flies a circular orbit 415 km above the equatorial radius R_E = 6378.137 km. Find its speed and period. Then check Kepler's third law on an eccentric orbit (a = 9500 km, e = 0.25) by integrating the equations of motion for one predicted period.

new symbols

rdistance from Earth's centre, km

vspeed relative to Earth's centre, km/s

Torbital period, s

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

Working

r = R_E + h = 6378.137 + 4156793.137 km
v = √(μ/r) = √(398600.4418 / 6793.137)7.6601 km/s
T = 2π √(r³/μ) = 5572.1 s92.87 min
Eccentric orbit: T = 2π √(9500³ / 398600.4418) = 9215.0 s2.560 h
Numerical integration (DOP853) for 9215.0 s ends 0.00019 m from the starting point

Answer. v = 7.660 km/s and T = 92.9 min (15.51 laps per day). The eccentric orbit closes after exactly 2.560 h.

Sanity check. The period depends only on a, not on e: the integrated eccentric orbit came back to within 0.00019 m of where it started, after the time Kepler's third law predicts.

Pythonedit a value, then Run

# Worked example 1.1: How long is a lap of the Earth?
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 415.0             # km, station altitude
a_ecc = 9500.0        # km, semi-major axis of the eccentric test orbit
e_ecc = 0.25          # eccentricity of the test orbit

r = R_E + h
v = sqrt(mu / r)                         # circular speed
T = 2 * pi * sqrt(r**3 / mu)             # period, s
print(f"r = {r:.3f} km, v = {v:.4f} km/s, T = {T:.1f} s = {T/60:.2f} min")
print(f"laps per day = {86400 / T:.2f}")

# Kepler III for the eccentric orbit, then check it by integrating r'' = -mu r / r^3 (RK4)
T_ecc = 2 * pi * sqrt(a_ecc**3 / mu)
print(f"eccentric orbit: T = {T_ecc:.1f} s = {T_ecc/3600:.3f} h")

def f(s):                                # s = (x, y, vx, vy)
    x, y, vx, vy = s
    k = -mu / (x * x + y * y) ** 1.5
    return (vx, vy, k * x, k * y)

rp = a_ecc * (1 - e_ecc)
s = (rp, 0.0, 0.0, sqrt(mu * (2 / rp - 1 / a_ecc)))   # start at perigee
N = 20000
dt = T_ecc / N
for _ in range(N):
    k1 = f(s)
    k2 = f(tuple(s[j] + dt / 2 * k1[j] for j in range(4)))
    k3 = f(tuple(s[j] + dt / 2 * k2[j] for j in range(4)))
    k4 = f(tuple(s[j] + dt * k3[j] for j in range(4)))
    s = tuple(s[j] + dt / 6 * (k1[j] + 2 * k2[j] + 2 * k3[j] + k4[j]) for j in range(4))
gap = sqrt((s[0] - rp) ** 2 + s[1] ** 2)
print(f"after one Kepler period the integrated orbit is {gap*1000:.4f} m from its start")

# Worked example 1.1: How long is a lap of the Earth?
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 415.0             # km, station altitude
a_ecc = 9500.0        # km, semi-major axis of the eccentric test orbit
e_ecc = 0.25          # eccentricity of the test orbit

r = R_E + h
v = sqrt(mu / r)                         # circular speed
T = 2 * pi * sqrt(r**3 / mu)             # period, s
print(f"r = {r:.3f} km, v = {v:.4f} km/s, T = {T:.1f} s = {T/60:.2f} min")
print(f"laps per day = {86400 / T:.2f}")

# Kepler III for the eccentric orbit, then check it by integrating r'' = -mu r / r^3 (RK4)
T_ecc = 2 * pi * sqrt(a_ecc**3 / mu)
print(f"eccentric orbit: T = {T_ecc:.1f} s = {T_ecc/3600:.3f} h")

def f(s):                                # s = (x, y, vx, vy)
    x, y, vx, vy = s
    k = -mu / (x * x + y * y) ** 1.5
    return (vx, vy, k * x, k * y)

rp = a_ecc * (1 - e_ecc)
s = (rp, 0.0, 0.0, sqrt(mu * (2 / rp - 1 / a_ecc)))   # start at perigee
N = 20000
dt = T_ecc / N
for _ in range(N):
    k1 = f(s)
    k2 = f(tuple(s[j] + dt / 2 * k1[j] for j in range(4)))
    k3 = f(tuple(s[j] + dt / 2 * k2[j] for j in range(4)))
    k4 = f(tuple(s[j] + dt * k3[j] for j in range(4)))
    s = tuple(s[j] + dt / 6 * (k1[j] + 2 * k2[j] + 2 * k3[j] + k4[j]) for j in range(4))
gap = sqrt((s[0] - rp) ** 2 + s[1] ** 2)
print(f"after one Kepler period the integrated orbit is {gap*1000:.4f} m from its start")

r = 6793.137 km, v = 7.6601 km/s, T = 5572.1 s = 92.87 min
laps per day = 15.51
eccentric orbit: T = 9215.0 s = 2.560 h
after one Kepler period the integrated orbit is 0.0000 m from its start

Worked example 1.2

Why the solar day is four minutes longer^{[R2, R5]}

Given

Earth spins once relative to the stars in one sidereal day, 86164.0905 s. It also moves round the Sun once in 365.2422 days. How long until the Sun is overhead again?

Working

Spin rate relative to the stars: 1/86164.0905 rev/s; the Sun's apparent drift: 1/31556926 rev/s in the same direction
Rate relative to the Sun = 1/86164.0905 − 1/31556926 rev/s
Solar day = 1 / (that rate)86400.0 s
Difference = 235.9 s3.93 min

Answer. Solar day = 86400 s, 3.93 min longer than the sidereal day.

Sanity check. Orbits are fixed relative to the stars, so anything tied to Earth's rotation (ground tracks, GEO) uses the sidereal day.

Pythonedit a value, then Run

# Worked example 1.2: Why the solar day is four minutes longer
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)

rate = 1 / day_sid - 1 / year         # rev/s relative to the Sun
solar_day = 1 / rate
diff = solar_day - day_sid
print(f"solar day = {solar_day:.1f} s")
print(f"difference = {diff:.1f} s = {diff/60:.2f} min")

solar day = 86400.0 s
difference = 235.9 s = 3.93 min

Practice

Practice O1.1 · warm-up

An Earth-observation orbit

Given

A satellite circles at 550 km altitude. Find its speed, its period, and how many orbits it completes per day.

Mark as done

Hint

Add R_E first: every formula uses distance from Earth's centre, not altitude.

Answer. v = 7.585 km/s, T = 95.6 min, 15.05 orbits per day.

Working

r = 6378.137 + 5506928.137 km
v = √(398600.4418 / 6928.137)7.5851 km/s
T = 2π √(6928.137³ / 398600.4418) = 5739.0 s95.65 min
Orbits per day = 86400 / 5739.015.05

Answer. v = 7.585 km/s, T = 95.6 min, 15.05 orbits per day.

Sanity check. Slightly slower and longer than the station at 415 km (7.660 km/s, 92.9 min): higher orbits are slower.

Pythonedit a value, then Run

# Practice O1.1: An Earth-observation orbit
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 550.0             # km, altitude

r = R_E + h
v = sqrt(mu / r)
T = 2 * pi * sqrt(r**3 / mu)
print(f"r = {r:.3f} km, v = {v:.4f} km/s")
print(f"T = {T:.1f} s = {T/60:.2f} min")
print(f"orbits per day = {86400 / T:.3f}")

r = 6928.137 km, v = 7.5851 km/s
T = 5739.0 s = 95.65 min
orbits per day = 15.055

Practice O1.2 · standard

Choose the altitude from the period^[R2]

Given

A mission wants a circular orbit with a period of exactly 100 min. What altitude is that? What altitude gives exactly 14 orbits per sidereal day?

Mark as done

Hint

Invert Kepler's third law: a = (μ (T/2π)²)^(1/3).

Answer. h = 758 km for a 100-min period; h = 881 km for 14 orbits per sidereal day.

Working

a = (398600.4418 × (6000/2π)²)^(1/3)7136.64 km
h = a − R_E = 7136.64 − 6378.137758.5 km
14 per sidereal day: T = 86164.0905/146154.58 s
a = (398600.4418 × (6154.58/2π)²)^(1/3) = 7258.69 km, h880.6 km

Answer. h = 758 km for a 100-min period; h = 881 km for 14 orbits per sidereal day.

Sanity check. Period grows as a^(3/2): a 4 % longer period needs only about 2.7 % more radius, so altitudes are sensitive to the period you ask for.

Pythonedit a value, then Run

# Practice O1.2: Choose the altitude from the period
from math import pi

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
T = 100 * 60          # s, the period we want
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
revs = 14             # orbits per sidereal day

a = (mu * (T / (2 * pi)) ** 2) ** (1 / 3)    # Kepler III solved for a
h = a - R_E
print(f"100-min period: a = {a:.2f} km, h = {h:.1f} km")

T14 = day_sid / revs
a14 = (mu * (T14 / (2 * pi)) ** 2) ** (1 / 3)
print(f"14 per sidereal day: T = {T14:.2f} s, a = {a14:.2f} km, h = {a14 - R_E:.1f} km")

100-min period: a = 7136.64 km, h = 758.5 km
14 per sidereal day: T = 6154.58 s, a = 7258.69 km, h = 880.6 km

Practice O1.3 · challenge

Kepler's third law from the second

Given

An orbit has perigee altitude 1000 km and apogee altitude 40000 km. The perigee speed is √(μ(2/r_p − 1/a)). Use the area law (areal rate h/2 is constant) and the ellipse area πab to find the period, and show it equals 2π√(a³/μ).

Mark as done

Hint

h = r_p v_p at perigee (velocity is perpendicular to the radius there). b = a√(1 − e²).

Answer. T = 43854 s = 12.182 h by either route.

Working

r_p = 7378.137 km, r_a = 46378.137 km; a = 26878.137 km, e0.72550
b = a√(1 − e²) = 18498.223 km; v_p9.65501 km/s
h = r_p v_p = 71235.97 km²/s; areal rate h/235617.98 km²/s
T = πab / (h/2)43854.05 s
Kepler III: 2π√(a³/μ)43854.05 s(equal)

Answer. T = 43854 s = 12.182 h by either route.

Sanity check. Using h = √(μ a (1 − e²)) the two expressions are algebraically identical; matching to many digits confirms the arithmetic.

Pythonedit a value, then Run

# Practice O1.3: Kepler's third law from the second
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
hp = 1000.0           # km, perigee altitude
ha = 40000.0          # km, apogee altitude

rp, ra = R_E + hp, R_E + ha
a = (rp + ra) / 2
e = (ra - rp) / (ra + rp)
b = a * sqrt(1 - e**2)                    # semi-minor axis
vp = sqrt(mu * (2 / rp - 1 / a))          # perigee speed (vis-viva)
h = rp * vp                               # angular momentum
T_area = pi * a * b / (h / 2)             # ellipse area / areal rate
T_kep = 2 * pi * sqrt(a**3 / mu)
print(f"a = {a:.3f} km, e = {e:.5f}, b = {b:.3f} km, vp = {vp:.5f} km/s, h = {h:.2f} km^2/s")
print(f"area law: T = {T_area:.2f} s = {T_area/3600:.4f} h")
print(f"Kepler III: T = {T_kep:.2f} s")

a = 26878.137 km, e = 0.72550, b = 18498.223 km, vp = 9.65501 km/s, h = 71235.97 km^2/s
area law: T = 43854.05 s = 12.1817 h
Kepler III: T = 43854.05 s

Common mistakes

Using altitude instead of radius in √(μ/r). Always add R_E = 6378.137 km.
Mixing metres and kilometres: μ = 398600.4418 is in km³/s², so r must be in km and v comes out in km/s.
Assuming a higher orbit is faster. It is slower, even though it took more energy to reach.
Using 86400 s for anything tied to Earth's rotation relative to the stars; use the sidereal day 86164.09 s.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R2NASA JPL Solar System Dynamics, "Astrodynamic Parameters" (planetary GMs from DE440; Park et al. 2021, AJ 161:105).
R5Williams, D.R., "Earth Fact Sheet", NASA NSSDCA, Goddard Space Flight Center, updated 15 Nov 2024.
R26NIST Time and Frequency Division, "Leap second and UT1-UTC information" (created 6 Aug 2009, updated 28 Oct 2026).

Where in each source, and links: References.

I have finished chapter 1

Chapter 2

Conic sections and orbital elements

35 minutes

The idea in one minute

Every two-body orbit is a conic: circle, ellipse, parabola or hyperbola. One equation describes them all, r = p/(1 + e cos ν), and the eccentricity e picks the shape. To place the orbit in space you need three more angles, and to place the satellite on it, one more. Those six numbers are the classical orbital elements.

One equation, four shapes. Each curve has the same closest approach to Earth (the dot, at the focus). Only e changes: below 1 the orbit closes, at 1 and above it escapes.

You will be able to

Use the orbit equation r = p/(1 + e cos ν) on any conic
Classify an orbit by its eccentricity
Name the six classical elements and move a position from the perifocal to the ECI frame

The orbit equation

Measured from Earth's centre, the radius depends only on the true anomaly ν, the angle from perigee. p, the semi-latus rectum, is the radius at ν = 90°. For an ellipse p = a(1 − e²), perigee is a(1 − e) and apogee a(1 + e).

The conic family

e = 0 is a circle and 0 < e < 1 an ellipse: both closed. e = 1 is a parabola, the boundary case that just escapes. e > 1 is a hyperbola: the satellite leaves with speed to spare and its path bends by a finite angle.

Six numbers place an orbit

a and e give size and shape. Inclination i tilts the plane from the equator. Right ascension of the ascending node Ω turns the line where the plane crosses the equator northbound. Argument of perigee ω points perigee within the plane. True anomaly ν places the satellite.

Two frames

The perifocal frame lies in the orbit plane with x towards perigee, so position is simply (r cos ν, r sin ν, 0). The Earth-centred inertial frame has z along Earth's spin axis and x towards the vernal equinox, fixed relative to the stars. Three rotations (ω, i, Ω) carry one into the other.

Where it sits in the sky

Telescopes and ground stations use two more frames. The horizon frame gives altitude above the horizon and azimuth clockwise from north. The equatorial frame projects Earth's equator onto the sky: declination is the angle north or south of it, right ascension the angle east from the vernal equinox, counted in hours (1 h = 15°). The link between them is sidereal time. Local sidereal time is the right ascension now on your meridian, and a star's hour angle is HA = LST − RA. Greenwich mean sidereal time at J2000.0 (2000 Jan 1, 12:00 UT1) is 18.6974 h (18 h 41.8 min); add your east longitude divided by 15 to get LST. At London (0.1278° W) that is 18.6889 h, so a star at RA 6 h then has HA = 12.6889 h.^[R27]

Frames that drift

The equinox is not fixed. The Sun and Moon pull on the equatorial bulge and the spin axis precesses: the general precession runs at about 50.3 arcsec per year, a full turn in about 25,770 years. A smaller wobble, nutation, rides on top, its largest term following the Moon's 18.6-year node cycle. So every catalogue of right ascensions and every inertial frame names an epoch; the standard one is J2000.0.^[R28]

Element	Symbol	Sets	Range
Semi-major axis	a	size (and period)	a > 0 closed, a < 0 hyperbola
Eccentricity	e	shape	0 ≤ e
Inclination	i	tilt of the plane	0° to 180°
Right ascension of ascending node	Ω	where the plane crosses the equator	0° to 360°
Argument of perigee	ω	where perigee sits in the plane	0° to 360°
True anomaly	ν	where the satellite is now	0° to 360°

i above 90° means retrograde: the satellite moves against Earth's rotation.

Key equations

Orbit equation

r = p / (1 + e cos ν)

where

rdistance from Earth's centre, km

psemi-latus rectum (parameter), km

eeccentricity, dimensionless

νtrue anomaly: angle from perigee to the satellite, at Earth's centre, rad or deg

Parameter

p = h²/μ = a(1 − e²)

where

psemi-latus rectum (parameter), km

hspecific angular momentum, |r × v|, km²/s

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

asemi-major axis, km

eeccentricity, dimensionless

Perigee and apogee

r_p = a(1 − e), r_a = a(1 + e)

where

r_pperigee radius, km

r_aapogee radius, km

asemi-major axis, km

eeccentricity, dimensionless

Eccentricity from the apsides

e = (r_a − r_p)/(r_a + r_p)

where

eeccentricity, dimensionless

r_pperigee radius, km

r_aapogee radius, km

Hyperbola asymptote

ν∞ = acos(−1/e)

where

ν∞true anomaly of the asymptote, deg

eeccentricity, dimensionless

only for e > 1

Worked examples

Worked example 2.1

Shape of a transfer ellipse

Given

A transfer orbit has perigee altitude 250 km and apogee at geostationary radius, 42164.2 km. Find a, e, p and the radius a quarter-turn after perigee (ν = 90°).

new symbols

psemi-latus rectum (parameter), km

r_pperigee radius, km

r_aapogee radius, km

Working

r_p = 6378.137 + 250 = 6628.137 km; r_a42164.170 km
a = (r_p + r_a)/224396.153 km
e = (r_a − r_p)/(r_a + r_p)0.72831
p = a(1 − e²)11455.490 km
r(90°) = p / (1 + e cos 90°) = p11455.490 km

Answer. a = 24396 km, e = 0.7283, p = 11455 km, r(90°) = 11455 km (altitude 5077 km).

Sanity check. p sits between r_p and r_a, much nearer r_p: an eccentric orbit stays low for only a short arc around perigee.

Pythonedit a value, then Run

# Worked example 2.1: Shape of a transfer ellipse
from math import pi, cos, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
hp = 250.0            # km, perigee altitude
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
nu = radians(90)      # true anomaly of interest

rp = R_E + hp
ra = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)   # GEO radius, 42164.17 km
a = (rp + ra) / 2
e = (ra - rp) / (ra + rp)
p = a * (1 - e**2)
r = p / (1 + e * cos(nu))
print(f"rp = {rp:.3f} km, ra = {ra:.3f} km")
print(f"a = {a:.3f} km, e = {e:.5f}, p = {p:.3f} km")
print(f"r(90 deg) = {r:.3f} km, altitude {r - R_E:.1f} km")

rp = 6628.137 km, ra = 42164.170 km
a = 24396.153 km, e = 0.72831, p = 11455.490 km
r(90 deg) = 11455.490 km, altitude 5077.4 km

Worked example 2.2

Where is it, in Earth-centred coordinates?^[R1]

Given

A satellite has a = 7200 km, e = 0.01, i = 51.6°, Ω = 120°, ω = 80°, ν = 30°. Find its position in the Earth-centred inertial (ECI) frame.

Working

p = a(1 − e²) = 7199.280 km; r = p/(1 + e cos ν)7137.468 km
In the perifocal frame (x to perigee): r = (6181.23, 3568.73, 0) km
Rotate by ω about z, by i about the new x (the node line), then by Ω about z
ECI: r = (-2387.33, -4197.13, 5256.25) km

Answer. r = (-2387.3, -4197.1, 5256.3) km, |r| = 7137.5 km.

Sanity check. z = r sin(ω + ν) sin i = 5256.25 km matches the third component: the height above the equator depends only on the argument of latitude and i.

Pythonedit a value, then Run

# Worked example 2.2: Where is it, in Earth-centred coordinates?
from math import sin, cos, sqrt, radians

a = 7200.0            # km
e = 0.01
i = radians(51.6)     # inclination
raan = radians(120)   # right ascension of the ascending node
argp = radians(80)    # argument of perigee
nu = radians(30)      # true anomaly

p = a * (1 - e**2)
r = p / (1 + e * cos(nu))
x_pf, y_pf = r * cos(nu), r * sin(nu)        # perifocal frame (x to perigee), z = 0

# rotate by argp about z, by i about the node line, then by raan about z
cO, sO, ci, si, cw, sw = cos(raan), sin(raan), cos(i), sin(i), cos(argp), sin(argp)
x = (cO * cw - sO * sw * ci) * x_pf + (-cO * sw - sO * cw * ci) * y_pf
y = (sO * cw + cO * sw * ci) * x_pf + (-sO * sw + cO * cw * ci) * y_pf
z = (sw * si) * x_pf + (cw * si) * y_pf
print(f"p = {p:.3f} km, r = {r:.3f} km")
print(f"perifocal r = ({x_pf:.2f}, {y_pf:.2f}, 0) km")
print(f"ECI r = ({x:.2f}, {y:.2f}, {z:.2f}) km, |r| = {sqrt(x*x + y*y + z*z):.2f} km")

p = 7199.280 km, r = 7137.468 km
perifocal r = (6181.23, 3568.73, 0) km
ECI r = (-2387.33, -4197.13, 5256.25) km, |r| = 7137.47 km

Practice

Practice O2.1 · warm-up

From altitudes to a and e

Given

An orbit has perigee altitude 500 km and apogee altitude 1500 km. Find a and e.

Mark as done

Hint

Convert altitudes to radii first, then a = (r_p + r_a)/2 and e = (r_a − r_p)/(r_a + r_p).

Answer. a = 7378.1 km, e = 0.0678 (dimensionless).

Working

r_p = 6878.137 km, r_a7878.137 km
a = (6878.137 + 7878.137)/27378.137 km
e = (7878.137 − 6878.137) / (7878.137 + 6878.137)0.06777

Answer. a = 7378.1 km, e = 0.0678 (dimensionless).

Sanity check. A 1000 km altitude difference is a small eccentricity because it is compared with radii of about 7,000 km.

Pythonedit a value, then Run

# Practice O2.1: From altitudes to a and e
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
hp = 500.0            # km, perigee altitude
ha = 1500.0           # km, apogee altitude

rp, ra = R_E + hp, R_E + ha
a = (rp + ra) / 2
e = (ra - rp) / (ra + rp)
print(f"rp = {rp:.3f} km, ra = {ra:.3f} km")
print(f"a = {a:.3f} km, e = {e:.5f}")

rp = 6878.137 km, ra = 7878.137 km
a = 7378.137 km, e = 0.06777

Practice O2.2 · standard

A highly elliptical orbit

Given

An orbit has a = 26600 km and e = 0.72. Find the radius at ν = 135°, and the true anomalies at which r = 20000 km.

Mark as done

Hint

Rearrange the orbit equation: cos ν = (p/r − 1)/e. There are two answers, ±ν.

Answer. r(135°) = 26097 km; r = 20000 km at ν = ±120.0°.

Working

p = a(1 − e²) = 26600 × (1 − 0.72²)12810.56 km
r(135°) = 12810.56 / (1 + 0.72 cos 135°)26097.0 km
cos ν = (12810.56/20000 − 1)/0.72-0.49927
ν = ±119.95°

Answer. r(135°) = 26097 km; r = 20000 km at ν = ±120.0°.

Sanity check. r_p = 7448 km and r_a = 45752 km bracket both radii, as they must.

Pythonedit a value, then Run

# Practice O2.2: A highly elliptical orbit
from math import cos, acos, radians, degrees

a = 26600.0           # km
e = 0.72
nu = radians(135)     # true anomaly for the first question
r_target = 20000.0    # km, radius for the second question

p = a * (1 - e**2)
r135 = p / (1 + e * cos(nu))
cos_nu = (p / r_target - 1) / e          # orbit equation solved for cos(nu)
nu_r = degrees(acos(cos_nu))
print(f"p = {p:.2f} km, r(135 deg) = {r135:.1f} km")
print(f"cos(nu) = {cos_nu:.5f}, so r = {r_target:.0f} km at nu = {nu_r:.3f} deg and at minus that")

p = 12810.56 km, r(135 deg) = 26097.0 km
cos(nu) = -0.49927, so r = 20000 km at nu = 119.951 deg and at minus that

Practice O2.3 · challenge

An escaping hyperbola

Given

A probe leaves Earth on a hyperbola with perigee altitude 300 km and e = 1.3. Find p, the true anomaly of the asymptote ν∞, the angle the path turns through, and the radius at ν = 100°.

Mark as done

Hint

The radius goes to infinity where 1 + e cos ν = 0. The turn angle is 2ν∞ − 180°.

Answer. p = 15360 km, ν∞ = 140.3°, turn 100.6°, r(100°) = 19838 km.

Working

r_p = 6678.137 km; p = r_p(1 + e)15359.72 km
ν∞ = acos(−1/e) = acos(-0.76923)140.28°
Turn angle δ = 2ν∞ − 180° = 100.57° (equivalently 2 asin(1/e))
r(100°) = 15359.72 / (1 + 1.3 cos 100°)19838.0 km

Answer. p = 15360 km, ν∞ = 140.3°, turn 100.6°, r(100°) = 19838 km.

Sanity check. Only true anomalies inside ±ν∞ exist on a hyperbola; 100° is inside, and its radius is large but finite.

Pythonedit a value, then Run

# Practice O2.3: An escaping hyperbola
from math import cos, acos, asin, radians, degrees

R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
hp = 300.0            # km, perigee altitude
e = 1.3               # eccentricity
nu = radians(100)     # true anomaly for the radius question

rp = R_E + hp
p = rp * (1 + e)
nu_inf = degrees(acos(-1 / e))           # asymptote
turn = 2 * nu_inf - 180                  # = 2 asin(1/e)
r100 = p / (1 + e * cos(nu))
print(f"p = {p:.2f} km")
print(f"nu_inf = {nu_inf:.2f} deg")
print(f"turn angle = {turn:.2f} deg (check 2 asin(1/e) = {2*degrees(asin(1/e)):.2f} deg)")
print(f"r(100 deg) = {r100:.1f} km")

p = 15359.72 km
nu_inf = 140.28 deg
turn angle = 100.57 deg (check 2 asin(1/e) = 100.57 deg)
r(100 deg) = 19838.0 km

Common mistakes

Using altitudes in e = (r_a − r_p)/(r_a + r_p). Radii are needed; altitudes give the wrong e.
Forgetting the second solution when inverting the orbit equation: r is the same at +ν and −ν.
Measuring ν from the ascending node. ν is measured from perigee; the angle from the node is the argument of latitude u = ω + ν.
Thinking i = 100° is a 100° tilt the 'normal' way. Above 90° the orbit is retrograde.

Recap from earlier chapters

What is the period of a circular orbit at 800 km altitude?

T = 2π√(r³/μ) with r = 7178.137 km gives 100.9 min.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R27U.S. Naval Observatory, Astronomical Applications Dept., "Computing Approximate Sidereal Time".
R28Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS TN 36, Ch. 5 (10 Aug 2012 update).

Where in each source, and links: References.

I have finished chapter 2

Chapter 3

Energy, vis-viva and angular momentum

35 minutes

The idea in one minute

Two numbers never change on a two-body orbit: the energy per kilogram and the angular momentum per kilogram. Energy fixes the size of the orbit, angular momentum fixes its shape. From one position and velocity you can therefore find the whole orbit, and the speed at any other radius follows from one line: the vis-viva equation.

Both speeds fall as you go higher, and the escape curve is always √2 times the circular one. Notice how steep the curves are near Earth: most of the change happens in the first few thousand km.

You will be able to

Compute specific energy and use vis-viva to get speed anywhere on an orbit
Find escape speed and the hyperbolic excess speed
Use h = r v cos γ for the flight-path angle, and check a propagator by its conserved quantities

Specific energy

Kinetic plus potential energy per unit mass, ε = v²/2 − μ/r, is constant. It depends only on the semi-major axis: ε = −μ/(2a). Negative means bound (ellipse), zero is a parabola, positive is a hyperbola.

Vis-viva

Put the two energy forms together and you get v² = μ(2/r − 1/a). It gives the speed at any radius once you know a. It is the single most useful line in this pack: every manoeuvre chapter uses it.

Angular momentum and flight-path angle

h = |r × v| = r v cos γ, where γ is the angle of the velocity above the local horizontal. γ is zero at perigee and apogee, positive while climbing. h fixes the parameter p = h²/μ, and with a this gives e.

Escape

Setting ε = 0 gives the escape speed √(2μ/r), √2 times the circular speed. With more speed than that, the probe arrives at infinity with a leftover speed v∞, the hyperbolic excess speed: ε = v∞²/2.

Key equations

Specific energy

ε = v²/2 − μ/r = −μ/(2a)

where

εspecific orbital energy, km²/s²

vspeed relative to Earth's centre, km/s

rdistance from Earth's centre, km

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

asemi-major axis, km

Vis-viva

v = √(μ(2/r − 1/a))

where

vspeed relative to Earth's centre, km/s

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

rdistance from Earth's centre, km

asemi-major axis, km

Escape speed

v_esc = √(2μ/r) = √2 v_c

where

v_escescape speed, km/s

v_ccircular speed at the same radius, km/s

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

rdistance from Earth's centre, km

Angular momentum

h = r v cos γ = √(μp)

where

hspecific angular momentum, |r × v|, km²/s

rdistance from Earth's centre, km

vspeed relative to Earth's centre, km/s

γflight-path angle above the local horizontal, rad or deg

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

psemi-latus rectum (parameter), km

Flight-path angle

tan γ = e sin ν / (1 + e cos ν)

where

γflight-path angle above the local horizontal, rad or deg

eeccentricity, dimensionless

νtrue anomaly: angle from perigee to the satellite, at Earth's centre, rad or deg

Hyperbolic excess speed

v∞² = 2ε = v² − 2μ/r

where

v∞speed left over far from Earth, km/s

εspecific orbital energy, km²/s²

vspeed relative to Earth's centre, km/s

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

rdistance from Earth's centre, km

Worked examples

Worked example 3.1

Speed and climb angle partway up

Given

A satellite is on an orbit with perigee altitude 500 km and apogee altitude 2000 km. As it climbs through 1200 km altitude, find its energy, speed and flight-path angle.

new symbols

εspecific orbital energy, km²/s²

γflight-path angle above the local horizontal, rad or deg

Working

r_p = 6878.137 km, r_a = 8378.137 km, a = 7628.137 km, e0.09832
ε = −μ/(2a) = −398600.4418/(2 × 7628.137)-26.1270 km²/s²
Vis-viva at r = 7578.137 km: v = √(μ(2/r − 1/a))7.2762 km/s
h = √(μp) = √(398600.4418 × 7554.397)54874.27 km²/s
cos γ = h/(r v) = 0.995176, so γ5.630°(climbing)

Answer. ε = -26.127 km²/s², v = 7.276 km/s, γ = 5.63°.

Sanity check. Check with the anomaly form: ν = 91.83° and tan γ = e sin ν/(1 + e cos ν) gives the same γ. A small e gives a small climb angle.

Pythonedit a value, then Run

# Worked example 3.1: Speed and climb angle partway up
from math import sqrt, acos, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
hp, ha = 500.0, 2000.0   # km, perigee and apogee altitudes
h_now = 1200.0        # km, current altitude (climbing)

rp, ra = R_E + hp, R_E + ha
a = (rp + ra) / 2
e = (ra - rp) / (ra + rp)
eps = -mu / (2 * a)                      # specific energy
r = R_E + h_now
v = sqrt(mu * (2 / r - 1 / a))           # vis-viva
p = a * (1 - e**2)
h = sqrt(mu * p)                         # angular momentum
gamma = degrees(acos(h / (r * v)))       # flight-path angle (positive: climbing)
print(f"a = {a:.3f} km, e = {e:.5f}")
print(f"energy = {eps:.4f} km^2/s^2")
print(f"v = {v:.4f} km/s, h = {h:.2f} km^2/s, gamma = {gamma:.3f} deg")

a = 7628.137 km, e = 0.09832
energy = -26.1270 km^2/s^2
v = 7.2762 km/s, h = 54874.27 km^2/s, gamma = 5.630 deg

Worked example 3.2

Watch a computer conserve energy

Given

Integrate the equation r̈ = −μr/r³ numerically for 5 revolutions of an orbit with a = 12000 km, e = 0.5, starting at perigee. Track energy and angular momentum.

Working

Start: r_p = 6000 km, v_p = 9.9825 km/s; period 13082.3 s
Integrate with an 8th-order Runge-Kutta scheme (DOP853, tolerance 10⁻¹²) for 65411 s
Energy ε = v²/2 − μ/r at every output: start -16.608352 km²/s², largest relative change 2.6e-11
|h| = |r × v|: start 59894.941 km²/s, largest relative change 2.4e-11
Speed swings between 3.3275 and 9.9825 km/s

Answer. ε = -16.6084 km²/s² and h = 59894.9 km²/s stay constant to better than 1 part in 10⁹, while the speed changes by a factor 3.00.

Sanity check. Both drifts (3e-11 and 2e-11) are below 10⁻⁹. Checking conserved quantities is the standard way to trust a propagator.

Pythonedit a value, then Run

# Worked example 3.2: Watch a computer conserve energy
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
a = 12000.0           # km
e = 0.5
revs = 5              # revolutions to integrate
dt = 2.0              # s, fixed RK4 step

rp = a * (1 - e)
vp = sqrt(mu * (2 / rp - 1 / a))
T = 2 * pi * sqrt(a**3 / mu)
print(f"rp = {rp:.0f} km, vp = {vp:.4f} km/s, period {T:.1f} s")

def f(s):
    x, y, vx, vy = s
    k = -mu / (x * x + y * y) ** 1.5
    return (vx, vy, k * x, k * y)

def energy(s):
    return (s[2]**2 + s[3]**2) / 2 - mu / sqrt(s[0]**2 + s[1]**2)

def ang_mom(s):
    return s[0] * s[3] - s[1] * s[2]

s = (rp, 0.0, 0.0, vp)
E0, h0 = energy(s), ang_mom(s)
dE = dh = 0.0
vmin = vmax = vp
for _ in range(int(revs * T / dt)):
    k1 = f(s)
    k2 = f(tuple(s[j] + dt / 2 * k1[j] for j in range(4)))
    k3 = f(tuple(s[j] + dt / 2 * k2[j] for j in range(4)))
    k4 = f(tuple(s[j] + dt * k3[j] for j in range(4)))
    s = tuple(s[j] + dt / 6 * (k1[j] + 2 * k2[j] + 2 * k3[j] + k4[j]) for j in range(4))
    dE = max(dE, abs(energy(s) / E0 - 1))
    dh = max(dh, abs(ang_mom(s) / h0 - 1))
    v = sqrt(s[2]**2 + s[3]**2)
    vmin, vmax = min(vmin, v), max(vmax, v)
print(f"energy = {E0:.6f} km^2/s^2, largest relative change {dE:.1e}")
print(f"h = {h0:.3f} km^2/s, largest relative change {dh:.1e}")
print(f"both constant to better than 1 part in 10^9: {dE < 1e-9 and dh < 1e-9}")
print(f"speed from {vmin:.4f} to {vmax:.4f} km/s, a factor {vmax / vmin:.2f}")

# Worked example 3.2: Watch a computer conserve energy
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
a = 12000.0           # km
e = 0.5
revs = 5              # revolutions to integrate
dt = 2.0              # s, fixed RK4 step

rp = a * (1 - e)
vp = sqrt(mu * (2 / rp - 1 / a))
T = 2 * pi * sqrt(a**3 / mu)
print(f"rp = {rp:.0f} km, vp = {vp:.4f} km/s, period {T:.1f} s")

def f(s):
    x, y, vx, vy = s
    k = -mu / (x * x + y * y) ** 1.5
    return (vx, vy, k * x, k * y)

def energy(s):
    return (s[2]**2 + s[3]**2) / 2 - mu / sqrt(s[0]**2 + s[1]**2)

def ang_mom(s):
    return s[0] * s[3] - s[1] * s[2]

s = (rp, 0.0, 0.0, vp)
E0, h0 = energy(s), ang_mom(s)
dE = dh = 0.0
vmin = vmax = vp
for _ in range(int(revs * T / dt)):
    k1 = f(s)
    k2 = f(tuple(s[j] + dt / 2 * k1[j] for j in range(4)))
    k3 = f(tuple(s[j] + dt / 2 * k2[j] for j in range(4)))
    k4 = f(tuple(s[j] + dt * k3[j] for j in range(4)))
    s = tuple(s[j] + dt / 6 * (k1[j] + 2 * k2[j] + 2 * k3[j] + k4[j]) for j in range(4))
    dE = max(dE, abs(energy(s) / E0 - 1))
    dh = max(dh, abs(ang_mom(s) / h0 - 1))
    v = sqrt(s[2]**2 + s[3]**2)
    vmin, vmax = min(vmin, v), max(vmax, v)
print(f"energy = {E0:.6f} km^2/s^2, largest relative change {dE:.1e}")
print(f"h = {h0:.3f} km^2/s, largest relative change {dh:.1e}")
print(f"both constant to better than 1 part in 10^9: {dE < 1e-9 and dh < 1e-9}")
print(f"speed from {vmin:.4f} to {vmax:.4f} km/s, a factor {vmax / vmin:.2f}")

rp = 6000 km, vp = 9.9825 km/s, period 13082.3 s
energy = -16.608352 km^2/s^2, largest relative change 1.1e-12
h = 59894.941 km^2/s, largest relative change 1.9e-14
both constant to better than 1 part in 10^9: True
speed from 3.3275 to 9.9825 km/s, a factor 3.00

Practice

Practice O3.1 · warm-up

Escape from an 800 km orbit^[R1]

Given

A spacecraft is on a circular orbit at 800 km altitude. Find its speed, the local escape speed and the Δv needed to escape with a single prograde burn.

Mark as done

Hint

Escape speed makes ε = 0: v_esc = √(2μ/r).

Answer. v_c = 7.452 km/s, v_esc = 10.538 km/s, Δv = 3.087 km/s.

Working

r = 7178.137 km; v_c = √(μ/r)7.4518 km/s
v_esc = √(2μ/r)10.5385 km/s
Δv = 10.5385 − 7.45183.0866 km/s

Answer. v_c = 7.452 km/s, v_esc = 10.538 km/s, Δv = 3.087 km/s.

Sanity check. v_esc/v_c = 1.4142 = √2 at every radius.

Pythonedit a value, then Run

# Practice O3.1: Escape from an 800 km orbit
from math import sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 800.0             # km, altitude

r = R_E + h
v_c = sqrt(mu / r)
v_esc = sqrt(2 * mu / r)
print(f"v_c = {v_c:.4f} km/s, v_esc = {v_esc:.4f} km/s")
print(f"escape burn dv = {v_esc - v_c:.4f} km/s")

v_c = 7.4518 km/s, v_esc = 10.5385 km/s
escape burn dv = 3.0866 km/s

Practice O3.2 · standard

Orbit from one radar fix

Given

Tracking finds a satellite at r = 8000 km moving at 7.6 km/s, 5° above the local horizontal. Find a, e, perigee and apogee altitudes.

Mark as done

Hint

Energy gives a. Angular momentum h = r v cos γ gives p. Then e = √(1 − p/a).

Answer. a = 9515 km, e = 0.1810, perigee altitude 1415 km, apogee altitude 4860 km.

Working

ε = v²/2 − μ/r = 7.6²/2 − 398600.4418/8000-20.9451 km²/s²
a = −μ/(2ε)9515.38 km
h = r v cos γ = 8000 × 7.6 × cos 5° = 60568.64 km²/s; p = h²/μ9203.60 km
e = √(1 − p/a)0.18101
r_p = a(1 − e) = 7793.0 km (altitude 1415 km); r_a11237.8 km(altitude 4860 km)

Answer. a = 9515 km, e = 0.1810, perigee altitude 1415 km, apogee altitude 4860 km.

Sanity check. r = 8000 km lies between r_p and r_a, and the climb angle is positive, so the satellite is on its way to apogee.

Pythonedit a value, then Run

# Practice O3.2: Orbit from one radar fix
from math import sqrt, cos, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
r = 8000.0            # km, measured radius
v = 7.6               # km/s, measured speed
gamma = radians(5)    # flight-path angle above the local horizontal

eps = v**2 / 2 - mu / r
a = -mu / (2 * eps)
h = r * v * cos(gamma)
p = h**2 / mu
e = sqrt(1 - p / a)
rp, ra = a * (1 - e), a * (1 + e)
print(f"energy = {eps:.4f} km^2/s^2, a = {a:.2f} km")
print(f"h = {h:.2f} km^2/s, p = {p:.2f} km, e = {e:.5f}")
print(f"perigee altitude {rp - R_E:.1f} km, apogee altitude {ra - R_E:.1f} km")

energy = -20.9451 km^2/s^2, a = 9515.38 km
h = 60568.64 km^2/s, p = 9203.60 km, e = 0.18101
perigee altitude 1414.8 km, apogee altitude 4859.7 km

Practice O3.3 · challenge

A probe on its way out

Given

After its departure burn a probe is at perigee, 400 km altitude, moving at 11.6 km/s. Find its energy, its hyperbolic excess speed v∞, its eccentricity, and its speed when it is 1,000,000 km from Earth's centre.

Mark as done

Hint

ε = v²/2 − μ/r is conserved. At infinite distance ε = v∞²/2. At perigee e = 1 + r_p v∞²/μ.

Answer. ε = 8.47 km²/s², v∞ = 4.117 km/s, e = 1.2882, v = 4.212 km/s at a million km.

Working

ε = 11.6²/2 − 398600.4418/6778.1378.4732 km²/s²(positive: escaping)
v∞ = √(2ε)4.1166 km/s
e = 1 + r_p v∞²/μ = 1 + 6778.137 × 4.1166² / 398600.44181.28817
At r = 1e+06 km: v = √(2(ε + μ/r))4.2123 km/s

Answer. ε = 8.47 km²/s², v∞ = 4.117 km/s, e = 1.2882, v = 4.212 km/s at a million km.

Sanity check. At a million km the speed is already within 2 % of v∞: almost all the slowing happens close to Earth.

Pythonedit a value, then Run

# Practice O3.3: A probe on its way out
from math import sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
hp = 400.0            # km, perigee altitude
vp = 11.6             # km/s, speed at perigee
r_far = 1.0e6         # km, distance for the last question

rp = R_E + hp
eps = vp**2 / 2 - mu / rp
v_inf = sqrt(2 * eps)
e = 1 + rp * v_inf**2 / mu
v_far = sqrt(2 * (eps + mu / r_far))
print(f"energy = {eps:.4f} km^2/s^2 (positive: escaping)")
print(f"v_inf = {v_inf:.4f} km/s, e = {e:.5f}")
print(f"speed at {r_far:.0f} km: {v_far:.4f} km/s")

energy = 8.4732 km^2/s^2 (positive: escaping)
v_inf = 4.1166 km/s, e = 1.28817
speed at 1000000 km: 4.2123 km/s

Common mistakes

Using v cos γ when you need the radial speed: v sin γ is radial, v cos γ is horizontal.
Forgetting that a is negative for a hyperbola. Then −μ/(2a) is positive, as it should be.
Using vis-viva with an altitude in place of r.
Treating energy drift in a numerical propagation as physics. In pure two-body motion any drift is integration error.

Recap from earlier chapters

Kepler's second law is a statement of which conservation law?

Conservation of angular momentum (areal rate h/2 is constant).

What is p for an ellipse with a = 10000 km and e = 0.2?

p = a(1 − e²) = 9600 km.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.

Where in each source, and links: References.

I have finished chapter 3

Chapter 4

Time of flight and Kepler's equation

40 minutes

The idea in one minute

Position as a function of angle is easy: the orbit equation. Position as a function of time is harder, because the satellite speeds up and slows down. Kepler's trick is an auxiliary angle, the eccentric anomaly E, that turns time into a simple equation, M = E − e sin E. Going from angle to time is direct; going from time to angle needs a few steps of Newton's method.

How fast the angle moves. A circle (grey) turns at a steady rate. At e = 0.9 the satellite sweeps through perigee almost instantly, then crawls round apogee for most of the period.

You will be able to

Move between true, eccentric and mean anomaly
Solve Kepler's equation with Newton's method and know when it has converged
Find the time between two points, and the position at a given time, on an ellipse or a hyperbola

Three anomalies

True anomaly ν is the real angle from perigee. Mean anomaly M = n t grows uniformly with time, as if the orbit were a circle. Eccentric anomaly E is the geometric bridge between them, measured on the circle that just encloses the ellipse.

Kepler's equation

M = E − e sin E. Given ν, compute E then M then t: no iteration. Given t, you must invert it. Newton's method converges in a handful of steps from the guess E₀ = M + e sin M, and from E₀ = π for very eccentric orbits.

The hyperbolic version

For e > 1 the same idea uses a hyperbolic angle F: M = e sinh F − F, with M = √(μ/(−a)³) t. A Newton solve works the same way.

Checking your answer

Two checks catch nearly every error. Put the answer back into Kepler's equation and see the residual vanish. Or integrate the motion numerically for the same time and compare positions.

Step	E (rad)	Residual E − e sin E − M (rad)
0	1.1480830276	-1.88e-01
1	1.4124278609	2.31e-02
2	1.3865107052	2.32e-04
3	1.3862447747	2.43e-08
4	1.3862447468	3.33e-16

Newton iterations for e = 0.7, M = 40°. The residual roughly squares at every step: quadratic convergence.

Key equations

Kepler's equation

M = E − e sin E

where

Mmean anomaly, rad

Eeccentric anomaly, rad

eeccentricity, dimensionless

Mean anomaly

M = n t, n = √(μ/a³)

where

Mmean anomaly, rad

nmean motion, 2π/T, rad/s

ttime since perigee passage, s

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

asemi-major axis, km

True from eccentric anomaly

tan(ν/2) = √((1+e)/(1−e)) tan(E/2)

where

νtrue anomaly: angle from perigee to the satellite, at Earth's centre, rad or deg

Eeccentric anomaly, rad

eeccentricity, dimensionless

Newton step

E_(k+1) = E_k − (E_k − e sin E_k − M)/(1 − e cos E_k)

where

E_keccentric anomaly after k Newton steps, rad

eeccentricity, dimensionless

Mmean anomaly, rad

Radius from E

r = a(1 − e cos E)

where

rdistance from Earth's centre, km

asemi-major axis, km

eeccentricity, dimensionless

Eeccentric anomaly, rad

Hyperbolic Kepler equation

M_h = e sinh F − F = √(μ/(−a)³) t

where

M_hhyperbolic mean anomaly, dimensionless

Fhyperbolic anomaly, dimensionless

eeccentricity, dimensionless

asemi-major axis, km

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

ttime since perigee passage, s

a < 0 for a hyperbola

Worked examples

Worked example 4.1

Solve Kepler's equation by Newton's method

Given

An orbit has e = 0.7. Find E and ν when the mean anomaly is M = 40°.

new symbols

Mmean anomaly, rad

Eeccentric anomaly, rad

nmean motion, 2π/T, rad/s

Working

M = 40° = 0.698132 rad. Start with E₀ = M + e sin M1.148083 rad
Update E ← E − (E − e sin E − M)/(1 − e cos E)
Iteration 1: E = 1.4124278609 rad, residual 2.31e-02 rad
Iteration 2: E = 1.3865107052 rad, residual 2.32e-04 rad
Iteration 3: E = 1.3862447747 rad, residual 2.43e-08 rad
Iteration 4: E = 1.3862447468 rad, residual 3.33e-16 rad
E = 79.4260°; ν = 2 atan(√((1+e)/(1−e)) tan(E/2))126.3430°

Answer. E = 79.426°, ν = 126.343°, in 5 Newton steps.

Sanity check. ν > E > M, as always between perigee and apogee: the satellite runs ahead of the uniform mean motion near perigee.

Pythonedit a value, then Run

# Worked example 4.1: Solve Kepler's equation by Newton's method
from math import sin, cos, tan, atan, sqrt, radians, degrees

e = 0.7               # eccentricity
M = radians(40)       # mean anomaly, rad

E = M + e * sin(M)    # starting guess
steps = 0
while True:
    dE = (E - e * sin(E) - M) / (1 - e * cos(E))
    E -= dE
    steps += 1
    print(f"step {steps}: E = {E:.10f} rad, residual {abs(E - e*sin(E) - M):.2e} rad")
    if abs(dE) < 1e-14:
        break
nu = 2 * atan(sqrt((1 + e) / (1 - e)) * tan(E / 2))
print(f"E = {degrees(E):.4f} deg, nu = {degrees(nu):.4f} deg, in {steps} Newton steps")

step 1: E = 1.4124278609 rad, residual 2.31e-02 rad
step 2: E = 1.3865107052 rad, residual 2.32e-04 rad
step 3: E = 1.3862447747 rad, residual 2.43e-08 rad
step 4: E = 1.3862447468 rad, residual 3.33e-16 rad
step 5: E = 1.3862447468 rad, residual 1.11e-16 rad
E = 79.4260 deg, nu = 126.3430 deg, in 5 Newton steps

Worked example 4.2

How long to climb to ν = 150°?

Given

A highly elliptical orbit has perigee altitude 600 km and apogee altitude 39000 km. How long after perigee does the satellite reach ν = 150°? Check by integrating the motion.

Working

a = 26178.137 km, e = 0.73344, n = √(μ/a³) = 1.490599e-04 rad/s, T11.709 h
E = 2 atan(√((1−e)/(1+e)) tan(ν/2))1.942745 rad
M = E − e sin E1.259460 rad
t = M/n = 8449.4 s2.347 h
Integrating from perigee for 8449.4 s lands 2.7e-05 m from the Kepler position

Answer. t = 2.35 h after perigee, 20.0 % of the period for 42 % of the angle.

Sanity check. The satellite covers the first 150° quickly and spends most of the period near apogee, which is the whole point of such an orbit.

Pythonedit a value, then Run

# Worked example 4.2: How long to climb to nu = 150 deg?
from math import pi, sin, tan, atan, sqrt, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
hp, ha = 600.0, 39000.0   # km, perigee and apogee altitudes
nu = radians(150)     # target true anomaly

rp, ra = R_E + hp, R_E + ha
a = (rp + ra) / 2
e = (ra - rp) / (ra + rp)
n = sqrt(mu / a**3)
T = 2 * pi / n
E = 2 * atan(sqrt((1 - e) / (1 + e)) * tan(nu / 2))
M = E - e * sin(E)
t = M / n
print(f"a = {a:.3f} km, e = {e:.5f}, T = {T/3600:.3f} h")
print(f"E = {E:.6f} rad, M = {M:.6f} rad")
print(f"t = {t:.1f} s = {t/3600:.3f} h after perigee")
print(f"that is {100*t/T:.1f} % of the period for {100*150/360:.0f} % of the angle")

a = 26178.137 km, e = 0.73344, T = 11.709 h
E = 1.942745 rad, M = 1.259460 rad
t = 8449.4 s = 2.347 h after perigee
that is 20.0 % of the period for 42 % of the angle

Practice

Practice O4.1 · warm-up

A quarter-turn on a transfer orbit

Given

On a transfer orbit from 250 km altitude to GEO radius (42164.2 km), how long after perigee is the satellite at ν = 90°?

Mark as done

Hint

ν → E → M → t. Use the half-angle formula for E.

Answer. t = 25.7 min after perigee.

Working

a = 24396.153 km, e0.72831
E = 2 atan(√((1−e)/(1+e)) tan 45°)0.754941 rad
M = E − e sin E0.255868 rad
t = M √(a³/μ) = 1544.3 s25.7 min

Answer. t = 25.7 min after perigee.

Sanity check. The whole climb to apogee takes 316 min; reaching 90° takes only 8 % of that.

Pythonedit a value, then Run

# Practice O4.1: A quarter-turn on a transfer orbit
from math import pi, sin, tan, atan, sqrt, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
hp = 250.0            # km, perigee altitude
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
nu = radians(90)

rp = R_E + hp
ra = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
a = (rp + ra) / 2
e = (ra - rp) / (ra + rp)
E = 2 * atan(sqrt((1 - e) / (1 + e)) * tan(nu / 2))
M = E - e * sin(E)
t = M * sqrt(a**3 / mu)
print(f"a = {a:.3f} km, e = {e:.5f}, E = {E:.6f} rad, M = {M:.6f} rad")
print(f"t = {t:.1f} s = {t/60:.2f} min after perigee")

a = 24396.153 km, e = 0.72831, E = 0.754941 rad, M = 0.255868 rad
t = 1544.3 s = 25.74 min after perigee

Practice O4.2 · standard

Where will it be in 40 minutes?

Given

A satellite passes perigee of an orbit with a = 9000 km, e = 0.3. Where is it (ν and r) 40 min later?

Mark as done

Hint

M = n t, then solve M = E − e sin E for E by Newton, then convert E to ν.

Answer. ν = 131.57°, r = 10226 km (altitude 3847 km).

Working

n = √(μ/a³) = 7.394437e-04 rad/s; M = n t1.774665 rad
Newton on M = E − e sin E: E = 2.041975 rad after 4 steps
ν = 2 atan(√((1+e)/(1−e)) tan(E/2))131.573°
r = a(1 − e cos E)10225.63 km

Answer. ν = 131.57°, r = 10226 km (altitude 3847 km).

Sanity check. 28 % of the period has passed, and ν = 132° is well over 102°: near perigee the satellite moves faster than average.

Pythonedit a value, then Run

# Practice O4.2: Where will it be in 40 minutes?
from math import sin, cos, tan, atan, sqrt, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
a = 9000.0            # km
e = 0.3
t = 40 * 60           # s after perigee

n = sqrt(mu / a**3)
M = n * t
E = M + e * sin(M)
max_steps = 20        # Newton iteration cap
for k in range(max_steps):               # Newton on M = E - e sin E
    dE = (E - e * sin(E) - M) / (1 - e * cos(E))
    E -= dE
    if abs(dE) < 1e-14:
        break
nu = 2 * atan(sqrt((1 + e) / (1 - e)) * tan(E / 2))
r = a * (1 - e * cos(E))
print(f"n = {n:.6e} rad/s, M = {M:.6f} rad, E = {E:.6f} rad")
print(f"nu = {degrees(nu):.3f} deg, r = {r:.2f} km (altitude {r - R_E:.1f} km)")

n = 7.394437e-04 rad/s, M = 1.774665 rad, E = 2.041975 rad
nu = 131.573 deg, r = 10225.63 km (altitude 3847.5 km)

Practice O4.3 · challenge

Leaving on a hyperbola

Given

A probe passes perigee at 300 km altitude on a hyperbola with e = 1.8. How long until it is 100,000 km from Earth's centre?

Mark as done

Hint

a = r_p/(1 − e) is negative. r = a(1 − e cosh F) gives F; then M = e sinh F − F and t = M √(−a³/μ).

Answer. t = 3.42 h after perigee, at ν = 116.9°.

Working

r_p = 6678.137 km; a = r_p/(1 − e)-8347.67 km
cosh F = (1 − r/a)/e = 7.210772, so F2.663880
M = e sinh F − F10.190090
t = M √((−a)³/μ) = 12310.0 s3.419 h
True anomaly there: ν = 2 atan(√((e+1)/(e−1)) tanh(F/2))116.85°(asymptote at 123.75°)

Answer. t = 3.42 h after perigee, at ν = 116.9°.

Sanity check. A universal-variable propagation of the perigee state for the same time lands at the same radius, and ν is below the asymptote angle as it must be.

Pythonedit a value, then Run

# Practice O4.3: Leaving on a hyperbola
from math import sqrt, sinh, acosh, tanh, atan, degrees, acos

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
hp = 300.0            # km, perigee altitude
e = 1.8               # eccentricity
r = 100000.0          # km, target distance

rp = R_E + hp
a = rp / (1 - e)                         # negative for a hyperbola
F = acosh((1 - r / a) / e)               # hyperbolic anomaly
M = e * sinh(F) - F
t = M * sqrt((-a) ** 3 / mu)
nu = 2 * atan(sqrt((e + 1) / (e - 1)) * tanh(F / 2))
print(f"a = {a:.2f} km, F = {F:.6f}, M = {M:.6f}")
print(f"t = {t:.1f} s = {t/3600:.3f} h")
print(f"nu = {degrees(nu):.2f} deg (asymptote at {degrees(acos(-1/e)):.2f} deg)")

a = -8347.67 km, F = 2.663880, M = 10.190090
t = 12310.0 s = 3.419 h
nu = 116.85 deg (asymptote at 123.75 deg)

Common mistakes

Working in degrees inside Kepler's equation. E − e sin E only makes sense with E in radians.
Getting the wrong quadrant from atan: use the half-angle formula (or atan2), which keeps E and ν in the same half of the orbit.
Stopping Newton after a fixed number of steps without checking the residual.
Using M = n t with t measured from the wrong event. t is time since perigee passage.

Recap from earlier chapters

What is the vis-viva equation?

v² = μ(2/r − 1/a).

What is the perigee radius of an ellipse in terms of a and e?

r_p = a(1 − e).

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.

Where in each source, and links: References.

I have finished chapter 4

Chapter 5

From state vectors to elements and back

40 minutes

The idea in one minute

Tracking gives you a position and a velocity at one instant: six numbers. The classical elements are six other numbers describing the same orbit in a more meaningful way. Going from r, v to elements is a short recipe built on three vectors (h, n, e) and a few arc cosines; going back is a rotation from the orbit's own frame into the Earth-centred one.

Three angles place the orbit in space. Ω turns the node line away from the equinox direction, i tilts the plane about that line, and ω points perigee within the plane. ν then says where the satellite is.

You will be able to

Compute the six classical elements from a position and velocity
Apply the three quadrant checks that every acos needs
Build r and v from elements, and handle the circular and equatorial cases

Three helper vectors

h = r × v is perpendicular to the orbit plane and gives i. The node vector n = k × h points to the ascending node and gives Ω. The eccentricity vector e points to perigee, its length is e, and it gives ω and ν.^[R11]

Quadrant checks

acos returns 0° to 180° only. Three sign tests fix this: if n_y < 0 then Ω = 360° − Ω; if e_z < 0 then ω = 360° − ω; if r·v < 0 (falling towards perigee) then ν = 360° − ν.

Back to r and v

In the perifocal frame the state is simple: r = (r cos ν, r sin ν, 0) and v = √(μ/p)(−sin ν, e + cos ν, 0). Rotate by ω, then i, then Ω and you are in ECI.

When elements break

On a circular orbit perigee does not exist, so ω and ν are undefined; use the argument of latitude u = ω + ν. On an equatorial orbit the node does not exist; use the longitude of perigee or true longitude. Numerically, near-circular orbits make ω and ν very sensitive to noise.

Key equations

Angular momentum and inclination

h = r × v, i = acos(h_z/h)

where

h (vector)specific angular momentum vector, km²/s

h_zz component of h, km²/s

iinclination, deg

r (vector)position in ECI, km

v (vector)velocity in ECI, km/s

Node vector and RAAN

n = k × h, Ω = acos(n_x/n) (360° − Ω if n_y < 0)

where

nnode vector, towards the ascending node, km²/s

kunit vector along ECI z, dimensionless

Ωright ascension of the ascending node, deg

n_xx component of n, km²/s

Eccentricity vector

e = ((v² − μ/r) r − (r·v) v)/μ

where

e (vector)eccentricity vector, towards perigee, dimensionless

vspeed relative to Earth's centre, km/s

rdistance from Earth's centre, km

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

Argument of perigee and true anomaly

ω = acos(n·e/(n e)), ν = acos(e·r/(e r))

where

ωargument of perigee (360° − ω if e_z < 0), deg

νtrue anomaly (360° − ν if r·v < 0), deg

Perifocal state

r_PQW = p/(1+e cos ν) (cos ν, sin ν, 0); v_PQW = √(μ/p) (−sin ν, e + cos ν, 0)

where

r_PQWposition in the perifocal frame (x to perigee), km

v_PQWvelocity in the perifocal frame, km/s

psemi-latus rectum (parameter), km

eeccentricity, dimensionless

νtrue anomaly: angle from perigee to the satellite, at Earth's centre, rad or deg

μgravitational parameter of Earth, G(M + m) ≈ GM = 398600.4418, km³/s²

Rotation to ECI

r_ECI = R₃(−Ω) R₁(−i) R₃(−ω) r_PQW

where

R₁, R₃rotation matrices about the x and z axes, dimensionless

r_ECIposition in the Earth-centred inertial frame, km

Worked examples

Worked example 5.1

Six elements from one state vector

Given

A tracking station gives r = (4506.847, 6754.624, 4127.819) km and v = (-1.220029, 3.493813, -5.021162) km/s in the ECI frame. Find the classical elements.

new symbols

nnode vector k × h, pointing to the ascending node, km²/s

e (vector)eccentricity vector, pointing to perigee, dimensionless

kunit vector along Earth's spin axis (ECI z), dimensionless

Working

|r| = 9109.089 km, |v| = 6.237569 km/s, r·v-2625.539 km²/s(negative: falling towards perigee)
h = r × v = (-48337.9, 17593.6, 23986.9) km²/s, |h|56757.88 km²/s
i = acos(h_z/|h|)65.000°
n = k × h = (-17593.6, -48337.9, 0); Ω = acos(n_x/|n|) = 110.000°, but n_y < 0 so Ω = 360° − that250.000°
e = ((v² − μ/r) r − (r·v) v)/μ: |e|0.120000
ω = acos(n·e/(|n||e|)) = 50.000°, but e_z < 0 so ω310.000°
ν = acos(e·r/(|e||r|)) = 160.000°, but r·v < 0 so ν200.000°
a = −μ/(2ε) with ε = v²/2 − μ/r = -24.30491 km²/s²: a8200.00 km

Answer. a = 8200.0 km, e = 0.1200, i = 65.00°, Ω = 250.00°, ω = 310.00°, ν = 200.00°.

Sanity check. Every acos returns 0° to 180°; each of the three sign tests moved an angle into the other half. Feeding the elements back through coe2rv returns the same state.

Pythonedit a value, then Run

# Worked example 5.1: Six elements from one state vector
from math import sqrt, acos, degrees, pi

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r = [4506.847, 6754.624, 4127.819]     # km, ECI position
v = [-1.220029, 3.493813, -5.021162]   # km/s, ECI velocity

def dot(a, b):
    return sum(x * y for x, y in zip(a, b))

def cross(a, b):
    return [a[1] * b[2] - a[2] * b[1], a[2] * b[0] - a[0] * b[2], a[0] * b[1] - a[1] * b[0]]

def norm(a):
    return sqrt(dot(a, a))

rm, vm, rv = norm(r), norm(v), dot(r, v)
h = cross(r, v); hm = norm(h)
i = acos(h[2] / hm)
n = cross([0, 0, 1], h); nm = norm(n)            # node vector
raan = acos(n[0] / nm)
if n[1] < 0:
    raan = 2 * pi - raan
evec = [((vm**2 - mu / rm) * r[k] - rv * v[k]) / mu for k in range(3)]
e = norm(evec)
argp = acos(dot(n, evec) / (nm * e))
if evec[2] < 0:
    argp = 2 * pi - argp
nu = acos(dot(evec, r) / (e * rm))
if rv < 0:
    nu = 2 * pi - nu
eps = vm**2 / 2 - mu / rm
a = -mu / (2 * eps)
print(f"|r| = {rm:.3f} km, |v| = {vm:.6f} km/s, r.v = {rv:.3f}, |h| = {hm:.2f} km^2/s")
print(f"a = {a:.2f} km, e = {e:.5f}, i = {degrees(i):.3f} deg")
print(f"RAAN = {degrees(raan):.3f} deg, argp = {degrees(argp):.3f} deg, nu = {degrees(nu):.3f} deg")

# Worked example 5.1: Six elements from one state vector
from math import sqrt, acos, degrees, pi

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r = [4506.847, 6754.624, 4127.819]     # km, ECI position
v = [-1.220029, 3.493813, -5.021162]   # km/s, ECI velocity

def dot(a, b):
    return sum(x * y for x, y in zip(a, b))

def cross(a, b):
    return [a[1] * b[2] - a[2] * b[1], a[2] * b[0] - a[0] * b[2], a[0] * b[1] - a[1] * b[0]]

def norm(a):
    return sqrt(dot(a, a))

rm, vm, rv = norm(r), norm(v), dot(r, v)
h = cross(r, v); hm = norm(h)
i = acos(h[2] / hm)
n = cross([0, 0, 1], h); nm = norm(n)            # node vector
raan = acos(n[0] / nm)
if n[1] < 0:
    raan = 2 * pi - raan
evec = [((vm**2 - mu / rm) * r[k] - rv * v[k]) / mu for k in range(3)]
e = norm(evec)
argp = acos(dot(n, evec) / (nm * e))
if evec[2] < 0:
    argp = 2 * pi - argp
nu = acos(dot(evec, r) / (e * rm))
if rv < 0:
    nu = 2 * pi - nu
eps = vm**2 / 2 - mu / rm
a = -mu / (2 * eps)
print(f"|r| = {rm:.3f} km, |v| = {vm:.6f} km/s, r.v = {rv:.3f}, |h| = {hm:.2f} km^2/s")
print(f"a = {a:.2f} km, e = {e:.5f}, i = {degrees(i):.3f} deg")
print(f"RAAN = {degrees(raan):.3f} deg, argp = {degrees(argp):.3f} deg, nu = {degrees(nu):.3f} deg")

|r| = 9109.089 km, |v| = 6.237569 km/s, r.v = -2625.539, |h| = 56757.88 km^2/s
a = 8200.00 km, e = 0.12000, i = 65.000 deg
RAAN = 250.000 deg, argp = 310.000 deg, nu = 200.000 deg

Worked example 5.2

Elements to a state vector

Given

An Earth-observation satellite has a = 7000 km, e = 0.0012, i = 97.8°, Ω = 30°, ω = 90°, ν = 45°. Find r and v in ECI.

Working

p = a(1 − e²) = 6999.990 km; r = p/(1 + e cos ν)6994.055 km
Perifocal: r = (4945.544, 4945.544, 0) km, v = √(μ/p)(−sin ν, e + cos ν, 0) = (-5.33587, 5.34492, 0) km/s
Multiply by Q = R₃(−Ω) R₁(−i) R₃(−ω)
ECI: r = (-3947.373, -3054.037, 4899.787) km
ECI: v = (-4.99092, -2.04532, -5.28650) km/s

Answer. r = (-3947.4, -3054.0, 4899.8) km, v = (-4.991, -2.045, -5.287) km/s.

Sanity check. A rotation never changes length: |r| and |v| are the same in both frames. With u = ω + ν = 135° the satellite has passed its most northerly point and z = 4900 km is still north of the equator.

Pythonedit a value, then Run

# Worked example 5.2: Elements to a state vector
from math import sin, cos, sqrt, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
a = 7000.0            # km
e = 0.0012
i = radians(97.8)
raan = radians(30)
argp = radians(90)
nu = radians(45)

p = a * (1 - e**2)
r = p / (1 + e * cos(nu))
r_pf = [r * cos(nu), r * sin(nu), 0.0]
v_pf = [-sqrt(mu / p) * sin(nu), sqrt(mu / p) * (e + cos(nu)), 0.0]

# Q = R3(-raan) R1(-i) R3(-argp): perifocal -> ECI
cO, sO, ci, si, cw, sw = cos(raan), sin(raan), cos(i), sin(i), cos(argp), sin(argp)
Q = [[cO * cw - sO * sw * ci, -cO * sw - sO * cw * ci, sO * si],
     [sO * cw + cO * sw * ci, -sO * sw + cO * cw * ci, -cO * si],
     [sw * si, cw * si, ci]]
r_eci = [sum(Q[j][k] * r_pf[k] for k in range(3)) for j in range(3)]
v_eci = [sum(Q[j][k] * v_pf[k] for k in range(3)) for j in range(3)]
print(f"p = {p:.3f} km, r = {r:.3f} km")
print("ECI r = (%.3f, %.3f, %.3f) km" % tuple(r_eci))
print("ECI v = (%.5f, %.5f, %.5f) km/s" % tuple(v_eci))

# Worked example 5.2: Elements to a state vector
from math import sin, cos, sqrt, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
a = 7000.0            # km
e = 0.0012
i = radians(97.8)
raan = radians(30)
argp = radians(90)
nu = radians(45)

p = a * (1 - e**2)
r = p / (1 + e * cos(nu))
r_pf = [r * cos(nu), r * sin(nu), 0.0]
v_pf = [-sqrt(mu / p) * sin(nu), sqrt(mu / p) * (e + cos(nu)), 0.0]

# Q = R3(-raan) R1(-i) R3(-argp): perifocal -> ECI
cO, sO, ci, si, cw, sw = cos(raan), sin(raan), cos(i), sin(i), cos(argp), sin(argp)
Q = [[cO * cw - sO * sw * ci, -cO * sw - sO * cw * ci, sO * si],
     [sO * cw + cO * sw * ci, -sO * sw + cO * cw * ci, -cO * si],
     [sw * si, cw * si, ci]]
r_eci = [sum(Q[j][k] * r_pf[k] for k in range(3)) for j in range(3)]
v_eci = [sum(Q[j][k] * v_pf[k] for k in range(3)) for j in range(3)]
print(f"p = {p:.3f} km, r = {r:.3f} km")
print("ECI r = (%.3f, %.3f, %.3f) km" % tuple(r_eci))
print("ECI v = (%.5f, %.5f, %.5f) km/s" % tuple(v_eci))

p = 6999.990 km, r = 6994.055 km
ECI r = (-3947.373, -3054.037, 4899.787) km
ECI v = (-4.99092, -2.04532, -5.28650) km/s

Practice

Practice O5.1 · warm-up

Inclination from a state

Given

A satellite is at r = (6800, 0, 0) km with v = (0, 5.2, 5.6) km/s. Find h and the inclination.

Mark as done

Hint

h = r × v; i = acos(h_z/|h|).

Answer. |h| = 51966 km²/s, i = 47.12°.

Working

h = r × v = (0·5.6 − 0·5.2, 0·0 − 6800·5.6, 6800·5.2 − 0) = (0, -38080, 35360) km²/s
|h|51965.53 km²/s
i = acos(35360/51965.53)47.121°

Answer. |h| = 51966 km²/s, i = 47.12°.

Sanity check. The satellite is on the x axis, in the equatorial plane, so it is at a node and i is simply the angle of v above the equator: atan(5.6/5.2) = 47.12°.

Pythonedit a value, then Run

# Practice O5.1: Inclination from a state
from math import sqrt, acos, degrees

r = [6800.0, 0.0, 0.0]   # km
v = [0.0, 5.2, 5.6]      # km/s

h = [r[1] * v[2] - r[2] * v[1], r[2] * v[0] - r[0] * v[2], r[0] * v[1] - r[1] * v[0]]   # h = r x v
hm = sqrt(h[0]**2 + h[1]**2 + h[2]**2)
i = degrees(acos(h[2] / hm))
print(f"h = ({h[0]:.0f}, {h[1]:.0f}, {h[2]:.0f}) km^2/s")
print(f"|h| = {hm:.2f} km^2/s, i = {i:.3f} deg")

h = (0, -38080, 35360) km^2/s
|h| = 51965.53 km^2/s, i = 47.121 deg

Practice O5.2 · standard

All six elements

Given

Find all six classical elements for r = (-7902.47, -296.38, 5413.27) km, v = (-1.29711, -6.84390, 0.06230) km/s.

Mark as done

Hint

Follow the recipe: h, i, n, Ω, e, ω, ν, a. Do the three sign checks (n_y, e_z, r·v) at the end.

Answer. a = 11500 km, e = 0.250, i = 35.00°, Ω = 80.00°, ω = 40.00°, ν = 60.00°.

Working

|r| = 9583.338 km, |v| = 6.96601 km/s, r·v12616.01 km²/s
h = (37029.4, -6529.3, 53699.3); i35.000°
n = (6529.3, 37029.4, 0); n_y ≥ 0 so Ω80.000°
e vector = (-0.09638, 0.21146, 0.09217); e = 0.25000; e_z ≥ 0 so ω40.000°
r·v > 0 so ν60.000°
ε = -17.33040 km²/s², a = −μ/(2ε)11500.04 km

Answer. a = 11500 km, e = 0.250, i = 35.00°, Ω = 80.00°, ω = 40.00°, ν = 60.00°.

Sanity check. Check: r_p = a(1 − e) must be above Earth's surface; here it is 2247 km altitude. Converting back with coe2rv reproduces the state to rounding.

Pythonedit a value, then Run

# Practice O5.2: All six elements
from math import sqrt, acos, degrees, pi

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r = [-7902.47, -296.38, 5413.27]       # km
v = [-1.29711, -6.84390, 0.06230]      # km/s

def dot(a, b):
    return sum(x * y for x, y in zip(a, b))

def cross(a, b):
    return [a[1] * b[2] - a[2] * b[1], a[2] * b[0] - a[0] * b[2], a[0] * b[1] - a[1] * b[0]]

def norm(a):
    return sqrt(dot(a, a))

rm, vm, rv = norm(r), norm(v), dot(r, v)
h = cross(r, v); hm = norm(h)
i = acos(h[2] / hm)
n = cross([0, 0, 1], h); nm = norm(n)
raan = acos(n[0] / nm)
if n[1] < 0:
    raan = 2 * pi - raan
evec = [((vm**2 - mu / rm) * r[k] - rv * v[k]) / mu for k in range(3)]
e = norm(evec)
argp = acos(dot(n, evec) / (nm * e))
if evec[2] < 0:
    argp = 2 * pi - argp
nu = acos(dot(evec, r) / (e * rm))
if rv < 0:
    nu = 2 * pi - nu
eps = vm**2 / 2 - mu / rm
a = -mu / (2 * eps)
print(f"energy = {eps:.5f} km^2/s^2")
print(f"a = {a:.2f} km, e = {e:.5f}, i = {degrees(i):.3f} deg")
print(f"RAAN = {degrees(raan):.3f} deg, argp = {degrees(argp):.3f} deg, nu = {degrees(nu):.3f} deg")

# Practice O5.2: All six elements
from math import sqrt, acos, degrees, pi

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r = [-7902.47, -296.38, 5413.27]       # km
v = [-1.29711, -6.84390, 0.06230]      # km/s

def dot(a, b):
    return sum(x * y for x, y in zip(a, b))

def cross(a, b):
    return [a[1] * b[2] - a[2] * b[1], a[2] * b[0] - a[0] * b[2], a[0] * b[1] - a[1] * b[0]]

def norm(a):
    return sqrt(dot(a, a))

rm, vm, rv = norm(r), norm(v), dot(r, v)
h = cross(r, v); hm = norm(h)
i = acos(h[2] / hm)
n = cross([0, 0, 1], h); nm = norm(n)
raan = acos(n[0] / nm)
if n[1] < 0:
    raan = 2 * pi - raan
evec = [((vm**2 - mu / rm) * r[k] - rv * v[k]) / mu for k in range(3)]
e = norm(evec)
argp = acos(dot(n, evec) / (nm * e))
if evec[2] < 0:
    argp = 2 * pi - argp
nu = acos(dot(evec, r) / (e * rm))
if rv < 0:
    nu = 2 * pi - nu
eps = vm**2 / 2 - mu / rm
a = -mu / (2 * eps)
print(f"energy = {eps:.5f} km^2/s^2")
print(f"a = {a:.2f} km, e = {e:.5f}, i = {degrees(i):.3f} deg")
print(f"RAAN = {degrees(raan):.3f} deg, argp = {degrees(argp):.3f} deg, nu = {degrees(nu):.3f} deg")

energy = -17.33040 km^2/s^2
a = 11500.04 km, e = 0.25000, i = 35.000 deg
RAAN = 80.000 deg, argp = 40.000 deg, nu = 60.000 deg

Practice O5.3 · challenge

When perigee is almost undefined

Given

A near-circular orbit has a = 6930 km, e = 2.0e-04, i = 97.5°, Ω = 300°, ω = 120°, ν = 250°. Convert to r, v and back. Which recovered quantity is reliable, and why? Find it directly from the node vector.

Mark as done

Hint

With e tiny, the direction of perigee is set by a vector only 10⁻⁴ long. The angle from the node to the satellite, u = ω + ν, does not need perigee at all.

Answer. u = 10.000° is reliable; ω and ν separately are ill-conditioned when e ≈ 0, though in exact arithmetic the round trip still closes (error about 10⁻¹² km, floating-point level).

Working

coe2rv: r = (3276.554, -5989.325, 1193.168) km, |r|6930.474 km
rv2coe gives ω = 120.0000° and ν250.0000°
u = ω + ν10.000000°(given 10.000000°)
Directly: u = acos(n·r/(|n||r|)) = 10.000000°, with z ≥ 0 deciding the half
Round trip back to r, v: position error 1.5e-12 km

Answer. u = 10.000° is reliable; ω and ν separately are ill-conditioned when e ≈ 0, though in exact arithmetic the round trip still closes (error about 10⁻¹² km, floating-point level).

Sanity check. Real tracking noise of a few metres changes the e vector's direction by tens of degrees at e = 10⁻⁴, but barely moves u. That is why near-circular orbits are described with u (or equinoctial elements).

Pythonedit a value, then Run

# Practice O5.3: When perigee is almost undefined
from math import sin, cos, sqrt, acos, radians, degrees, pi

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
a, e = 6930.0, 2.0e-4                  # km, near-circular
i, raan, argp, nu = map(radians, (97.5, 300.0, 120.0, 250.0))

def dot(a, b):
    return sum(x * y for x, y in zip(a, b))

def cross(a, b):
    return [a[1] * b[2] - a[2] * b[1], a[2] * b[0] - a[0] * b[2], a[0] * b[1] - a[1] * b[0]]

def norm(a):
    return sqrt(dot(a, a))

def coe2rv(a, e, i, raan, argp, nu):
    p = a * (1 - e * e)
    rpf = [cos(nu) * p / (1 + e * cos(nu)), sin(nu) * p / (1 + e * cos(nu)), 0.0]
    vpf = [-sin(nu) * sqrt(mu / p), (e + cos(nu)) * sqrt(mu / p), 0.0]
    cO, sO, ci, si, cw, sw = cos(raan), sin(raan), cos(i), sin(i), cos(argp), sin(argp)
    Q = [[cO * cw - sO * sw * ci, -cO * sw - sO * cw * ci, sO * si],
         [sO * cw + cO * sw * ci, -sO * sw + cO * cw * ci, -cO * si],
         [sw * si, cw * si, ci]]
    return [dot(Q[j], rpf) for j in range(3)], [dot(Q[j], vpf) for j in range(3)]

def rv2coe(r, v):
    rm, vm = norm(r), norm(v)
    h = cross(r, v); n = cross([0, 0, 1], h); nm = norm(n)
    ev = [((vm * vm - mu / rm) * r[k] - dot(r, v) * v[k]) / mu for k in range(3)]; ee = norm(ev)
    aa = -mu / (2 * (vm * vm / 2 - mu / rm))
    ii = acos(h[2] / norm(h))
    O = acos(max(-1, min(1, n[0] / nm)));  O = 2 * pi - O if n[1] < 0 else O
    w = acos(max(-1, min(1, dot(n, ev) / (nm * ee))));  w = 2 * pi - w if ev[2] < 0 else w
    f = acos(max(-1, min(1, dot(ev, r) / (ee * rm))));  f = 2 * pi - f if dot(r, v) < 0 else f
    return aa, ee, ii, O, w, f, n

r, v = coe2rv(a, e, i, raan, argp, nu)
aa, ee, ii, O, w, f, n = rv2coe(r, v)
u_sum = degrees(w + f) % 360
u_dir = degrees(acos(dot(n, r) / (norm(n) * norm(r))))   # z >= 0: first half of the orbit
if r[2] < 0:
    u_dir = 360 - u_dir
r2, _ = coe2rv(aa, ee, ii, O, w, f)
err = norm([r2[k] - r[k] for k in range(3)])
print("r = (%.3f, %.3f, %.3f) km" % tuple(r))
print(f"recovered argp = {degrees(w):.4f} deg, nu = {degrees(f):.4f} deg, e = {ee:.1e}")
print(f"u = argp + nu = {u_sum:.6f} deg; directly from the node vector u = {u_dir:.6f} deg")
print(f"round-trip position error {err:.1e} km")

# Practice O5.3: When perigee is almost undefined
from math import sin, cos, sqrt, acos, radians, degrees, pi

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
a, e = 6930.0, 2.0e-4                  # km, near-circular
i, raan, argp, nu = map(radians, (97.5, 300.0, 120.0, 250.0))

def dot(a, b):
    return sum(x * y for x, y in zip(a, b))

def cross(a, b):
    return [a[1] * b[2] - a[2] * b[1], a[2] * b[0] - a[0] * b[2], a[0] * b[1] - a[1] * b[0]]

def norm(a):
    return sqrt(dot(a, a))

def coe2rv(a, e, i, raan, argp, nu):
    p = a * (1 - e * e)
    rpf = [cos(nu) * p / (1 + e * cos(nu)), sin(nu) * p / (1 + e * cos(nu)), 0.0]
    vpf = [-sin(nu) * sqrt(mu / p), (e + cos(nu)) * sqrt(mu / p), 0.0]
    cO, sO, ci, si, cw, sw = cos(raan), sin(raan), cos(i), sin(i), cos(argp), sin(argp)
    Q = [[cO * cw - sO * sw * ci, -cO * sw - sO * cw * ci, sO * si],
         [sO * cw + cO * sw * ci, -sO * sw + cO * cw * ci, -cO * si],
         [sw * si, cw * si, ci]]
    return [dot(Q[j], rpf) for j in range(3)], [dot(Q[j], vpf) for j in range(3)]

def rv2coe(r, v):
    rm, vm = norm(r), norm(v)
    h = cross(r, v); n = cross([0, 0, 1], h); nm = norm(n)
    ev = [((vm * vm - mu / rm) * r[k] - dot(r, v) * v[k]) / mu for k in range(3)]; ee = norm(ev)
    aa = -mu / (2 * (vm * vm / 2 - mu / rm))
    ii = acos(h[2] / norm(h))
    O = acos(max(-1, min(1, n[0] / nm)));  O = 2 * pi - O if n[1] < 0 else O
    w = acos(max(-1, min(1, dot(n, ev) / (nm * ee))));  w = 2 * pi - w if ev[2] < 0 else w
    f = acos(max(-1, min(1, dot(ev, r) / (ee * rm))));  f = 2 * pi - f if dot(r, v) < 0 else f
    return aa, ee, ii, O, w, f, n

r, v = coe2rv(a, e, i, raan, argp, nu)
aa, ee, ii, O, w, f, n = rv2coe(r, v)
u_sum = degrees(w + f) % 360
u_dir = degrees(acos(dot(n, r) / (norm(n) * norm(r))))   # z >= 0: first half of the orbit
if r[2] < 0:
    u_dir = 360 - u_dir
r2, _ = coe2rv(aa, ee, ii, O, w, f)
err = norm([r2[k] - r[k] for k in range(3)])
print("r = (%.3f, %.3f, %.3f) km" % tuple(r))
print(f"recovered argp = {degrees(w):.4f} deg, nu = {degrees(f):.4f} deg, e = {ee:.1e}")
print(f"u = argp + nu = {u_sum:.6f} deg; directly from the node vector u = {u_dir:.6f} deg")
print(f"round-trip position error {err:.1e} km")

r = (3276.554, -5989.325, 1193.168) km
recovered argp = 120.0000 deg, nu = 250.0000 deg, e = 2.0e-04
u = argp + nu = 10.000000 deg; directly from the node vector u = 10.000000 deg
round-trip position error 6.4e-13 km

Common mistakes

Skipping the quadrant checks. Half of all random states give a wrong Ω, ω or ν without them.
Using the r·v test for ω or the e_z test for ν: each angle has its own test.
Trusting ω and ν on a near-circular orbit. Report u = ω + ν instead.
Applying the rotations in the wrong order. The perifocal-to-ECI matrix is R₃(−Ω) R₁(−i) R₃(−ω).

Recap from earlier chapters

Write the orbit equation.

r = p/(1 + e cos ν), with p = a(1 − e²).

How many Newton steps does Kepler's equation usually need at e = 0.7?

Very few: 5 for M = 40° to machine precision.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R11Vallado, D.A., companion software to Fundamentals of Astrodynamics and Applications (CelesTrak), GitHub repo CelesTrak/fundamentals-of-astrodynamics, commit 49df047 (AGPL-3.0; values read, no code copied).

Where in each source, and links: References.

I have finished chapter 5

Chapter 6

Perturbations: J₂, drag, third body and pressure from sunlight

40 minutes

The idea in one minute

A real orbit is a Kepler ellipse that slowly bends. Earth's equatorial bulge (J₂) turns the orbit plane and the ellipse inside it, steadily, by degrees per day. Air drag, thin as it is, takes energy and lowers the orbit. The Moon and the Sun pull a little differently on the satellite than on Earth, and sunlight itself pushes. Each is tiny compared with Earth's central pull, but they add up over days and years, and each one wins at a different height.

Read straight up from your altitude. J₂ is the biggest push everywhere below GEO; drag matters only low down (it falls below sunlight pressure near 612 km for A/m = 0.01 m²/kg); at GEO the Moon and Sun catch up with J₂.

You will be able to

Compute the J₂ drift of the node and the perigee and say which way they turn
Estimate drag decay per orbit and a rough orbital lifetime
Rank J₂, drag, the Moon, the Sun and sunlight pressure for any altitude

Earth's bulge: J₂

Earth is wider at the equator by about 21 km. The extra mass pulls an inclined orbit towards the equator, and the orbit responds like a gyroscope: the plane precesses. The node moves west for a prograde orbit and east for a retrograde one. Inside the plane the ellipse rotates too, except at 63.4°, the critical inclination.^[R1]

Air drag

Drag removes energy, so the orbit shrinks and, oddly, the satellite speeds up. Density falls roughly exponentially with height, so the decay rate depends hugely on altitude and on the Sun's activity, which heats and swells the upper atmosphere.^{[R14, R15]}

The Moon, the Sun and sunlight

A third body matters through the difference in its pull on the satellite and on Earth: a tidal acceleration that grows with distance from Earth. Sunlight carries momentum and pushes on every lit surface, by an amount set by the area-to-mass ratio, not by altitude.^{[R2, R8]}

What a tracker sees: the period shrinks

Radar measures the period more easily than the height. From T = 2π√(a³/μ), dT/da = 3T/(2a), and with the drag decay per revolution this gives dT/dt = −3π C_D (A/m) ρ a (SI units, circular orbit, density held fixed). For the small satellite of worked example 6.2 at 400 km (ρ = 2.8e-12 kg/m³) the period shortens by 0.34 s per day. Run the same relation backwards and a measured period decay gives the density along the orbit.^[R14]

Key equations

Node regression (J₂)

Ω̇ = −(3/2) n J₂ (R_E/p)² cos i

where

Ω̇rate of change of the right ascension of the ascending node, rad/s

nmean motion, √(μ/a³), rad/s

J₂Earth's oblateness coefficient, 1.0826 × 10⁻³, dimensionless

R_EEarth's equatorial radius, 6378.137, km

psemi-latus rectum, a(1 − e²), km

iinclination, deg

Apsidal rotation (J₂)

ω̇ = (3/4) n J₂ (R_E/p)² (5 cos² i − 1)

where

ω̇rate of change of the argument of perigee, rad/s

nmean motion, √(μ/a³), rad/s

J₂Earth's oblateness coefficient, 1.0826 × 10⁻³, dimensionless

R_EEarth's equatorial radius, 6378.137, km

psemi-latus rectum, a(1 − e²), km

iinclination, deg

zero at the critical inclination 63.4°

Drag decay per orbit

Δa_rev = −2π C_D (A/m) ρ a²

where

Δa_revchange of semi-major axis in one revolution, m

C_Ddrag coefficient, dimensionless

A/marea-to-mass ratio, m²/kg

ρatmospheric density at the orbit, kg/m³

asemi-major axis (in metres here), m

circular orbit, SI units

Exponential atmosphere

ρ = ρ₀ exp(−(h − h₀)/H)

where

ρ₀density at the reference height, kg/m³

haltitude, km

h₀reference altitude, km

Hscale height, km

Third-body tidal pull

a₃ ≈ 2 μ₃ r / d³

where

a₃perturbing acceleration (worst direction), km/s²

μ₃gravitational parameter of the Moon or Sun, km³/s²

rsatellite distance from Earth's centre, km

ddistance from Earth to the third body, km

valid when r ≪ d

Sunlight pressure

a_SRP = (S/c) C_R (A/m)

where

a_SRPacceleration from sunlight, m/s²

Ssolar flux, 1361 at 1 AU, W/m²

cspeed of light, 299,792,458, m/s

C_Rradiation-pressure coefficient (1 to 2), dimensionless

A/marea-to-mass ratio, m²/kg

Worked examples

Worked example 6.1

How fast does the orbit plane turn?^[R1]

Given

A circular orbit has altitude 800 km and inclination 70°. Find the J₂ node rate and the perigee rate in degrees per day.

Working

a = R_E + h = 6378.137 + 800 = 7178.137 km; n = √(μ/a³)1.038129e-03 rad/s
Common factor n J₂ (R_E/p)² = 1.038129e-03 × 0.00108262668 × (6378.137/7178.137)²8.8735e-07 rad/s
Ω̇ = −1.5 × that × cos 70° = -4.5524e-07 rad/s-2.254 °/day
ω̇ = 0.75 × that × (5 cos² 70° − 1) = -2.7626e-07 rad/s-1.368 °/day

Answer. Ω̇ = -2.25 °/day (westward), ω̇ = -1.37 °/day.

Sanity check. The node moves about 823° in a year: J₂ is not a small effect over a mission. A negative ω̇ is expected above the critical inclination of 63.4°.

Pythonedit a value, then Run

# Worked example 6.1: How fast does the orbit plane turn?
from math import sqrt, cos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
h = 800.0             # km, altitude (circular, so p = a)
i = radians(70)       # inclination

a = R_E + h
n = sqrt(mu / a**3)
k = n * J2 * (R_E / a) ** 2              # common factor
raan_dot = -1.5 * k * cos(i)
argp_dot = 0.75 * k * (5 * cos(i) ** 2 - 1)
day = 86400.0
print(f"n = {n:.6e} rad/s, factor = {k:.4e} rad/s")
print(f"node rate    = {raan_dot:.4e} rad/s = {degrees(raan_dot) * day:.3f} deg/day")
print(f"perigee rate = {argp_dot:.4e} rad/s = {degrees(argp_dot) * day:.3f} deg/day")

n = 1.038129e-03 rad/s, factor = 8.8735e-07 rad/s
node rate    = -4.5524e-07 rad/s = -2.254 deg/day
perigee rate = -2.7626e-07 rad/s = -1.368 deg/day

Worked example 6.2

How fast does drag pull a small satellite down?^{[R14, R15]}

Given

A small satellite (C_D = 2.2, A/m = 0.01 m²/kg) is in a circular orbit at 400 km, where the density is about 2.8e-12 kg/m³. Find the height lost per orbit and per day, then estimate the time to fall to 200 km.

new symbols

C_Ddrag coefficient, dimensionless

A/marea-to-mass ratio, m²/kg

ρatmospheric density, kg/m³

Hdensity scale height, km

Working

a = 6778.137 km6.7781e+06 m
Δa per rev = −2π C_D (A/m) ρ a² = −2π × 2.2 × 0.01 × 2.80e-12 × (6.7781e+06)²-17.78 m
Revolutions per day = 86,400 / 5553.6 s = 15.56, so Δa per day-277 m
Local scale height from the density table: H = 54.6 km. Density grows quickly as the orbit drops, so the decay speeds up.
Integrating da/dt = −√(μa) ρ(h) C_D (A/m) with the table down to 200 km gives 185 days

Answer. About 18 m per orbit (0.28 km per day) at first; roughly 185 days (6 months) to reach 200 km.

Sanity check. The decay is not linear: at the start the satellite would need 723 days at the initial rate, yet it falls in 185. Halve the density (solar minimum) and the lifetime roughly doubles.

Pythonedit a value, then Run

# Worked example 6.2: How fast does drag pull a small satellite down?
from math import pi, sqrt, log, exp

mu = 398600.4418 * 1e9  # m^3/s^2, Earth's gravitational parameter in SI (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
C_D = 2.2             # drag coefficient
Am = 0.01             # m^2/kg, area-to-mass ratio
h0, h_end = 400.0, 200.0   # km, start and end altitudes
# density table (altitude km, density kg/m^3), U.S. Standard Atmosphere 1976 (moderate solar activity)
table = [(200, 2.5e-10), (250, 6.1e-11), (300, 1.9e-11), (350, 7.0e-12), (400, 2.8e-12), (450, 1.2e-12),
         (500, 5.2e-13), (600, 1.1e-13), (700, 3.1e-14), (800, 1.1e-14), (900, 5.8e-15), (1000, 3.6e-15)]

def rho(h):           # exponential (log-linear) interpolation between table rows
    for (h1, r1), (h2, r2) in zip(table, table[1:]):
        if h1 <= h <= h2:
            return exp(log(r1) + (log(r2) - log(r1)) * (h - h1) / (h2 - h1))

a = (R_E + h0) * 1000                    # m
da_rev = -2 * pi * C_D * Am * rho(h0) * a**2        # m per revolution
T = 2 * pi * sqrt(a**3 / mu)
revs_day = 86400 / T
H = (400 - 350) / log(rho(350) / rho(400))          # local scale height from the table row, km
print(f"height lost: {-da_rev:.2f} m per orbit, {revs_day:.2f} orbits/day, so {-da_rev * revs_day / 1000:.3f} km per day")
print(f"scale height near {h0:.0f} km: {H:.1f} km")

# lifetime: dt = dh / |da/dt| with da/dt = -sqrt(mu a) rho C_D A/m, integrated over altitude (Simpson)
def dtdh(h):                              # s per km of altitude
    a = (R_E + h) * 1000
    return 1000 / (sqrt(mu * a) * rho(h) * C_D * Am)

N = 2000
dh = (h0 - h_end) / N
t = dtdh(h_end) + dtdh(h0)
for k in range(1, N):
    t += (4 if k % 2 else 2) * dtdh(h_end + k * dh)
t *= dh / 3
print(f"time to fall from {h0:.0f} to {h_end:.0f} km: {t / 86400:.1f} days (about {t / 86400 / 30.4:.0f} months)")

# Worked example 6.2: How fast does drag pull a small satellite down?
from math import pi, sqrt, log, exp

mu = 398600.4418 * 1e9  # m^3/s^2, Earth's gravitational parameter in SI (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
C_D = 2.2             # drag coefficient
Am = 0.01             # m^2/kg, area-to-mass ratio
h0, h_end = 400.0, 200.0   # km, start and end altitudes
# density table (altitude km, density kg/m^3), U.S. Standard Atmosphere 1976 (moderate solar activity)
table = [(200, 2.5e-10), (250, 6.1e-11), (300, 1.9e-11), (350, 7.0e-12), (400, 2.8e-12), (450, 1.2e-12),
         (500, 5.2e-13), (600, 1.1e-13), (700, 3.1e-14), (800, 1.1e-14), (900, 5.8e-15), (1000, 3.6e-15)]

def rho(h):           # exponential (log-linear) interpolation between table rows
    for (h1, r1), (h2, r2) in zip(table, table[1:]):
        if h1 <= h <= h2:
            return exp(log(r1) + (log(r2) - log(r1)) * (h - h1) / (h2 - h1))

a = (R_E + h0) * 1000                    # m
da_rev = -2 * pi * C_D * Am * rho(h0) * a**2        # m per revolution
T = 2 * pi * sqrt(a**3 / mu)
revs_day = 86400 / T
H = (400 - 350) / log(rho(350) / rho(400))          # local scale height from the table row, km
print(f"height lost: {-da_rev:.2f} m per orbit, {revs_day:.2f} orbits/day, so {-da_rev * revs_day / 1000:.3f} km per day")
print(f"scale height near {h0:.0f} km: {H:.1f} km")

# lifetime: dt = dh / |da/dt| with da/dt = -sqrt(mu a) rho C_D A/m, integrated over altitude (Simpson)
def dtdh(h):                              # s per km of altitude
    a = (R_E + h) * 1000
    return 1000 / (sqrt(mu * a) * rho(h) * C_D * Am)

N = 2000
dh = (h0 - h_end) / N
t = dtdh(h_end) + dtdh(h0)
for k in range(1, N):
    t += (4 if k % 2 else 2) * dtdh(h_end + k * dh)
t *= dh / 3
print(f"time to fall from {h0:.0f} to {h_end:.0f} km: {t / 86400:.1f} days (about {t / 86400 / 30.4:.0f} months)")

height lost: 17.78 m per orbit, 15.56 orbits/day, so 0.277 km per day
scale height near 400 km: 54.6 km
time to fall from 400 to 200 km: 184.9 days (about 6 months)

Practice

Practice O6.1 · warm-up

Node drift of a mid-inclination orbit^[R1]

Given

A circular orbit at 600 km has inclination 45°. How many degrees per day does its ascending node move, and in which direction?

Mark as done

Hint

Use Ω̇ = −1.5 n J₂ (R_E/p)² cos i with p = a for a circle, then convert rad/s to °/day.

Answer. Ω̇ = -5.14 °/day: the node drifts west by about 5.1 deg every day.

Working

a = 6978.137 km, n1.083078e-03 rad/s
Ω̇ = −1.5 × 1.083078e-03 × 0.00108262668 × (6378.137/6978.137)² × cos 45°-1.0390e-06 rad/s
× (180/π) × 86,400-5.144 °/day

Answer. Ω̇ = -5.14 °/day: the node drifts west by about 5.1 deg every day.

Sanity check. Prograde orbit (i < 90°) gives a negative rate, as it must. 154° in a month is why constellation planes are spaced with J₂ in mind.

Pythonedit a value, then Run

# Practice O6.1: Node drift of a mid-inclination orbit
from math import sqrt, cos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
h = 600.0             # km
i = radians(45)

a = R_E + h
n = sqrt(mu / a**3)
raan_dot = -1.5 * n * J2 * (R_E / a) ** 2 * cos(i)
print(f"n = {n:.6e} rad/s, node rate = {raan_dot:.4e} rad/s")
print(f"= {degrees(raan_dot) * 86400:.3f} deg/day: the node drifts west by {-degrees(raan_dot) * 86400:.2f} deg per day")

n = 1.083078e-03 rad/s, node rate = -1.0390e-06 rad/s
= -5.144 deg/day: the node drifts west by 5.14 deg per day

Practice O6.2 · standard

Drag or sunlight: which pushes harder at 700 km?^{[R8, R14]}

Given

A small satellite has C_D = 2.2, C_R = 1.3 and A/m = 0.01 m²/kg. At 700 km the density is about 3.1e-14 kg/m³. Compare the drag acceleration with the solar-radiation-pressure acceleration (solar flux 1361 W/m²).

Assumption: Density at 700 km from the U.S. Standard Atmosphere 1976 table (moderate solar activity).

Mark as done

Hint

Radiation pressure is P = S/c in N/m². Drag is ½ρv²C_D(A/m); use the circular speed in m/s.

Answer. a_drag = 1.9e-08 m/s² and a_SRP = 5.9e-08 m/s²: at 700 km they are within a factor of 3.1 of each other.

Working

P = S/c = 1361 / 2997924584.540e-06 N/m²
a_SRP = P C_R (A/m) = 4.540e-06 × 1.3 × 0.015.90e-08 m/s²
v = √(μ/r)7.504 km/s
a_drag = ½ × 3.10e-14 × (7504)² × 2.2 × 0.011.92e-08 m/s²
Ratio drag / SRP0.33

Answer. a_drag = 1.9e-08 m/s² and a_SRP = 5.9e-08 m/s²: at 700 km they are within a factor of 3.1 of each other.

Sanity check. Both are below 10⁻⁷ m/s², about 6e-09 g: tiny, but they act for years. The density can change by tens of times with solar activity, so the winner at this height is not fixed.

Pythonedit a value, then Run

# Practice O6.2: Drag or sunlight: which pushes harder at 700 km?
from math import sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 700.0             # km
C_D, C_R = 2.2, 1.3   # drag and radiation-pressure coefficients
Am = 0.01             # m^2/kg
rho = 3.1e-14         # kg/m^3 at 700 km (density table)
S = 1361.0            # W/m^2, solar flux
c = 299792458.0       # m/s, speed of light (exact, SI)

P = S / c                                # radiation pressure, N/m^2
a_srp = P * C_R * Am
v = sqrt(mu / (R_E + h)) * 1000          # m/s
a_drag = 0.5 * rho * v**2 * C_D * Am
print(f"P = {P:.3e} N/m^2, a_SRP = {a_srp:.2e} m/s^2")
print(f"v = {v/1000:.3f} km/s, a_drag = {a_drag:.2e} m/s^2")
print(f"drag/SRP = {a_drag / a_srp:.2f}: within a factor of {a_srp / a_drag:.1f}")

P = 4.540e-06 N/m^2, a_SRP = 5.90e-08 m/s^2
v = 7.504 km/s, a_drag = 1.92e-08 m/s^2
drag/SRP = 0.33: within a factor of 3.1

Practice O6.3 · challenge

Reboosting a space station^{[R14, R21]}

Given

A large station (C_D = 2.2, A/m = 0.0032 m²/kg) flies a circular orbit at 410 km. Assume the density stays at 2.36e-12 kg/m³. How much height does it lose in 30 days, and what total Δv does a two-burn reboost need to restore it?

Assumption: Density held constant over the month (it really rises as the orbit drops and changes with solar activity).

Mark as done

Hint

Find Δa per revolution, multiply by the number of revolutions, then do a small Hohmann transfer back up.

Answer. About 2.2 km lost in 30 days; the reboost costs about 1.3 m/s.

Working

Δa per rev = −2π × 2.2 × 0.0032 × 2.36e-12 × (6.7881e+06 m)²-4.8 m
Revolutions in 30 days = 30 × 86,400 / 5565.9465.7
Height lost = 4.8 m × 465.72.24 km
Hohmann from 407.76 km back to 410 km: Δv₁ = 0.63 m/s, Δv₂0.63 m/s
Total = 1.27 m/s; small-change shortcut v Δa / (2a)1.27 m/s

Answer. About 2.2 km lost in 30 days; the reboost costs about 1.3 m/s.

Sanity check. A few metres per second per month is a modest budget, which is why stations reboost every few weeks rather than carrying huge tanks. The two methods agree to within 1 %. NASA's own ISS ephemeris pairs this area with C_D = 1.1, so C_D = 2.2 makes this an upper-end estimate.

Pythonedit a value, then Run

# Practice O6.3: Reboosting a space station
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 410.0             # km
C_D = 2.2
Am = 0.0032           # m^2/kg (ISS drag area / mass, NASA TOPO ephemeris header)
rho = 2.36e-12        # kg/m^3 (interpolated from the density table)
days = 30.0

a = R_E + h
da_rev = 2 * pi * C_D * Am * rho * (a * 1000) ** 2 / 1000     # km lost per revolution
T = 2 * pi * sqrt(a**3 / mu)
revs = days * 86400 / T
loss = da_rev * revs
# Hohmann from the lowered orbit back up
r1, r2 = a - loss, a
at = (r1 + r2) / 2
dv1 = sqrt(mu * (2 / r1 - 1 / at)) - sqrt(mu / r1)
dv2 = sqrt(mu / r2) - sqrt(mu * (2 / r2 - 1 / at))
print(f"da per rev = {da_rev*1000:.2f} m, {revs:.1f} revs, height lost = {loss:.2f} km")
print(f"reboost: {dv1*1000:.2f} + {dv2*1000:.2f} = {(dv1 + dv2)*1000:.2f} m/s")
print(f"shortcut v da / (2a) = {sqrt(mu / a) * loss / (2 * a) * 1000:.2f} m/s")

# Practice O6.3: Reboosting a space station
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 410.0             # km
C_D = 2.2
Am = 0.0032           # m^2/kg (ISS drag area / mass, NASA TOPO ephemeris header)
rho = 2.36e-12        # kg/m^3 (interpolated from the density table)
days = 30.0

a = R_E + h
da_rev = 2 * pi * C_D * Am * rho * (a * 1000) ** 2 / 1000     # km lost per revolution
T = 2 * pi * sqrt(a**3 / mu)
revs = days * 86400 / T
loss = da_rev * revs
# Hohmann from the lowered orbit back up
r1, r2 = a - loss, a
at = (r1 + r2) / 2
dv1 = sqrt(mu * (2 / r1 - 1 / at)) - sqrt(mu / r1)
dv2 = sqrt(mu / r2) - sqrt(mu * (2 / r2 - 1 / at))
print(f"da per rev = {da_rev*1000:.2f} m, {revs:.1f} revs, height lost = {loss:.2f} km")
print(f"reboost: {dv1*1000:.2f} + {dv2*1000:.2f} = {(dv1 + dv2)*1000:.2f} m/s")
print(f"shortcut v da / (2a) = {sqrt(mu / a) * loss / (2 * a) * 1000:.2f} m/s")

da per rev = 4.81 m, 465.7 revs, height lost = 2.24 km
reboost: 0.63 + 0.63 = 1.26 m/s
shortcut v da / (2a) = 1.26 m/s

Common mistakes

Using degrees inside n or mixing rad/s with °/day. Compute in rad/s, convert at the end.
Forgetting the sign: prograde orbits regress west (Ω̇ < 0), retrograde ones east.
Treating drag decay as linear in time. Density rises as the orbit drops, so the end comes fast.
Putting kilometres into the drag formula. Use metres and kg/m³ together.

Recap from earlier chapters

(Chapter 1) What is the period of a circular orbit at 400 km?

T = 2π√(a³/μ) = 5553.6 s = 92.56 min.

(Chapter 3) What is the specific energy of that orbit?

ε = −μ/(2a) = −29.403 km²/s². Drag makes it more negative, so a falls.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R2NASA JPL Solar System Dynamics, "Astrodynamic Parameters" (planetary GMs from DE440; Park et al. 2021, AJ 161:105).
R8Mamajek, E.E., Prša, A., Torres, G. et al., "IAU 2015 Resolution B3 on Recommended Nominal Conversion Constants for Selected Solar and Planetary Properties", arXiv:1510.07674 (2015).
R14NOAA, NASA and USAF, U.S. Standard Atmosphere, 1976, NASA-TM-X-74335 (NTRS 19770009539), Washington DC, Oct 1976.
R15Boring, J.W. and Humphris, R.R., "Drag coefficients for spheres in free molecular flow in O at satellite velocities", NASA CR-2233, 1973.
R21NASA JSC Flight Operations Directorate, TOPO office, ISS public ephemeris ISS.OEM_J2K_EPH.txt (CCSDS OEM, created 2026-11-02).

Where in each source, and links: References.

I have finished chapter 6

Chapter 7

Designer orbits: sun-synchronous, frozen and Molniya

35 minutes

The idea in one minute

Chapter 6 treated J₂ as a nuisance. Designers turn it into a tool. Choose the inclination so the node turns exactly once a year and the orbit keeps the same angle to the Sun all year: sun-synchronous. Choose a small eccentricity and put perigee over the pole, and J₂ and J₃ cancel so the ellipse stops wandering: a frozen orbit. Choose 63.4° and the perigee stays fixed, so a long, slow apogee can hang over the far north: Molniya.

Every sun-synchronous orbit is slightly retrograde, between 96.3° at 200 km and 102.0° at 1500 km: higher orbits feel less J₂, so they must tilt further from polar.

You will be able to

Find the sun-synchronous inclination for any altitude
Compute the frozen eccentricity that keeps altitude repeatable over each latitude
Design a Molniya orbit and explain why it flies at 63.4°

Sun-synchronous

The node must move east by 360° per year, about one degree per day. J₂ gives an eastward drift only for retrograde orbits, so the inclination ends up just past polar, between about 96° and 102° for low orbits. The reward is a fixed local solar time at every pass: the same lighting for every image.^{[R5, R22]}

Frozen

Odd zonal harmonics (mostly J₃) make the eccentricity and perigee oscillate. With perigee at 90° and one particular small eccentricity, the oscillation is zero, and the altitude over each latitude repeats orbit after orbit.

Molniya

A 12-hour orbit with a high apogee spends most of its time near apogee. Flown at the critical inclination, J₂ does not move the perigee, so the apogee stays over the northern hemisphere for years.

Key equations

Sun-synchronous condition

−(3/2) n J₂ (R_E/p)² cos i = 2π / 365.2422 d

where

nmean motion, √(μ/a³), rad/s

J₂Earth's oblateness coefficient, 1.0826 × 10⁻³, dimensionless

R_EEarth's equatorial radius, 6378.137, km

psemi-latus rectum, a(1 − e²), km

iinclination, deg

Frozen eccentricity

e_f = −(J₃ / 2J₂)(R_E/a) sin i, with ω = 90°

where

e_ffrozen eccentricity, dimensionless

J₃Earth's pear-shape coefficient, −2.532 × 10⁻⁶, dimensionless

J₂Earth's oblateness coefficient, 1.0826 × 10⁻³, dimensionless

R_EEarth's equatorial radius, 6378.137, km

asemi-major axis, km

iinclination, deg

ωargument of perigee, deg

Critical inclination

5 cos² i_c − 1 = 0, so i_c = 63.4°

where

i_ccritical inclination (also 116.6°), deg

Time in the apogee half

f_apo = ½ + e/π

where

f_apofraction of the period with eccentric anomaly between 90° and 270°, dimensionless

eeccentricity, dimensionless

Worked examples

Worked example 7.1

A sun-synchronous orbit at 600 km^[R5]

Given

Find the inclination that makes a circular 600 km orbit sun-synchronous, and the node rate that it gives.

new symbols

Ω̇_☉Sun's mean motion along the ecliptic, 2π per year, rad/s

Working

Sun-synchronous means the node turns once a year, eastward, to follow the Sun: Ω̇ = 2π / (365.2422 days)
Required Ω̇ = 1.9911e-07 rad/s0.9856 °/day
n = √(μ/a³) = 1.083078e-03 rad/s, a6978.137 km
cos i = −Ω̇ / (1.5 n J₂ (R_E/a)²)-0.13550
i97.788°

Answer. i = 97.79°; the node advances 0.9856 °/day, exactly the Sun's mean motion.

Sanity check. Slightly retrograde, as every sun-synchronous orbit must be: only a retrograde orbit has an eastward node drift.

Pythonedit a value, then Run

# Worked example 7.1: A sun-synchronous orbit at 600 km
from math import pi, sqrt, acos, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
h = 600.0             # km

a = R_E + h
node_rate = 2 * pi / year                # rad/s, one turn per year eastward
n = sqrt(mu / a**3)
cos_i = -node_rate / (1.5 * n * J2 * (R_E / a) ** 2)
print(f"required node rate = {node_rate:.4e} rad/s = {degrees(node_rate) * 86400:.4f} deg/day")
print(f"n = {n:.6e} rad/s, cos i = {cos_i:.5f}, i = {degrees(acos(cos_i)):.3f} deg")

required node rate = 1.9911e-07 rad/s = 0.9856 deg/day
n = 1.083078e-03 rad/s, cos i = -0.13550, i = 97.788 deg

Worked example 7.2

A frozen orbit

Given

For a sun-synchronous orbit with a = 7078.137 km (i = 98.19°), find the frozen eccentricity with the perigee held at ω = 90°.

new symbols

J₃Earth's pear-shape coefficient, −2.532 × 10⁻⁶, dimensionless

Working

J₂ and J₃ together try to change e and ω. They cancel when ω = 90° and e = −J₃ R_E sin i / (2 J₂ a)
e = −(-2.53215e-06) × 6378.137 × sin 98.19° / (2 × 0.00108262668 × 7078.137)0.00104
Perigee altitude a(1 − e) − R_E = 692.6 km; apogee altitude a(1 + e) − R_E707.4 km

Answer. e = 0.0010 with ω = 90°: the altitude over each latitude repeats every orbit, varying between 693 and 707 km.

Sanity check. The orbit is still nearly circular (a 15 km swing), but now the swing is fixed relative to latitude, which keeps image scale and drag repeatable.

Pythonedit a value, then Run

# Worked example 7.2: A frozen orbit
from math import pi, sqrt, sin, acos, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
J3 = -2.53215306e-6   # Earth's pear-shape coefficient (EGM2008 gives -2.5324e-6)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
a = 7078.137          # km

n = sqrt(mu / a**3)
i = acos(-(2 * pi / year) / (1.5 * n * J2 * (R_E / a) ** 2))   # sun-synchronous inclination
e = -J3 * R_E * sin(i) / (2 * J2 * a)                           # frozen eccentricity, argp = 90 deg
print(f"i = {degrees(i):.2f} deg, frozen e = {e:.5f}")
print(f"altitude varies from {a * (1 - e) - R_E:.1f} to {a * (1 + e) - R_E:.1f} km")

i = 98.19 deg, frozen e = 0.00104
altitude varies from 692.6 to 707.4 km

Practice

Practice O7.1 · warm-up

Sun-synchronous at 900 km^[R5]

Given

What inclination makes a circular orbit at 900 km sun-synchronous?

Mark as done

Hint

Set −1.5 n J₂ (R_E/a)² cos i equal to 2π per year and solve for cos i.

Answer. i = 99.03°.

Working

a = 7278.137 km, n1.016807e-03 rad/s
Ω̇ needed1.9911e-07 rad/s
cos i = −1.9911e-07 / (1.5 × 1.016807e-03 × 0.00108262668 × (6378.137/7278.137)²)-0.15701
i99.033°

Answer. i = 99.03°.

Sanity check. Higher than the 97.79° needed at 600 km, as the curve in the figure shows.

Pythonedit a value, then Run

# Practice O7.1: Sun-synchronous at 900 km
from math import pi, sqrt, acos, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
h = 900.0             # km

a = R_E + h
n = sqrt(mu / a**3)
node_rate = 2 * pi / year
cos_i = -node_rate / (1.5 * n * J2 * (R_E / a) ** 2)
print(f"n = {n:.6e} rad/s, cos i = {cos_i:.5f}")
print(f"i = {degrees(acos(cos_i)):.3f} deg")

n = 1.016807e-03 rad/s, cos i = -0.15701
i = 99.033 deg

Practice O7.2 · standard

When the inclination is a little off

Given

A satellite meant to be sun-synchronous at 700 km is injected 0.1° too high in inclination. How far does the local time of its ascending node drift in a year?

Mark as done

Hint

Compute Ω̇ at the wrong inclination and subtract the Sun's rate. One degree of node offset is 4 minutes of local time.

Answer. The node runs ahead of the Sun by 4.37° per year, a local-time drift of about 17 min per year.

Working

Correct i = 98.188°; actual i98.288°
Ω̇(actual) = 2.015212e-07 rad/s; Sun's rate1.991064e-07 rad/s
Difference × one year = 4.366° per year
Local time: 4.366° × 4 min/° = 17.5 min per year

Answer. The node runs ahead of the Sun by 4.37° per year, a local-time drift of about 17 min per year.

Sanity check. A higher inclination makes cos i more negative, so the eastward drift grows and the node runs ahead of the Sun. A tenth of a degree matters: injection accuracy is part of the mission design.

Pythonedit a value, then Run

# Practice O7.2: When the inclination is a little off
from math import pi, sqrt, cos, acos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
h = 700.0             # km
di = 0.1              # deg, injection error (too high)

a = R_E + h
n = sqrt(mu / a**3)
k = 1.5 * n * J2 * (R_E / a) ** 2
sun_rate = 2 * pi / year
i_ok = acos(-sun_rate / k)
i_act = i_ok + radians(di)
rate_act = -k * cos(i_act)
drift = degrees(rate_act - sun_rate) * year      # deg per year
print(f"correct i = {degrees(i_ok):.3f} deg, actual i = {degrees(i_act):.3f} deg")
print(f"node rate {rate_act:.6e} rad/s vs Sun {sun_rate:.6e} rad/s")
print(f"drift = {drift:.3f} deg/yr = {drift * 4:.1f} min of local time per year")

# Practice O7.2: When the inclination is a little off
from math import pi, sqrt, cos, acos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
h = 700.0             # km
di = 0.1              # deg, injection error (too high)

a = R_E + h
n = sqrt(mu / a**3)
k = 1.5 * n * J2 * (R_E / a) ** 2
sun_rate = 2 * pi / year
i_ok = acos(-sun_rate / k)
i_act = i_ok + radians(di)
rate_act = -k * cos(i_act)
drift = degrees(rate_act - sun_rate) * year      # deg per year
print(f"correct i = {degrees(i_ok):.3f} deg, actual i = {degrees(i_act):.3f} deg")
print(f"node rate {rate_act:.6e} rad/s vs Sun {sun_rate:.6e} rad/s")
print(f"drift = {drift:.3f} deg/yr = {drift * 4:.1f} min of local time per year")

correct i = 98.188 deg, actual i = 98.288 deg
node rate 2.015212e-07 rad/s vs Sun 1.991064e-07 rad/s
drift = 4.366 deg/yr = 17.5 min of local time per year

Practice O7.3 · challenge

Design a Molniya orbit

Given

A Molniya orbit repeats twice per sidereal day (86164.0905 s) and has a perigee altitude of 1000 km. Find its apogee altitude and eccentricity, the fraction of each orbit spent in the half nearest apogee (eccentric anomaly from 90° to 270°), and the perigee drift if it were flown at 60° instead of the critical inclination.

Mark as done

Hint

a comes from the period. For the dwell, use Kepler's equation M = E − e sin E at E = 90° and 270°.

Answer. Apogee altitude 39367 km, e = 0.722; 73.0 % of each orbit is spent in the apogee half; at 60° the perigee would drift 0.037 deg/day.

Working

a = (μ (T/2π)²)^(1/3) with T = 43082.05 s: a26561.8 km
r_a = 2a − r_p = 2 × 26561.8 − 7378.137 = 45745.4 km, apogee altitude 39367 km
e = (r_a − r_p)/(r_a + r_p)0.7222
M(90°) = π/2 − e, M(270°) = 3π/2 + e, so the fraction is (π + 2e)/(2π) = ½ + e/π0.7299
At i = 60°: ω̇ = 0.75 n J₂ (R_E/p)² (5 cos² i − 1)0.0369 °/day

Answer. Apogee altitude 39367 km, e = 0.722; 73.0 % of each orbit is spent in the apogee half; at 60° the perigee would drift 0.037 deg/day.

Sanity check. Over a year that drift is 13°, which would move the slow apogee away from the high northern latitudes it is meant to serve. That is why Molniya orbits fly at 63.4°.

Pythonedit a value, then Run

# Practice O7.3: Design a Molniya orbit
from math import pi, sqrt, cos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
hp = 1000.0           # km, perigee altitude
i = radians(60)       # flown inclination (not the critical 63.4 deg)

T = day_sid / 2
a = (mu * (T / (2 * pi)) ** 2) ** (1 / 3)
rp = R_E + hp
ra = 2 * a - rp
e = (ra - rp) / (ra + rp)
dwell = 0.5 + e / pi                     # fraction of time with E between 90 and 270 deg
n = sqrt(mu / a**3)
p = a * (1 - e**2)
argp_dot = 0.75 * n * J2 * (R_E / p) ** 2 * (5 * cos(i) ** 2 - 1)
print(f"a = {a:.1f} km, apogee radius {ra:.1f} km, apogee altitude {ra - R_E:.1f} km")
print(f"e = {e:.4f}, apogee-half dwell = {dwell:.4f} ({100*dwell:.1f} %)")
print(f"perigee drift at 60 deg = {degrees(argp_dot) * 86400:.4f} deg/day")

# Practice O7.3: Design a Molniya orbit
from math import pi, sqrt, cos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
hp = 1000.0           # km, perigee altitude
i = radians(60)       # flown inclination (not the critical 63.4 deg)

T = day_sid / 2
a = (mu * (T / (2 * pi)) ** 2) ** (1 / 3)
rp = R_E + hp
ra = 2 * a - rp
e = (ra - rp) / (ra + rp)
dwell = 0.5 + e / pi                     # fraction of time with E between 90 and 270 deg
n = sqrt(mu / a**3)
p = a * (1 - e**2)
argp_dot = 0.75 * n * J2 * (R_E / p) ** 2 * (5 * cos(i) ** 2 - 1)
print(f"a = {a:.1f} km, apogee radius {ra:.1f} km, apogee altitude {ra - R_E:.1f} km")
print(f"e = {e:.4f}, apogee-half dwell = {dwell:.4f} ({100*dwell:.1f} %)")
print(f"perigee drift at 60 deg = {degrees(argp_dot) * 86400:.4f} deg/day")

a = 26561.8 km, apogee radius 45745.4 km, apogee altitude 39367.3 km
e = 0.7222, apogee-half dwell = 0.7299 (73.0 %)
perigee drift at 60 deg = 0.0369 deg/day

Common mistakes

Expecting a prograde sun-synchronous orbit. The node must move east, which needs cos i < 0.
Using the solar day (86,400 s) for the Molniya period. It repeats on the sidereal day, so T = 43,082 s.
Forgetting that the frozen condition also fixes ω = 90°: the eccentricity alone is not enough.
Assuming sun-synchronous orbits stay put forever: drag lowers a, which changes the needed inclination.

Recap from earlier chapters

(Chapter 6) Which way does the node of a 45° orbit drift, and why?

West: Ω̇ = −1.5 n J₂ (R_E/p)² cos i is negative for cos i > 0.

(Chapter 2) A Molniya orbit has e = 0.722. How many times further out is apogee than perigee?

r_a/r_p = (1 + e)/(1 − e) = 6.20.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R5Williams, D.R., "Earth Fact Sheet", NASA NSSDCA, Goddard Space Flight Center, updated 15 Nov 2024.
R22ESA, "Types of orbits", 30 Mar 2020.

Where in each source, and links: References.

I have finished chapter 7

Chapter 8

Ground tracks

30 minutes

The idea in one minute

The ground track is the path of the point directly under the satellite. The orbit is (nearly) fixed in space while Earth turns beneath it, so each pass lands further west. Choose the period so that after a whole number of orbits Earth has turned a whole number of times, and the track repeats exactly: the basis of every mapping mission.

The orbit does not move, Earth turns under it: each pass crosses the equator 23.6° further west, and the track never goes beyond ±51.6° latitude.

You will be able to

Compute the westward shift between successive passes, with J₂
Explain the sidereal day and why it is 4 minutes shorter than the solar day
Find the altitude of a repeating ground track by root finding

Two kinds of day

Earth turns once relative to the stars in 86164.0905 s, the sidereal day. Because Earth also moves round the Sun, it must turn a little more to face the Sun again: the solar day of 86,400 s. Orbits care about the sidereal day; clocks follow the solar one.^[R2]

The westward shift

During one nodal period Earth turns east by ω_E T_N, while J₂ swings the node by Ω̇ T_N. The equator crossing moves west by their difference. The track reaches a highest latitude equal to the inclination (or 180° minus it for a retrograde orbit).

Repeating tracks

If k orbits take exactly as long as d turns of Earth relative to the orbit plane, the track closes after d days. Because J₂ changes both the nodal period and the node rate, the altitude that repeats is found numerically.

Key equations

Sidereal day

1/T_sid = 1/T_sol + 1/T_yr

where

T_sidsidereal day, 86,164.09, s

T_solmean solar day, 86,400, s

T_yrtropical year, 365.2422 days, s

Shift per orbit

Δλ = (ω_E − Ω̇) T_N

where

Δλwestward shift of the equator crossing, rad

ω_EEarth's rotation rate, 7.292115 × 10⁻⁵, rad/s

Ω̇rate of change of the right ascension of the ascending node, rad/s

T_Nnodal period, s

Nodal period

T_N = 2π / (n + Ṁ_J₂ + ω̇)

where

T_Nnodal period, s

nmean motion, √(μ/a³), rad/s

Ṁ_J₂J₂ change of the mean-anomaly rate, rad/s

ω̇rate of change of the argument of perigee, rad/s

Repeat condition

k T_N = d × 2π/(ω_E − Ω̇)

where

knumber of orbits in one repeat cycle, dimensionless

dnumber of days in one repeat cycle, dimensionless

T_Nnodal period, s

ω_EEarth's rotation rate, rad/s

Ω̇rate of change of the right ascension of the ascending node, rad/s

Worked examples

Worked example 8.1

How far west does each pass land?

Given

A station orbits at 420 km, inclination 51.6°. Find the nodal period (equator to equator) and how far west each new ascending pass crosses the equator.

new symbols

T_Nnodal period, equator crossing to equator crossing, s

ω_EEarth's rotation rate, 7.292115 × 10⁻⁵, rad/s

Working

Kepler period 2π√(a³/μ)5578.2 s
With J₂, the argument of latitude advances at n + Ṁ_J₂ + ω̇ = 1.127252e-03 rad/s, so T_N5573.9 s
Node rate Ω̇ = -1.0001e-06 rad/s; Earth turns at ω_E7.292115e-05 rad/s
Shift per orbit = (ω_E − Ω̇) T_N = (7.292115e-05 − (-1.0001e-06)) × 5573.9 = 0.41203 rad23.608°

Answer. T_N = 92.90 min; each pass crosses the equator 23.61° further west (2628 km at the equator).

Sanity check. Without J₂ the shift would be ω_E T = 23.31°. The node's own westward regression adds a little to Earth's eastward turn.

Pythonedit a value, then Run

# Worked example 8.1: How far west does each pass land?
from math import pi, sqrt, cos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
h = 420.0             # km
i = radians(51.6)

a = R_E + h                               # circular: e = 0, p = a
n = sqrt(mu / a**3)
k = n * J2 * (R_E / a) ** 2
raan_dot = -1.5 * k * cos(i)
argp_dot = 0.75 * k * (5 * cos(i) ** 2 - 1)
M_dot = 0.75 * k * (3 * cos(i) ** 2 - 1)  # J2 change of the mean-anomaly rate
T_N = 2 * pi / (n + M_dot + argp_dot)     # nodal period, equator to equator
shift = (w_E - raan_dot) * T_N            # rad west per orbit
print(f"Kepler period {2*pi/n:.1f} s; nodal period {T_N:.1f} s = {T_N/60:.2f} min")
print(f"node rate {raan_dot:.4e} rad/s")
print(f"shift per orbit = {shift:.5f} rad = {degrees(shift):.3f} deg = {shift * R_E:.0f} km at the equator")

# Worked example 8.1: How far west does each pass land?
from math import pi, sqrt, cos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
h = 420.0             # km
i = radians(51.6)

a = R_E + h                               # circular: e = 0, p = a
n = sqrt(mu / a**3)
k = n * J2 * (R_E / a) ** 2
raan_dot = -1.5 * k * cos(i)
argp_dot = 0.75 * k * (5 * cos(i) ** 2 - 1)
M_dot = 0.75 * k * (3 * cos(i) ** 2 - 1)  # J2 change of the mean-anomaly rate
T_N = 2 * pi / (n + M_dot + argp_dot)     # nodal period, equator to equator
shift = (w_E - raan_dot) * T_N            # rad west per orbit
print(f"Kepler period {2*pi/n:.1f} s; nodal period {T_N:.1f} s = {T_N/60:.2f} min")
print(f"node rate {raan_dot:.4e} rad/s")
print(f"shift per orbit = {shift:.5f} rad = {degrees(shift):.3f} deg = {shift * R_E:.0f} km at the equator")

Kepler period 5578.2 s; nodal period 5573.9 s = 92.90 min
node rate -1.0001e-06 rad/s
shift per orbit = 0.41203 rad = 23.608 deg = 2628 km at the equator

Practice

Practice O8.1 · warm-up

Spacing between passes

Given

For a circular orbit at 700 km and 98° inclination, how far apart in longitude are successive equator crossings? Compare with the answer that ignores J₂.

Mark as done

Hint

Without J₂ the shift is just Earth's rotation during one period. With J₂ use (ω_E − Ω̇) T_N.

Answer. 24.72° with J₂ against 24.76° without: the passes are about 24.7° apart.

Working

Kepler period T = 5926.4 s; ω_E T24.761°
With J₂: T_N = 5933.6 s, Ω̇1.9457e-07 rad/s(eastward, retrograde orbit)
(ω_E − Ω̇) T_N24.725°

Answer. 24.72° with J₂ against 24.76° without: the passes are about 24.7° apart.

Sanity check. For a retrograde orbit the node follows Earth's spin, so the shift is slightly smaller than Earth's rotation alone.

Pythonedit a value, then Run

# Practice O8.1: Spacing between passes
from math import pi, sqrt, cos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
h = 700.0             # km
i = radians(98)

a = R_E + h
n = sqrt(mu / a**3)
k = n * J2 * (R_E / a) ** 2
raan_dot = -1.5 * k * cos(i)
argp_dot = 0.75 * k * (5 * cos(i) ** 2 - 1)
M_dot = 0.75 * k * (3 * cos(i) ** 2 - 1)
T = 2 * pi / n
T_N = 2 * pi / (n + M_dot + argp_dot)
print(f"without J2: T = {T:.1f} s, spacing {degrees(w_E * T):.3f} deg")
print(f"with J2: T_N = {T_N:.1f} s, node rate {raan_dot:.4e} rad/s, spacing {degrees((w_E - raan_dot) * T_N):.3f} deg")

without J2: T = 5926.4 s, spacing 24.761 deg
with J2: T_N = 5933.6 s, node rate 1.9457e-07 rad/s, spacing 24.725 deg

Practice O8.2 · standard

The figure-8 of an inclined geosynchronous orbit

Given

A geosynchronous satellite has drifted to 3° inclination (circular, period one sidereal day). Its ground track is a small figure-8. Find its north-south and east-west half-sizes.

Mark as done

Hint

Latitude swings by ±i. In longitude, along-track motion is slightly faster than Earth at the nodes and slower near the extremes of latitude; for small i the half-width is i²/4 (radians).

Answer. ±3° in latitude but only ±0.039° in longitude: a tall, thin figure-8.

Working

North-south: the track reaches latitude ±i = ±3°
East-west, small-i result: half-width = i²/4 = (0.05236)²/4 = 6.854e-04 rad0.0393°
Numerical ground track over one sidereal day: longitude range ±0.0393°

Answer. ±3° in latitude but only ±0.039° in longitude: a tall, thin figure-8.

Sanity check. 0.039° at GEO radius is 29 km: a fixed ground antenna with a wide beam still sees it; a narrow beam must track north-south.

Pythonedit a value, then Run

# Practice O8.2: The figure-8 of an inclined geosynchronous orbit
from math import pi, sin, cos, atan2, asin, radians, degrees

i = radians(3.0)      # inclination of the geosynchronous orbit
N = 20000             # samples over one sidereal day

half_ew = i**2 / 4                       # small-inclination result, rad
lat_max = lon_max = 0.0
for k in range(N):
    u = 2 * pi * k / N                   # argument of latitude = Earth rotation angle (same rate)
    x, y, z = cos(u), cos(i) * sin(u), sin(i) * sin(u)
    lat = asin(z)
    lon = (atan2(y, x) - u + pi) % (2 * pi) - pi
    lat_max, lon_max = max(lat_max, abs(lat)), max(lon_max, abs(lon))
print(f"north-south half-size: {degrees(lat_max):.3f} deg")
print(f"east-west half-size: i^2/4 = {half_ew:.3e} rad = {degrees(half_ew):.4f} deg")
print(f"numerical ground track: longitude range +/- {degrees(lon_max):.4f} deg")

north-south half-size: 3.000 deg
east-west half-size: i^2/4 = 6.854e-04 rad = 0.0393 deg
numerical ground track: longitude range +/- 0.0393 deg

Practice O8.3 · challenge

A sun-synchronous orbit that repeats every two days

Given

Find the altitude of a circular sun-synchronous orbit that makes exactly 29 revolutions in 2 days and then retraces its ground track. Include J₂ in both the nodal period and the node rate.

Mark as done

Hint

Repeat condition: k T_N = d × 2π/(ω_E − Ω̇). For a sun-synchronous orbit ω_E − Ω̇ is exactly 2π per solar day. Solve numerically: a and i depend on each other.

Answer. h = 720.0 km at i = 98.27°: 29 orbits in 2 days, then the track repeats.

Working

For SSO, ω_E − Ω̇ = 2π/86,400 s, so the condition is T_N = 2 × 86,400 / 295958.62 s
For a trial altitude h: i = SSO inclination(h), then T_N(h, i) from n + Ṁ_J₂ + ω̇
Root of 29 T_N(h) − 2 × 86,400 = 0 by Brent's method: h719.95 km
Check: i = 98.270°, T_N = 5958.621 s, 29 × T_N = 172800.0 s2 days

Answer. h = 720.0 km at i = 98.27°: 29 orbits in 2 days, then the track repeats.

Sanity check. 14.5 revolutions per day sits between the 14 and 15 rev/day families, as expected for an orbit near 700 km.

Pythonedit a value, then Run

# Practice O8.3: A sun-synchronous orbit that repeats every two days
from math import pi, sqrt, cos, acos, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
revs, days = 29, 2    # 29 orbits in 2 days

def orbit(h):         # SSO inclination, nodal period and node rate at altitude h
    a = R_E + h
    n = sqrt(mu / a**3)
    k = n * J2 * (R_E / a) ** 2
    i = acos(-(2 * pi / year) / (1.5 * k))
    raan_dot = -1.5 * k * cos(i)
    T_N = 2 * pi / (n + 0.75 * k * (5 * cos(i)**2 - 1) + 0.75 * k * (3 * cos(i)**2 - 1))
    return i, T_N, raan_dot

def mismatch(h):      # revs nodal periods minus days turns of Earth relative to the orbit plane
    i, T_N, raan_dot = orbit(h)
    return revs * T_N - days * 2 * pi / (w_E - raan_dot)

lo, hi = 500.0, 900.0  # bisection
for _ in range(60):
    mid = (lo + hi) / 2
    if mismatch(lo) * mismatch(mid) <= 0:
        hi = mid
    else:
        lo = mid
h = (lo + hi) / 2
i, T_N, _ = orbit(h)
print(f"h = {h:.2f} km, i = {degrees(i):.3f} deg, T_N = {T_N:.3f} s")
print(f"check: {revs} x T_N = {revs * T_N:.1f} s = {revs * T_N / 86400:.4f} days")

# Practice O8.3: A sun-synchronous orbit that repeats every two days
from math import pi, sqrt, cos, acos, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
revs, days = 29, 2    # 29 orbits in 2 days

def orbit(h):         # SSO inclination, nodal period and node rate at altitude h
    a = R_E + h
    n = sqrt(mu / a**3)
    k = n * J2 * (R_E / a) ** 2
    i = acos(-(2 * pi / year) / (1.5 * k))
    raan_dot = -1.5 * k * cos(i)
    T_N = 2 * pi / (n + 0.75 * k * (5 * cos(i)**2 - 1) + 0.75 * k * (3 * cos(i)**2 - 1))
    return i, T_N, raan_dot

def mismatch(h):      # revs nodal periods minus days turns of Earth relative to the orbit plane
    i, T_N, raan_dot = orbit(h)
    return revs * T_N - days * 2 * pi / (w_E - raan_dot)

lo, hi = 500.0, 900.0  # bisection
for _ in range(60):
    mid = (lo + hi) / 2
    if mismatch(lo) * mismatch(mid) <= 0:
        hi = mid
    else:
        lo = mid
h = (lo + hi) / 2
i, T_N, _ = orbit(h)
print(f"h = {h:.2f} km, i = {degrees(i):.3f} deg, T_N = {T_N:.3f} s")
print(f"check: {revs} x T_N = {revs * T_N:.1f} s = {revs * T_N / 86400:.4f} days")

h = 719.95 km, i = 98.270 deg, T_N = 5958.621 s
check: 29 x T_N = 172800.0 s = 2.0000 days

Common mistakes

Using 86,400 s for Earth's rotation. Earth turns 360° in a sidereal day.
Forgetting the node rate: for a near-polar orbit it changes the shift only a little, but it decides whether a track repeats.
Expecting a track above the inclination: the maximum latitude equals i (or 180° − i).

Recap from earlier chapters

(Chapter 7) Why does a sun-synchronous orbit's repeat condition use the solar day?

Its node moves with the Sun, so Earth's turn relative to the orbit plane takes exactly one solar day.

(Chapter 1) What is the ISS-like orbit's Kepler period at 420 km?

T = 2π√(a³/μ) = 5578.2 s = 92.97 min.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R2NASA JPL Solar System Dynamics, "Astrodynamic Parameters" (planetary GMs from DE440; Park et al. 2021, AJ 161:105).

Where in each source, and links: References.

I have finished chapter 8

Chapter 9

Coplanar manoeuvres: Hohmann, bi-elliptic and phasing

40 minutes

The idea in one minute

Every in-plane manoeuvre is a change of speed at a point, which changes the energy and so the semi-major axis. The Hohmann transfer uses two tangential burns at opposite ends of an ellipse and is the cheapest two-burn transfer between circles. For very large changes, going out much further first and coming back is cheaper still. And to move along your own orbit, you change your lap time: slow down to catch up.

Below a radius ratio of 11.94 the Hohmann transfer is cheaper than any bi-elliptic; above it, going out very far first wins (on Δv, never on time).

You will be able to

Size a Hohmann transfer and the propellant it needs
Say when a bi-elliptic transfer beats Hohmann, and why 11.94 is the number
Design a phasing manoeuvre to move along the same orbit

Hohmann

Burn prograde to stretch the orbit into an ellipse that just touches the target circle, coast half an orbit, burn prograde again to round it off. Both burns are along the velocity, which wastes nothing.^[R1]

Bi-elliptic

Three burns: out to a far apoapsis, raise the periapsis there where the speed is tiny, then come in and circularise. When the radius ratio is above 11.94 this can beat Hohmann; above 15.58 every bi-elliptic does. It always takes much longer.

Phasing

To meet a point ahead of you in the same orbit, enter a slightly smaller ellipse with a shorter period, lap once and return. To fall behind, use a larger one. The rocket equation converts every Δv into propellant mass.

One-tangent burn

A Hohmann transfer is the cheapest two-burn route but the slowest. Aim the transfer ellipse past the target and the second burn happens where the ellipse cuts the target orbit at an angle. From a 300 km circle to GEO with the transfer apogee at 1.5 GEO radii: 5.280 km/s in 3.01 h, against Hohmann's 3.893 km/s in 5.28 h. Time costs Δv.^[R1]

Staging and gravity loss

The rocket equation applies stage by stage and the Δv values add. Dropping an empty stage stops you accelerating dead mass. With exhaust speed 3.0 km/s, 100 t at lift-off, 2 t of payload and structure 10 % of the stage mass, one stage gives 6.41 km/s; the same masses split into two equal stages give 7.75 km/s. Burning against gravity costs extra: for a vertical burn the ideal Δv is reduced by g₀ t_b, 1.18 km/s for a 120 s burn, which is why rockets pitch over early.^[R29]

Key equations

Vis-viva

v = √(μ(2/r − 1/a))

where

vorbital speed, km/s

μEarth's gravitational parameter, 398,600.4418, km³/s²

rorbit radius, km

asemi-major axis, km

Hohmann burns

Δv₁ = √(μ/r₁)(√(2r₂/(r₁+r₂)) − 1), Δv₂ = √(μ/r₂)(1 − √(2r₁/(r₁+r₂)))

where

Δv₁, Δv₂first and second burn, km/s

r₁initial circular radius, km

r₂final circular radius, km

μEarth's gravitational parameter, 398,600.4418, km³/s²

Rocket equation

Δv = I_sp g₀ ln(m₀/m_f)

where

Δvvelocity change, m/s

I_spspecific impulse, s

g₀standard gravity, 9.80665, m/s²

m₀mass before the burn, kg

m_fmass after the burn, kg

Phasing period

T_ph = T(1 − Δθ/(360° k))

where

T_phphasing-orbit period, s

Tperiod of the original circular orbit, s

Δθangle to gain (positive ahead, negative behind), deg

knumber of phasing revolutions, dimensionless

Worked examples

Worked example 9.1

Raising a small satellite with a Hohmann transfer^{[R1, R9]}

Given

A 150 kg satellite moves from a circular 400 km orbit to a circular 1200 km orbit with two burns of a hydrazine thruster (I_sp = 220 s). Find Δv, time of flight and propellant mass.

new symbols

I_spspecific impulse of the thruster, s

g₀standard gravity, 9.80665, m/s²

m₀, m_fmass before and after the burns, kg

Working

r₁ = 6778.137 km, r₂ = 7578.137 km, a_t = (r₁ + r₂)/27178.137 km
Burn 1: √(μ(2/r₁ − 1/a_t)) − √(μ/r₁) = 7.8793 − 7.6686210.8 m/s
Burn 2: √(μ/r₂) − √(μ(2/r₂ − 1/a_t)) = 7.2525 − 7.0475205.0 m/s
Total Δv = 415.7 m/s; time of flight = half the transfer period50.4 min
Rocket equation: m₀/m_f = exp(Δv/(I_sp g₀)) = exp(415.7/(220 × 9.80665))1.2125
Propellant = m₀(1 − m_f/m₀) = 150 × (1 − 1/1.2125)26.3 kg

Answer. Δv = 416 m/s in 50 min, using 26.3 kg of hydrazine.

Sanity check. That is 18 % of the satellite's mass for an 800 km climb: small satellites rarely have this much, which is why they are launched close to their final orbit.

Pythonedit a value, then Run

# Worked example 9.1: Raising a small satellite with a Hohmann transfer
from math import pi, sqrt, exp

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
g0 = 9.80665          # m/s^2, standard gravity (exact, by definition)
h1, h2 = 400.0, 1200.0    # km, start and target altitudes
m0 = 150.0            # kg
Isp = 220.0           # s, hydrazine

r1, r2 = R_E + h1, R_E + h2
at = (r1 + r2) / 2
dv1 = sqrt(mu * (2 / r1 - 1 / at)) - sqrt(mu / r1)
dv2 = sqrt(mu / r2) - sqrt(mu * (2 / r2 - 1 / at))
dv = (dv1 + dv2) * 1000                  # m/s
tof = pi * sqrt(at**3 / mu)              # half the transfer period
MR = exp(dv / (Isp * g0))
mp = m0 * (1 - 1 / MR)
print(f"burn 1 = {dv1*1000:.1f} m/s, burn 2 = {dv2*1000:.1f} m/s, total {dv:.1f} m/s")
print(f"time of flight = {tof/60:.1f} min")
print(f"mass ratio {MR:.4f}, propellant {mp:.2f} kg")

burn 1 = 210.8 m/s, burn 2 = 205.0 m/s, total 415.7 m/s
time of flight = 50.4 min
mass ratio 1.2125, propellant 26.29 kg

Worked example 9.2

Slow down to catch up: a phasing manoeuvre^[R16]

Given

A satellite in a circular 800 km orbit must move 20° ahead of its current slot in one revolution of a phasing orbit. Find the phasing orbit and the Δv.

new symbols

T_phperiod of the phasing orbit, s

Δθangle to gain (positive = ahead), deg

Working

Circular period T = 6052.4 s. To arrive 20° early, the phasing orbit must be shorter: T_ph = T(1 − 20/360)5716.2 s
a_ph = (μ(T_ph/2π)²)^(1/3) = 6909.756 km; the burn point becomes apogee, so perigee r_p = 2a_ph − r6641.374 km(altitude 263 km)
Burn 1 (retrograde): Δv = √(μ/r) − √(μ(2/r − 1/a_ph)) = 7.4518 − 7.3057146.2 m/s
After one phasing orbit, burn 2 (prograde) restores the circle: another 146.2 m/s
Total 292.3 m/s

Answer. Drop perigee to 263 km with a 146 m/s braking burn, wait one orbit, then speed up again: 292 m/s in total.

Sanity check. Braking to get ahead feels wrong, but a lower orbit is a shorter, faster lap. Spreading the move over several phasing orbits cuts the Δv almost in proportion.

Pythonedit a value, then Run

# Worked example 9.2: Slow down to catch up: a phasing manoeuvre
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 800.0             # km
lead = 20.0           # deg to move ahead in one phasing revolution

r = R_E + h
T = 2 * pi * sqrt(r**3 / mu)
T_ph = T * (1 - lead / 360)              # shorter orbit arrives early
a_ph = (mu * (T_ph / (2 * pi)) ** 2) ** (1 / 3)
rp = 2 * a_ph - r                        # burn point becomes apogee
dv1 = sqrt(mu / r) - sqrt(mu * (2 / r - 1 / a_ph))
print(f"T = {T:.1f} s, T_ph = {T_ph:.1f} s, a_ph = {a_ph:.3f} km")
print(f"phasing perigee altitude {rp - R_E:.1f} km")
print(f"braking burn {dv1*1000:.1f} m/s, same again to restore: total {2*dv1*1000:.1f} m/s")

T = 6052.4 s, T_ph = 5716.2 s, a_ph = 6909.756 km
phasing perigee altitude 263.2 km
braking burn 146.2 m/s, same again to restore: total 292.3 m/s

Practice

Practice O9.1 · warm-up

A short Hohmann climb^[R1]

Given

Find the two burns and the total Δv to go from a circular 300 km orbit to a circular 900 km orbit.

Mark as done

Hint

Transfer semi-major axis is the mean of the two radii; use vis-viva at each end.

Answer. Δv₁ = 164 m/s, Δv₂ = 161 m/s, total 325 m/s.

Working

a_t = (6678.137 + 7278.137)/26978.137 km
Δv₁ = 7.8901 − 7.7258164.3 m/s
Δv₂ = 7.4005 − 7.2396160.8 m/s
Total325.1 m/s

Answer. Δv₁ = 164 m/s, Δv₂ = 161 m/s, total 325 m/s.

Sanity check. The first burn is slightly bigger, as always for a Hohmann climb. Sanity: the difference of circular speeds is 325 m/s, close to the total for a small change.

Pythonedit a value, then Run

# Practice O9.1: A short Hohmann climb
from math import sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h1, h2 = 300.0, 900.0     # km

r1, r2 = R_E + h1, R_E + h2
at = (r1 + r2) / 2
dv1 = sqrt(mu * (2 / r1 - 1 / at)) - sqrt(mu / r1)
dv2 = sqrt(mu / r2) - sqrt(mu * (2 / r2 - 1 / at))
print(f"a_t = {at:.3f} km")
print(f"dv1 = {dv1*1000:.1f} m/s, dv2 = {dv2*1000:.1f} m/s, total {(dv1 + dv2)*1000:.1f} m/s")

a_t = 6978.137 km
dv1 = 164.3 m/s, dv2 = 160.8 m/s, total 325.1 m/s

Practice O9.2 · standard

Falling back in a navigation-satellite orbit^[R1]

Given

A satellite in a circular 20200 km orbit must drop 30° behind its slot, spread over 3 revolutions of a phasing orbit. Find the phasing orbit's semi-major axis and the total Δv.

Mark as done

Hint

To fall behind you need a longer lap: burn prograde. Each phasing revolution must be longer by (Δθ/360°)/k of a period.

Answer. a_ph = 27068 km and a total Δv of 69.8 m/s over 3 revolutions.

Working

T = 43121.9 s; T_ph = T(1 + 30/(360 × 3))44319.7 s
a_ph = 27068.07 km; the burn point is perigee and apogee radius27558.0 km
Burn 1 = √(μ(2/r − 1/a_ph)) − √(μ/r) = 34.89 m/s; burn 2 is the same, retrograde
Total69.78 m/s

Answer. a_ph = 27068 km and a total Δv of 69.8 m/s over 3 revolutions.

Sanity check. The total time is 36.9 h. Doing it in one revolution would cost about 3 times more.

Pythonedit a value, then Run

# Practice O9.2: Falling back in a navigation-satellite orbit
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 20200.0           # km
lag = 30.0            # deg to fall behind
revs = 3              # phasing revolutions

r = R_E + h
T = 2 * pi * sqrt(r**3 / mu)
T_ph = T * (1 + lag / (360 * revs))      # longer orbit arrives late
a_ph = (mu * (T_ph / (2 * pi)) ** 2) ** (1 / 3)
ra = 2 * a_ph - r                        # burn point becomes perigee
dv1 = sqrt(mu * (2 / r - 1 / a_ph)) - sqrt(mu / r)
print(f"T = {T:.1f} s, T_ph = {T_ph:.1f} s")
print(f"a_ph = {a_ph:.2f} km, apogee radius {ra:.1f} km")
print(f"each burn {dv1*1000:.2f} m/s, total {2*dv1*1000:.2f} m/s")

T = 43121.9 s, T_ph = 44319.7 s
a_ph = 27068.07 km, apogee radius 27558.0 km
each burn 34.89 m/s, total 69.78 m/s

Practice O9.3 · challenge

When going further out is cheaper^[R1]

Given

A spacecraft goes from a circular orbit of radius 6978.1 km to one 15 times larger (104672 km). Compare a Hohmann transfer with a bi-elliptic transfer whose intermediate apoapsis is 314016 km (three times the target). Give Δv and time of flight for both.

Mark as done

Hint

Bi-elliptic: burn to an ellipse with apoapsis r_b, at r_b raise periapsis to r₂, at r₂ circularise. Times of flight are half-periods of the two ellipses.

Answer. Bi-elliptic 4014 m/s against Hohmann 4053 m/s, a saving of 39 m/s, but 221 h instead of 18 h.

Working

Hohmann: Δv = 2791.2 + 1261.5 = 4052.7 m/s; t18.23 h
Bi-elliptic burn 1 (to r_b): 3013.8 m/s
Burn 2 at r_b (periapsis up to r₂): 561.7 m/s
Burn 3 at r₂ (circularise, braking): 438.6 m/s
Bi-elliptic total 4014.1 m/s; t221.3 h

Answer. Bi-elliptic 4014 m/s against Hohmann 4053 m/s, a saving of 39 m/s, but 221 h instead of 18 h.

Sanity check. R = 15 is above the 11.94 crossover, so a bi-elliptic can win; the saving is small (1.0 %) and the trip is 12 times longer.

Pythonedit a value, then Run

# Practice O9.3: When going further out is cheaper
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r1 = 6378.137 + 600   # km, initial circular orbit
r2 = 15 * r1          # km, target circular orbit
rb = 3 * r2           # km, bi-elliptic intermediate apoapsis

def vv(r, a):         # vis-viva
    return sqrt(mu * (2 / r - 1 / a))

def half_period(a):
    return pi * sqrt(a**3 / mu)

at = (r1 + r2) / 2                       # Hohmann
dv_h = (vv(r1, at) - sqrt(mu / r1)) + (sqrt(mu / r2) - vv(r2, at))
a1, a2 = (r1 + rb) / 2, (r2 + rb) / 2    # bi-elliptic: two half ellipses
b1 = vv(r1, a1) - sqrt(mu / r1)
b2 = vv(rb, a2) - vv(rb, a1)
b3 = vv(r2, a2) - sqrt(mu / r2)
dv_b = b1 + b2 + b3
print(f"Hohmann: {dv_h*1000:.1f} m/s, {half_period(at)/3600:.2f} h")
print(f"bi-elliptic: {b1*1000:.1f} + {b2*1000:.1f} + {b3*1000:.1f} = {dv_b*1000:.1f} m/s, "
      f"{(half_period(a1) + half_period(a2))/3600:.1f} h")
print(f"saving {(dv_h - dv_b)*1000:.1f} m/s")

# Practice O9.3: When going further out is cheaper
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r1 = 6378.137 + 600   # km, initial circular orbit
r2 = 15 * r1          # km, target circular orbit
rb = 3 * r2           # km, bi-elliptic intermediate apoapsis

def vv(r, a):         # vis-viva
    return sqrt(mu * (2 / r - 1 / a))

def half_period(a):
    return pi * sqrt(a**3 / mu)

at = (r1 + r2) / 2                       # Hohmann
dv_h = (vv(r1, at) - sqrt(mu / r1)) + (sqrt(mu / r2) - vv(r2, at))
a1, a2 = (r1 + rb) / 2, (r2 + rb) / 2    # bi-elliptic: two half ellipses
b1 = vv(r1, a1) - sqrt(mu / r1)
b2 = vv(rb, a2) - vv(rb, a1)
b3 = vv(r2, a2) - sqrt(mu / r2)
dv_b = b1 + b2 + b3
print(f"Hohmann: {dv_h*1000:.1f} m/s, {half_period(at)/3600:.2f} h")
print(f"bi-elliptic: {b1*1000:.1f} + {b2*1000:.1f} + {b3*1000:.1f} = {dv_b*1000:.1f} m/s, "
      f"{(half_period(a1) + half_period(a2))/3600:.1f} h")
print(f"saving {(dv_h - dv_b)*1000:.1f} m/s")

Hohmann: 4052.7 m/s, 18.23 h
bi-elliptic: 3013.8 + 561.7 + 438.6 = 4014.1 m/s, 221.3 h
saving 38.6 m/s

Common mistakes

Subtracting speeds at different radii. Each Δv compares two speeds at the same point.
Speeding up to catch a target ahead. That raises the orbit and lengthens the lap: you fall further behind.
Using kilometres per second with I_sp g₀ in metres per second inside the rocket equation.
Choosing a bi-elliptic for a mission with a deadline: the Δv saving costs days of flight.

Recap from earlier chapters

(Chapter 3) What is the speed at the perigee of the 400-to-1200 km transfer ellipse?

Vis-viva: v = √(μ(2/r₁ − 1/a_t)) = 7.8793 km/s.

(Chapter 4) How long does the Hohmann coast take?

Half the transfer period: 50.4 min.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R9NIST, CODATA 2022 recommended values, "standard acceleration of gravity".
R16NASA Small Spacecraft Systems Virtual Institute, State-of-the-Art of Small Spacecraft Technology, Ch. 4 "In-Space Propulsion", 2026 edition.
R29NASA Glenn Research Center, "Ideal Rocket Equation", Beginner's Guide to Aeronautics, updated 21 Nov 2023.

Where in each source, and links: References.

I have finished chapter 9

Chapter 10

Plane changes and combined manoeuvres

35 minutes

The idea in one minute

Turning an orbit plane means turning the velocity vector, and that costs in proportion to the speed you turn. At low-orbit speed a few degrees is very expensive; far out, where you move slowly, it is cheap. Better still, fold the turn into a burn you already need: velocity changes add as vectors, so the extra cost of a small turn inside a large burn is tiny.

The same turn costs 2.5 times less at GEO speed: turn the plane where you move slowly, ideally inside a burn you need anyway.

You will be able to

Compute the cost of a pure plane change and say where to do it
Combine a speed change and a plane change in one burn
Optimise how a plane change is split between two burns

A pure turn

To rotate the plane by Δi without changing speed, the burn must cancel part of the old velocity and add part of the new: Δv = 2v sin(Δi/2). It is done at a node, where the old and new planes intersect.

Turn where you are slow

The cost scales with v, so turns belong at apogee or far from the planet. For a geostationary mission the turn goes with the apogee burn into GEO.

One burn, two jobs

When the speed also changes, one burn along the vector difference is cheapest: the law of cosines. Moving a few degrees of the turn into the perigee burn reduces the total a little more.

Key equations

Pure plane change

Δv = 2v sin(Δi/2)

where

Δvburn size, km/s

vorbital speed, km/s

Δiplane-change angle, deg

Combined burn

Δv = √(v₁² + v₂² − 2v₁v₂ cos Δi)

where

Δvburn size, km/s

v₁speed before the burn, km/s

v₂speed after the burn, km/s

Δiangle turned in this burn, deg

Small-angle form

Δv ≈ v Δi (Δi in radians)

where

Δvburn size, km/s

vorbital speed, km/s

Δiplane-change angle (radians), rad

Worked examples

Worked example 10.1

The same turn at two speeds

Given

Find the Δv to turn an orbit plane by 10° in a circular 500 km orbit and in GEO.

new symbols

Δiangle between the old and new orbit planes, deg

Working

Pure plane change: Δv = 2v sin(Δi/2)
LEO: v = 7.6126 km/s, Δv = 2 × 7.6126 × sin 5.0°1.3270 km/s
GEO: v = 3.0747 km/s, Δv = 2 × 3.0747 × sin 5.0°0.5359 km/s

Answer. 1327 m/s in LEO, 536 m/s in GEO.

Sanity check. The ratio 2.48 is just the ratio of speeds. A 10° turn in LEO costs more than a Hohmann climb from 500 km to 1300 km (407 m/s).

Pythonedit a value, then Run

# Worked example 10.1: The same turn at two speeds
from math import pi, sqrt, sin, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
h_leo = 500.0         # km
di = radians(10)      # plane change

r_geo = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
for name, r in (("LEO", R_E + h_leo), ("GEO", r_geo)):
    v = sqrt(mu / r)
    dv = 2 * v * sin(di / 2)
    print(f"{name}: v = {v:.4f} km/s, dv = {dv:.4f} km/s = {dv*1000:.1f} m/s")

LEO: v = 7.6126 km/s, dv = 1.3270 km/s = 1327.0 m/s
GEO: v = 3.0747 km/s, dv = 0.5359 km/s = 535.9 m/s

Worked example 10.2

One burn at apogee: circularise and turn together

Given

A transfer orbit from a 28.5° latitude site has perigee altitude 250 km, apogee at GEO radius and inclination 28.5°. Compare one combined burn at apogee with circularising first and then turning the plane.

new symbols

v_aspeed at apogee of the transfer orbit, km/s

v_GEOcircular speed at GEO radius, km/s

Working

a = (6628.137 + 42164.17)/2 = 24396.15 km; apogee speed v_a = √(μ(2/r_a − 1/a))1.6026 km/s
GEO speed v_GEO3.0747 km/s
Combined: Δv = √(v_a² + v_GEO² − 2 v_a v_GEO cos Δi) = √(1.6026² + 3.0747² − 2 × 1.6026 × 3.0747 × cos 28.5°)1.8333 km/s
Separate: (3.0747 − 1.6026) + 2 × 3.0747 × sin 14.25°2.9857 km/s

Answer. Combined 1833 m/s against 2986 m/s for two separate burns: a saving of 1152 m/s.

Sanity check. Velocity changes are vectors. One burn along the vector difference is always shorter than two burns along two sides of the triangle.

Pythonedit a value, then Run

# Worked example 10.2: One burn at apogee: circularise and turn together
from math import pi, sqrt, sin, cos, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
hp = 250.0            # km, transfer perigee altitude
di = radians(28.5)    # plane change = site latitude

rp = R_E + hp
ra = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)   # GEO radius
a = (rp + ra) / 2
va = sqrt(mu * (2 / ra - 1 / a))
vg = sqrt(mu / ra)
dv_comb = sqrt(va**2 + vg**2 - 2 * va * vg * cos(di))
dv_sep = (vg - va) + 2 * vg * sin(di / 2)
print(f"apogee speed {va:.4f} km/s, GEO speed {vg:.4f} km/s")
print(f"combined burn {dv_comb*1000:.1f} m/s, separate burns {dv_sep*1000:.1f} m/s")
print(f"saving {(dv_sep - dv_comb)*1000:.1f} m/s")

apogee speed 1.6026 km/s, GEO speed 3.0747 km/s
combined burn 1833.3 m/s, separate burns 2985.7 m/s
saving 1152.4 m/s

Practice

Practice O10.1 · warm-up

A small plane trim

Given

A satellite in a circular 700 km orbit must change its inclination by 5°. What Δv does that cost?

Mark as done

Hint

Δv = 2v sin(Δi/2) with the circular speed.

Answer. Δv = 655 m/s.

Working

v = √(μ/r) = √(398600.4418/7078.137)7.5043 km/s
Δv = 2 × 7.5043 × sin 2.5°0.6547 km/s

Answer. Δv = 655 m/s.

Sanity check. About 8.7 % of the orbital speed for a 5° turn; for small angles Δv ≈ v Δi = 655 m/s.

Pythonedit a value, then Run

# Practice O10.1: A small plane trim
from math import sqrt, sin, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 700.0             # km
di = radians(5)

v = sqrt(mu / (R_E + h))
dv = 2 * v * sin(di / 2)
print(f"v = {v:.4f} km/s, dv = {dv:.4f} km/s = {dv*1000:.1f} m/s")

v = 7.5043 km/s, dv = 0.6547 km/s = 654.7 m/s

Practice O10.2 · standard

Why launch sites near the equator win for GEO^[R25]

Given

Using the transfer orbit of worked example 10.2 (apogee speed 1.6026 km/s), find the combined apogee burn into GEO for transfer orbits inclined at 5.2° and at 45° (the site latitudes).

Mark as done

Hint

Same law of cosines, only the angle changes.

Answer. 1486 m/s from 5.2° and 2248 m/s from 45°, a difference of 762 m/s.

Working

Δv = √(v_a² + v_GEO² − 2 v_a v_GEO cos Δi), v_a = 1.6026, v_GEO3.0747 km/s
Δi = 5.2°: Δv1.4857 km/s
Δi = 45°: Δv2.2480 km/s

Answer. 1486 m/s from 5.2° and 2248 m/s from 45°, a difference of 762 m/s.

Sanity check. The 5.2° case is close to the pure circularisation cost of 1472 m/s, as it should be for a small angle.

Pythonedit a value, then Run

# Practice O10.2: Why launch sites near the equator win for GEO
from math import pi, sqrt, cos, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
hp = 250.0            # km, transfer perigee altitude (worked example 10.2)

rp = R_E + hp
ra = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
va = sqrt(mu * (2 / ra - 1 / ((rp + ra) / 2)))     # 1.6026 km/s
vg = sqrt(mu / ra)
dv = {}
for lat in (5.2, 45.0):
    di = radians(lat)
    dv[lat] = sqrt(va**2 + vg**2 - 2 * va * vg * cos(di))
    print(f"di = {lat} deg: combined apogee burn {dv[lat]*1000:.1f} m/s")
print(f"difference {(dv[45.0] - dv[5.2])*1000:.1f} m/s")

di = 5.2 deg: combined apogee burn 1485.7 m/s
di = 45.0 deg: combined apogee burn 2248.0 m/s
difference 762.2 m/s

Practice O10.3 · challenge

Where to do the plane change: an optimisation

Given

A spacecraft goes from a circular 300 km orbit at 40° inclination to an equatorial GEO with a Hohmann transfer. A share s of the plane change is done in the perigee burn and the rest in the apogee burn. Find the best s and the saving compared with doing it all at apogee.

Mark as done

Hint

Write the total as two law-of-cosines burns, one at each end, as a function of s. Minimise numerically (bounded scalar search).

Answer. Do about 2.6° of the turn at perigee (s = 0.066); total 4504 m/s, saving 34 m/s.

Working

Perigee: v_c1 = 7.7258, v_p = 10.1515 km/s. Apogee: v_a = 1.6078, v_GEO3.0747 km/s
Δv(s) = √(v_c1² + v_p² − 2v_c1v_p cos(sΔi)) + √(v_a² + v_GEO² − 2v_av_GEO cos((1 − s)Δi))
All at apogee, s = 0: Δv4.5387 km/s
Bounded minimisation: s = 0.0659 (2.64° at perigee), Δv4.5043 km/s

Answer. Do about 2.6° of the turn at perigee (s = 0.066); total 4504 m/s, saving 34 m/s.

Sanity check. A small turn added to a big burn costs almost nothing (the vector barely gets longer), so the optimum always shifts a few degrees to perigee.

Pythonedit a value, then Run

# Practice O10.3: Where to do the plane change: an optimisation
from math import pi, sqrt, cos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
h1 = 300.0            # km, circular start orbit
di = radians(40)      # total plane change

r1 = R_E + h1
r2 = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
at = (r1 + r2) / 2
vc1, vp = sqrt(mu / r1), sqrt(mu * (2 / r1 - 1 / at))
va, vg = sqrt(mu * (2 / r2 - 1 / at)), sqrt(mu / r2)

def burn(v1, v2, angle):              # one burn that changes speed and turns by angle
    return sqrt(v1**2 + v2**2 - 2 * v1 * v2 * cos(angle))

def total(s):                         # share s of the turn at perigee
    return burn(vc1, vp, s * di) + burn(va, vg, (1 - s) * di)

lo, hi = 0.0, 1.0                     # golden-section search
g = (sqrt(5) - 1) / 2
for _ in range(100):
    x1, x2 = hi - g * (hi - lo), lo + g * (hi - lo)
    if total(x1) < total(x2):
        hi = x2
    else:
        lo = x1
s = (lo + hi) / 2
print(f"vc1 = {vc1:.4f}, vp = {vp:.4f}, va = {va:.4f}, vGEO = {vg:.4f} km/s")
print(f"all at apogee: {total(0)*1000:.1f} m/s")
print(f"best s = {s:.4f} ({degrees(s * di):.2f} deg at perigee): {total(s)*1000:.1f} m/s")
print(f"saving {(total(0) - total(s))*1000:.1f} m/s")

# Practice O10.3: Where to do the plane change: an optimisation
from math import pi, sqrt, cos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
h1 = 300.0            # km, circular start orbit
di = radians(40)      # total plane change

r1 = R_E + h1
r2 = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
at = (r1 + r2) / 2
vc1, vp = sqrt(mu / r1), sqrt(mu * (2 / r1 - 1 / at))
va, vg = sqrt(mu * (2 / r2 - 1 / at)), sqrt(mu / r2)

def burn(v1, v2, angle):              # one burn that changes speed and turns by angle
    return sqrt(v1**2 + v2**2 - 2 * v1 * v2 * cos(angle))

def total(s):                         # share s of the turn at perigee
    return burn(vc1, vp, s * di) + burn(va, vg, (1 - s) * di)

lo, hi = 0.0, 1.0                     # golden-section search
g = (sqrt(5) - 1) / 2
for _ in range(100):
    x1, x2 = hi - g * (hi - lo), lo + g * (hi - lo)
    if total(x1) < total(x2):
        hi = x2
    else:
        lo = x1
s = (lo + hi) / 2
print(f"vc1 = {vc1:.4f}, vp = {vp:.4f}, va = {va:.4f}, vGEO = {vg:.4f} km/s")
print(f"all at apogee: {total(0)*1000:.1f} m/s")
print(f"best s = {s:.4f} ({degrees(s * di):.2f} deg at perigee): {total(s)*1000:.1f} m/s")
print(f"saving {(total(0) - total(s))*1000:.1f} m/s")

vc1 = 7.7258, vp = 10.1515, va = 1.6078, vGEO = 3.0747 km/s
all at apogee: 4538.7 m/s
best s = 0.0659 (2.64 deg at perigee): 4504.3 m/s
saving 34.4 m/s

Common mistakes

Using degrees inside the small-angle form. Δv ≈ v Δi needs Δi in radians.
Doing the plane change in LEO because it is the first burn. It is the most expensive place.
Adding the speed change and the plane change as scalars: they are vectors.
Burning away from a node: a plane change has to happen where the two planes cross.

Recap from earlier chapters

(Chapter 9) What is the circularisation Δv at GEO for the 250 km-perigee transfer, with no plane change?

v_GEO − v_a = 3.0747 − 1.6026 = 1472 m/s.

(Chapter 2) Which element does a pure plane change at the node alter?

The inclination (and, if not done at a node, the RAAN too). Shape and size stay the same.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R25ESA, "Europe's Spaceport".

Where in each source, and links: References.

I have finished chapter 10

Chapter 11

Rendezvous: Clohessy-Wiltshire

40 minutes

The idea in one minute

Rendezvous is orbital mechanics seen from the target. Put the target at the origin and ride along with it: radial up, along-track forward, cross-track out of the plane. Close to the target the equations become linear and have a neat closed-form solution, the Clohessy-Wiltshire (CW) equations. They contain the great paradox of orbits: to catch up with something ahead of you, you slow down. A slower craft drops to a lower orbit, a lower orbit is faster, and it overtakes.

A burn backwards drops the chaser below the target, where it moves faster and gains 16.7 km per orbit: slow down to catch up.

You will be able to

Describe relative motion in the target's LVLH frame (radial, along-track, cross-track)
Explain why a retrograde burn makes a chaser move ahead
Plan a two-burn approach with the Clohessy-Wiltshire solution and check it against the full orbit

The LVLH frame

Local vertical, local horizontal: x points radially out from Earth through the target, y points along the target's velocity, z completes the right-handed set along the orbit normal. The line along y through the target is the V-bar; the line along x is the R-bar. Approaches to a space station come in along one of them.

The CW equations

For a target on a circular orbit and a chaser a few kilometres away, gravity differences and the rotating frame give three linear equations. The radial and along-track motions are coupled through Coriolis terms 2n; the cross-track motion is a simple oscillation at the orbit rate. Because they are linear, the motion after any burn is a matrix times the state: the state transition matrix.

Slow down to catch up

A burn along +y raises the opposite side of the orbit; the craft climbs, its period grows and it falls behind. A burn along −y does the reverse: it drops, speeds up relative to the target and drifts ahead by 6π Δv/n per orbit. The relative path is a looping cycloid, as in the figure.

Checking the linear model

CW assumes a circular target orbit, small separations and no perturbations. Measure x as the difference in radius and y as the distance along the curved orbit, and for a 10 km separation at 400 km altitude the linear prediction is within a few metres of the full two-body answer; at hundreds of kilometres it breaks down, and you go back to Lambert's problem (Chapter 12).

Key equations

Clohessy-Wiltshire equations

ẍ − 3n²x − 2nẏ = 0, ÿ + 2nẋ = 0, z̈ + n²z = 0

where

xradial offset of the chaser from the target (up is positive), km

yalong-track offset (ahead is positive), km

zcross-track offset (along the orbit normal), km

ẋ, ẏrelative velocity components in the rotating frame, km/s

ẍ, ÿ, z̈relative acceleration components, km/s²

nmean motion of the target's circular orbit, √(μ/a³), rad/s

valid for a circular target orbit and separations much smaller than the orbit radius

Mean motion

n = √(μ/a³)

where

nmean motion of the target's circular orbit, √(μ/a³), rad/s

μgravitational parameter of the central body, km³/s²

asemi-major axis (radius, for a circular orbit), km

Drift per orbit after an along-track burn

Δy = −6π Δv_y / n

where

Δyalong-track displacement after one orbit, km

Δv_yalong-track burn (negative = backwards), km/s

nmean motion of the target's circular orbit, √(μ/a³), rad/s

Natural drift of a lower orbit

ẏ = −(3/2) n x

where

ẏalong-track drift rate, km/s

xradial offset of a circular orbit (below is negative), km

nmean motion of the target's circular orbit, √(μ/a³), rad/s

Two-impulse transfer

v₀⁺ = −Φ_rv⁻¹(t) Φ_rr(t) r₀, Δv₁ = v₀⁺ − v₀

where

r₀initial relative position, km

v₀relative velocity before the burn, km/s

v₀⁺relative velocity needed just after the burn, km/s

Φ_rr, Φ_rvCW state transition blocks evaluated at the transfer time, dimensionless, s

Δv₁first burn, km/s

ttransfer time, s

Worked examples

Worked example 11.1

Slow down to catch up

Given

A target is on a circular orbit at 400 km. A chaser sits 10 km behind it on the same orbit (on the V-bar). It wants to arrive at the target after exactly one orbit with two along-track burns. Find the burn, the total Δv and how low the chaser dips.

new symbols

Δyalong-track gain over one orbit, km

Working

Mean motion n = √(μ/a³) = √(398600.4418 / 6778.137³) = 1.131367e-03 rad/s; period T = 2π/n92.56 min
An along-track burn Δv_y moves the chaser by Δy = −6π Δv_y / n per orbit (CW solution at t = T)
We need Δy = +10 km, so Δv_y = −10 × 1.131367e-03 / (6π)-0.600 m/s(a retrograde burn)
After one orbit the CW solution gives position (0.0e+00, 1.8e-15) km and along-track rate -0.600 m/s, so a second burn of +0.600 m/s stops it
Total Δv = 2 × 0.600 = 1.200 m/s; the lowest point is x = 4Δv_y/n = -2122 m, half an orbit after the burn
Full two-body check: the chaser arrives 3.1 m from the target, 0.03 % of the gap

Answer. Burn 0.60 m/s retrograde, wait one orbit (92.6 min), burn 0.60 m/s prograde: 1.20 m/s in total. The chaser dips 2122 m below the target's orbit.

Sanity check. Slowing down lowers the orbit, a lower orbit has a shorter period, and so the chaser laps ahead. The tiny dip (hundreds of metres for a 10 km gap) is why phasing near a station is so cheap.

Pythonedit a value, then Run

# Worked example 11.1: Slow down to catch up
from math import pi, sqrt, sin, cos

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 400.0             # km, target's circular orbit
gap = 10.0            # km, chaser behind the target on the V-bar

a = R_E + h
n = sqrt(mu / a**3)
T = 2 * pi / n
dvy = -gap * n / (6 * pi)                # along-track burn so that dy = -6 pi dvy / n = +gap
# CW solution (x radial, y along-track) from x0 = 0, y0 = -gap, xdot0 = 0, ydot0 = dvy
t = T
s, c = sin(n * t), cos(n * t)
x = 2 * (1 - c) / n * dvy
y = -gap + (4 * s - 3 * n * t) / n * dvy
ydot = (4 * c - 3) * dvy
x_low = 4 * dvy / n                      # lowest point, half an orbit after the burn
print(f"n = {n:.6e} rad/s, T = {T/60:.2f} min")
print(f"burn 1 = {dvy*1000:.3f} m/s (retrograde), i.e. {abs(dvy)*1000:.2f} m/s")
print(f"after one orbit: x = {x:.1e} km, y = {y:.1e} km, ydot = {ydot*1000:.3f} m/s")
print(f"burn 2 = {-ydot*1000:.2f} m/s prograde; total {2*abs(dvy)*1000:.2f} m/s")
print(f"the chaser dips {abs(x_low)*1000:.0f} m below the target's orbit")

# Worked example 11.1: Slow down to catch up
from math import pi, sqrt, sin, cos

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 400.0             # km, target's circular orbit
gap = 10.0            # km, chaser behind the target on the V-bar

a = R_E + h
n = sqrt(mu / a**3)
T = 2 * pi / n
dvy = -gap * n / (6 * pi)                # along-track burn so that dy = -6 pi dvy / n = +gap
# CW solution (x radial, y along-track) from x0 = 0, y0 = -gap, xdot0 = 0, ydot0 = dvy
t = T
s, c = sin(n * t), cos(n * t)
x = 2 * (1 - c) / n * dvy
y = -gap + (4 * s - 3 * n * t) / n * dvy
ydot = (4 * c - 3) * dvy
x_low = 4 * dvy / n                      # lowest point, half an orbit after the burn
print(f"n = {n:.6e} rad/s, T = {T/60:.2f} min")
print(f"burn 1 = {dvy*1000:.3f} m/s (retrograde), i.e. {abs(dvy)*1000:.2f} m/s")
print(f"after one orbit: x = {x:.1e} km, y = {y:.1e} km, ydot = {ydot*1000:.3f} m/s")
print(f"burn 2 = {-ydot*1000:.2f} m/s prograde; total {2*abs(dvy)*1000:.2f} m/s")
print(f"the chaser dips {abs(x_low)*1000:.0f} m below the target's orbit")

n = 1.131367e-03 rad/s, T = 92.56 min
burn 1 = -0.600 m/s (retrograde), i.e. 0.60 m/s
after one orbit: x = -0.0e+00 km, y = 1.8e-15 km, ydot = -0.600 m/s
burn 2 = 0.60 m/s prograde; total 1.20 m/s
the chaser dips 2122 m below the target's orbit

Worked example 11.2

A two-burn approach in half an orbit

Given

The chaser is 1 km below and 8 km behind the target (same 400 km orbit for the target). Being on a lower circular orbit it already drifts forward at ẏ = −(3/2)n x = 1.70 m/s. Plan a two-burn CW transfer that reaches the target in half an orbit.

new symbols

Φ_rr, Φ_rvblocks of the CW state transition matrix (position from position, position from velocity), dimensionless, s

Working

Transfer time t = T/2 = 46.28 min, so nt = π
Required velocity just after burn 1: v₀⁺ = −Φ_rv⁻¹ Φ_rr r₀ = (-1.596, 1.980, 0) m/s
Burn 1: Δv₁ = v₀⁺ − v₀ = (-1.596, 0.283, 0) m/s, size 1.621 m/s
Arrival velocity from the CW solution; burn 2 cancels it: |Δv₂|1.621 m/s
Total = 3.242 m/s. Full two-body propagation of the same burn misses by 0.8 m

Answer. Δv₁ = 1.62 m/s, Δv₂ = 1.62 m/s, total 3.24 m/s, arriving 46.3 min later.

Sanity check. Metres per second, not kilometres per second: relative motion near a target is a game of very small burns. The linear CW model is good to a few metres here because the separation is tiny compared with the orbit radius.

Pythonedit a value, then Run

# Worked example 11.2: A two-burn approach in half an orbit
from math import pi, sqrt, sin, cos

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 400.0             # km, target orbit
x0, y0 = -1.0, -8.0   # km: 1 km below, 8 km behind
a = R_E + h
n = sqrt(mu / a**3)
t = pi / n            # half an orbit
vx0, vy0 = 0.0, -1.5 * n * x0            # natural drift of a lower circular orbit, km/s

s, c, nt = sin(n * t), cos(n * t), n * t
# in-plane CW state-transition blocks
Prr = [[4 - 3 * c, 0], [6 * (s - nt), 1]]
Prv = [[s / n, 2 * (1 - c) / n], [2 * (c - 1) / n, (4 * s - 3 * nt) / n]]
Pvr = [[3 * n * s, 0], [6 * n * (c - 1), 0]]
Pvv = [[c, 2 * s], [-2 * s, 4 * c - 3]]
# velocity after burn 1 so that r(t) = 0:  Prv v = -Prr r0  (2x2 solve)
bx = -(Prr[0][0] * x0 + Prr[0][1] * y0)
by = -(Prr[1][0] * x0 + Prr[1][1] * y0)
det = Prv[0][0] * Prv[1][1] - Prv[0][1] * Prv[1][0]
vx = (bx * Prv[1][1] - Prv[0][1] * by) / det
vy = (Prv[0][0] * by - Prv[1][0] * bx) / det
dv1 = sqrt((vx - vx0)**2 + (vy - vy0)**2)
vfx = Pvr[0][0] * x0 + Pvr[0][1] * y0 + Pvv[0][0] * vx + Pvv[0][1] * vy
vfy = Pvr[1][0] * x0 + Pvr[1][1] * y0 + Pvv[1][0] * vx + Pvv[1][1] * vy
dv2 = sqrt(vfx**2 + vfy**2)             # burn 2 cancels the arrival velocity
print(f"drift rate before burn 1: {vy0*1000:.2f} m/s")
print(f"velocity after burn 1: ({vx*1000:.3f}, {vy*1000:.3f}) m/s")
print(f"dv1 = {dv1*1000:.3f} m/s, dv2 = {dv2*1000:.3f} m/s, total {(dv1 + dv2)*1000:.3f} m/s")
print(f"arrival {t/60:.2f} min after burn 1")

# Worked example 11.2: A two-burn approach in half an orbit
from math import pi, sqrt, sin, cos

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 400.0             # km, target orbit
x0, y0 = -1.0, -8.0   # km: 1 km below, 8 km behind
a = R_E + h
n = sqrt(mu / a**3)
t = pi / n            # half an orbit
vx0, vy0 = 0.0, -1.5 * n * x0            # natural drift of a lower circular orbit, km/s

s, c, nt = sin(n * t), cos(n * t), n * t
# in-plane CW state-transition blocks
Prr = [[4 - 3 * c, 0], [6 * (s - nt), 1]]
Prv = [[s / n, 2 * (1 - c) / n], [2 * (c - 1) / n, (4 * s - 3 * nt) / n]]
Pvr = [[3 * n * s, 0], [6 * n * (c - 1), 0]]
Pvv = [[c, 2 * s], [-2 * s, 4 * c - 3]]
# velocity after burn 1 so that r(t) = 0:  Prv v = -Prr r0  (2x2 solve)
bx = -(Prr[0][0] * x0 + Prr[0][1] * y0)
by = -(Prr[1][0] * x0 + Prr[1][1] * y0)
det = Prv[0][0] * Prv[1][1] - Prv[0][1] * Prv[1][0]
vx = (bx * Prv[1][1] - Prv[0][1] * by) / det
vy = (Prv[0][0] * by - Prv[1][0] * bx) / det
dv1 = sqrt((vx - vx0)**2 + (vy - vy0)**2)
vfx = Pvr[0][0] * x0 + Pvr[0][1] * y0 + Pvv[0][0] * vx + Pvv[0][1] * vy
vfy = Pvr[1][0] * x0 + Pvr[1][1] * y0 + Pvv[1][0] * vx + Pvv[1][1] * vy
dv2 = sqrt(vfx**2 + vfy**2)             # burn 2 cancels the arrival velocity
print(f"drift rate before burn 1: {vy0*1000:.2f} m/s")
print(f"velocity after burn 1: ({vx*1000:.3f}, {vy*1000:.3f}) m/s")
print(f"dv1 = {dv1*1000:.3f} m/s, dv2 = {dv2*1000:.3f} m/s, total {(dv1 + dv2)*1000:.3f} m/s")
print(f"arrival {t/60:.2f} min after burn 1")

drift rate before burn 1: 1.70 m/s
velocity after burn 1: (-1.596, 1.980) m/s
dv1 = 1.621 m/s, dv2 = 1.621 m/s, total 3.242 m/s
arrival 46.28 min after burn 1

Practice

Practice O11.1 · warm-up

How far does a tiny burn move you?

Given

A satellite on a circular orbit at 500 km makes an along-track burn of -0.2 m/s (retrograde). How far, and in which direction, has it moved along-track relative to where it would have been, after one orbit?

Mark as done

Hint

Use Δy = −6π Δv_y / n with n in rad/s and Δv in km/s.

Answer. Δy = +3.41 km: about 3.4 km ahead after one orbit.

Working

a = 6378.137 + 500 = 6878.137 km; n = √(μ/a³)1.106783e-03 rad/s
Δy = −6π × (-2.0e-04) / 1.106783e-033.406 km
Positive y is ahead of the reference point

Answer. Δy = +3.41 km: about 3.4 km ahead after one orbit.

Sanity check. 0.2 m/s is about 1/38,063 of the orbital speed, yet the effect grows every orbit: drift is how operators move satellites along a constellation.

Pythonedit a value, then Run

# Practice O11.1: How far does a tiny burn move you?
from math import pi, sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 500.0             # km
dvy = -0.2e-3         # km/s, along-track burn (retrograde)

a = R_E + h
n = sqrt(mu / a**3)
dy = -6 * pi * dvy / n                   # CW along-track shift after one orbit
print(f"n = {n:.6e} rad/s")
print(f"dy = {dy:+.3f} km ({'ahead' if dy > 0 else 'behind'})")

n = 1.106783e-03 rad/s
dy = +3.406 km (ahead)

Practice O11.2 · standard

Drifting in from below

Given

A chaser is on a circular orbit 2 km below a target at 450 km, 20 km behind it. How fast does it close the gap, and how long until it reaches the target's along-track position?

Mark as done

Hint

A slightly lower circular orbit drifts forward at ẏ = −(3/2) n x.

Answer. ẏ = 3.36 m/s, so the gap closes in 1.7 h.

Working

n = √(μ/(6378.137 + 450)³)1.118963e-03 rad/s
ẏ = −1.5 × 1.118963e-03 × (-2)3.357 m/s
Time = 20 km / 3.357 m/s = 5958 s1.65 h
Exact check: (n_chaser − n_target) × a3.358 m/s

Answer. ẏ = 3.36 m/s, so the gap closes in 1.7 h.

Sanity check. The linear rate and the exact difference of angular rates agree to 0.04 %, as they should for a 2 km offset.

Pythonedit a value, then Run

# Practice O11.2: Drifting in from below
from math import sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 450.0             # km, target orbit
x = -2.0              # km, chaser 2 km below
gap = 20.0            # km behind

a = R_E + h
n = sqrt(mu / a**3)
ydot = -1.5 * n * x                      # km/s, CW drift of a lower circular orbit
t = gap / ydot
exact = (sqrt(mu / (a + x)**3) - n) * a  # difference in mean motion times radius
print(f"n = {n:.6e} rad/s, drift {ydot*1000:.3f} m/s (exact {exact*1000:.3f} m/s)")
print(f"time to close {gap:.0f} km: {t:.0f} s = {t/3600:.3f} h")

n = 1.118963e-03 rad/s, drift 3.357 m/s (exact 3.358 m/s)
time to close 20 km: 5958 s = 1.655 h

Practice O11.3 · challenge

Coming down the R-bar

Given

A chaser is released 3 km directly above a target (on the R-bar) at 400 km, with zero velocity relative to the LVLH frame. Plan a two-burn CW transfer that reaches the target in a quarter of an orbit. Give both burns and the total.

Mark as done

Hint

Use the CW state transition matrix at nt = π/2: v₀⁺ = −Φ_rv⁻¹ Φ_rr r₀. The second burn cancels the arrival velocity.

Answer. Δv₁ = 6.27 m/s, Δv₂ = 2.19 m/s, total 8.47 m/s.

Working

n = 1.131367e-03 rad/s, t = T/4 = 23.14 min, nt = π/2
v₀⁺ = (-4.130, -4.723, 0) m/s, so Δv₁6.274 m/s
Arrival velocity cancelled by Δv₂2.192 m/s
Total8.466 m/s
Two-body check: the chaser arrives 6.0 m from the target

Answer. Δv₁ = 6.27 m/s, Δv₂ = 2.19 m/s, total 8.47 m/s.

Sanity check. The first burn has a large along-track part, not just a radial push down: in orbit, radial and along-track motion are coupled by the Coriolis terms.

Pythonedit a value, then Run

# Practice O11.3: Coming down the R-bar
from math import pi, sqrt, sin, cos

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 400.0             # km, target orbit
x0, y0 = 3.0, 0.0     # km, 3 km above on the R-bar
vx0, vy0 = 0.0, 0.0   # at rest in the LVLH frame

a = R_E + h
n = sqrt(mu / a**3)
t = (pi / 2) / n      # a quarter of an orbit
s, c, nt = sin(n * t), cos(n * t), n * t
Prr = [[4 - 3 * c, 0], [6 * (s - nt), 1]]
Prv = [[s / n, 2 * (1 - c) / n], [2 * (c - 1) / n, (4 * s - 3 * nt) / n]]
Pvr = [[3 * n * s, 0], [6 * n * (c - 1), 0]]
Pvv = [[c, 2 * s], [-2 * s, 4 * c - 3]]
bx = -(Prr[0][0] * x0 + Prr[0][1] * y0)
by = -(Prr[1][0] * x0 + Prr[1][1] * y0)
det = Prv[0][0] * Prv[1][1] - Prv[0][1] * Prv[1][0]
vx = (bx * Prv[1][1] - Prv[0][1] * by) / det
vy = (Prv[0][0] * by - Prv[1][0] * bx) / det
dv1 = sqrt((vx - vx0)**2 + (vy - vy0)**2)
vfx = Pvr[0][0] * x0 + Pvr[0][1] * y0 + Pvv[0][0] * vx + Pvv[0][1] * vy
vfy = Pvr[1][0] * x0 + Pvr[1][1] * y0 + Pvv[1][0] * vx + Pvv[1][1] * vy
dv2 = sqrt(vfx**2 + vfy**2)
print(f"transfer time {t/60:.2f} min, velocity after burn 1 ({vx*1000:.3f}, {vy*1000:.3f}) m/s")
print(f"dv1 = {dv1*1000:.3f} m/s, dv2 = {dv2*1000:.3f} m/s, total {(dv1 + dv2)*1000:.3f} m/s")

# Practice O11.3: Coming down the R-bar
from math import pi, sqrt, sin, cos

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 400.0             # km, target orbit
x0, y0 = 3.0, 0.0     # km, 3 km above on the R-bar
vx0, vy0 = 0.0, 0.0   # at rest in the LVLH frame

a = R_E + h
n = sqrt(mu / a**3)
t = (pi / 2) / n      # a quarter of an orbit
s, c, nt = sin(n * t), cos(n * t), n * t
Prr = [[4 - 3 * c, 0], [6 * (s - nt), 1]]
Prv = [[s / n, 2 * (1 - c) / n], [2 * (c - 1) / n, (4 * s - 3 * nt) / n]]
Pvr = [[3 * n * s, 0], [6 * n * (c - 1), 0]]
Pvv = [[c, 2 * s], [-2 * s, 4 * c - 3]]
bx = -(Prr[0][0] * x0 + Prr[0][1] * y0)
by = -(Prr[1][0] * x0 + Prr[1][1] * y0)
det = Prv[0][0] * Prv[1][1] - Prv[0][1] * Prv[1][0]
vx = (bx * Prv[1][1] - Prv[0][1] * by) / det
vy = (Prv[0][0] * by - Prv[1][0] * bx) / det
dv1 = sqrt((vx - vx0)**2 + (vy - vy0)**2)
vfx = Pvr[0][0] * x0 + Pvr[0][1] * y0 + Pvv[0][0] * vx + Pvv[0][1] * vy
vfy = Pvr[1][0] * x0 + Pvr[1][1] * y0 + Pvv[1][0] * vx + Pvv[1][1] * vy
dv2 = sqrt(vfx**2 + vfy**2)
print(f"transfer time {t/60:.2f} min, velocity after burn 1 ({vx*1000:.3f}, {vy*1000:.3f}) m/s")
print(f"dv1 = {dv1*1000:.3f} m/s, dv2 = {dv2*1000:.3f} m/s, total {(dv1 + dv2)*1000:.3f} m/s")

transfer time 23.14 min, velocity after burn 1 (-4.130, -4.723) m/s
dv1 = 6.274 m/s, dv2 = 2.192 m/s, total 8.466 m/s

Common mistakes

Burning towards the target to catch it. A forward burn raises the orbit and the chaser falls behind.
Using CW for large separations or an elliptical target orbit. Check the separation is a small fraction of the radius.
Mixing the inertial and rotating-frame velocities. CW velocities are measured in the rotating LVLH frame.
Forgetting the second burn. A two-impulse transfer arrives with a relative velocity that must be cancelled.

Recap from earlier chapters

(Chapter 3) What is the circular speed at 400 km altitude?

v = √(μ/r) = 7.669 km/s.

(Chapter 1) What is the orbital period at 400 km?

T = 2π√(a³/μ) = 92.56 min.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.

Where in each source, and links: References.

I have finished chapter 11

Chapter 12

Lambert's problem

45 minutes

The idea in one minute

Most orbit questions are 'given where I am and how fast I'm going, where will I be?'. Mission design asks the opposite: I am here, I must be there at that time; what velocity do I need? That is Lambert's problem. Two position vectors and a flight time define exactly one conic (for less than one revolution, going the short or the long way). Every launch window, every interplanetary porkchop plot and every long-range rendezvous is a grid of Lambert solves.

Same two points, different flight times: the shorter the time, the faster and flatter the arc Lambert's solver returns.

You will be able to

Say why Lambert's problem sits behind launch windows, targeting and rendezvous
Solve it with the universal-variable method and Stumpff functions
Verify a solver against a Hohmann case and by propagating the answer

Why it matters

Targeting a planet, choosing a launch date, intercepting a satellite, or picking up a sample at a given time are all boundary-value problems: the end points and the time are fixed, the orbit is unknown. Lambert's problem answers them, and its output (v₁ and v₂) gives the burns at both ends by simple subtraction.^[R13]

The universal-variable method

Write the unknown orbit with the universal variable z, which is positive for ellipses, zero for parabolas and negative for hyperbolas. The flight time becomes a single monotonic function of z built from the Stumpff functions C(z) and S(z). Solve F(z) = 0 with a bracketed root finder, then the Lagrange coefficients f, g and ġ give both velocities.^[R13]

Short way, long way, and 180°

For a given pair of points you can go the short way (Δθ < 180°) or the long way (Δθ > 180°); the direction of motion decides which. At exactly 180° the two vectors do not define a plane, so the problem is singular; real tools handle it with a small plane offset. Allowing extra full revolutions adds further solutions, each pair with its own minimum time.

Trust, but verify

A Lambert solver should pass three tests: it rebuilds a known ellipse from two of its points, its answer propagated forward lands on r₂, and the orbit's energy is the same at both ends. This chapter's solver passes all three, and reproduces a standard textbook-style test case to four decimals.^{[R12, R13]}

Key equations

Lambert's problem

(r₁, r₂, Δt) → (v₁, v₂)

where

r₁, r₂position vectors at departure and arrival, km

Δtflight time, s

v₁, v₂velocities on the transfer orbit at the two points, km/s

Geometry constant

A = sin Δθ √(r₁r₂/(1 − cos Δθ))

where

Aconstant fixed by the geometry, km

Δθtransfer angle swept between r₁ and r₂, rad

r₁, r₂lengths of the two position vectors, km

A = 0 at Δθ = 180°: the singular case

Time of flight in z

√μ Δt = (y/C)^3/2 S + A√y, y = r₁ + r₂ + A (zS − 1)/√C

where

μgravitational parameter of the central body, km³/s²

Δtflight time, s

zuniversal variable (z > 0 ellipse, z < 0 hyperbola), dimensionless

yauxiliary function of z, km

C, SStumpff functions evaluated at z, dimensionless

Ageometry constant, km

r₁, r₂radii of the two points, km

Stumpff functions (z > 0)

C(z) = (1 − cos√z)/z, S(z) = (√z − sin√z)/√z³

where

zuniversal variable, dimensionless

C, SStumpff functions (cosh and sinh forms for z < 0; 1/2 and 1/6 at z = 0), dimensionless

Velocities from f and g

f = 1 − y/r₁, g = A√(y/μ), ġ = 1 − y/r₂; v₁ = (r₂ − f r₁)/g, v₂ = (ġ r₂ − r₁)/g

where

f, ġLagrange coefficients, dimensionless

gLagrange coefficient, s

yauxiliary function at the root z, km

μgravitational parameter of the central body, km³/s²

r₁, r₂position vectors, km

v₁, v₂transfer velocities, km/s

Ageometry constant, km

Minimum-energy transfer

c = |r₂ − r₁|, s = (r₁ + r₂ + c)/2, a_min = s/2

where

cchord between the two points, km

ssemi-perimeter of the triangle Earth-r₁-r₂, km

a_minsemi-major axis of the lowest-energy transfer, km

r₁, r₂radii, km

Worked examples

Worked example 12.1

Lambert finds the Hohmann ellipse^[R1]

Given

Take the Hohmann transfer from a 300 km circular orbit to GEO (r = 42164.2 km). Pick the departure point (perigee) and the point 170° further round the ellipse, and the flight time between them. Lambert's solver, given only the two positions and the time, must rebuild the same ellipse.

Working

Transfer ellipse: a_t = (6678.137 + 42164.2)/2 = 24421.2 km, e0.72654
Time from ν = 0 to ν = 170° by Kepler's equation (Chapter 4): Δt = 14532.2 s4.037 h
Lambert(r₁, r₂, Δt) returns |v₁|10.151493 km/s
Vis-viva at perigee: √(μ(2/r_p − 1/a_t))10.151493 km/s
Orbit from Lambert's v₁: a = 24421.2 km, e0.72654

Answer. Lambert gives 10.1515 km/s, exactly the Hohmann perigee speed, and the same ellipse (a = 24421 km).

Sanity check. Two positions plus a time define one conic. If you feed in points from a known ellipse, you must get that ellipse back: this is the first test any Lambert solver should pass.

Pythonedit a value, then Run

# Worked example 12.1: Lambert finds the Hohmann ellipse
from math import pi, sqrt, sin, cos, sinh, cosh, acos, atan2, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
hp = 300.0            # km, parking orbit
nu2 = radians(170)    # arrival point on the ellipse

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

    def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

    def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

    lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

rp = R_E + hp
ra = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
at = (rp + ra) / 2
et = (ra - rp) / (ra + rp)
E2 = 2 * atan2(sqrt(1 - et) * sin(nu2 / 2), sqrt(1 + et) * cos(nu2 / 2))
tof = (E2 - et * sin(E2)) * sqrt(at**3 / mu)       # Kepler's equation from perigee
r2m = at * (1 - et**2) / (1 + et * cos(nu2))
r1 = [rp, 0.0, 0.0]
r2 = [r2m * cos(nu2), r2m * sin(nu2), 0.0]
v1, v2, z, A, dth = lambert(r1, r2, tof, mu)
v1m = sqrt(sum(c * c for c in v1))
vp = sqrt(mu * (2 / rp - 1 / at))
a = 1 / (2 / rp - v1m**2 / mu)           # orbit from Lambert's v1 (vis-viva)
e = 1 - rp / a                           # r1 is the perigee
print(f"transfer ellipse a = {at:.1f} km, e = {et:.5f}, flight time {tof:.1f} s = {tof/3600:.3f} h")
print(f"Lambert |v1| = {v1m:.6f} km/s, vis-viva perigee speed {vp:.6f} km/s")
print(f"orbit from Lambert: a = {a:.1f} km, e = {e:.5f}")

# Worked example 12.1: Lambert finds the Hohmann ellipse
from math import pi, sqrt, sin, cos, sinh, cosh, acos, atan2, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
hp = 300.0            # km, parking orbit
nu2 = radians(170)    # arrival point on the ellipse

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

rp = R_E + hp
ra = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
at = (rp + ra) / 2
et = (ra - rp) / (ra + rp)
E2 = 2 * atan2(sqrt(1 - et) * sin(nu2 / 2), sqrt(1 + et) * cos(nu2 / 2))
tof = (E2 - et * sin(E2)) * sqrt(at**3 / mu)       # Kepler's equation from perigee
r2m = at * (1 - et**2) / (1 + et * cos(nu2))
r1 = [rp, 0.0, 0.0]
r2 = [r2m * cos(nu2), r2m * sin(nu2), 0.0]
v1, v2, z, A, dth = lambert(r1, r2, tof, mu)
v1m = sqrt(sum(c * c for c in v1))
vp = sqrt(mu * (2 / rp - 1 / at))
a = 1 / (2 / rp - v1m**2 / mu)           # orbit from Lambert's v1 (vis-viva)
e = 1 - rp / a                           # r1 is the perigee
print(f"transfer ellipse a = {at:.1f} km, e = {et:.5f}, flight time {tof:.1f} s = {tof/3600:.3f} h")
print(f"Lambert |v1| = {v1m:.6f} km/s, vis-viva perigee speed {vp:.6f} km/s")
print(f"orbit from Lambert: a = {a:.1f} km, e = {e:.5f}")

transfer ellipse a = 24421.2 km, e = 0.72654, flight time 14532.2 s = 4.037 h
Lambert |v1| = 10.151493 km/s, vis-viva perigee speed 10.151493 km/s
orbit from Lambert: a = 24421.2 km, e = 0.72654

Worked example 12.2

Aim for a point, arrive on time

Given

A spacecraft on a circular orbit of radius 7000 km wants to reach the point at radius 12000 km that lies 110° ahead in the same plane, in exactly 60 min. Find the departure velocity, the transfer orbit and the departure burn.

Working

Inputs: r₁ = 7000 km, r₂ = 12000 km, Δθ = 110°, Δt3600 s(prograde, short way)
A = sin Δθ √(r₁r₂/(1 − cos Δθ))7434.4 km
Solve √μ Δt = (y/C)^3/2 S + A√y for z: z5.57899(z > 0, so an ellipse)
Lagrange coefficients give v₁ = (2.7041, 7.8177, 0) km/s, |v₁|8.2721 km/s
Transfer orbit from (r₁, v₁): a = 8768.6 km, e0.3784
Circular speed at r₁ is 7.5461 km/s along +y, so the burn is |v₁ − v_c|2.7177 km/s
Check: numerical propagation of v₁ for 3600 s misses r₂ by 7.0e-10 km

Answer. v₁ = 8.272 km/s on an ellipse with a = 8769 km and e = 0.378; departure burn 2.718 km/s.

Sanity check. The burn has a radial part as well as an along-track part: to arrive at a fixed time you generally leave at a flight-path angle, not tangentially.

Pythonedit a value, then Run

# Worked example 12.2: Aim for a point, arrive on time
from math import pi, sqrt, sin, cos, sinh, cosh, acos, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r1m, r2m = 7000.0, 12000.0   # km
dtheta = radians(110) # transfer angle (prograde, short way)
tof = 3600.0          # s

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

    def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

    def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

    lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

r1 = [r1m, 0.0, 0.0]
r2 = [r2m * cos(dtheta), r2m * sin(dtheta), 0.0]
v1, v2, z, A, dth = lambert(r1, r2, tof, mu)
v1m = sqrt(sum(c * c for c in v1))
a = 1 / (2 / r1m - v1m**2 / mu)
h = r1[0] * v1[1] - r1[1] * v1[0]        # angular momentum (planar)
e = sqrt(1 - h**2 / (mu * a))
vc = sqrt(mu / r1m)                      # circular speed along +y
dv = sqrt(v1[0]**2 + (v1[1] - vc)**2)
print(f"A = {A:.1f} km, z = {z:.5f}")
print(f"v1 = ({v1[0]:.4f}, {v1[1]:.4f}, 0) km/s, |v1| = {v1m:.4f} km/s")
print(f"transfer orbit a = {a:.1f} km, e = {e:.4f}")
print(f"departure burn {dv:.4f} km/s")

# Worked example 12.2: Aim for a point, arrive on time
from math import pi, sqrt, sin, cos, sinh, cosh, acos, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r1m, r2m = 7000.0, 12000.0   # km
dtheta = radians(110) # transfer angle (prograde, short way)
tof = 3600.0          # s

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

r1 = [r1m, 0.0, 0.0]
r2 = [r2m * cos(dtheta), r2m * sin(dtheta), 0.0]
v1, v2, z, A, dth = lambert(r1, r2, tof, mu)
v1m = sqrt(sum(c * c for c in v1))
a = 1 / (2 / r1m - v1m**2 / mu)
h = r1[0] * v1[1] - r1[1] * v1[0]        # angular momentum (planar)
e = sqrt(1 - h**2 / (mu * a))
vc = sqrt(mu / r1m)                      # circular speed along +y
dv = sqrt(v1[0]**2 + (v1[1] - vc)**2)
print(f"A = {A:.1f} km, z = {z:.5f}")
print(f"v1 = ({v1[0]:.4f}, {v1[1]:.4f}, 0) km/s, |v1| = {v1m:.4f} km/s")
print(f"transfer orbit a = {a:.1f} km, e = {e:.4f}")
print(f"departure burn {dv:.4f} km/s")

A = 7434.4 km, z = 5.57899
v1 = (2.7041, 7.8177, 0) km/s, |v1| = 8.2721 km/s
transfer orbit a = 8768.6 km, e = 0.3784
departure burn 2.7177 km/s

Practice

Practice O12.1 · warm-up

The cheapest-energy ellipse

Given

Two points at r₁ = 8000 km and r₂ = 10000 km are 90° apart. Find the chord, the semi-perimeter and the semi-major axis of the minimum-energy transfer ellipse between them.

Mark as done

Hint

Chord by the law of cosines; semi-perimeter s = (r₁ + r₂ + c)/2; the minimum-energy ellipse has a = s/2.

Answer. c = 12806 km, s = 15403 km, a_min = 7702 km.

Working

c = √(r₁² + r₂² − 2r₁r₂ cos Δθ)12806.2 km
s = (8000 + 10000 + 12806.2)/215403.1 km
a_min = s/27701.6 km
Its flight time (Lambert's theorem) is 3259 s; a Lambert solve at that time returns a7701.6 km

Answer. c = 12806 km, s = 15403 km, a_min = 7702 km.

Sanity check. a_min must exceed half the larger radius (5000 km) and is less than the mean radius of a Hohmann-like path: 7702 km fits.

Pythonedit a value, then Run

# Practice O12.1: The cheapest-energy ellipse
from math import pi, sqrt, cos, radians

r1, r2 = 8000.0, 10000.0   # km
dtheta = radians(90)

c = sqrt(r1**2 + r2**2 - 2 * r1 * r2 * cos(dtheta))   # chord
s = (r1 + r2 + c) / 2                                  # semi-perimeter
a_min = s / 2
print(f"chord c = {c:.1f} km, semi-perimeter s = {s:.1f} km")
print(f"minimum-energy a = {a_min:.1f} km")

chord c = 12806.2 km, semi-perimeter s = 15403.1 km
minimum-energy a = 7701.6 km

Practice O12.2 · standard

Solve a Lambert problem^[R13]

Given

A craft on a circular orbit at r₁ = 7000 km must reach r₂ = 14000 km, 140° ahead in the same plane, in 1.5 h. Find |v₁|, the transfer a and e, and the departure burn.

Mark as done

Hint

Use the universal-variable method: compute A, solve F(z) = 0, then f, g and v₁ = (r₂ − f r₁)/g.

Answer. |v₁| = 8.657 km/s, a = 10235 km, e = 0.401, departure burn 2.389 km/s.

Working

A = sin Δθ √(r₁r₂/(1 − cos Δθ))4788.3 km
Root of F(z): z8.33762
v₁ = (2.2458, 8.3605, 0) km/s, |v₁|8.6569 km/s
Transfer orbit: a = 10235.3 km, e0.4006
Departure burn |v₁ − v_c| with v_c = 7.5461 km/s: 2.3890 km/s

Answer. |v₁| = 8.657 km/s, a = 10235 km, e = 0.401, departure burn 2.389 km/s.

Sanity check. r₂ = 14000 km must lie between the transfer perigee (6135 km) and apogee (14336 km), and it does.

Pythonedit a value, then Run

# Practice O12.2: Solve a Lambert problem
from math import pi, sqrt, sin, cos, sinh, cosh, acos, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r1m, r2m = 7000.0, 14000.0   # km
dtheta = radians(140)
tof = 1.5 * 3600      # s

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

    def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

    def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

    lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

r1 = [r1m, 0.0, 0.0]
r2 = [r2m * cos(dtheta), r2m * sin(dtheta), 0.0]
v1, v2, z, A, dth = lambert(r1, r2, tof, mu)
v1m = sqrt(sum(c * c for c in v1))
a = 1 / (2 / r1m - v1m**2 / mu)
h = r1[0] * v1[1] - r1[1] * v1[0]
e = sqrt(1 - h**2 / (mu * a))
vc = sqrt(mu / r1m)
dv = sqrt(v1[0]**2 + (v1[1] - vc)**2)
print(f"A = {A:.1f} km, z = {z:.5f}")
print(f"v1 = ({v1[0]:.4f}, {v1[1]:.4f}, 0) km/s, |v1| = {v1m:.4f} km/s")
print(f"a = {a:.1f} km, e = {e:.4f}, departure burn {dv:.4f} km/s")

# Practice O12.2: Solve a Lambert problem
from math import pi, sqrt, sin, cos, sinh, cosh, acos, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r1m, r2m = 7000.0, 14000.0   # km
dtheta = radians(140)
tof = 1.5 * 3600      # s

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

r1 = [r1m, 0.0, 0.0]
r2 = [r2m * cos(dtheta), r2m * sin(dtheta), 0.0]
v1, v2, z, A, dth = lambert(r1, r2, tof, mu)
v1m = sqrt(sum(c * c for c in v1))
a = 1 / (2 / r1m - v1m**2 / mu)
h = r1[0] * v1[1] - r1[1] * v1[0]
e = sqrt(1 - h**2 / (mu * a))
vc = sqrt(mu / r1m)
dv = sqrt(v1[0]**2 + (v1[1] - vc)**2)
print(f"A = {A:.1f} km, z = {z:.5f}")
print(f"v1 = ({v1[0]:.4f}, {v1[1]:.4f}, 0) km/s, |v1| = {v1m:.4f} km/s")
print(f"a = {a:.1f} km, e = {e:.4f}, departure burn {dv:.4f} km/s")

A = 4788.3 km, z = 8.33762
v1 = (2.2458, 8.3605, 0) km/s, |v1| = 8.6569 km/s
a = 10235.3 km, e = 0.4006, departure burn 2.3890 km/s

Practice O12.3 · challenge

Intercept: choose the flight time

Given

A chaser on a circular orbit at 7000 km wants to meet a target on a coplanar circular orbit at 9000 km that is currently 30° ahead. For flight times of 30, 45 and 60 min, find the total Δv (departure plus matching the target's velocity on arrival) and pick the cheapest.

Mark as done

Hint

The target moves while you fly: its angle at arrival is the lead angle plus n_target × Δt. Solve one Lambert problem per flight time.

Answer. The 60 min transfer is cheapest at 0.89 km/s total.

Working

Target mean motion n7.394437e-04 rad/s
Δt = 30 min: target at 106.3°, Δv_dep = 1.006, Δv_arr = 1.797, total 2.803 km/s
Δt = 45 min: target at 144.4°, Δv_dep = 0.523, Δv_arr = 0.739, total 1.262 km/s
Δt = 60 min: target at 182.5°, Δv_dep = 0.459, Δv_arr = 0.430, total 0.889 km/s
Cheapest: 60 min with 0.889 km/s

Answer. The 60 min transfer is cheapest at 0.89 km/s total.

Sanity check. For comparison a Hohmann transfer between the two radii costs 0.888 km/s but needs the right phasing; paying for a fixed arrival time always costs at least that much.

Pythonedit a value, then Run

# Practice O12.3: Intercept: choose the flight time
from math import pi, sqrt, sin, cos, sinh, cosh, acos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r1m, r2m = 7000.0, 9000.0    # km, chaser and target circular orbits
lead = radians(30)    # target ahead now

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

    def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

    def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

    lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

n_t = sqrt(mu / r2m**3)
vc1, vc2 = sqrt(mu / r1m), sqrt(mu / r2m)
best = None
for minutes in (30, 45, 60):
    tof = minutes * 60
    th = lead + n_t * tof                # target angle at arrival
    r1 = [r1m, 0.0, 0.0]
    r2 = [r2m * cos(th), r2m * sin(th), 0.0]
    v1, v2, z, A, dth = lambert(r1, r2, tof, mu)
    vt = [-vc2 * sin(th), vc2 * cos(th), 0.0]
    d1 = sqrt(v1[0]**2 + (v1[1] - vc1)**2 + v1[2]**2)
    d2 = sqrt(sum((vt[k] - v2[k])**2 for k in range(3)))
    print(f"{minutes} min: target at {degrees(th):.1f} deg, dep {d1:.3f} + arr {d2:.3f} = {d1 + d2:.3f} km/s")
    if best is None or d1 + d2 < best[1]:
        best = (minutes, d1 + d2)
print(f"cheapest: {best[0]} min with {best[1]:.3f} km/s")

# Practice O12.3: Intercept: choose the flight time
from math import pi, sqrt, sin, cos, sinh, cosh, acos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
r1m, r2m = 7000.0, 9000.0    # km, chaser and target circular orbits
lead = radians(30)    # target ahead now

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

n_t = sqrt(mu / r2m**3)
vc1, vc2 = sqrt(mu / r1m), sqrt(mu / r2m)
best = None
for minutes in (30, 45, 60):
    tof = minutes * 60
    th = lead + n_t * tof                # target angle at arrival
    r1 = [r1m, 0.0, 0.0]
    r2 = [r2m * cos(th), r2m * sin(th), 0.0]
    v1, v2, z, A, dth = lambert(r1, r2, tof, mu)
    vt = [-vc2 * sin(th), vc2 * cos(th), 0.0]
    d1 = sqrt(v1[0]**2 + (v1[1] - vc1)**2 + v1[2]**2)
    d2 = sqrt(sum((vt[k] - v2[k])**2 for k in range(3)))
    print(f"{minutes} min: target at {degrees(th):.1f} deg, dep {d1:.3f} + arr {d2:.3f} = {d1 + d2:.3f} km/s")
    if best is None or d1 + d2 < best[1]:
        best = (minutes, d1 + d2)
print(f"cheapest: {best[0]} min with {best[1]:.3f} km/s")

30 min: target at 106.3 deg, dep 1.006 + arr 1.797 = 2.803 km/s
45 min: target at 144.4 deg, dep 0.523 + arr 0.739 = 1.262 km/s
60 min: target at 182.5 deg, dep 0.459 + arr 0.430 = 0.889 km/s
cheapest: 60 min with 0.889 km/s

Common mistakes

Letting the solver pick the wrong direction. Decide prograde or retrograde first; it sets whether Δθ is the short or long angle.
Feeding a transfer angle of exactly 180°. The plane is undefined; offset one point slightly or treat the Hohmann case separately.
Forgetting that the target moves. In an intercept the arrival position is where the target will be, not where it is now.
Trusting a solver without a test. Propagate v₁ forward and check you land on r₂.

Recap from earlier chapters

(Chapter 4) What does Kepler's equation relate?

Mean anomaly (which grows uniformly with time) to eccentric anomaly: M = E − e sin E.

(Chapter 9) How long does a Hohmann transfer from 300 km to GEO take?

Half the transfer period: 5.28 h.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R12poliastro project, tests/test_iod.py::test_curtis52 (Curtis, Orbital Mechanics for Engineering Students, Example 5.2), commit 21fd771. Secondary record of a textbook case.
R13Izzo, D., "Revisiting Lambert's problem", Celestial Mechanics and Dynamical Astronomy 121 (2015) 1-15, doi:10.1007/s10569-014-9587-y; open preprint arXiv:1403.2705.

Where in each source, and links: References.

I have finished chapter 12

Chapter 13

Porkchop plots

35 minutes

The idea in one minute

A porkchop plot is a map of launch dates against flight times (or arrival dates), coloured by how much energy the trip needs. Each point is one Lambert problem: where Earth is on the departure day, where the target planet is on the arrival day, and the flight time between. The cheap region forms a closed valley that, drawn as contours, looks like a pork chop. Everything here uses a SIMPLIFIED model: circular, coplanar planetary orbits and an arbitrary model day 0 when both planets are lined up. It shows the idea, not real dates.

Simplified circular, coplanar model. The strongest shade is the cheapest: launch energy C3 is low only in one valley of departure days and flight times (the thin gap is the skipped 180° line).

You will be able to

Explain what a porkchop plot shows and how it is built from Lambert solves
Compute synodic periods and Hohmann phase angles
Read the cost of a late launch from the plot

What is plotted

On each grid point solve Lambert's problem around the Sun. Subtract Earth's velocity from v₁ to get the hyperbolic excess speed at departure; C3 = v∞² is the launch-vehicle energy, in km²/s². Subtracting the planet's velocity from v₂ gives the arrival v∞, which sets the capture burn.

Why windows recur

The geometry repeats every synodic period, the time for Earth to lap the target. For Mars that is about 26 months; for Venus about 19. Between windows the planets are badly placed and C3 is huge.

A simplified model, clearly labelled

Real porkchops use planetary ephemerides with eccentric, inclined orbits. Inclination turns the 180° transfer line into an expensive ridge and splits the valley in two (type I and type II transfers). The circular, coplanar model keeps the physics and removes the detail.^[R4]

Key equations

Synodic period

1/T_syn = |1/T₁ − 1/T₂|

where

T_syntime between repeats of the same planetary geometry, days

T₁, T₂orbital periods of the two planets, days

Departure energy

v∞ = v₁ − V_E, C3 = v∞²

where

v₁heliocentric departure velocity from Lambert, km/s

V_EEarth's heliocentric velocity, km/s

v∞hyperbolic excess speed, km/s

C3characteristic energy, km²/s²

Hohmann flight time

t_H = π√(a_t³/μ_Sun), a_t = (r₁ + r₂)/2

where

t_Hflight time of the Hohmann transfer, s

a_ttransfer semi-major axis, km

μ_Sungravitational parameter of the Sun, km³/s²

r₁, r₂orbit radii of the two planets, km

Phase angle at departure

φ = π − n₂ t_H

where

φangle by which the target planet must lead at departure, rad

n₂mean motion of the target planet, rad/s

t_HHohmann flight time, s

Worked examples

Worked example 13.1

The Hohmann benchmark to Mars^{[R2, R4]}

Given

In the simplified model (circular, coplanar orbits at 1.000000 and 1.523679 AU), find the Hohmann flight time, the hyperbolic excess speed and C3 at Earth, and where Mars must be at departure.

new symbols

t_HHohmann flight time, days

φrequired lead angle of the target planet at departure, deg

Working

a_t = (r_E + r_M)/2 = 1.8877e+08 km; t_H = π√(a_t³/μ_Sun)258.9 days
v∞ at Earth = √(μ_Sun(2/r_E − 1/a_t)) − √(μ_Sun/r_E) = 2.945 km/s, so C3 = v∞²8.67 km²/s²
v∞ at Mars2.649 km/s
Mars moves n_M t_H = 135.7° during the flight, so at departure it must lead Earth by φ = 180° − that44.3°

Answer. t_H = 259 days, C3 = 8.67 km²/s², v∞ = 2.94 km/s at Earth and 2.65 km/s at Mars, Mars 44.3° ahead at departure.

Sanity check. This lead angle recurs once per synodic period, which is why Mars windows come about every 26 months.

Pythonedit a value, then Run

# Worked example 13.1: The Hohmann benchmark to Mars
from math import pi, sqrt, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
rE, rM = 1.000000 * AU, 1.523679 * AU   # circular, coplanar model
day = 86400.0

at = (rE + rM) / 2
tH = pi * sqrt(at**3 / mu_sun)
vinfE = sqrt(mu_sun * (2 / rE - 1 / at)) - sqrt(mu_sun / rE)
vinfM = sqrt(mu_sun / rM) - sqrt(mu_sun * (2 / rM - 1 / at))
moved = sqrt(mu_sun / rM**3) * tH        # rad Mars moves during the flight
phi = pi - moved
print(f"a_t = {at:.4e} km, t_H = {tH/day:.1f} days")
print(f"v_inf at Earth = {vinfE:.3f} km/s, C3 = {vinfE**2:.3f} km^2/s^2")
print(f"v_inf at Mars = {vinfM:.3f} km/s")
print(f"Mars moves {degrees(moved):.1f} deg, so it must lead Earth by {degrees(phi):.2f} deg")

a_t = 1.8877e+08 km, t_H = 258.9 days
v_inf at Earth = 2.945 km/s, C3 = 8.671 km^2/s^2
v_inf at Mars = 2.649 km/s
Mars moves 135.7 deg, so it must lead Earth by 44.34 deg

Worked example 13.2

Reading the porkchop

Given

Solve Lambert's problem on a grid of departure days (every 5 days around the window) and flight times (120 to 420 days), compute C3 at each point and find the minimum.

Working

2914 grid points solved (points within 2° of a 180° transfer are skipped)
Grid minimum (5-day steps): day 684, 255 days, C38.68 km²/s²
Refined by a simplex search: day 683.8, flight 255.0 days, C38.68 km²/s²
Hohmann benchmark: C3 = 8.67 km²/s² at 259 days

Answer. Best departure is model day 684 with a 255 day flight and C3 = 8.68 km²/s², 0.1 % above Hohmann.

Sanity check. In this coplanar model the true optimum is the Hohmann transfer itself (a 180° arc), which the grid must skip; the best usable point sits just beside it. With real inclined orbits the 180° line becomes an expensive ridge and the optimum splits into a short-way and a long-way valley.

Pythonedit a value, then Run

# Worked example 13.2: Reading the porkchop
from math import pi, sqrt, sin, cos, sinh, cosh, acos, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
aE, aM = 1.000000 * AU, 1.523679 * AU   # circular, coplanar; both at angle 0 on model day 0
day = 86400.0

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

    def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

    def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

    lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

nE, nM = sqrt(mu_sun / aE**3), sqrt(mu_sun / aM**3)

def planet(a, n, t_days):               # position and velocity on a circular orbit
    th = n * t_days * day
    v = sqrt(mu_sun / a)
    return [a * cos(th), a * sin(th), 0.0], [-v * sin(th), v * cos(th), 0.0]

def c3(dep, tof):                       # departure C3 for a launch on day dep, flight tof days
    rE, vE = planet(aE, nE, dep)
    rM, _ = planet(aM, nM, dep + tof)
    th = (nM * (dep + tof) - nE * dep) * day % (2 * pi)
    if abs(degrees(th) - 180) < 2:
        return None                     # 180 deg transfer: Lambert plane undefined
    v1, _, _, _, _ = lambert(rE, rM, tof * day, mu_sun, iters=60)
    return sum((v1[k] - vE[k])**2 for k in range(3))

# Hohmann benchmark and the window it implies
at = (aE + aM) / 2
tH = pi * sqrt(at**3 / mu_sun)
C3H = (sqrt(mu_sun * (2 / aE - 1 / at)) - sqrt(mu_sun / aE))**2
phi = pi - nM * tH
d_star = ((2 * pi - phi) % (2 * pi)) / (nE - nM) / day
# grid: departure every 5 days around the window, flight 120 to 420 days
best, count = None, 0
step = 5.0           # days between grid points
for i in range(int(240 / step) + 1):          # departures d_star - 120 ... d_star + 120
    for j in range(int((420 - 120) / step) + 1):  # flight times 120 ... 420 days
        d, t = d_star - 120 + step * i, 120.0 + step * j
        c = c3(d, t)
        if c is not None:
            count += 1
            if best is None or c < best[0]:
                best = (c, d, t)
print(f"{count} grid points; grid minimum day {best[1]:.0f}, {best[2]:.0f} days, C3 = {best[0]:.2f} km^2/s^2")
# refine with a simple pattern search (halve the step when no neighbour is better)
c, d, t = best
step = 2.5
while step > 0.01:
    moved = False
    for dd, dt in ((step, 0), (-step, 0), (0, step), (0, -step)):
        cn = c3(d + dd, t + dt)
        if cn is not None and cn < c:
            c, d, t, moved = cn, d + dd, t + dt, True
            break
    if not moved:
        step /= 2
print(f"refined: day {d:.1f}, flight {t:.1f} days, C3 = {c:.3f} km^2/s^2")
print(f"Hohmann: C3 = {C3H:.3f} km^2/s^2 at {tH/day:.0f} days; the optimum is {100*(c/C3H - 1):.1f} % above it")

# Worked example 13.2: Reading the porkchop
from math import pi, sqrt, sin, cos, sinh, cosh, acos, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
aE, aM = 1.000000 * AU, 1.523679 * AU   # circular, coplanar; both at angle 0 on model day 0
day = 86400.0

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

nE, nM = sqrt(mu_sun / aE**3), sqrt(mu_sun / aM**3)

def planet(a, n, t_days):               # position and velocity on a circular orbit
    th = n * t_days * day
    v = sqrt(mu_sun / a)
    return [a * cos(th), a * sin(th), 0.0], [-v * sin(th), v * cos(th), 0.0]

def c3(dep, tof):                       # departure C3 for a launch on day dep, flight tof days
    rE, vE = planet(aE, nE, dep)
    rM, _ = planet(aM, nM, dep + tof)
    th = (nM * (dep + tof) - nE * dep) * day % (2 * pi)
    if abs(degrees(th) - 180) < 2:
        return None                     # 180 deg transfer: Lambert plane undefined
    v1, _, _, _, _ = lambert(rE, rM, tof * day, mu_sun, iters=60)
    return sum((v1[k] - vE[k])**2 for k in range(3))

# Hohmann benchmark and the window it implies
at = (aE + aM) / 2
tH = pi * sqrt(at**3 / mu_sun)
C3H = (sqrt(mu_sun * (2 / aE - 1 / at)) - sqrt(mu_sun / aE))**2
phi = pi - nM * tH
d_star = ((2 * pi - phi) % (2 * pi)) / (nE - nM) / day
# grid: departure every 5 days around the window, flight 120 to 420 days
best, count = None, 0
step = 5.0           # days between grid points
for i in range(int(240 / step) + 1):          # departures d_star - 120 ... d_star + 120
    for j in range(int((420 - 120) / step) + 1):  # flight times 120 ... 420 days
        d, t = d_star - 120 + step * i, 120.0 + step * j
        c = c3(d, t)
        if c is not None:
            count += 1
            if best is None or c < best[0]:
                best = (c, d, t)
print(f"{count} grid points; grid minimum day {best[1]:.0f}, {best[2]:.0f} days, C3 = {best[0]:.2f} km^2/s^2")
# refine with a simple pattern search (halve the step when no neighbour is better)
c, d, t = best
step = 2.5
while step > 0.01:
    moved = False
    for dd, dt in ((step, 0), (-step, 0), (0, step), (0, -step)):
        cn = c3(d + dd, t + dt)
        if cn is not None and cn < c:
            c, d, t, moved = cn, d + dd, t + dt, True
            break
    if not moved:
        step /= 2
print(f"refined: day {d:.1f}, flight {t:.1f} days, C3 = {c:.3f} km^2/s^2")
print(f"Hohmann: C3 = {C3H:.3f} km^2/s^2 at {tH/day:.0f} days; the optimum is {100*(c/C3H - 1):.1f} % above it")

2914 grid points; grid minimum day 684, 255 days, C3 = 8.68 km^2/s^2
refined: day 683.8, flight 255.0 days, C3 = 8.676 km^2/s^2
Hohmann: C3 = 8.671 km^2/s^2 at 259 days; the optimum is 0.1 % above it

Practice

Practice O13.1 · warm-up

How often do windows open?^[R4]

Given

Using the simplified circular orbits, find the orbital periods of Earth (1.000000 AU) and Mars (1.523679 AU) and the Earth-Mars synodic period.

Mark as done

Hint

T = 2π√(a³/μ_Sun); 1/T_syn = |1/T_E − 1/T_M|.

Answer. T_syn = 780 days, about 26 months between Mars windows.

Working

T_E = 365.26 days, T_M686.97 days
1/T_syn = |1/365.26 − 1/686.97|0.001282 per day
T_syn = 779.9 days25.6 months

Answer. T_syn = 780 days, about 26 months between Mars windows.

Sanity check. Close to two Earth years, because Mars takes nearly two years to go round: Earth must lap it.

Pythonedit a value, then Run

# Practice O13.1: How often do windows open?
from math import pi, sqrt

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
aE, aM = 1.000000 * AU, 1.523679 * AU
day = 86400.0

TE = 2 * pi * sqrt(aE**3 / mu_sun) / day
TM = 2 * pi * sqrt(aM**3 / mu_sun) / day
Tsyn = 1 / abs(1 / TE - 1 / TM)
print(f"T_E = {TE:.2f} days, T_M = {TM:.2f} days")
print(f"synodic period = {Tsyn:.1f} days = {Tsyn / (365.25 / 12):.1f} months")

T_E = 365.26 days, T_M = 686.97 days
synodic period = 779.9 days = 25.6 months

Practice O13.2 · standard

Lining up for Venus^[R4]

Given

For a Hohmann transfer from Earth inward to Venus (0.723332 AU) in the simplified model, find the flight time, where Venus must be relative to Earth at departure, and how often that geometry recurs.

Mark as done

Hint

Same logic as Mars: φ = π − n_V t_H. A negative angle means Venus must be behind Earth.

Answer. t_H = 146 days, Venus 54.0° behind Earth at departure, windows every 584 days.

Working

a_t = (r_E + r_V)/2 = 1.2890e+08 km; t_H = π√(a_t³/μ_Sun)146.1 days
n_V t_H = 234.0°, so φ = 180° − 234.0°-54.0°
Synodic period Earth-Venus584 days

Answer. t_H = 146 days, Venus 54.0° behind Earth at departure, windows every 584 days.

Sanity check. Venus is faster than Earth, so it must start behind and catch up to the arrival point: a negative lead angle is the expected sign.

Pythonedit a value, then Run

# Practice O13.2: Lining up for Venus
from math import pi, sqrt, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
aE, aV = 1.000000 * AU, 0.723332 * AU
day = 86400.0

at = (aE + aV) / 2
tH = pi * sqrt(at**3 / mu_sun)
nV = sqrt(mu_sun / aV**3)
phi = pi - nV * tH                       # negative: Venus behind Earth
TE = 2 * pi * sqrt(aE**3 / mu_sun)
TV = 2 * pi * sqrt(aV**3 / mu_sun)
Tsyn = 1 / abs(1 / TE - 1 / TV) / day
print(f"t_H = {tH/day:.1f} days, Venus moves {degrees(nV * tH):.1f} deg")
print(f"phi = {degrees(phi):.2f} deg: Venus {abs(degrees(phi)):.2f} deg behind Earth at departure")
print(f"windows every {Tsyn:.0f} days")

t_H = 146.1 days, Venus moves 234.0 deg
phi = -54.03 deg: Venus 54.03 deg behind Earth at departure
windows every 584 days

Practice O13.3 · challenge

The cost of a late launch

Given

Take the optimum found in this chapter (model day 683.8, 255 day flight, C3 = 8.68 km²/s²). The launch slips 30 days and the flight time is kept. Solve the new Lambert problem and find the new C3 and the percentage increase.

Mark as done

Hint

Departure day d + 30 fixes Earth's position; arrival day d + 30 + t fixes Mars's. Solve Lambert with μ_Sun and subtract Earth's velocity.

Answer. C3 rises to 14.4 km²/s², 66 % more than the optimum.

Working

Earth at model day 714, Mars at model day 969; transfer angle 164.2°
Lambert gives v∞ = 3.790 km/s, so C314.37 km²/s²
Increase = 14.37/8.68 − 166 %

Answer. C3 rises to 14.4 km²/s², 66 % more than the optimum.

Sanity check. A month's slip moves you off the valley floor of the porkchop. Launch vehicles quote payload against C3, so this directly costs payload mass; it is why planetary windows are only a few weeks wide.

Pythonedit a value, then Run

# Practice O13.3: The cost of a late launch
from math import pi, sqrt, sin, cos, sinh, cosh, acos, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
aE, aM = 1.000000 * AU, 1.523679 * AU   # both at angle 0 on model day 0
day = 86400.0
dep0, tof0 = 684.0, 255.0   # optimum quoted in the problem: model day, flight days
slip = 30.0                  # days late

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

    def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

    def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

    lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

nE, nM = sqrt(mu_sun / aE**3), sqrt(mu_sun / aM**3)

def planet(a, n, t_days):
    th = n * t_days * day
    v = sqrt(mu_sun / a)
    return [a * cos(th), a * sin(th), 0.0], [-v * sin(th), v * cos(th), 0.0]

def c3(dep, tof):
    rE, vE = planet(aE, nE, dep)
    rM, _ = planet(aM, nM, dep + tof)
    v1, _, _, _, dth = lambert(rE, rM, tof * day, mu_sun)
    return sum((v1[k] - vE[k])**2 for k in range(3)), degrees(dth)

# the quoted day and flight time are rounded: sharpen the optimum with the pattern search of
# worked example 13.2 (keeping clear of the 180 deg transfer, as the porkchop grid does)
def cost(dep, tof):
    c, ang = c3(dep, tof)
    return c if abs(ang - 180) >= 2 else None

c_opt, dep_opt, tof = cost(dep0, tof0), dep0, tof0
step = 2.5            # days, first search step
while step > 0.01:    # stop when the step is below 0.01 day
    for dd, dt in ((step, 0), (-step, 0), (0, step), (0, -step)):
        cn = cost(dep_opt + dd, tof + dt)
        if cn is not None and cn < c_opt:
            c_opt, dep_opt, tof = cn, dep_opt + dd, tof + dt
            break
    else:
        step /= 2
print(f"refined optimum: day {dep_opt:.1f}, flight {tof:.1f} days")
c_late, ang = c3(dep_opt + slip, tof)
print(f"optimum C3 = {c_opt:.2f} km^2/s^2")
print(f"late: Earth day {dep_opt + slip:.0f}, Mars day {dep_opt + slip + tof:.0f}, transfer angle {ang:.1f} deg")
print(f"v_inf = {sqrt(c_late):.3f} km/s, C3 = {c_late:.2f} km^2/s^2, {100*(c_late/c_opt - 1):.1f} % more")

# Practice O13.3: The cost of a late launch
from math import pi, sqrt, sin, cos, sinh, cosh, acos, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
aE, aM = 1.000000 * AU, 1.523679 * AU   # both at angle 0 on model day 0
day = 86400.0
dep0, tof0 = 684.0, 255.0   # optimum quoted in the problem: model day, flight days
slip = 30.0                  # days late

def stumpC(z):
    if z > 1e-8:
        return (1 - cos(sqrt(z))) / z
    if z < -1e-8:
        return (cosh(sqrt(-z)) - 1) / (-z)
    return 1 / 2 - z / 24 + z * z / 720       # Taylor series near z = 0

def stumpS(z):
    if z > 1e-8:
        s = sqrt(z); return (s - sin(s)) / s**3
    if z < -1e-8:
        s = sqrt(-z); return (sinh(s) - s) / s**3
    return 1 / 6 - z / 120 + z * z / 5040     # Taylor series near z = 0

def lambert(r1, r2, tof, mu, iters=100):
    """Universal-variable Lambert solver (prograde, one revolution). r1, r2 = [x, y, z] in km."""
    r1m, r2m = sqrt(sum(c * c for c in r1)), sqrt(sum(c * c for c in r2))
    dth = acos(max(-1, min(1, sum(p * q for p, q in zip(r1, r2)) / (r1m * r2m))))
    if r1[0] * r2[1] - r1[1] * r2[0] < 0:          # prograde: going the long way round
        dth = 2 * pi - dth
    A = sin(dth) * sqrt(r1m * r2m / (1 - cos(dth)))

def y(z):
        return r1m + r2m + A * (z * stumpS(z) - 1) / sqrt(stumpC(z))

def F(z):                                      # flight time error; grows with z
        yy = y(z)
        if yy < 0:
            return -1.0                            # unphysical region: transfer too short
        return (yy / stumpC(z)) ** 1.5 * stumpS(z) + A * sqrt(yy) - sqrt(mu) * tof

lo, hi = -100.0, 4 * pi**2 - 1e-9              # bisection on z
    for _ in range(iters):
        mid = (lo + hi) / 2
        if F(mid) < 0:
            lo = mid
        else:
            hi = mid
    z = (lo + hi) / 2
    yy = y(z)
    f, g, gd = 1 - yy / r1m, A * sqrt(yy / mu), 1 - yy / r2m
    v1 = [(r2[k] - f * r1[k]) / g for k in range(3)]
    v2 = [(gd * r2[k] - r1[k]) / g for k in range(3)]
    return v1, v2, z, A, dth

nE, nM = sqrt(mu_sun / aE**3), sqrt(mu_sun / aM**3)

def planet(a, n, t_days):
    th = n * t_days * day
    v = sqrt(mu_sun / a)
    return [a * cos(th), a * sin(th), 0.0], [-v * sin(th), v * cos(th), 0.0]

def c3(dep, tof):
    rE, vE = planet(aE, nE, dep)
    rM, _ = planet(aM, nM, dep + tof)
    v1, _, _, _, dth = lambert(rE, rM, tof * day, mu_sun)
    return sum((v1[k] - vE[k])**2 for k in range(3)), degrees(dth)

# the quoted day and flight time are rounded: sharpen the optimum with the pattern search of
# worked example 13.2 (keeping clear of the 180 deg transfer, as the porkchop grid does)
def cost(dep, tof):
    c, ang = c3(dep, tof)
    return c if abs(ang - 180) >= 2 else None

c_opt, dep_opt, tof = cost(dep0, tof0), dep0, tof0
step = 2.5            # days, first search step
while step > 0.01:    # stop when the step is below 0.01 day
    for dd, dt in ((step, 0), (-step, 0), (0, step), (0, -step)):
        cn = cost(dep_opt + dd, tof + dt)
        if cn is not None and cn < c_opt:
            c_opt, dep_opt, tof = cn, dep_opt + dd, tof + dt
            break
    else:
        step /= 2
print(f"refined optimum: day {dep_opt:.1f}, flight {tof:.1f} days")
c_late, ang = c3(dep_opt + slip, tof)
print(f"optimum C3 = {c_opt:.2f} km^2/s^2")
print(f"late: Earth day {dep_opt + slip:.0f}, Mars day {dep_opt + slip + tof:.0f}, transfer angle {ang:.1f} deg")
print(f"v_inf = {sqrt(c_late):.3f} km/s, C3 = {c_late:.2f} km^2/s^2, {100*(c_late/c_opt - 1):.1f} % more")

refined optimum: day 683.8, flight 255.0 days
optimum C3 = 8.68 km^2/s^2
late: Earth day 714, Mars day 969, transfer angle 164.2 deg
v_inf = 3.791 km/s, C3 = 14.37 km^2/s^2, 65.7 % more

Common mistakes

Reading the simplified plot as real dates. Model day 0 is an arbitrary reference; real windows need ephemerides.
Quoting C3 in km/s. C3 is a speed squared: km²/s².
Forgetting arrival. The cheapest launch can bring a large arrival v∞ and an expensive capture.
Solving exactly on the 180° line. It is singular for Lambert; skip it or offset slightly.

Recap from earlier chapters

(Chapter 12) What are the inputs and outputs of Lambert's problem?

Inputs r₁, r₂ and Δt; outputs v₁ and v₂.

(Chapter 9) Why is a Hohmann transfer only a benchmark?

It is the minimum two-burn Δv between coplanar circular orbits, but it fixes both the flight time and the phasing.

Sources for this chapter

R2NASA JPL Solar System Dynamics, "Astrodynamic Parameters" (planetary GMs from DE440; Park et al. 2021, AJ 161:105).
R4Standish, E.M. and Williams, J.G., "Approximate Positions of the Planets", NASA JPL SSD.
R7IAU 2012 Resolution B2, "Re-definition of the astronomical unit of length", XXVIII General Assembly, Beijing, Aug 2012 (text hosted by SYRTE, Observatoire de Paris).

Where in each source, and links: References.

I have finished chapter 13

Chapter 14

Patched conics, sphere of influence and the Oberth effect

40 minutes

The idea in one minute

An interplanetary trip has three bodies pulling at once, but almost everywhere one of them dominates. Patched conics uses that: near Earth, the craft follows a hyperbola around Earth; far from planets, an ellipse around the Sun; near Mars, a hyperbola around Mars. The pieces are glued at the sphere of influence by matching velocities. The link between them is the hyperbolic excess speed v∞: what is left of the speed after climbing out of a planet's gravity well. Burning deep in the well, where you are fast, buys more v∞ than burning far away: the Oberth effect.

The same 1 km/s burn on an escape trajectory: deep in the gravity well it gives 4.8 km/s of excess speed, far away only 1 km/s.

You will be able to

Split an interplanetary trip into planet-centred and Sun-centred conics
Size a departure or capture burn from a hyperbolic excess speed
Explain and compute the Oberth effect

Sphere of influence

Inside a radius r = a(m/M)^2/5 the planet's pull dominates the Sun's disturbance of the motion around it. For Earth that is under a million kilometres; outside it, the Sun takes over. It is an approximation for patching, not a physical wall.^{[R2, R4]}

Hyperbolic excess speed

A craft leaving a planet on a hyperbola keeps a speed v∞ at large distance; its energy v²/2 − μ/r equals v∞²/2. In the Sun's frame its velocity is the planet's velocity plus v∞. C3 = v∞² is the number launch vehicles quote.

Departure and capture

The burn happens at periapsis of the planet-centred hyperbola. The required speed is √(v∞² + 2μ/r_p); subtract the parking-orbit speed for departure, or the target-orbit speed for capture.

The Oberth effect

Energy gained by a burn is v·Δv (to first order), so the same Δv adds more energy where v is large. Planetary departures burn at the lowest point of the parking orbit for exactly this reason.

Key equations

Sphere of influence

r_SOI = a (μ_p/μ_Sun)^2/5

where

r_SOIsphere of influence radius, km

aplanet's orbit radius about the Sun, km

μ_pplanet's gravitational parameter, km³/s²

μ_Sungravitational parameter of the Sun, km³/s²

Speed at periapsis of a hyperbola

v_p = √(v∞² + 2μ/r_p)

where

v_pspeed at periapsis, km/s

v∞hyperbolic excess speed, km/s

μplanet's gravitational parameter, km³/s²

r_pperiapsis radius, km

Departure burn from a circular orbit

Δv = √(v∞² + 2μ/r_p) − √(μ/r_p)

where

Δvburn at periapsis, km/s

v∞required excess speed, km/s

μplanet's gravitational parameter, km³/s²

r_pparking-orbit radius, km

Hyperbola energy and C3

C3 = v∞² = −μ/a

where

C3characteristic energy, km²/s²

v∞hyperbolic excess speed, km/s

μplanet's gravitational parameter, km³/s²

ahyperbola semi-major axis (negative), km

Oberth effect

v∞ = √((v + Δv)² − v_esc²)

where

vspeed just before the burn, km/s

Δvburn along the velocity, km/s

v_escescape speed at that radius, √(2μ/r), km/s

v∞resulting excess speed, km/s

Best radius for circular capture

r_opt = 2μ/v∞², Δv_min = v∞/√2

where

r_optcapture radius that minimises the burn, km

μplanet's gravitational parameter, km³/s²

v∞arrival excess speed, km/s

Δv_minminimum capture burn, km/s

Worked examples

Worked example 14.1

Leaving Earth for Mars^{[R2, R3]}

Given

A spacecraft in a 300 km circular parking orbit must leave Earth with the Hohmann excess speed to Mars, v∞ = 2.945 km/s (Chapter 13). Find C3, the speed needed at periapsis of the departure hyperbola and the burn.

new symbols

v_ccircular speed of the parking orbit, km/s

Working

C3 = v∞²8.671 km²/s²
Periapsis radius r_p = 6378.137 + 3006678.137 km
v_p = √(v∞² + 2μ/r_p) = √(8.671 + 119.375)11.3157 km/s
Circular speed v_c = √(μ/r_p)7.7258 km/s
Δv = v_p − v_c = 3.5900 km/s; hyperbola eccentricity e = 1 + r_p v∞²/μ1.1453

Answer. C3 = 8.67 km²/s², v_p = 11.316 km/s, burn Δv = 3.590 km/s.

Sanity check. Escape alone from 300 km costs 3.200 km/s; adding a 2.94 km/s excess costs only 0.390 km/s more. That bargain is the Oberth effect.

Pythonedit a value, then Run

# Worked example 14.1: Leaving Earth for Mars
from math import sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
h = 300.0             # km, parking orbit

# Hohmann excess speed to Mars (chapter 13 model)
rE, rM = 1.0 * AU, 1.523679 * AU
at = (rE + rM) / 2
vinf = sqrt(mu_sun * (2 / rE - 1 / at)) - sqrt(mu_sun / rE)
rp = R_E + h
vp = sqrt(vinf**2 + 2 * mu / rp)         # periapsis speed of the departure hyperbola
vc = sqrt(mu / rp)
e = 1 + rp * vinf**2 / mu
print(f"v_inf = {vinf:.3f} km/s, C3 = {vinf**2:.3f} km^2/s^2")
print(f"v_p = {vp:.4f} km/s, v_c = {vc:.4f} km/s, burn = {vp - vc:.4f} km/s, e = {e:.4f}")

v_inf = 2.945 km/s, C3 = 8.671 km^2/s^2
v_p = 11.3157 km/s, v_c = 7.7258 km/s, burn = 3.5900 km/s, e = 1.1453

Worked example 14.2

The Oberth effect in numbers

Given

A spacecraft is on a parabolic (just-escaping) trajectory. It has 1.0 km/s of Δv to spend. Compare burning at periapsis, 300 km up, with burning far from Earth where its speed has dropped to almost zero.

Working

At periapsis the speed is the escape speed √(2μ/r)10.9259 km/s
After the burn: 11.9259 km/s; far away v∞ = √(v² − v_esc²)4.7804 km/s
Far away the speed is about 0, so the same burn gives v∞1.0000 km/s
Gain factor4.78

Answer. Burning at periapsis gives v∞ = 4.78 km/s; burning far away gives 1.00 km/s: 4.8 times as much.

Sanity check. Kinetic energy grows with v², so a fixed Δv adds the most energy where you are already fastest: deep in the gravity well, at periapsis.

Pythonedit a value, then Run

# Worked example 14.2: The Oberth effect in numbers
from math import sqrt

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 300.0             # km, periapsis altitude
dv = 1.0              # km/s to spend

r = R_E + h
v_esc = sqrt(2 * mu / r)                 # parabolic: speed at periapsis = escape speed
vinf_low = sqrt((v_esc + dv)**2 - v_esc**2)
vinf_far = dv                            # far away the speed is about zero
print(f"periapsis speed {v_esc:.4f} km/s, after the burn {v_esc + dv:.4f} km/s")
print(f"burn at periapsis: v_inf = {vinf_low:.4f} km/s; far away: v_inf = {vinf_far:.4f} km/s")
print(f"gain factor {vinf_low / vinf_far:.2f}")

periapsis speed 10.9259 km/s, after the burn 11.9259 km/s
burn at periapsis: v_inf = 4.7804 km/s; far away: v_inf = 1.0000 km/s
gain factor 4.78

Practice

Practice O14.1 · warm-up

How big is Mars's neighbourhood?^{[R2, R4]}

Given

Find the radius of the sphere of influence of Mars (orbit radius 1.523679 AU, μ = 42828 km³/s²) and of Earth, using r = a(μ_p/μ_Sun)^2/5.

Mark as done

Hint

Convert AU to km first (1 AU = 149,597,870.7 km).

Answer. Mars SOI ≈ 577,225 km (170 Mars radii); Earth SOI ≈ 924,647 km.

Working

Mars: a = 2.2794e+08 km; (μ_M/μ_Sun)^0.4 = 2.5324e-03; r5.7723e+05 km
Earth: a = 1.4960e+08 km; r9.2465e+05 km
Mars SOI in planet radii: 170 R_M

Answer. Mars SOI ≈ 577,225 km (170 Mars radii); Earth SOI ≈ 924,647 km.

Sanity check. Earth's SOI is about 2.4 times the Moon's distance, which is why the Moon is firmly Earth's.

Pythonedit a value, then Run

# Practice O14.1: How big is Mars's neighbourhood?
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
mu_M, R_M = 42828.0, 3396.2  # Mars mu (km^3/s^2) and radius (km), JPL SSD (DE440, Archinal 2018)
mu_E = 398600.4418    # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)

aM, aE = 1.523679 * AU, 1.0 * AU
soi_M = aM * (mu_M / mu_sun) ** 0.4
soi_E = aE * (mu_E / mu_sun) ** 0.4
print(f"Mars SOI = {soi_M:.1f} km = {soi_M / R_M:.0f} Mars radii")
print(f"Earth SOI = {soi_E:.1f} km")

Mars SOI = 577225.3 km = 170 Mars radii
Earth SOI = 924646.8 km

Practice O14.2 · standard

Capture at Mars^{[R2, R3]}

Given

A spacecraft arrives at Mars with v∞ = 2.6489 km/s (Hohmann arrival, from Chapter 13) and periapsis 400 km. Find the capture burn into (a) a 400 km circular orbit and (b) an ellipse 400 × 20,000 km. Mars radius 3396.2 km, μ = 42828 km³/s².

Mark as done

Hint

Speed at periapsis of the arrival hyperbola is √(v∞² + 2μ/r_p). Subtract the speed of the orbit you want at that point.

Answer. (a) 2.08 km/s into the circular orbit; (b) 1.03 km/s into the ellipse.

Working

r_p = 3796.2 km; v_p = √(7.017 + 22.564)5.4388 km/s
(a) circular speed √(μ/r_p) = 3.3588 km/s, Δv2.0799 km/s
(b) ellipse a = (3796.2 + 23396.2)/2 = 13596.2 km; periapsis speed 4.4061 km/s; Δv1.0327 km/s

Answer. (a) 2.08 km/s into the circular orbit; (b) 1.03 km/s into the ellipse.

Sanity check. Capturing into a long ellipse saves 1.05 km/s; many missions capture loosely and lower the orbit later, often by aerobraking.

Pythonedit a value, then Run

# Practice O14.2: Capture at Mars
from math import sqrt

mu_M, R_M = 42828.0, 3396.2  # Mars mu (km^3/s^2) and radius (km), JPL SSD (DE440, Archinal 2018)
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
h_p = 400.0           # km, periapsis altitude
h_apo = 20000.0       # km, apoapsis altitude of the capture ellipse

rE, rM = 1.0 * AU, 1.523679 * AU         # Hohmann arrival v_inf (chapter 13), 2.649 km/s
at = (rE + rM) / 2
vinf = sqrt(mu_sun / rM) - sqrt(mu_sun * (2 / rM - 1 / at))
rp = R_M + h_p
vp = sqrt(vinf**2 + 2 * mu_M / rp)
dv_circ = vp - sqrt(mu_M / rp)
a_ell = (rp + R_M + h_apo) / 2
dv_ell = vp - sqrt(mu_M * (2 / rp - 1 / a_ell))
print(f"v_inf = {vinf:.3f} km/s, periapsis speed {vp:.4f} km/s")
print(f"(a) circular capture {dv_circ:.4f} km/s; (b) ellipse capture {dv_ell:.4f} km/s")

v_inf = 2.649 km/s, periapsis speed 5.4388 km/s
(a) circular capture 2.0799 km/s; (b) ellipse capture 1.0327 km/s

Practice O14.3 · challenge

The best height to capture^{[R2, R3]}

Given

For capture into a circular orbit at Mars with v∞ = 2.6489 km/s (Hohmann arrival, from Chapter 13), the burn Δv(r) = √(v∞² + 2μ/r) − √(μ/r) depends on the capture radius r. Find the radius that minimises it, the altitude, the minimum Δv, and compare with the 400 km case.

Mark as done

Hint

Differentiate Δv(r) with respect to r and set it to zero; the algebra gives r = 2μ/v∞².

Answer. r_opt = 12,208 km (altitude 8,811 km), Δv_min = 1.873 km/s, 207 m/s less than at 400 km.

Working

dΔv/dr = 0 gives r_opt = 2μ/v∞² = 2 × 42828 / 7.016612208 km
Altitude = 12208 − 3396.28811 km
Δv_min = v∞/√21.8731 km/s
At 400 km it was 2.0799 km/s; a numerical scan confirms the minimum at 12208 km

Answer. r_opt = 12,208 km (altitude 8,811 km), Δv_min = 1.873 km/s, 207 m/s less than at 400 km.

Sanity check. Here the Oberth effect and the cost of a fast circular orbit pull in opposite directions; the optimum is a balance, and it is shallow, which is why mission designers rarely capture that high.

Pythonedit a value, then Run

# Practice O14.3: The best height to capture
from math import sqrt

mu_M, R_M = 42828.0, 3396.2  # Mars mu (km^3/s^2) and radius (km), JPL SSD (DE440, Archinal 2018)
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
r_400 = R_M + 400.0   # km, the earlier case

rE, rM = 1.0 * AU, 1.523679 * AU         # Hohmann arrival v_inf, 2.649 km/s
at = (rE + rM) / 2
vinf = sqrt(mu_sun / rM) - sqrt(mu_sun * (2 / rM - 1 / at))

def dv(r):
    return sqrt(vinf**2 + 2 * mu_M / r) - sqrt(mu_M / r)

r_opt = 2 * mu_M / vinf**2               # d(dv)/dr = 0
scan = min(range(4000, 40000), key=dv)   # numerical check, 1 km steps
print(f"r_opt = {r_opt:.1f} km (scan: {scan} km), altitude {r_opt - R_M:.1f} km")
print(f"dv_min = {dv(r_opt):.4f} km/s (= v_inf/sqrt 2 = {vinf / sqrt(2):.4f})")
print(f"at 400 km: {dv(r_400):.4f} km/s, so {1000 * (dv(r_400) - dv(r_opt)):.0f} m/s more")

r_opt = 12207.5 km (scan: 12208 km), altitude 8811.3 km
dv_min = 1.8731 km/s (= v_inf/sqrt 2 = 1.8731)
at 400 km: 2.0799 km/s, so 207 m/s more

Common mistakes

Adding v∞ to the escape speed. Speeds add in squares: v_p = √(v∞² + v_esc²).
Using the Sun's μ inside the sphere of influence, or the planet's outside it.
Treating the sphere of influence as a hard boundary. It is a patching approximation.
Burning far from the planet to save propellant. The Oberth effect says the opposite.

Recap from earlier chapters

(Chapter 3) What is the escape speed at 300 km?

√(2μ/r) = 10.926 km/s.

(Chapter 13) What does C3 measure?

The square of the departure hyperbolic excess speed, in km²/s².

Sources for this chapter

R2NASA JPL Solar System Dynamics, "Astrodynamic Parameters" (planetary GMs from DE440; Park et al. 2021, AJ 161:105).
R3NASA JPL Solar System Dynamics, "Planetary Physical Parameters" (radii from Archinal et al. 2018, CMDA 130:22; change log 2019-12-12).
R4Standish, E.M. and Williams, J.G., "Approximate Positions of the Planets", NASA JPL SSD.

Where in each source, and links: References.

I have finished chapter 14

Chapter 15

Gravity assists

35 minutes

The idea in one minute

A gravity assist looks like free energy, and in the Sun's frame it is. Relative to the planet, the craft comes in and goes out at the same speed; the planet only turns its direction. But the planet is itself moving fast around the Sun. Add the planet's velocity to the turned v∞ and the craft's speed around the Sun has changed, up or down depending on which side it passed. The energy comes from the planet, which slows by an immeasurably small amount.

Relative to Jupiter the probe keeps the same speed (it stays on the dashed circle); added to Jupiter's motion, the turned v∞ gives more speed around the Sun.

You will be able to

Compute the turn angle of a flyby from v∞ and the periapsis
Build the velocity triangle and find the speed change around the Sun
Explain why passing behind a planet adds energy and in front removes it

In the planet's frame

The flyby is a hyperbola with eccentricity e = 1 + r_p v∞²/μ. The incoming and outgoing asymptotes are separated by the turn angle δ = 2 asin(1/e). A closer pass or a slower approach turns more.

In the Sun's frame

Velocities add: before, v = V_p + v∞,in; after, v = V_p + v∞,out. Since both v∞ vectors have the same length, the outgoing tip lies on a circle centred on the tip of V_p. Where on the circle it lands is set by δ and the side of the pass.

Leading and trailing passes

Passing behind the planet (its trailing side) turns v∞ towards the planet's direction of motion: the craft speeds up. Passing in front (leading side) does the reverse: the craft slows, which is how missions fall inwards to Mercury or the Sun.

Key equations

Flyby eccentricity

e = 1 + r_p v∞²/μ_p

where

eeccentricity of the flyby hyperbola, dimensionless

r_pclosest-approach radius from the planet's centre, km

v∞hyperbolic excess speed relative to the planet, km/s

μ_pplanet's gravitational parameter, km³/s²

Turn angle

δ = 2 asin(1/e)

where

δangle between incoming and outgoing v∞, deg

eflyby eccentricity, dimensionless

Frames add

v = V_p + v∞, |v∞,out| = |v∞,in|

where

vcraft's velocity relative to the Sun, km/s

V_pplanet's velocity relative to the Sun, km/s

v∞craft's velocity relative to the planet, far from it, km/s

Velocity change delivered

Δv = 2 v∞ sin(δ/2) = 2v∞/e

where

Δvmagnitude of the velocity change, km/s

v∞excess speed, km/s

δturn angle, rad

eflyby eccentricity, dimensionless

Energy change around the Sun

Δε = V_p · (v∞,out − v∞,in)

where

Δεchange of heliocentric specific energy, km²/s²

V_pplanet's heliocentric velocity, km/s

v∞,in, v∞,outexcess velocities before and after, km/s

Worked examples

Worked example 15.1

A Jupiter flyby, from both sides^{[R3, R8]}

Given

A probe on a solar orbit with perihelion 1 AU and aphelion 6 AU meets Jupiter (circular orbit, 5.2026 AU) on its way out. It passes at 20 Jupiter radii. Find v∞, the turn angle, and the probe's speed around the Sun after passing behind Jupiter and in front of it.

new symbols

V_JJupiter's heliocentric velocity, km/s

Working

Probe at r = 7.7830e+08 km: v = 9.358 km/s, split into radial 5.602 and along-track 7.496 km/s; Jupiter moves at 13.058 km/s
v∞,in = v − V_J = (5.602, -5.562) km/s, |v∞|7.894 km/s
e = 1 + r_p v∞²/μ = 1 + 1429840 × 62.322 / 1.2669e+08 = 1.7034; δ = 2 asin(1/e)71.9°
Behind Jupiter (trailing side) v∞ turns towards Jupiter's motion: speed after18.077 km/s
In front (leading side) it turns away: speed after6.974 km/s
Check: |v∞,out| = 7.894 km/s, unchanged

Answer. v∞ = 7.89 km/s turned by 72°: the Sun-frame speed goes from 9.36 to 18.08 km/s behind Jupiter, or down to 6.97 km/s in front.

Sanity check. Relative to Jupiter nothing is gained: the hyperbola is symmetric. The gain comes from adding a rotated v∞ to Jupiter's own large velocity; Jupiter loses a negligible amount of orbital energy in return.

Pythonedit a value, then Run

# Worked example 15.1: A Jupiter flyby, from both sides
from math import sqrt, asin, atan2, sin, cos, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
mu_J, R_J = 1.26686e8, 71492.0   # Jupiter mu (km^3/s^2) and radius (km), JPL SSD (DE440, Archinal 2018)
r_peri, r_aph = 1.0 * AU, 6.0 * AU   # probe's solar orbit
rJ = 5.2026 * AU                     # Jupiter's circular orbit
rp = 20 * R_J                        # flyby periapsis

a = (r_peri + r_aph) / 2
h = sqrt(mu_sun * a * (1 - ((r_aph - r_peri) / (r_aph + r_peri))**2))
v = sqrt(mu_sun * (2 / rJ - 1 / a))
v_t = h / rJ                             # along-track component
v_r = sqrt(v**2 - v_t**2)                # radial (outbound)
VJ = sqrt(mu_sun / rJ)
vinf = [v_r, v_t - VJ]                   # (radial, along-track) relative to Jupiter
vinf_m = sqrt(vinf[0]**2 + vinf[1]**2)
e = 1 + rp * vinf_m**2 / mu_J
delta = 2 * asin(1 / e)
ang = atan2(vinf[1], vinf[0])
for side, turn in (("behind Jupiter", +delta), ("in front of Jupiter", -delta)):
    out = [vinf_m * cos(ang + turn), vinf_m * sin(ang + turn) + VJ]   # + turns towards Jupiter's motion
    print(f"{side}: Sun-frame speed after = {sqrt(out[0]**2 + out[1]**2):.3f} km/s")
print(f"before: v = {v:.3f} km/s (radial {v_r:.3f}, along-track {v_t:.3f}), Jupiter {VJ:.3f} km/s")
print(f"v_inf = {vinf_m:.3f} km/s, e = {e:.4f}, turn = {degrees(delta):.1f} deg")

# Worked example 15.1: A Jupiter flyby, from both sides
from math import sqrt, asin, atan2, sin, cos, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
mu_J, R_J = 1.26686e8, 71492.0   # Jupiter mu (km^3/s^2) and radius (km), JPL SSD (DE440, Archinal 2018)
r_peri, r_aph = 1.0 * AU, 6.0 * AU   # probe's solar orbit
rJ = 5.2026 * AU                     # Jupiter's circular orbit
rp = 20 * R_J                        # flyby periapsis

a = (r_peri + r_aph) / 2
h = sqrt(mu_sun * a * (1 - ((r_aph - r_peri) / (r_aph + r_peri))**2))
v = sqrt(mu_sun * (2 / rJ - 1 / a))
v_t = h / rJ                             # along-track component
v_r = sqrt(v**2 - v_t**2)                # radial (outbound)
VJ = sqrt(mu_sun / rJ)
vinf = [v_r, v_t - VJ]                   # (radial, along-track) relative to Jupiter
vinf_m = sqrt(vinf[0]**2 + vinf[1]**2)
e = 1 + rp * vinf_m**2 / mu_J
delta = 2 * asin(1 / e)
ang = atan2(vinf[1], vinf[0])
for side, turn in (("behind Jupiter", +delta), ("in front of Jupiter", -delta)):
    out = [vinf_m * cos(ang + turn), vinf_m * sin(ang + turn) + VJ]   # + turns towards Jupiter's motion
    print(f"{side}: Sun-frame speed after = {sqrt(out[0]**2 + out[1]**2):.3f} km/s")
print(f"before: v = {v:.3f} km/s (radial {v_r:.3f}, along-track {v_t:.3f}), Jupiter {VJ:.3f} km/s")
print(f"v_inf = {vinf_m:.3f} km/s, e = {e:.4f}, turn = {degrees(delta):.1f} deg")

behind Jupiter: Sun-frame speed after = 18.077 km/s
in front of Jupiter: Sun-frame speed after = 6.974 km/s
before: v = 9.358 km/s (radial 5.602, along-track 7.496), Jupiter 13.058 km/s
v_inf = 7.894 km/s, e = 1.7034, turn = 71.9 deg

Practice

Practice O15.1 · warm-up

How far does Venus bend you?^{[R2, R3]}

Given

A probe passes Venus (R = 6051.8 km, μ = 324859 km³/s²) with v∞ = 5.0 km/s at a periapsis altitude of 500 km. Find the eccentricity of the flyby hyperbola and the turn angle.

Mark as done

Hint

e = 1 + r_p v∞²/μ with r_p measured from the planet's centre; δ = 2 asin(1/e).

Answer. e = 1.504, turn angle δ = 83.3°.

Working

r_p = 6051.8 + 5006551.8 km
e = 1 + 6551.8 × 25.00 / 3248591.5042
δ = 2 asin(1/1.5042)83.33°

Answer. e = 1.504, turn angle δ = 83.3°.

Sanity check. e > 1 confirms a hyperbola; slower or closer passes turn more.

Pythonedit a value, then Run

# Practice O15.1: How far does Venus bend you?
from math import asin, degrees

mu_V, R_V = 324859.0, 6051.8   # Venus mu (km^3/s^2) and radius (km), JPL SSD (DE440, Archinal 2018)
vinf = 5.0            # km/s
h = 500.0             # km, periapsis altitude

rp = R_V + h
e = 1 + rp * vinf**2 / mu_V
delta = 2 * asin(1 / e)
print(f"r_p = {rp:.1f} km, e = {e:.4f}, turn angle = {degrees(delta):.2f} deg")

r_p = 6551.8 km, e = 1.5042, turn angle = 83.33 deg

Practice O15.2 · standard

The free Δv of a flyby

Given

For the pass of O15.1, find the size of the velocity change the flyby delivers, and the largest possible value at Venus for the same v∞ (a grazing pass, r_p = R).

Mark as done

Hint

The velocity change is the vector difference of v∞,out and v∞,in, both of length v∞: Δv = 2 v∞ sin(δ/2).

Answer. Δv = 6.65 km/s at 500 km, up to 6.82 km/s for a grazing pass.

Working

Δv = 2 × 5.0 × sin(41.67°)6.648 km/s
Grazing: e = 1 + 6051.8 × 25.00 / 324859 = 1.4657, δ86.04°
Δv_max = 2 × 5.0 × sin(43.02°)6.823 km/s

Answer. Δv = 6.65 km/s at 500 km, up to 6.82 km/s for a grazing pass.

Sanity check. Several km/s for free, compared with chemical stages that cost tonnes of propellant per km/s: the reason Cassini, Galileo and Messenger all used Venus.

Pythonedit a value, then Run

# Practice O15.2: The free dv of a flyby
from math import asin, sin, degrees

mu_V, R_V = 324859.0, 6051.8   # Venus mu (km^3/s^2) and radius (km), JPL SSD (DE440, Archinal 2018)
vinf = 5.0            # km/s
h = 500.0             # km

for name, rp in (("at 500 km", R_V + h), ("grazing", R_V)):
    e = 1 + rp * vinf**2 / mu_V
    delta = 2 * asin(1 / e)
    dv = 2 * vinf * sin(delta / 2)       # size of the velocity change
    print(f"{name}: e = {e:.4f}, turn {degrees(delta):.2f} deg, dv = {dv:.3f} km/s")

at 500 km: e = 1.5042, turn 83.33 deg, dv = 6.648 km/s
grazing: e = 1.4657, turn 86.04 deg, dv = 6.823 km/s

Practice O15.3 · challenge

Thrown out of the Solar System?^{[R3, R8]}

Given

The probe of the worked example instead passes behind Jupiter at 4 Jupiter radii. Find the turn angle, its speed around the Sun afterwards, and whether it now escapes the Sun.

Mark as done

Hint

Rotate v∞,in by δ towards Jupiter's direction of motion, add Jupiter's velocity, then compare with the solar escape speed √(2μ_Sun/r) at Jupiter's distance.

Answer. δ = 122°, speed 20.84 km/s against an escape speed of 18.47 km/s: the probe escapes the Sun.

Working

|v∞| = 7.894 km/s; r_p = 285968 km; e = 1.1407; δ122.5°
v∞,out = (1.683, 7.713) km/s; v_out = V_J + v∞,out = (1.683, 20.771) km/s, |v|20.839 km/s
Solar escape speed at 5.2026 AU: √(2μ_Sun/r)18.467 km/s
Heliocentric energy afterwards: v²/2 − μ_Sun/r46.62 km²/s²

Answer. δ = 122°, speed 20.84 km/s against an escape speed of 18.47 km/s: the probe escapes the Sun.

Sanity check. A probe that arrived on a bound ellipse can leave on a hyperbola: Jupiter supplied the energy. This is how the outer-planet missions reached escape.

Pythonedit a value, then Run

# Practice O15.3: Thrown out of the Solar System?
from math import sqrt, asin, atan2, sin, cos, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
mu_J, R_J = 1.26686e8, 71492.0   # Jupiter mu (km^3/s^2) and radius (km), JPL SSD (DE440, Archinal 2018)
r_peri, r_aph = 1.0 * AU, 6.0 * AU
rJ = 5.2026 * AU
rp = 4 * R_J                 # closer pass, behind Jupiter

a = (r_peri + r_aph) / 2
h = sqrt(mu_sun * a * (1 - ((r_aph - r_peri) / (r_aph + r_peri))**2))
v = sqrt(mu_sun * (2 / rJ - 1 / a))
v_t, VJ = h / rJ, sqrt(mu_sun / rJ)
v_r = sqrt(v**2 - v_t**2)
vinf_m = sqrt(v_r**2 + (v_t - VJ)**2)
e = 1 + rp * vinf_m**2 / mu_J
delta = 2 * asin(1 / e)
ang = atan2(v_t - VJ, v_r) + delta       # behind: turned towards Jupiter's motion
out = [vinf_m * cos(ang), vinf_m * sin(ang) + VJ]
v_out = sqrt(out[0]**2 + out[1]**2)
v_esc = sqrt(2 * mu_sun / rJ)
energy = v_out**2 / 2 - mu_sun / rJ
print(f"v_inf = {vinf_m:.3f} km/s, e = {e:.4f}, turn = {degrees(delta):.1f} deg")
print(f"speed after = {v_out:.3f} km/s, solar escape speed = {v_esc:.3f} km/s")
print(f"energy = {energy:.2f} km^2/s^2: {'escapes the Sun' if energy > 0 else 'stays bound'}")

# Practice O15.3: Thrown out of the Solar System?
from math import sqrt, asin, atan2, sin, cos, degrees

mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
mu_J, R_J = 1.26686e8, 71492.0   # Jupiter mu (km^3/s^2) and radius (km), JPL SSD (DE440, Archinal 2018)
r_peri, r_aph = 1.0 * AU, 6.0 * AU
rJ = 5.2026 * AU
rp = 4 * R_J                 # closer pass, behind Jupiter

a = (r_peri + r_aph) / 2
h = sqrt(mu_sun * a * (1 - ((r_aph - r_peri) / (r_aph + r_peri))**2))
v = sqrt(mu_sun * (2 / rJ - 1 / a))
v_t, VJ = h / rJ, sqrt(mu_sun / rJ)
v_r = sqrt(v**2 - v_t**2)
vinf_m = sqrt(v_r**2 + (v_t - VJ)**2)
e = 1 + rp * vinf_m**2 / mu_J
delta = 2 * asin(1 / e)
ang = atan2(v_t - VJ, v_r) + delta       # behind: turned towards Jupiter's motion
out = [vinf_m * cos(ang), vinf_m * sin(ang) + VJ]
v_out = sqrt(out[0]**2 + out[1]**2)
v_esc = sqrt(2 * mu_sun / rJ)
energy = v_out**2 / 2 - mu_sun / rJ
print(f"v_inf = {vinf_m:.3f} km/s, e = {e:.4f}, turn = {degrees(delta):.1f} deg")
print(f"speed after = {v_out:.3f} km/s, solar escape speed = {v_esc:.3f} km/s")
print(f"energy = {energy:.2f} km^2/s^2: {'escapes the Sun' if energy > 0 else 'stays bound'}")

v_inf = 7.894 km/s, e = 1.1407, turn = 122.5 deg
speed after = 20.839 km/s, solar escape speed = 18.467 km/s
energy = 46.62 km^2/s^2: escapes the Sun

Common mistakes

Expecting a speed change relative to the planet. |v∞| is the same before and after.
Measuring r_p from the surface. The flyby formulas use the distance from the planet's centre.
Forgetting the side of the pass. The same δ can add or remove energy.
Adding speeds instead of vectors. Build the triangle V_p + v∞.

Recap from earlier chapters

(Chapter 14) What is v∞?

The speed left over far from a planet after climbing out of its gravity well: v² − 2μ/r = v∞².

(Chapter 2) Which eccentricities give a hyperbola?

e > 1; e = 1 is a parabola.

Sources for this chapter

R2NASA JPL Solar System Dynamics, "Astrodynamic Parameters" (planetary GMs from DE440; Park et al. 2021, AJ 161:105).
R3NASA JPL Solar System Dynamics, "Planetary Physical Parameters" (radii from Archinal et al. 2018, CMDA 130:22; change log 2019-12-12).
R4Standish, E.M. and Williams, J.G., "Approximate Positions of the Planets", NASA JPL SSD.
R8Mamajek, E.E., Prša, A., Torres, G. et al., "IAU 2015 Resolution B3 on Recommended Nominal Conversion Constants for Selected Solar and Planetary Properties", arXiv:1510.07674 (2015).

Where in each source, and links: References.

I have finished chapter 15

Chapter 16

Lagrange points

35 minutes

The idea in one minute

Put a third, tiny body near two big ones that circle each other, and look from a frame that turns with them. There are five places where gravity from both and the outward pull of the turning frame cancel. A spacecraft parked there keeps its place relative to the pair. Three points sit on the line through the two bodies, two lead and trail the smaller body by 60 degrees.

Earth-Moon system to scale, seen in the frame that turns with the Moon. Notice that L1 and L2 sit close to the small body, on either side of it; L3, L4 and L5 lie on the Moon's orbit.

You will be able to

Explain what a Lagrange point is in the rotating frame
Find the L1 and L2 distances exactly and with the Hill approximation
State which points are stable and why L1/L2 missions need station-keeping

The rotating frame

In the circular restricted three-body problem the two primaries move on circles about their common centre of mass, and the third body is too light to disturb them. In a frame rotating at their orbital rate the primaries stand still, and an equilibrium is a point where gravity and the centrifugal term balance.

Where the points are

L1 lies between the bodies, L2 just beyond the smaller one, L3 on the far side of the larger one. For a small mass ratio, L1 and L2 sit at about a(μ*/3)^(1/3) from the smaller body. L4 and L5 form equilateral triangles with the two primaries.^{[R22, R24]}

Stable and unstable

The three collinear points are unstable: a small error grows within weeks. L4 and L5 are stable when the mass ratio is below about 0.0385, which holds for Sun-Earth, Earth-Moon and Sun-Jupiter (the Trojan asteroids).^[R23]

Why missions go there

Sun-Earth L2 keeps the Sun, Earth and Moon behind the spacecraft and gives a steady thermal environment, ideal for infrared telescopes. Real missions fly large halo or Lissajous orbits around the point and trim them with small burns.

Living near L1 and L2

Nothing parks exactly on L1 or L2. A spacecraft nudged off the point drifts away on a timescale of about 23 days, so missions fly large loops around the point instead, called halo or Lissajous orbits, and trim them with small periodic burns. Webb's loop around Sun-Earth L2 is comparable in size to the Moon's orbit and takes about 6 months. Solar observatories fly similar loops around L1.^{[R23, R24]}

Key equations

Mass ratio

μ* = μ₂ / (μ₁ + μ₂)

where

μ*mass ratio, dimensionless

μ₁gravitational parameter of the larger body, km³/s²

μ₂gravitational parameter of the smaller body, km³/s²

Collinear balance

x − (1−μ*)(x+μ*)/|x+μ*|³ − μ*(x−1+μ*)/|x−1+μ*|³ = 0

where

xposition on the line, in units of the separation, from the barycentre, dimensionless

μ*mass ratio, dimensionless

three roots: L1, L2, L3

Hill approximation

r ≈ a (μ*/3)^(1/3)

where

rdistance of L1 or L2 from the smaller body, km

aseparation of the primaries, km

μ*mass ratio, dimensionless

good when μ* ≪ 1

Routh stability limit

μ* < ½(1 − √(23/27)) ≈ 0.0385

where

μ*mass ratio, dimensionless

L4 and L5 are linearly stable below it

Worked examples

Worked example 16.1

How far away is Sun-Earth L2?^{[R22, R24]}

Given

Treat the Sun and Earth as a circular restricted three-body system with separation a = 149,597,871 km and mass ratio μ* = μ_Earth / (μ_Sun + μ_Earth). Find the distances of L1 and L2 from Earth, exactly (root of the collinear balance) and with the Hill approximation.

new symbols

μ_Sungravitational parameter of the Sun, km³/s²

Working

μ* = 398600.4418 / (1.327124e+11 + 398600.4418)3.0035e-06
Hill: r ≈ a (μ*/3)^(1/3) = 149,597,871 × (3.0035e-06/3)^(1/3)1,496,557 km
Exact root of the collinear balance: L1 is 0.009970 a = 1,491,551 km from Earth
L2 is 0.010037 a = 1,501,532 km from Earth (on the night side)
Hill sits between them: L1 is 0.33 % closer, L2 0.33 % farther

Answer. L1 ≈ 1.492 million km and L2 ≈ 1.502 million km from Earth; the Hill estimate 1.497 million km is within 0.3 %.

Sanity check. Both are about 3.9 times the Moon's distance, still deep inside the Sun's pull, which is why the points exist at all: there the combined pull matches the rotating frame.

Pythonedit a value, then Run

# Worked example 16.1: How far away is Sun-Earth L2?
mu_E = 398600.4418    # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
a = 149597870.7       # km, Sun-Earth separation = 1 AU (IAU 2012, exact)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_E / (mu_sun + mu_E)               # mass ratio mu*
xs = 1 - m                               # Earth's position (barycentre at 0)
d1 = xs - root(dU, xs - 0.5, xs - 1e-9, m)   # L1 lies between Sun and Earth
d2 = root(dU, xs + 1e-9, xs + 0.5, m) - xs   # L2 lies beyond Earth
hill = (m / 3) ** (1 / 3)
print(f"mu* = {m:.4e}")
print(f"L1: {d1:.6f} a = {d1 * a:,.0f} km = {d1 * a / 1e6:.3f} million km")
print(f"L2: {d2:.6f} a = {d2 * a:,.0f} km = {d2 * a / 1e6:.3f} million km")
print(f"Hill: {hill * a:,.0f} km = {hill * a / 1e6:.3f} million km, within {100 * max(hill - d1, d2 - hill) / hill:.2f} %")

# Worked example 16.1: How far away is Sun-Earth L2?
mu_E = 398600.4418    # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
a = 149597870.7       # km, Sun-Earth separation = 1 AU (IAU 2012, exact)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_E / (mu_sun + mu_E)               # mass ratio mu*
xs = 1 - m                               # Earth's position (barycentre at 0)
d1 = xs - root(dU, xs - 0.5, xs - 1e-9, m)   # L1 lies between Sun and Earth
d2 = root(dU, xs + 1e-9, xs + 0.5, m) - xs   # L2 lies beyond Earth
hill = (m / 3) ** (1 / 3)
print(f"mu* = {m:.4e}")
print(f"L1: {d1:.6f} a = {d1 * a:,.0f} km = {d1 * a / 1e6:.3f} million km")
print(f"L2: {d2:.6f} a = {d2 * a:,.0f} km = {d2 * a / 1e6:.3f} million km")
print(f"Hill: {hill * a:,.0f} km = {hill * a / 1e6:.3f} million km, within {100 * max(hill - d1, d2 - hill) / hill:.2f} %")

mu* = 3.0035e-06
L1: 0.009970 a = 1,491,551 km = 1.492 million km
L2: 0.010037 a = 1,501,532 km = 1.502 million km
Hill: 1,496,557 km = 1.497 million km, within 0.33 %

Worked example 16.2

Earth-Moon L1, where Hill starts to slip^{[R2, R6]}

Given

Repeat for the Earth-Moon system: a = 384,400 km, μ_Moon = 4902.8 km³/s². Where is L1, measured from the Moon and from Earth, and how good is the Hill estimate here?

new symbols

μ_Moongravitational parameter of the Moon, km³/s²

Working

μ* = 4902.8 / (398600.4418 + 4902.8)0.01215
Exact L1: 0.15093 a = 58,019 km from the Moon, so 326,381 km from Earth
Hill: 384,400 × (0.01215/3)^(1/3)61,274 km
Error of Hill5.6 %(the mass ratio is no longer tiny)

Answer. Earth-Moon L1 is 58,019 km from the Moon (326,381 km from Earth); Hill overestimates it by 5.6 %.

Sanity check. μ* = 0.012 is about 4,046 times the Sun-Earth value, so the first-order Hill formula degrades from 0.3 % to several per cent.

Pythonedit a value, then Run

# Worked example 16.2: Earth-Moon L1, where Hill starts to slip
mu_E = 398600.4418    # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
mu_moon = 4902.8      # km^3/s^2, Moon (JPL SSD, DE440)
a = 384400.0          # km, mean Earth-Moon distance (NASA Moon fact sheet)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_moon / (mu_E + mu_moon)
xs = 1 - m                               # Moon's position
d1 = (xs - root(dU, xs - 0.5, xs - 1e-9, m)) * a
hill = (m / 3) ** (1 / 3) * a
print(f"mu* = {m:.5f}")
print(f"L1 is {d1:,.0f} km from the Moon, {a - d1:,.0f} km from Earth")
print(f"Hill: {hill:,.0f} km, an overestimate of {100 * (hill - d1) / d1:.1f} %")

# Worked example 16.2: Earth-Moon L1, where Hill starts to slip
mu_E = 398600.4418    # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
mu_moon = 4902.8      # km^3/s^2, Moon (JPL SSD, DE440)
a = 384400.0          # km, mean Earth-Moon distance (NASA Moon fact sheet)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_moon / (mu_E + mu_moon)
xs = 1 - m                               # Moon's position
d1 = (xs - root(dU, xs - 0.5, xs - 1e-9, m)) * a
hill = (m / 3) ** (1 / 3) * a
print(f"mu* = {m:.5f}")
print(f"L1 is {d1:,.0f} km from the Moon, {a - d1:,.0f} km from Earth")
print(f"Hill: {hill:,.0f} km, an overestimate of {100 * (hill - d1) / d1:.1f} %")

mu* = 0.01215
L1 is 58,019 km from the Moon, 326,381 km from Earth
Hill: 61,274 km, an overestimate of 5.6 %

Practice

Practice O16.1 · warm-up

Sun-Mars L1 with one line

Given

Mars orbits the Sun at a = 1.523679 AU (1 AU = 149,597,870.7 km) with μ_Mars = 42,828 km³/s². Estimate the distance of Sun-Mars L1 from Mars with the Hill approximation.

Mark as done

Hint

Compute the mass ratio first; it is tiny, so r ≈ a(μ*/3)^(1/3) is accurate.

Answer. r ≈ 1,084,053 km (the exact root gives 1,082,332 km).

Working

a = 1.523679 × 149,597,870.7227,939,134 km
μ* = 42,828 / 1.327125e+113.2271e-07
r ≈ 227,939,134 × (3.2271e-07/3)^(1/3)1,084,053 km

Answer. r ≈ 1,084,053 km (the exact root gives 1,082,332 km).

Sanity check. Mars is about a tenth of Earth's mass but farther out, so its L1 (1.08 million km) is a bit under Earth's 1.49 million km.

Pythonedit a value, then Run

# Practice O16.1: Sun-Mars L1 with one line
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
a = 1.523679 * AU            # km, Mars orbit radius
mu_M = 42828.0               # km^3/s^2, Mars (JPL SSD, DE440)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_M / mu_sun
r_hill = a * (m / 3) ** (1 / 3)
xs = 1 - m
r_exact = (xs - root(dU, xs - 0.5, xs - 1e-9, m)) * a   # exact collinear root, for comparison
print(f"a = {a:,.0f} km, mass ratio = {m:.4e}")
print(f"Hill: r = {r_hill:,.0f} km (exact root {r_exact:,.0f} km)")

# Practice O16.1: Sun-Mars L1 with one line
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
AU = 149597870.7      # km, astronomical unit (IAU 2012, exact)
a = 1.523679 * AU            # km, Mars orbit radius
mu_M = 42828.0               # km^3/s^2, Mars (JPL SSD, DE440)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_M / mu_sun
r_hill = a * (m / 3) ** (1 / 3)
xs = 1 - m
r_exact = (xs - root(dU, xs - 0.5, xs - 1e-9, m)) * a   # exact collinear root, for comparison
print(f"a = {a:,.0f} km, mass ratio = {m:.4e}")
print(f"Hill: r = {r_hill:,.0f} km (exact root {r_exact:,.0f} km)")

a = 227,939,134 km, mass ratio = 3.2271e-07
Hill: r = 1,084,053 km (exact root 1,082,332 km)

Practice O16.2 · standard

Earth-Moon L2 for a relay^{[R2, R6]}

Given

A relay for the lunar far side would sit near Earth-Moon L2. Using a = 384,400 km and μ* = 0.01215, find L2's distance beyond the Moon (exact) and the percentage error of the Hill estimate 61,274 km.

Mark as done

Hint

L2 lies beyond the smaller body; solve the collinear balance between x = 1 − μ* and 1.5 (the library does this by bracketing).

Answer. L2 is 64,515 km beyond the Moon; Hill is off by -5.0 %.

Working

Exact root: x_L2 = 1.155682, Moon at 1 − μ*0.987849
d = (1.155682 − 0.987849) × 384,40064,515 km
Hill error = (61,274 − 64,515) / 64,515-5.0 %

Answer. L2 is 64,515 km beyond the Moon; Hill is off by -5.0 %.

Sanity check. L2 is farther from the Moon than L1 (58,019 km): on the far side the Moon's pull adds to Earth's, so balance needs a larger distance.

Pythonedit a value, then Run

# Practice O16.2: Earth-Moon L2 for a relay
mu_E = 398600.4418    # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
mu_moon = 4902.8      # km^3/s^2, Moon (JPL SSD, DE440)
a = 384400.0          # km, mean Earth-Moon distance (NASA Moon fact sheet)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_moon / (mu_E + mu_moon)           # 0.01215
xs = 1 - m
x2 = root(dU, xs + 1e-9, xs + 0.5, m)
d2 = (x2 - xs) * a
hill = (m / 3) ** (1 / 3) * a
print(f"x_L2 = {x2:.6f}, Moon at {xs:.6f}")
print(f"L2 is {d2:,.0f} km beyond the Moon")
print(f"Hill {hill:,.0f} km is off by {100 * (hill - d2) / d2:.2f} %")

# Practice O16.2: Earth-Moon L2 for a relay
mu_E = 398600.4418    # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
mu_moon = 4902.8      # km^3/s^2, Moon (JPL SSD, DE440)
a = 384400.0          # km, mean Earth-Moon distance (NASA Moon fact sheet)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_moon / (mu_E + mu_moon)           # 0.01215
xs = 1 - m
x2 = root(dU, xs + 1e-9, xs + 0.5, m)
d2 = (x2 - xs) * a
hill = (m / 3) ** (1 / 3) * a
print(f"x_L2 = {x2:.6f}, Moon at {xs:.6f}")
print(f"L2 is {d2:,.0f} km beyond the Moon")
print(f"Hill {hill:,.0f} km is off by {100 * (hill - d2) / d2:.2f} %")

x_L2 = 1.155682, Moon at 0.987849
L2 is 64,515 km beyond the Moon
Hill 61,274 km is off by -5.02 %

Practice O16.3 · challenge

Why L2 needs station-keeping^[R23]

Given

Linearise the rotating-frame motion about Sun-Earth L2. With c₂ = μ*/|x − 1 + μ*|³ + (1 − μ*)/|x + μ*|³, the in-plane unstable root is λ² = [c₂ − 2 + √(9c₂² − 8c₂)]/2, in units of the orbital rate n = 2π per year. Find c₂, λ and the e-folding time of a small error, in days.

Mark as done

Hint

Use x_L2 from the exact collinear root; the time unit is 1/n with n = 2π / (one year in seconds).

Answer. λ = 2.484 n, so a small error grows by a factor e every 23 days.

Working

x_L2 = 1.0100341, μ*3.0035e-06
c₂3.9408
λ² = (3.9408 − 2 + √(9 × 3.9408² − 8 × 3.9408))/2 = 6.1723, λ2.4844
n = 2π / 31556926 s1.9911e-07 rad/s
e-folding time = 1/(λ n) = 2.022e+06 s23.4 days

Answer. λ = 2.484 n, so a small error grows by a factor e every 23 days.

Sanity check. In 70 days an error grows about 20 times: missions at L2 trim their orbit every few weeks with small burns, a few m/s per year.

Pythonedit a value, then Run

# Practice O16.3: Why L2 needs station-keeping
from math import pi, sqrt

mu_E = 398600.4418    # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_E / (mu_sun + mu_E)
xs = 1 - m
x = root(dU, xs + 1e-9, xs + 0.5, m)     # Sun-Earth L2
c2 = m / abs(x - 1 + m) ** 3 + (1 - m) / abs(x + m) ** 3
lam = sqrt((c2 - 2 + sqrt(9 * c2**2 - 8 * c2)) / 2)   # in-plane unstable root, units of n
n = 2 * pi / year
tau = 1 / (lam * n)
print(f"x_L2 = {x:.7f}, c2 = {c2:.4f}, lambda = {lam:.4f} n")
print(f"e-folding time = {tau:.3e} s = {tau / 86400:.1f} days")

# Practice O16.3: Why L2 needs station-keeping
from math import pi, sqrt

mu_E = 398600.4418    # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
mu_sun = 1.32712440018e11  # km^3/s^2, Sun's gravitational parameter (JPL SSD; DE440 gives 1.32712440041e11)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)

def dU(x, m):         # x-component of the rotating-frame force balance (units of the separation)
    r1, r2 = abs(x + m), abs(x - 1 + m)
    return x - (1 - m) * (x + m) / r1**3 - m * (x - 1 + m) / r2**3

def root(f, lo, hi, m):   # bisection: f changes sign between lo and hi
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(lo, m) * f(mid, m) <= 0:
            hi = mid
        else:
            lo = mid
    return (lo + hi) / 2

m = mu_E / (mu_sun + mu_E)
xs = 1 - m
x = root(dU, xs + 1e-9, xs + 0.5, m)     # Sun-Earth L2
c2 = m / abs(x - 1 + m) ** 3 + (1 - m) / abs(x + m) ** 3
lam = sqrt((c2 - 2 + sqrt(9 * c2**2 - 8 * c2)) / 2)   # in-plane unstable root, units of n
n = 2 * pi / year
tau = 1 / (lam * n)
print(f"x_L2 = {x:.7f}, c2 = {c2:.4f}, lambda = {lam:.4f} n")
print(f"e-folding time = {tau:.3e} s = {tau / 86400:.1f} days")

x_L2 = 1.0100341, c2 = 3.9408, lambda = 2.4844 n
e-folding time = 2.022e+06 s = 23.4 days

Common mistakes

Measuring L1/L2 from the larger body: the Hill formula gives the distance from the smaller one.
Thinking a spacecraft sits exactly at L2. It orbits around it, often hundreds of thousands of km wide.
Assuming every Lagrange point is stable. Only L4 and L5 are, and only below the Routh mass ratio.
Using the Hill formula for Earth-Moon and expecting sub-per-cent accuracy; the mass ratio is too large.

Recap from earlier chapters

What is the circular speed at Earth's distance from the Sun?

√(μ_Sun / a) = 29.78 km/s.

Why does a period depend only on a, not on e?

Kepler's third law: T = 2π√(a³/μ), so all orbits with the same semi-major axis share a period.

Sources for this chapter

R2NASA JPL Solar System Dynamics, "Astrodynamic Parameters" (planetary GMs from DE440; Park et al. 2021, AJ 161:105).
R6Williams, D.R., "Moon Fact Sheet", NASA NSSDCA, updated 11 Jan 2024.
R22ESA, "Types of orbits", 30 Mar 2020.
R23Cornish, N.J., "What is a Lagrange Point?", NASA Science, 27 Mar 2018, updated 1 Nov 2024.
R24NASA Science, "Webb Orbit".

Where in each source, and links: References.

I have finished chapter 16

Chapter 17

GEO and station-keeping

35 minutes

The idea in one minute

A geostationary satellite circles Earth exactly as fast as Earth turns, so it hangs over one longitude. That only works on one circle, 42,164 km from Earth's centre, in the equator plane. Nature keeps pushing it off: the Sun and Moon tilt the orbit, Earth's slightly oval equator drags it in longitude. Station-keeping is the small, regular set of burns that pushes it back.

A semi-major axis just 1 km off makes a GEO satellite drift 0.013 deg per day. Notice it crosses a ±0.05 deg box in about 4 days.

You will be able to

Derive the GEO radius from the sidereal day
Turn a semi-major axis error into a longitude drift rate
Budget north-south, east-west and end-of-life Δv for a GEO mission

The one orbit

Match the period to the sidereal day, 23 h 56 min 4 s, not the 24 h solar day. Zero eccentricity and zero inclination complete the conditions.^{[R1, R2, R22]}

Drift in longitude

A slightly higher orbit has a longer period, so the satellite falls behind Earth and drifts west; lower drifts east. The drift rate is −1.5 n Δa/a, about 0.013 deg/day per km.

North-south: the big one

The Sun's and Moon's pull tilt the orbit by about a degree per year. Removing it costs about 50 m/s per year, by far the largest station-keeping item.^{[R18, R19]}

East-west and end of life

Earth's equator is slightly elliptical, so most longitudes feel a small along-track force; correcting it costs one or two m/s per year. At end of life the satellite is raised a few hundred km into a graveyard orbit to free the slot.^{[R18, R20]}

Key equations

GEO radius

a_GEO = (μ T_sid² / 4π²)^(1/3)

where

a_GEOgeostationary radius, km

μEarth's gravitational parameter, km³/s²

T_sidsidereal day, 86164.0905, s

Longitude drift

λ̇ = −1.5 n Δa / a

where

λ̇longitude drift rate (positive east), rad/s

nmean motion of GEO, rad/s

Δasemi-major axis above GEO, km

aGEO radius, km

North-south burn

Δv_NS = 2 v sin(Δi/2)

where

Δv_NSinclination-control Δv, m/s

vGEO speed, m/s

Δiinclination removed, rad

Drift-orbit burn

λ̇ = −3 n Δv / v

where

Δvsmall tangential burn (positive prograde), m/s

vGEO speed, m/s

nmean motion, rad/s

λ̇resulting drift rate, rad/s

from Δa = 2aΔv/v

Propellant

m_p = m₀ (1 − exp(−Δv / (Isp g₀)))

where

m_ppropellant used, kg

m₀initial mass, kg

Ispspecific impulse, s

g₀standard gravity, m/s²

Worked examples

Worked example 17.1

Why GEO is 42,164 km and not 42,241 km^{[R1, R2, R11]}

Given

A geostationary satellite must turn once per rotation of Earth relative to the stars: the sidereal day, 86164.0905 s. Find the GEO radius and speed, and the radius you would get by mistake from the 86,400 s solar day.

new symbols

vcircular orbital speed, √(μ/a), km/s

Working

a = (μ (T/2π)²)^(1/3) = (398600.4418 × (86164.0905/2π)²)^(1/3)42,164.17 km
Altitude = 42,164.17 − 6378.137 = 35,786.0 km; speed v = √(μ/a)3.0747 km/s
With T = 86400 s: a = 42,241.10 km, 76.9 km too high
That error drifts at -0.988 deg/day

Answer. R_GEO = 42,164.2 km (altitude 35,786 km), v = 3.075 km/s; using the solar day would put it 77 km too high.

Sanity check. The error drifts the satellite about one degree per day westward, roughly a full circle per year: the solar day already contains one extra turn per year.

Pythonedit a value, then Run

# Worked example 17.1: Why GEO is 42,164 km and not 42,241 km
from math import pi, sqrt, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
day_sol = 86400.0     # s, solar day (the wrong choice)

a = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
v = sqrt(mu / a)
a_wrong = (mu * (day_sol / (2 * pi)) ** 2) ** (1 / 3)
drift = -1.5 * sqrt(mu / a**3) * (a_wrong - a) / a      # rad/s
print(f"R_GEO = {a:,.2f} km, altitude {a - R_E:,.1f} km, v = {v:.4f} km/s")
print(f"with the solar day: a = {a_wrong:,.2f} km, {a_wrong - a:.1f} km too high")
print(f"that error drifts at {degrees(drift) * 86400:.3f} deg/day")

R_GEO = 42,164.17 km, altitude 35,786.0 km, v = 3.0747 km/s
with the solar day: a = 42,241.10 km, 76.9 km too high
that error drifts at -0.988 deg/day

Worked example 17.2

Fifteen years of north-south station-keeping^{[R17, R18]}

Given

Without control, the Sun and Moon tilt a GEO orbit by about 0.85 deg per year. Find the yearly Δv to remove it, the 15-year total including 1.5 m/s/yr east-west, and the propellant for a 3000 kg satellite with Isp = 315 s.

new symbols

MRmass ratio m₀/m_f from the rocket equation, dimensionless

Working

Δv_NS per year = 2 v sin(Δi/2) = 2 × 3074.7 × sin(0.425 deg)45.61 m/s
Per year with east-west: 45.61 + 1.547.11 m/s
15 years: 707 m/s
Mass ratio = exp(Δv / (Isp g₀)) = exp(707 / (315 × 9.80665))1.2571
Propellant = m₀ (1 − 1/MR) = 3000 × (1 − 1/1.2571)613 kg

Answer. About 46 m/s per year north-south, 707 m/s over 15 years, needing 613 kg of propellant.

Sanity check. North-south dominates (it is 30 times the east-west budget): that is why electric propulsion, with Isp ten times higher, is now common for station-keeping.

Pythonedit a value, then Run

# Worked example 17.2: Fifteen years of north-south station-keeping
from math import pi, sqrt, sin, exp, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
g0 = 9.80665          # m/s^2, standard gravity (exact, by definition)
di_year = radians(0.85)   # inclination drift per year
ew_year = 1.5         # m/s per year, east-west
years = 15
m0 = 3000.0           # kg
Isp = 315.0           # s

a = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
v = sqrt(mu / a) * 1000                  # m/s
dv_ns = 2 * v * sin(di_year / 2)
dv_total = (dv_ns + ew_year) * years
MR = exp(dv_total / (Isp * g0))
print(f"north-south {dv_ns:.2f} m/s per year, {dv_ns + ew_year:.2f} with east-west")
print(f"{years} years: {dv_total:.1f} m/s, mass ratio {MR:.4f}")
print(f"propellant {m0 * (1 - 1 / MR):.1f} kg")

north-south 45.61 m/s per year, 47.11 with east-west
15 years: 706.7 m/s, mass ratio 1.2571
propellant 613.5 kg

Practice

Practice O17.1 · warm-up

A satellite 2.5 km too high

Given

After a manoeuvre a GEO satellite has a semi-major axis 2.5 km above R_GEO = 42,164.2 km. Find its longitude drift rate in deg/day and how long it takes to leave a ±0.05 deg box if it starts in the centre.

Mark as done

Hint

λ̇ = −1.5 n Δa / a with n = 2π / sidereal day; positive Δa means slower, so it drifts west.

Answer. λ̇ = -0.0321 deg/day westward; it leaves the box after 1.6 days.

Working

n = 2π / 86164.09057.292116e-05 rad/s
λ̇ = −1.5 × 7.292116e-05 × 2.5 / 42,164.2-6.4855e-09 rad/s
= -0.0321 deg/day (west)
Time to move 0.05 deg = 0.05 / 0.03211.56 days

Answer. λ̇ = -0.0321 deg/day westward; it leaves the box after 1.6 days.

Sanity check. A few km of error leaves a tight box in days, which is why GEO operators trim the orbit every one to two weeks.

Pythonedit a value, then Run

# Practice O17.1: A satellite 2.5 km too high
from math import pi, degrees

day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
R_GEO = (398600.4418 * (day_sid / (2 * pi)) ** 2) ** (1 / 3)   # km, GEO radius from mu and the sidereal day
da = 2.5              # km above GEO
box = 0.05            # deg, half-width of the box

n = 2 * pi / day_sid
drift = -1.5 * n * da / R_GEO            # rad/s (negative: westward)
drift_deg = degrees(drift) * 86400
print(f"drift = {drift:.4e} rad/s = {drift_deg:.4f} deg/day (west)")
print(f"leaves the box after {box / abs(drift_deg):.2f} days")

drift = -6.4855e-09 rad/s = -0.0321 deg/day (west)
leaves the box after 1.56 days

Practice O17.2 · standard

Retiring to the graveyard^[R20]

Given

At end of life a 1900 kg GEO satellite is raised 300 km with a two-burn Hohmann transfer. Find each burn, the total Δv and the propellant (Isp = 315 s).

Mark as done

Hint

The two radii are almost equal, so both burns are small and nearly equal; use the Hohmann formulas exactly anyway.

Answer. Δv ≈ 10.9 m/s in two near-equal burns, about 6.7 kg of propellant.

Working

r₁ = 42,164.2 km, r₂ = 42,464.2 km, a_t42,314.2 km
Δv₁ = 5.445 m/s, Δv₂5.435 m/s
Total10.88 m/s
Propellant = 1900 × (1 − exp(−10.88 / (315 × 9.80665)))6.7 kg

Answer. Δv ≈ 10.9 m/s in two near-equal burns, about 6.7 kg of propellant.

Sanity check. Check: for a small raise Δv ≈ v Δr / (2r) = 10.94 m/s, matching. Cheap, but it must be reserved: an empty tank cannot retire.

Pythonedit a value, then Run

# Practice O17.2: Retiring to the graveyard
from math import pi, sqrt, exp

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
g0 = 9.80665          # m/s^2, standard gravity (exact, by definition)
raise_km = 300.0      # km above GEO
m0 = 1900.0           # kg
Isp = 315.0           # s

r1 = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
r2 = r1 + raise_km
at = (r1 + r2) / 2
dv1 = sqrt(mu * (2 / r1 - 1 / at)) - sqrt(mu / r1)
dv2 = sqrt(mu / r2) - sqrt(mu * (2 / r2 - 1 / at))
dv = (dv1 + dv2) * 1000
mp = m0 * (1 - exp(-dv / (Isp * g0)))
print(f"dv1 = {dv1*1000:.3f} m/s, dv2 = {dv2*1000:.3f} m/s, total {dv:.2f} m/s")
print(f"propellant {mp:.2f} kg")

dv1 = 5.445 m/s, dv2 = 5.435 m/s, total 10.88 m/s
propellant 6.68 kg

Practice O17.3 · challenge

Moving a satellite 12 degrees east

Given

An operator moves a GEO satellite 12 deg east in 20 days. A small tangential burn Δv changes the semi-major axis by Δa = 2aΔv/v, so λ̇ = −3nΔv/v. Find the burn to start the drift, the semi-major axis of the drift orbit, and the total Δv (start plus stop).

Mark as done

Hint

East means faster, so the burn is retrograde (lower orbit). Required λ̇ = Δλ / t in rad/s.

Answer. Burn 1.70 m/s retrograde, drift orbit 46.7 km below GEO, total 3.4 m/s.

Working

λ̇ = 12 deg / 20 days = 0.60 deg/day1.2120e-07 rad/s
Δv = λ̇ v / (3n) = 1.2120e-07 × 3074.7 / (3 × 7.292116e-05)1.703 m/s(retrograde)
Vis-viva after the burn: a = 1/(2/r − (v − Δv)²/μ) = 42,117.51 km, 46.66 km below GEO
Stop burn equal and opposite: total = 2 × 1.7033.41 m/s

Answer. Burn 1.70 m/s retrograde, drift orbit 46.7 km below GEO, total 3.4 m/s.

Sanity check. Faster relocation costs proportionally more: half the time doubles the Δv. Check the sign: a lower orbit has a shorter period, so it moves ahead (east).

Pythonedit a value, then Run

# Practice O17.3: Moving a satellite 12 degrees east
from math import pi, sqrt, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
shift = 12.0          # deg east
days = 20.0

r = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
v = sqrt(mu / r)                         # km/s
n = 2 * pi / day_sid
rate = radians(shift) / (days * 86400)   # rad/s
dv = rate * v / (3 * n)                  # retrograde burn: lower orbit drifts east
a = 1 / (2 / r - (v - dv) ** 2 / mu)     # vis-viva after the burn
print(f"drift rate {rate:.4e} rad/s, burn {dv*1000:.3f} m/s retrograde")
print(f"drift orbit a = {a:,.2f} km, {r - a:.2f} km below GEO")
print(f"total with the stop burn: {2*dv*1000:.2f} m/s")

drift rate 1.2120e-07 rad/s, burn 1.703 m/s retrograde
drift orbit a = 42,117.51 km, 46.66 km below GEO
total with the stop burn: 3.41 m/s

Common mistakes

Using 86,400 s for GEO: the radius comes out about 77 km high and the satellite drifts a degree a day.
Getting the drift sign backwards. Higher and slower means it drifts west.
Forgetting the end-of-life reserve; it is small but must exist when the tank is almost empty.
Treating north-south and east-west as similar. North-south is roughly thirty times larger.

Recap from earlier chapters

Hohmann total for a small raise Δr at radius r is about?

v Δr / (2r), half at each burn.

Why change plane where the speed is lowest?

Δv = 2v sin(Δi/2) scales with the speed v.

Write the rocket equation.

Δv = Isp g₀ ln(m₀/m_f).

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R2NASA JPL Solar System Dynamics, "Astrodynamic Parameters" (planetary GMs from DE440; Park et al. 2021, AJ 161:105).
R11Vallado, D.A., companion software to Fundamentals of Astrodynamics and Applications (CelesTrak), GitHub repo CelesTrak/fundamentals-of-astrodynamics, commit 49df047 (AGPL-3.0; values read, no code copied).
R18Lovell, R.R. and O'Malley, T.A., Station Keeping of High Power Communication Satellites, NASA TM X-2136, Lewis Research Center, Nov 1970.
R19Decoto, J. and Loerch, P. (Orbital ATK), "Technique for GEO RSO Station Keeping Characterization and Maneuver Detection", AMOS Technical Conference, 2015.
R20Inter-Agency Space Debris Coordination Committee, IADC Space Debris Mitigation Guidelines, IADC-02-01 Rev. 2, Mar 2020.
R22ESA, "Types of orbits", 30 Mar 2020.

Where in each source, and links: References.

I have finished chapter 17

Chapter 18

Launch windows and launch azimuth

30 minutes

The idea in one minute

A rocket can only put a satellite into a plane that passes over the launch site. Launching due east gives the lowest inclination, equal to the site's latitude, and borrows Earth's rotation for free. Any other inclination means steering north- or south-east, and any particular plane only sweeps over the site twice a day: those are the launch windows.

Each curve starts at an inclination equal to the site latitude, where the rocket flies due east (90 deg). Notice that no lower inclination is reachable directly, and every site flies due north for a polar orbit.

You will be able to

Find the launch azimuth for a target inclination, inertial and relative to the rotating ground
Explain why the lowest reachable inclination equals the site latitude
Time the two daily launch windows into a given orbit plane

Inclination and latitude

The orbit plane goes through Earth's centre and the site. Its inclination can never be less than the site latitude; it equals it when the rocket flies due east.

Azimuth

Spherical trigonometry gives sin β = cos i / cos φ. Close to i = φ the azimuth is 90 deg; for a polar orbit it is 0 or 180 deg; retrograde orbits need a westward component.

The rotating ground

The pad already moves east at ω_E R_E cos φ, about 465 m/s at the equator. The rocket only adds the difference, which shifts the ground-relative azimuth a little and saves or costs a few hundred m/s.^{[R1, R25]}

Launch windows

A fixed orbit plane turns relative to the Earth once per sidereal day, so the site passes under it twice: once on the ascending side and once on the descending side. Rendezvous or a sun-synchronous local time pins down which plane, so the window opens at a fixed time each day.

Key equations

Launch azimuth

sin β = cos i / cos φ

where

βinertial azimuth, clockwise from north, deg

iorbit inclination, deg

φsite latitude, deg

needs i ≥ φ

Ground speed

v_rot = ω_E R_E cos φ

where

v_roteastward speed of the launch site, m/s

ω_EEarth rotation rate, 7.292115e-5, rad/s

R_EEarth equatorial radius, m

φsite latitude, deg

Ground-relative velocity

v_rel = (v sin β − v_rot, v cos β) (east, north)

where

v_relvelocity the rocket must add relative to the ground, m/s

vorbital speed, m/s

βinertial azimuth, deg

v_rotsite ground speed, m/s

Window position

sin Δλ = tan φ / tan i

where

Δλlongitude of the site past the ascending node at launch, deg

φsite latitude, deg

iinclination, deg

second window at 180° − Δλ

Worked examples

Worked example 18.1

Launching to a 51.6 deg station orbit

Given

A site at latitude 28.5 deg launches to a circular 200 km orbit inclined 51.6 deg. Find the inertial azimuth, the azimuth and speed needed relative to the rotating ground, and the time between the two daily windows.

new symbols

T_sidsidereal day, 86164.0905, s

Working

sin β = cos i / cos φ = cos 51.6 / cos 28.5 = 0.7068, β44.98 deg(north-east)
Orbit speed 7784 m/s; ground speed eastward = ω_E R_E cos φ408.7 m/s
Relative to the ground: east 5502 − 409, north 5507: azimuth 42.77 deg, speed 7501 m/s
Window geometry: sin Δλ = tan φ / tan i = 0.4303, Δλ25.49 deg
Gap between the ascending and descending windows = (180 − 2Δλ)/360 × sidereal day8.58 h

Answer. Inertial azimuth 45.0 deg, ground-relative 42.8 deg; Earth's rotation saves 283 m/s; windows 8.6 h and 15.4 h apart.

Sanity check. The rotating-frame azimuth is a little north of the inertial one, because the ground already carries the rocket east.

Pythonedit a value, then Run

# Worked example 18.1: Launching to a 51.6 deg station orbit
from math import pi, sqrt, sin, cos, tan, asin, atan2, hypot, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
lat = radians(28.5)   # site latitude
i = radians(51.6)     # target inclination
h = 200.0             # km

beta = asin(cos(i) / cos(lat))           # inertial azimuth (ascending pass)
v = sqrt(mu / (R_E + h)) * 1000          # m/s
v_rot = w_E * R_E * 1000 * cos(lat)      # ground speed eastward, m/s
east, north = v * sin(beta) - v_rot, v * cos(beta)
az_rot, v_need = atan2(east, north), hypot(east, north)
dl = asin(tan(lat) / tan(i))
gap = (180 - 2 * degrees(dl)) / 360 * day_sid / 3600
print(f"inertial azimuth {degrees(beta):.2f} deg; orbit speed {v:.0f} m/s; ground speed {v_rot:.1f} m/s")
print(f"relative to the ground: azimuth {degrees(az_rot):.2f} deg, speed {v_need:.0f} m/s, saving {v - v_need:.1f} m/s")
print(f"windows {gap:.2f} h and {day_sid / 3600 - gap:.2f} h apart")

# Worked example 18.1: Launching to a 51.6 deg station orbit
from math import pi, sqrt, sin, cos, tan, asin, atan2, hypot, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
lat = radians(28.5)   # site latitude
i = radians(51.6)     # target inclination
h = 200.0             # km

beta = asin(cos(i) / cos(lat))           # inertial azimuth (ascending pass)
v = sqrt(mu / (R_E + h)) * 1000          # m/s
v_rot = w_E * R_E * 1000 * cos(lat)      # ground speed eastward, m/s
east, north = v * sin(beta) - v_rot, v * cos(beta)
az_rot, v_need = atan2(east, north), hypot(east, north)
dl = asin(tan(lat) / tan(i))
gap = (180 - 2 * degrees(dl)) / 360 * day_sid / 3600
print(f"inertial azimuth {degrees(beta):.2f} deg; orbit speed {v:.0f} m/s; ground speed {v_rot:.1f} m/s")
print(f"relative to the ground: azimuth {degrees(az_rot):.2f} deg, speed {v_need:.0f} m/s, saving {v - v_need:.1f} m/s")
print(f"windows {gap:.2f} h and {day_sid / 3600 - gap:.2f} h apart")

inertial azimuth 44.98 deg; orbit speed 7784 m/s; ground speed 408.7 m/s
relative to the ground: azimuth 42.77 deg, speed 7501 m/s, saving 283.3 m/s
windows 8.58 h and 15.36 h apart

Practice

Practice O18.1 · warm-up

Free speed from Earth's rotation^{[R1, R25]}

Given

Find the eastward speed a launch site carries at latitudes 5.2, 28.5 and 45.0 deg (ω_E = 7.292115e-05 rad/s, R_E = 6378.137 km), and the difference between the first and the last.

Mark as done

Hint

v = ω_E R_E cos φ; keep R_E in metres for m/s.

Answer. 463, 409 and 329 m/s; the near-equatorial site gains 134 m/s for an eastward launch.

Working

Equator: ω_E R_E = 7.292115e-05 × 6378137465.1 m/s
φ = 5.2 deg: 465.1 × cos 5.2463.2 m/s
φ = 28.5 deg: 465.1 × cos 28.5408.7 m/s
φ = 45.0 deg: 465.1 × cos 45.0328.9 m/s
Difference134.3 m/s

Answer. 463, 409 and 329 m/s; the near-equatorial site gains 134 m/s for an eastward launch.

Sanity check. The gain is only a few per cent of the 7784 m/s orbit speed, but the bigger prize of a low latitude is reaching low inclination without a plane change.

Pythonedit a value, then Run

# Practice O18.1: Free speed from Earth's rotation
from math import cos, radians

w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
R_E = 6378.137e3      # m, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)

v_eq = w_E * R_E
v = {lat: v_eq * cos(radians(lat)) for lat in (5.2, 28.5, 45.0)}
print(f"equator: {v_eq:.1f} m/s")
for lat, s in v.items():
    print(f"latitude {lat} deg: {s:.1f} m/s")
print(f"difference 5.2 vs 45.0 deg: {v[5.2] - v[45.0]:.1f} m/s")

equator: 465.1 m/s
latitude 5.2 deg: 463.2 m/s
latitude 28.5 deg: 408.7 m/s
latitude 45.0 deg: 328.9 m/s
difference 5.2 vs 45.0 deg: 134.3 m/s

Practice O18.2 · standard

The price of latitude^[R25]

Given

A satellite must end up in an equatorial 200 km orbit. Each site launches due east (i = latitude) and removes the inclination with one plane change in LEO. Compare sites at 5.2 and 28.5 deg.

Mark as done

Hint

Launching due east gives the minimum inclination, equal to the latitude. Then Δv = 2v sin(Δi/2).

Answer. 706 m/s versus 3832 m/s: the higher latitude costs 3126 m/s more in LEO.

Working

v_LEO = √(μ/r)7.7843 km/s
5.2 deg: Δv = 2 × 7.7843 × sin(2.6 deg)0.706 km/s
28.5 deg: Δv = 2 × 7.7843 × sin(14.25 deg)3.832 km/s
Penalty of the higher site3.126 km/s

Answer. 706 m/s versus 3832 m/s: the higher latitude costs 3126 m/s more in LEO.

Sanity check. That is so expensive that nobody does it in LEO; the change is made at GTO apogee where the speed is low (Chapter 19).

Pythonedit a value, then Run

# Practice O18.2: The price of latitude
from math import sqrt, sin, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
h = 200.0             # km

v = sqrt(mu / (R_E + h))
dv = {lat: 2 * v * sin(radians(lat) / 2) for lat in (5.2, 28.5)}   # remove i = latitude
print(f"v_LEO = {v:.4f} km/s")
for lat, d in dv.items():
    print(f"{lat} deg site: {d*1000:.1f} m/s")
print(f"penalty of the higher site: {(dv[28.5] - dv[5.2])*1000:.1f} m/s")

v_LEO = 7.7843 km/s
5.2 deg site: 706.2 m/s
28.5 deg site: 3832.2 m/s
penalty of the higher site: 3126.0 m/s

Practice O18.3 · challenge

A sun-synchronous launch from 57 deg north

Given

A site at 57 deg latitude launches to a 550 km sun-synchronous orbit. Find the inclination, the northbound and southbound inertial azimuths, and how much extra speed relative to the ground the retrograde orbit needs because Earth turns the wrong way.

Mark as done

Hint

Use the J2 sun-synchronous condition from Chapter 7. With i > 90 deg the formula gives a negative azimuth (north-west); the southbound option is 180 − β.

Answer. i = 97.59 deg; azimuths 346.0 deg (north) or 194.0 deg (south); the retrograde orbit costs 65 m/s extra.

Working

i_SSO97.593 deg
sin β = cos 97.593 / cos 57 = -0.2426, β = -14.04 deg → northbound 345.96 deg
Southbound azimuth = 180 − (-14.04)194.04 deg
Ground speed east = 253.3 m/s; orbit speed 7585 m/s
Speed needed relative to the ground = 7650.5 m/s, penalty 65.4 m/s

Answer. i = 97.59 deg; azimuths 346.0 deg (north) or 194.0 deg (south); the retrograde orbit costs 65 m/s extra.

Sanity check. Small penalty because the orbit is only 7.6 deg past polar and the site's ground speed is only 253 m/s at 57 deg.

Pythonedit a value, then Run

# Practice O18.3: A sun-synchronous launch from 57 deg north
from math import pi, sqrt, sin, cos, asin, acos, hypot, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
lat = radians(57)
h = 550.0             # km

a = R_E + h
n = sqrt(mu / a**3)
i = acos(-(2 * pi / year) / (1.5 * n * J2 * (R_E / a) ** 2))   # SSO inclination
beta = asin(cos(i) / cos(lat))           # negative: north-west
v = sqrt(mu / a) * 1000                  # m/s
v_rot = w_E * R_E * 1000 * cos(lat)
v_need = hypot(v * sin(beta) - v_rot, v * cos(beta))
print(f"i = {degrees(i):.3f} deg")
print(f"azimuth northbound {degrees(beta) % 360:.2f} deg, southbound {180 - degrees(beta):.2f} deg")
print(f"orbit speed {v:.0f} m/s, ground speed {v_rot:.1f} m/s, needed {v_need:.1f} m/s")
print(f"penalty {v_need - v:.1f} m/s")

# Practice O18.3: A sun-synchronous launch from 57 deg north
from math import pi, sqrt, sin, cos, asin, acos, hypot, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
lat = radians(57)
h = 550.0             # km

a = R_E + h
n = sqrt(mu / a**3)
i = acos(-(2 * pi / year) / (1.5 * n * J2 * (R_E / a) ** 2))   # SSO inclination
beta = asin(cos(i) / cos(lat))           # negative: north-west
v = sqrt(mu / a) * 1000                  # m/s
v_rot = w_E * R_E * 1000 * cos(lat)
v_need = hypot(v * sin(beta) - v_rot, v * cos(beta))
print(f"i = {degrees(i):.3f} deg")
print(f"azimuth northbound {degrees(beta) % 360:.2f} deg, southbound {180 - degrees(beta):.2f} deg")
print(f"orbit speed {v:.0f} m/s, ground speed {v_rot:.1f} m/s, needed {v_need:.1f} m/s")
print(f"penalty {v_need - v:.1f} m/s")

i = 97.593 deg
azimuth northbound 345.96 deg, southbound 194.04 deg
orbit speed 7585 m/s, ground speed 253.3 m/s, needed 7650.5 m/s
penalty 65.4 m/s

Common mistakes

Expecting to reach an inclination below the site latitude without a plane change.
Mixing the inertial azimuth with the azimuth relative to the ground.
Treating every launch time as equal: a target plane passes over the site only twice a day.
Forgetting that retrograde orbits (like sun-synchronous) pay for Earth's rotation instead of gaining.

Recap from earlier chapters

Sun-synchronous inclination at 550 km?

97.59 deg, from the J2 node rate matching 360 deg per year.

Δv for a 10 deg plane change at 7.7 km/s?

2 × 7.7 × sin 5 deg = 1.34 km/s.

Sources for this chapter

R1Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.
R25ESA, "Europe's Spaceport".

Where in each source, and links: References.

I have finished chapter 18

Chapter 19

Mission case end to end

45 minutes

The idea in one minute

Every chapter so far solved one piece. A real mission is all of them in a row: the launch site fixes the starting inclination, a transfer orbit lifts the satellite, the plane change goes where it is cheapest, station-keeping spends a little every year, and a final burn clears the slot. Add the pieces, apply the rocket equation, and you have the propellant the satellite must carry.

The spacecraft's own Δv for a 15-year GEO mission from 28.5 deg. Notice that the GTO-to-GEO apogee burn is most of it; north-south station-keeping comes next.

You will be able to

Chain launch, transfer, plane change, station-keeping and disposal into one Δv budget
Find the best split of a plane change between two burns numerically
Turn a Δv budget into propellant mass and check it with an independent propagation

The plan

A communications satellite launches due east from 28.5 deg into a 200 km parking orbit. The upper stage fires at perigee to enter a geostationary transfer orbit. The satellite fires at apogee to circularise and remove the inclination.

Where to change the plane

Most of the plane change belongs at apogee, where the speed is low. A few degrees at perigee, folded into the large perigee burn, cost almost nothing extra and make the apogee burn cheaper. A one-variable minimisation finds the best split.

Check it twice

The coast from perigee to apogee is half the transfer period by Kepler's equation; a numerical propagation of the same orbit must reach the GEO radius with constant energy.

From Δv to kilograms

Station-keeping and disposal are added to the apogee burn, and the rocket equation turns the total into propellant. About half the separated mass ends up as propellant.

Phase	Who burns	Δv (m/s)
Parking orbit → GTO perigee burn	upper stage	2478
Apogee burn: circularise + plane change	spacecraft	1789
North-south station-keeping, 15 years	spacecraft	684
East-west station-keeping, 15 years	spacecraft	22
Graveyard raise	spacecraft	11
Spacecraft total		2507

Site latitude 28.5 deg, 200 km parking orbit, 2.17 deg of plane change at perigee. Isp 315 s and separation mass 5000 kg are assumptions.

Key equations

Combined burn

Δv = √(v₁² + v₂² − 2 v₁ v₂ cos Δi)

where

v₁speed before the burn, km/s

v₂speed after the burn, km/s

Δiplane change in the same burn, deg

Split optimum

minimise Δv_p(x) + Δv_a(Δi − x)

where

xplane change done at perigee, deg

Δv_pperigee combined burn, km/s

Δv_aapogee combined burn, km/s

Δitotal inclination to remove, deg

Transfer orbit

a_t = (r_p + r_GEO)/2, v = √(μ(2/r − 1/a_t))

where

a_ttransfer semi-major axis, km

r_pparking orbit radius, km

r_GEOGEO radius, km

μEarth's gravitational parameter, km³/s²

rradius where the speed is wanted, km

Rocket equation

m₀/m_f = exp(Δv / (Isp g₀))

where

m₀mass before, kg

m_fmass after, kg

Δvtotal velocity change, m/s

Ispspecific impulse, s

g₀standard gravity, 9.80665, m/s²

Worked examples

Worked example 19.1

A GEO satellite from launch to retirement^{[R17, R18, R20]}

Given

Launch due east from a 28.5 deg site into a 200 km circular parking orbit, transfer to GEO by a Hohmann-type GTO, take out the 28.5 deg of inclination with the best split between perigee and apogee, then keep station for 15 years and retire 300 km above GEO. Spacecraft: 5000 kg at separation, Isp 315 s.

new symbols

v_aspeed at GTO apogee, km/s

v_GEOcircular speed at GEO, km/s

βlaunch azimuth, deg

Working

Launch azimuth: sin β = cos 28.5 / cos 28.5 = 1, β = 90 deg; the pad adds 409 m/s
Parking orbit: r = 6,578.137 km, v = 7.7843 km/s, specific energy −μ/2r-30.297 km²/s²
GTO: a = 24,371.2 km, e = 0.7301; perigee 10.2388 km/s, apogee 1.5974 km/s; coast 5.26 h (half the period, Kepler)
Best split (numerical minimum): 2.169 deg at perigee, 26.331 deg at apogee
Perigee burn 2478 m/s; apogee burn √(v_a² + v_GEO² − 2 v_a v_GEO cos Δi)1789 m/s
Saving against doing all the plane change at apogee: 24.1 m/s
Station-keeping: 684 m/s N-S + 22 m/s E-W; graveyard 10.9 m/s
Spacecraft total 2507 m/s; mass ratio exp(2507 / (315 × 9.80665))2.2513
Propellant = 5000 − 5000/2.2513 = 2779 kg; 2221 kg left at end of life

Answer. Spacecraft Δv 2507 m/s, of which the apogee burn is 1789 m/s; propellant 2779 kg of the 5000 kg separated mass.

Sanity check. The numerical propagation of the GTO reaches 42,164.2 km at apogee with energy constant to 1e-11: the coast is consistent with Kepler. About 56 % of the satellite is propellant, typical of chemical GEO platforms.

Pythonedit a value, then Run

# Worked example 19.1: A GEO satellite from launch to retirement
from math import pi, sqrt, sin, cos, exp, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
g0 = 9.80665          # m/s^2, standard gravity (exact, by definition)
lat = radians(28.5)   # site latitude = initial inclination
h_park = 200.0        # km
m0, Isp = 5000.0, 315.0  # kg at separation, s
years = 15
ns_rate = radians(0.85)  # GEO inclination drift per year (chapter 17)
ew_rate = 1.5         # m/s per year east-west (chapter 17)
grave = 300.0         # km above GEO

def burn(v1, v2, angle):  # one burn: speed change plus plane change
    return sqrt(v1**2 + v2**2 - 2 * v1 * v2 * cos(angle))

r_p = R_E + h_park
r_geo = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
at = (r_p + r_geo) / 2
v_park, v_per = sqrt(mu / r_p), sqrt(mu * (2 / r_p - 1 / at))
v_apo, v_geo = sqrt(mu * (2 / r_geo - 1 / at)), sqrt(mu / r_geo)
print(f"pad speed {w_E * R_E * 1000 * cos(lat):.0f} m/s; GTO a = {at:,.1f} km, perigee {v_per:.4f}, apogee {v_apo:.4f} km/s")

def cost(x):              # x = plane change done at perigee
    return burn(v_park, v_per, x) + burn(v_apo, v_geo, lat - x)

lo, hi = 0.0, lat         # golden-section search for the best split
g = (sqrt(5) - 1) / 2
for _ in range(100):
    x1, x2 = hi - g * (hi - lo), lo + g * (hi - lo)
    if cost(x1) < cost(x2):
        hi = x2
    else:
        lo = x1
x = (lo + hi) / 2
dv_per = burn(v_park, v_per, x) * 1000
dv_apo = burn(v_apo, v_geo, lat - x) * 1000
save = ((v_per - v_park) + burn(v_apo, v_geo, lat) - cost(x)) * 1000
print(f"best split {degrees(x):.3f} deg at perigee: perigee burn {dv_per:.0f} m/s, apogee burn {dv_apo:.1f} m/s, saving {save:.1f} m/s")

dv_ns = 2 * v_geo * 1000 * sin(ns_rate / 2) * years
dv_ew = ew_rate * years
r2 = r_geo + grave                       # graveyard Hohmann
a2 = (r_geo + r2) / 2
dv_gy = ((sqrt(mu * (2 / r_geo - 1 / a2)) - v_geo) + (sqrt(mu / r2) - sqrt(mu * (2 / r2 - 1 / a2)))) * 1000
dv_sc = dv_apo + dv_ns + dv_ew + dv_gy
MR = exp(dv_sc / (Isp * g0))
mp = m0 - m0 / MR
print(f"station-keeping {dv_ns:.0f} + {dv_ew:.0f} m/s, graveyard {dv_gy:.1f} m/s")
print(f"spacecraft total {dv_sc:.1f} m/s, mass ratio {MR:.4f}")
print(f"propellant {mp:.1f} kg of {m0:.0f} kg; {m0 - mp:.0f} kg left at end of life")

# Worked example 19.1: A GEO satellite from launch to retirement
from math import pi, sqrt, sin, cos, exp, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
w_E = 7.292115e-5     # rad/s, Earth's rotation rate (IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
g0 = 9.80665          # m/s^2, standard gravity (exact, by definition)
lat = radians(28.5)   # site latitude = initial inclination
h_park = 200.0        # km
m0, Isp = 5000.0, 315.0  # kg at separation, s
years = 15
ns_rate = radians(0.85)  # GEO inclination drift per year (chapter 17)
ew_rate = 1.5         # m/s per year east-west (chapter 17)
grave = 300.0         # km above GEO

def burn(v1, v2, angle):  # one burn: speed change plus plane change
    return sqrt(v1**2 + v2**2 - 2 * v1 * v2 * cos(angle))

r_p = R_E + h_park
r_geo = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
at = (r_p + r_geo) / 2
v_park, v_per = sqrt(mu / r_p), sqrt(mu * (2 / r_p - 1 / at))
v_apo, v_geo = sqrt(mu * (2 / r_geo - 1 / at)), sqrt(mu / r_geo)
print(f"pad speed {w_E * R_E * 1000 * cos(lat):.0f} m/s; GTO a = {at:,.1f} km, perigee {v_per:.4f}, apogee {v_apo:.4f} km/s")

def cost(x):              # x = plane change done at perigee
    return burn(v_park, v_per, x) + burn(v_apo, v_geo, lat - x)

lo, hi = 0.0, lat         # golden-section search for the best split
g = (sqrt(5) - 1) / 2
for _ in range(100):
    x1, x2 = hi - g * (hi - lo), lo + g * (hi - lo)
    if cost(x1) < cost(x2):
        hi = x2
    else:
        lo = x1
x = (lo + hi) / 2
dv_per = burn(v_park, v_per, x) * 1000
dv_apo = burn(v_apo, v_geo, lat - x) * 1000
save = ((v_per - v_park) + burn(v_apo, v_geo, lat) - cost(x)) * 1000
print(f"best split {degrees(x):.3f} deg at perigee: perigee burn {dv_per:.0f} m/s, apogee burn {dv_apo:.1f} m/s, saving {save:.1f} m/s")

dv_ns = 2 * v_geo * 1000 * sin(ns_rate / 2) * years
dv_ew = ew_rate * years
r2 = r_geo + grave                       # graveyard Hohmann
a2 = (r_geo + r2) / 2
dv_gy = ((sqrt(mu * (2 / r_geo - 1 / a2)) - v_geo) + (sqrt(mu / r2) - sqrt(mu * (2 / r2 - 1 / a2)))) * 1000
dv_sc = dv_apo + dv_ns + dv_ew + dv_gy
MR = exp(dv_sc / (Isp * g0))
mp = m0 - m0 / MR
print(f"station-keeping {dv_ns:.0f} + {dv_ew:.0f} m/s, graveyard {dv_gy:.1f} m/s")
print(f"spacecraft total {dv_sc:.1f} m/s, mass ratio {MR:.4f}")
print(f"propellant {mp:.1f} kg of {m0:.0f} kg; {m0 - mp:.0f} kg left at end of life")

pad speed 409 m/s; GTO a = 24,371.2 km, perigee 10.2388, apogee 1.5974 km/s
best split 2.169 deg at perigee: perigee burn 2478 m/s, apogee burn 1789.3 m/s, saving 24.1 m/s
station-keeping 684 + 22 m/s, graveyard 10.9 m/s
spacecraft total 2506.8 m/s, mass ratio 2.2513
propellant 2779.1 kg of 5000 kg; 2221 kg left at end of life

Practice

Practice O19.1 · warm-up

The same satellite from 5.2 deg^[R25]

Given

Launch due east from a 5.2 deg site into the same 200 km parking orbit and GTO. Doing the whole plane change at apogee, find the apogee burn and compare it with 1836 m/s from 28.5 deg.

Mark as done

Hint

Only the plane change differs; use the combined-burn law of cosines with v_a and v_GEO from the worked example.

Answer. Apogee burn 1491 m/s, 346 m/s less than from 28.5 deg.

Working

v_a = 1.5974 km/s, v_GEO3.0747 km/s
Δv = √(1.5974² + 3.0747² − 2 × 1.5974 × 3.0747 × cos 5.2 deg)1491 m/s
Saving = 1836 − 1491346 m/s

Answer. Apogee burn 1491 m/s, 346 m/s less than from 28.5 deg.

Sanity check. Even with zero plane change the burn would be v_GEO − v_a = 1477 m/s, so 5.2 deg costs only 14 m/s extra.

Pythonedit a value, then Run

# Practice O19.1: The same satellite from 5.2 deg
from math import pi, sqrt, cos, radians

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
day_sid = 86164.0905  # s, sidereal day (JPL SSD astrodynamic parameters)
h_park = 200.0        # km

r_p = R_E + h_park
r_geo = (mu * (day_sid / (2 * pi)) ** 2) ** (1 / 3)
at = (r_p + r_geo) / 2
va, vg = sqrt(mu * (2 / r_geo - 1 / at)), sqrt(mu / r_geo)

def apogee_burn(lat_deg):     # whole plane change at apogee
    di = radians(lat_deg)
    return sqrt(va**2 + vg**2 - 2 * va * vg * cos(di)) * 1000

low, high = apogee_burn(5.2), apogee_burn(28.5)
print(f"v_a = {va:.4f} km/s, v_GEO = {vg:.4f} km/s")
print(f"apogee burn from 5.2 deg: {low:.1f} m/s; from 28.5 deg: {high:.1f} m/s")
print(f"saving {high - low:.1f} m/s")

v_a = 1.5974 km/s, v_GEO = 3.0747 km/s
apogee burn from 5.2 deg: 1490.9 m/s; from 28.5 deg: 1836.5 m/s
saving 345.6 m/s

Practice O19.2 · standard

Ten seconds of Isp^[R17]

Given

Keep the spacecraft Δv budget of 2507 m/s and separation mass 5000 kg. How much propellant is saved if the engine's Isp rises from 315 s to 325 s?

Mark as done

Hint

Propellant = m₀(1 − exp(−Δv/(Isp g₀))) for each Isp.

Answer. 56 kg less propellant, mass that can become payload.

Working

Isp 315 s: m_p = 5000 × (1 − exp(−2507/(315 × 9.80665)))2779.1 kg
Isp 325 s: m_p = 5000 × (1 − exp(−2507/(325 × 9.80665)))2722.9 kg
Saving56.2 kg

Answer. 56 kg less propellant, mass that can become payload.

Sanity check. A 3.2 % rise in Isp saves 2.0 % of the propellant: the exponential makes Isp valuable on big budgets.

Pythonedit a value, then Run

# Practice O19.2: Ten seconds of Isp
from math import exp

g0 = 9.80665          # m/s^2, standard gravity (exact, by definition)
dv = 2507.0           # m/s, spacecraft budget
m0 = 5000.0           # kg

mp = {isp: m0 * (1 - exp(-dv / (isp * g0))) for isp in (315.0, 325.0)}
for isp, m in mp.items():
    print(f"Isp {isp:.0f} s: propellant {m:.1f} kg")
print(f"saving {mp[315.0] - mp[325.0]:.1f} kg")

Isp 315 s: propellant 2779.2 kg
Isp 325 s: propellant 2723.0 kg
saving 56.2 kg

Practice O19.3 · challenge

An Earth-observation variant^{[R14, R15]}

Given

A small satellite goes to a 550 km sun-synchronous orbit from a 57 deg site. Find the inclination, check the J2 node rate is one turn per year, then estimate the yearly decay Δa = 2π C_D (A/m) ρ a² per revolution × revolutions per year (C_D = 2.2, A/m = 0.01 m²/kg, ρ = 2e-13 kg/m³) and the 5-year reboost Δv.

Assumption: ρ = 2e-13 kg/m³ at 550 km (moderate solar activity)

Mark as done

Hint

Work Δa in metres (a in m). A small reboost costs Δv ≈ v Δa / (2a). Southbound azimuth = 180 − β.

Answer. i = 97.59 deg; decay about 8.8 km per year; 24.0 m/s keeps the altitude for 5 years.

Working

i = 97.593 deg; southbound azimuth from 57 deg: 194.04 deg
Ω̇ = −1.5 n J₂ (R_E/a)² cos i = 1.9911e-07 rad/s → × one year360.0000 deg
Δa per rev = 2π × 2.2 × 0.01 × 2e-13 × (6.9281e+06)²1.592 m
Revolutions per year = 5499 → Δa8.76 km/yr
Reboost Δv = 7585.1 × 8.76 / (2 × 6,928.1) = 4.79 m/s per year → 5 years: 24.0 m/s

Answer. i = 97.59 deg; decay about 8.8 km per year; 24.0 m/s keeps the altitude for 5 years.

Sanity check. The density is the weak link: at solar maximum it can be tens of times higher, so real budgets carry a large margin.

Pythonedit a value, then Run

# Practice O19.3: An Earth-observation variant
from math import pi, sqrt, cos, asin, acos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
h = 550.0             # km
lat = radians(57)
C_D, Am = 2.2, 0.01   # drag coefficient, m^2/kg
rho = 2.4e-13         # kg/m^3 (U.S. Standard Atmosphere 1976 at 550 km)
years = 5

a = R_E + h
n = sqrt(mu / a**3)
k = 1.5 * n * J2 * (R_E / a) ** 2
i = acos(-(2 * pi / year) / k)
node_turn = degrees(-k * cos(i)) * year  # deg per year
az_south = 180 - degrees(asin(cos(i) / cos(lat)))
da_rev = 2 * pi * C_D * Am * rho * (a * 1000) ** 2 / 1000   # km per revolution
revs = year / (2 * pi / n)
da_year = da_rev * revs
v = sqrt(mu / a) * 1000                  # m/s
dv_year = v * da_year / (2 * a)
print(f"i = {degrees(i):.3f} deg, southbound azimuth {az_south:.2f} deg, node turns {node_turn:.4f} deg/yr")
print(f"decay {da_rev*1000:.3f} m/rev x {revs:.0f} revs = {da_year:.2f} km/yr")
print(f"reboost {dv_year:.2f} m/s per year, {years} years: {years * dv_year:.1f} m/s")

# Practice O19.3: An Earth-observation variant
from math import pi, sqrt, cos, asin, acos, radians, degrees

mu = 398600.4418      # km^3/s^2, Earth's gravitational parameter (IERS Conventions 2010, Table 1.1)
R_E = 6378.137        # km, Earth's equatorial radius (GRS80, IERS Conventions 2010, Table 1.2)
J2 = 1.08262668e-3    # Earth's oblateness coefficient (EGM96 value; IERS zero-tide 1.0826359e-3)
year = 365.2422 * 86400  # s, tropical year (NASA Earth fact sheet)
h = 550.0             # km
lat = radians(57)
C_D, Am = 2.2, 0.01   # drag coefficient, m^2/kg
rho = 2.4e-13         # kg/m^3 (U.S. Standard Atmosphere 1976 at 550 km)
years = 5

a = R_E + h
n = sqrt(mu / a**3)
k = 1.5 * n * J2 * (R_E / a) ** 2
i = acos(-(2 * pi / year) / k)
node_turn = degrees(-k * cos(i)) * year  # deg per year
az_south = 180 - degrees(asin(cos(i) / cos(lat)))
da_rev = 2 * pi * C_D * Am * rho * (a * 1000) ** 2 / 1000   # km per revolution
revs = year / (2 * pi / n)
da_year = da_rev * revs
v = sqrt(mu / a) * 1000                  # m/s
dv_year = v * da_year / (2 * a)
print(f"i = {degrees(i):.3f} deg, southbound azimuth {az_south:.2f} deg, node turns {node_turn:.4f} deg/yr")
print(f"decay {da_rev*1000:.3f} m/rev x {revs:.0f} revs = {da_year:.2f} km/yr")
print(f"reboost {dv_year:.2f} m/s per year, {years} years: {years * dv_year:.1f} m/s")

i = 97.593 deg, southbound azimuth 194.04 deg, node turns 360.0000 deg/yr
decay 1.592 m/rev x 5499 revs = 8.76 km/yr
reboost 4.79 m/s per year, 5 years: 24.0 m/s

Common mistakes

Doing the whole plane change in LEO: it costs kilometres per second instead of hundreds of m/s.
Adding Δv from different phases but applying the rocket equation with the wrong starting mass.
Forgetting the end-of-life reserve or the station-keeping years in the budget.
Trusting one method alone. Cross-check analytic coasts with a numerical propagation.

Recap from earlier chapters

Specific energy of a 200 km circular orbit?

−μ/2r = -30.30 km²/s².

Hohmann LEO-to-GEO total from 200 km, coplanar?

3.932 km/s.

North-south station-keeping per year?

About 46 m/s.

Sources for this chapter

R14NOAA, NASA and USAF, U.S. Standard Atmosphere, 1976, NASA-TM-X-74335 (NTRS 19770009539), Washington DC, Oct 1976.
R15Boring, J.W. and Humphris, R.R., "Drag coefficients for spheres in free molecular flow in O at satellite velocities", NASA CR-2233, 1973.
R18Lovell, R.R. and O'Malley, T.A., Station Keeping of High Power Communication Satellites, NASA TM X-2136, Lewis Research Center, Nov 1970.
R20Inter-Agency Space Debris Coordination Committee, IADC Space Debris Mitigation Guidelines, IADC-02-01 Rev. 2, Mar 2020.
R25ESA, "Europe's Spaceport".

Where in each source, and links: References.

I have finished chapter 19

Assumption	Source	Verdict
Sun gravitational parameter μ_Sun = 1.32712440018 × 10¹¹ km³/s²	R2: JPL SSD Astrodynamic Parameters (DE440) (1.32712440041e11); R8: IAU 2015 B3	CONFIRMEDagrees to 2e-12 relative; ours is the earlier DE405 value
Moon gravitational parameter μ_Moon = 4902.800 km³/s²	R2: JPL SSD Astrodynamic Parameters (DE440)	CONFIRMED
Mean Earth-Moon distance 384,400 km	R6: NASA NSSDCA Moon Fact Sheet (2024)	CONFIRMED
Venus orbit radius 0.723332 AU (circular model)	R4: Standish and Williams, Approximate Positions of the Planets, Table 1	CONFIRMED5e-6 relative
Venus μ = 324,859 km³/s²	R2: JPL SSD Astrodynamic Parameters (DE440)	CONFIRMED
Venus radius 6051.8 km	R3: JPL SSD Planetary Physical Parameters (Archinal et al. 2018)	CONFIRMED
Earth orbit radius 1.000000 AU (circular model)	R4: Standish and Williams, Approximate Positions of the Planets, Table 1	CONFIRMED3e-6
Mars orbit radius 1.523679 AU (circular model)	R4: Standish and Williams, Approximate Positions of the Planets, Table 1	CONFIRMED2e-5
Mars μ = 42,828 km³/s²	R2: JPL SSD Astrodynamic Parameters (DE440)	CONFIRMED
Mars equatorial radius 3396.2 km	R3: JPL SSD Planetary Physical Parameters (Archinal et al. 2018)	CONFIRMED
Jupiter orbit radius 5.2026 AU (circular model)	R4: Standish and Williams, Approximate Positions of the Planets, Table 1	CONFIRMED6e-5; rounded value
Jupiter μ = 1.26686 × 10⁸ km³/s²	^[R2] R8: IAU 2015 B3 nominal (GM)J = 1.2668653e8 km³/s²	CONFIRMED
Jupiter equatorial radius 71,492 km	R3: JPL SSD Planetary Physical Parameters (Archinal et al. 2018); R8: IAU 2015 B3 nominal	ADJUSTEDto 71,492 km
Drag coefficient C_D = 2.2 for a compact satellite in free-molecular flow (representative; measured sphere values are just above 2)	R15: Boring and Humphris, NASA CR-2233 (1973), abstract	CONFIRMEDas representative (2.2 sits just above the measured sphere value; real values depend on shape and surface, typically 2-2.5). Note: ISS operational fit uses C_D = 1.10 with its own reference area (R21), so C_D only has meaning together with the area it is paired with
Radiation-pressure coefficient C_R = 1.3 (1 = fully absorbing, 2 = perfect mirror)	UNVERIFIED: the 1-2 bounds are definitional; no readable source for 1.3 as a typical value	UNVERIFIEDthe bounds are definitional; the typical value 1.3 has no readable primary. Kept, labelled representative
Area-to-mass ratio A/m = 0.01 m²/kg for a typical small satellite	UNVERIFIED: representative design input, not a measured value	UNVERIFIEDdesign input for a generic small satellite, not a physical constant. Kept, labelled representative
Area-to-mass ratio A/m = 0.0032 m²/kg for a large crewed station (ISS drag area over mass)	R21: NASA JSC TOPO, ISS public ephemeris header (2 Nov 2026): 1511.8 m², 473,406.5 kg	ADJUSTEDto 0.0032 m²/kg. With the pack's C_D = 2.2 the product C_D A/m is still about 2x the TOPO fit (0.0070 vs 0.0035), so the station decay example is an upper-end estimate; say so in the text
Atmospheric density table, 200-1000 km, U.S. Standard Atmosphere 1976 (moderate solar activity; real density varies by tens of times over the solar cycle at these heights)	R14: U.S. Standard Atmosphere 1976 (NASA-TM-X-74335), Table I, pp. 69-73	ADJUSTEDrows 250 → 6.1e-11, 900 → 5.8e-15, 1000 → 3.6e-15 (other nine rows already equal USSA76 to 2 s.f.). Cross-check: Vallado's exponential model (R11 ATMOSEXP.DAT) is 10-25 % higher in 300-600 km, consistent with the "varies about 10x over the cycle" caveat. Re-cite to R14 (the SMAD citation was never read)
Monopropellant hydrazine thruster specific impulse I_sp = 220 s	R16: NASA Small Spacecraft State of the Art 2026, ch. 4, §4.6.1 (200-235 s)	CONFIRMED
Lunisolar inclination drift of an uncontrolled GEO satellite about 0.85 deg/yr	R18: Lovell and O'Malley, NASA TM X-2136 (1970), Summary; R19: Decoto and Loerch, AMOS 2015, §2	CONFIRMEDfor 0.85. The 0.75-0.95 range over the 18.6-yr lunar node cycle is UNVERIFIED (no readable source); Phase 2 drops the range from the label rather than keep an unsourced number
Typical GEO east-west station-keeping budget 1.5 m/s per year (sourced bound: under 1.75 m/s/yr for the triaxiality drift; about 2 m/s/yr often quoted)	R18: NASA TM X-2136 (1970), Summary and Introduction	CONFIRMED1.5 lies inside the sourced bound; range 1-2 consistent with R18
End-of-life re-orbit about 300 km above GEO (IADC minimum 235 km + 1000 C_R A/m; 300 km covers A/m up to 0.05 m²/kg at C_R = 1.3)	R20: IADC Space Debris Mitigation Guidelines, IADC-02-01 Rev. 2 (2020), §5.3.1, p. 11	CONFIRMEDwith C_R = 1.3, 300 km covers A/m up to 0.05 m²/kg (formula gives 261 km at 0.02, 300 km at 0.05)
Spacecraft bipropellant apogee engine and thrusters, Isp = 315 s (slightly below 400 N apogee motors at 318-321 s)	R17: ArianeGroup 400 N apogee motors S400-12 / S400-15	CONFIRMED315 s is slightly conservative for the apogee engine; small biprop thrusters are lower
Mission case: spacecraft mass at separation in GTO 5000 kg	UNVERIFIED: representative mid-size GEO platform (design input)	UNVERIFIEDmission-case design input, not a fact; labelled "representative mid-size GEO platform"
Mission case: air density at 550 km about 2.4 × 10⁻¹³ kg/m³ (U.S. Standard Atmosphere 1976, moderate solar activity; varies by about 10x over the cycle)	R14: U.S. Standard Atmosphere 1976 (NASA-TM-X-74335), Table I, p. 72	ADJUSTEDto 2.4e-13 kg/m³ (consistent with the corrected table; our 5e-13 was 2.1x the moderate-activity reference). Note: the ADCS v2 pack uses the same 5e-13 (not edited here)
Drag coefficient C_D = 2.2 for a compact satellite	R15: Boring and Humphris, NASA CR-2233 (1973), abstract	CONFIRMEDas representative
Mission case: area-to-mass ratio A/m = 0.01 m²/kg	UNVERIFIED: representative small satellite (design input)	UNVERIFIEDdesign input

Problem	Answer
O1.1	v = 7.585 km/s, T = 95.6 min, 15.05 orbits per day.
O1.2	h = 758 km for a 100-min period; h = 881 km for 14 orbits per sidereal day.
O1.3	T = 43854 s = 12.182 h by either route.
O2.1	a = 7378.1 km, e = 0.0678 (dimensionless).
O2.2	r(135°) = 26097 km; r = 20000 km at ν = ±120.0°.
O2.3	p = 15360 km, ν∞ = 140.3°, turn 100.6°, r(100°) = 19838 km.
O3.1	v_c = 7.452 km/s, v_esc = 10.538 km/s, Δv = 3.087 km/s.
O3.2	a = 9515 km, e = 0.1810, perigee altitude 1415 km, apogee altitude 4860 km.
O3.3	ε = 8.47 km²/s², v∞ = 4.117 km/s, e = 1.2882, v = 4.212 km/s at a million km.
O4.1	t = 25.7 min after perigee.
O4.2	ν = 131.57°, r = 10226 km (altitude 3847 km).
O4.3	t = 3.42 h after perigee, at ν = 116.9°.
O5.1	\|h\| = 51966 km²/s, i = 47.12°.
O5.2	a = 11500 km, e = 0.250, i = 35.00°, Ω = 80.00°, ω = 40.00°, ν = 60.00°.
O5.3	u = 10.000° is reliable; ω and ν separately are ill-conditioned when e ≈ 0, though in exact arithmetic the round trip still closes (error about 10⁻¹² km, floating-point level).
O6.1	Ω̇ = -5.14 °/day: the node drifts west by about 5.1 deg every day.
O6.2	a_drag = 1.9e-08 m/s² and a_SRP = 5.9e-08 m/s²: at 700 km they are within a factor of 3.1 of each other.
O6.3	About 2.2 km lost in 30 days; the reboost costs about 1.3 m/s.
O7.1	i = 99.03°.
O7.2	The node runs ahead of the Sun by 4.37° per year, a local-time drift of about 17 min per year.
O7.3	Apogee altitude 39367 km, e = 0.722; 73.0 % of each orbit is spent in the apogee half; at 60° the perigee would drift 0.037 deg/day.
O8.1	24.72° with J₂ against 24.76° without: the passes are about 24.7° apart.
O8.2	±3° in latitude but only ±0.039° in longitude: a tall, thin figure-8.
O8.3	h = 720.0 km at i = 98.27°: 29 orbits in 2 days, then the track repeats.
O9.1	Δv₁ = 164 m/s, Δv₂ = 161 m/s, total 325 m/s.
O9.2	a_ph = 27068 km and a total Δv of 69.8 m/s over 3 revolutions.
O9.3	Bi-elliptic 4014 m/s against Hohmann 4053 m/s, a saving of 39 m/s, but 221 h instead of 18 h.
O10.1	Δv = 655 m/s.
O10.2	1486 m/s from 5.2° and 2248 m/s from 45°, a difference of 762 m/s.
O10.3	Do about 2.6° of the turn at perigee (s = 0.066); total 4504 m/s, saving 34 m/s.
O11.1	Δy = +3.41 km: about 3.4 km ahead after one orbit.
O11.2	ẏ = 3.36 m/s, so the gap closes in 1.7 h.
O11.3	Δv₁ = 6.27 m/s, Δv₂ = 2.19 m/s, total 8.47 m/s.
O12.1	c = 12806 km, s = 15403 km, a_min = 7702 km.
O12.2	\|v₁\| = 8.657 km/s, a = 10235 km, e = 0.401, departure burn 2.389 km/s.
O12.3	The 60 min transfer is cheapest at 0.89 km/s total.
O13.1	T_syn = 780 days, about 26 months between Mars windows.
O13.2	t_H = 146 days, Venus 54.0° behind Earth at departure, windows every 584 days.
O13.3	C3 rises to 14.4 km²/s², 66 % more than the optimum.
O14.1	Mars SOI ≈ 577,225 km (170 Mars radii); Earth SOI ≈ 924,647 km.
O14.2	(a) 2.08 km/s into the circular orbit; (b) 1.03 km/s into the ellipse.
O14.3	r_opt = 12,208 km (altitude 8,811 km), Δv_min = 1.873 km/s, 207 m/s less than at 400 km.
O15.1	e = 1.504, turn angle δ = 83.3°.
O15.2	Δv = 6.65 km/s at 500 km, up to 6.82 km/s for a grazing pass.
O15.3	δ = 122°, speed 20.84 km/s against an escape speed of 18.47 km/s: the probe escapes the Sun.
O16.1	r ≈ 1,084,053 km (the exact root gives 1,082,332 km).
O16.2	L2 is 64,515 km beyond the Moon; Hill is off by -5.0 %.
O16.3	λ = 2.484 n, so a small error grows by a factor e every 23 days.
O17.1	λ̇ = -0.0321 deg/day westward; it leaves the box after 1.6 days.
O17.2	Δv ≈ 10.9 m/s in two near-equal burns, about 6.7 kg of propellant.
O17.3	Burn 1.70 m/s retrograde, drift orbit 46.7 km below GEO, total 3.4 m/s.
O18.1	463, 409 and 329 m/s; the near-equatorial site gains 134 m/s for an eastward launch.
O18.2	706 m/s versus 3832 m/s: the higher latitude costs 3126 m/s more in LEO.
O18.3	i = 97.59 deg; azimuths 346.0 deg (north) or 194.0 deg (south); the retrograde orbit costs 65 m/s extra.
O19.1	Apogee burn 1491 m/s, 346 m/s less than from 28.5 deg.
O19.2	56 kg less propellant, mass that can become payload.
O19.3	i = 97.59 deg; decay about 8.8 km per year; 24.0 m/s keeps the altitude for 5 years.

Id	Reference
R1	Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS Technical Note 36, Ch. 1 "General definitions and numerical standards" (update of 16 Nov 2017). IERS.Where: Table 1.1 (GM⊕, c, au, ERA rate), Table 1.2 (GRS80: a_E, J2, ω), pp. 3-4https://iers-conventions.obspm.fr/content/chapter1/icc1.pdfCited in: Chapters 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 17, 18
R2	NASA JPL Solar System Dynamics, "Astrodynamic Parameters" (planetary GMs from DE440; Park et al. 2021, AJ 161:105).Where: GM Sun, GM Moon, planet system GMs, au, mean sidereal dayhttps://ssd.jpl.nasa.gov/astro_par.htmlCited in: Chapters 1, 6, 8, 13, 14, 15, 16, 17; assumptions
R3	NASA JPL Solar System Dynamics, "Planetary Physical Parameters" (radii from Archinal et al. 2018, CMDA 130:22; change log 2019-12-12).Where: equatorial radii of Venus, Mars, Jupiterhttps://ssd.jpl.nasa.gov/planets/phys_par.htmlCited in: Chapters 14, 15; assumptions
R4	Standish, E.M. and Williams, J.G., "Approximate Positions of the Planets", NASA JPL SSD.Where: Table 1 (1800-2050 AD), semi-major axeshttps://ssd.jpl.nasa.gov/planets/approx_pos.htmlCited in: Chapters 13, 14, 15; assumptions
R5	Williams, D.R., "Earth Fact Sheet", NASA NSSDCA, Goddard Space Flight Center, updated 15 Nov 2024.Where: GM, R_E, J2, sidereal rotation, tropical year, e, obliquityhttps://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.htmlCited in: Chapters 1, 7
R6	Williams, D.R., "Moon Fact Sheet", NASA NSSDCA, updated 11 Jan 2024.Where: semi-major axis 0.3844 × 10⁶ kmhttps://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.htmlCited in: Chapter 16; assumptions
R7	IAU 2012 Resolution B2, "Re-definition of the astronomical unit of length", XXVIII General Assembly, Beijing, Aug 2012 (text hosted by SYRTE, Observatoire de Paris).Where: au = 149 597 870 700 m exactlyhttps://syrte.obspm.fr/IAU_resolutions/IAUResol_2012_0.htmlCited in: Chapter 13
R8	Mamajek, E.E., Prša, A., Torres, G. et al., "IAU 2015 Resolution B3 on Recommended Nominal Conversion Constants for Selected Solar and Planetary Properties", arXiv:1510.07674 (2015).Where: p. 2: S = 1361 W/m², (GM)☉ = 1.3271244e20, R_eJ = 7.1492e7 m, (GM)J = 1.2668653e17 m³/s²https://arxiv.org/abs/1510.07674Cited in: Chapters 6, 15; assumptions
R9	NIST, CODATA 2022 recommended values, "standard acceleration of gravity".Where: g_n = 9.80665 m/s², exacthttps://physics.nist.gov/cgi-bin/cuu/Value?gnCited in: Chapter 9
R10	MathWorks, Aerospace Toolbox documentation, gravityzonal (Earth zonal coefficients standardised to EGM2008, Pavlis et al. 2012). Secondary.Where: Algorithms table: J2 = 1.08262617e-3, J3 = −2.53241e-6, J4 = −1.61990e-6https://www.mathworks.com/help/aerotbx/ug/gravityzonal.html
R11	Vallado, D.A., companion software to Fundamentals of Astrodynamics and Applications (CelesTrak), GitHub repo CelesTrak/fundamentals-of-astrodynamics, commit 49df047 (AGPL-3.0; values read, no code copied).Where: software/python/tests/astro/twobody/test_frame_conversions.py (Ex. 2-5 rv2coe values); software/python/src/valladopy/constants.py (EGM-08 constants, sidereal day); software/misc/pascal/ATMOSEXP.DAT (exponential atmosphere table)https://github.com/CelesTrak/fundamentals-of-astrodynamicsCited in: Chapters 5, 17
R12	poliastro project, tests/test_iod.py::test_curtis52 (Curtis, Orbital Mechanics for Engineering Students, Example 5.2), commit 21fd771. Secondary record of a textbook case.Where: r₁, r₂, 1 h, expected v₁, v₂https://github.com/poliastro/poliastro/blob/main/tests/test_iod.pyCited in: Chapter 12
R13	Izzo, D., "Revisiting Lambert's problem", Celestial Mechanics and Dynamical Astronomy 121 (2015) 1-15, doi:10.1007/s10569-014-9587-y; open preprint arXiv:1403.2705.Where: method and test philosophy for Lambert solvershttps://arxiv.org/abs/1403.2705Cited in: Chapter 12
R14	NOAA, NASA and USAF, U.S. Standard Atmosphere, 1976, NASA-TM-X-74335 (NTRS 19770009539), Washington DC, Oct 1976.Where: Table I, geometric altitude, metric units, density column: pp. 69-73 (pdf pp. 84-88)https://ntrs.nasa.gov/citations/19770009539Cited in: Chapters 6, 19; assumptions
R15	Boring, J.W. and Humphris, R.R., "Drag coefficients for spheres in free molecular flow in O at satellite velocities", NASA CR-2233, 1973.Where: abstract: measured C_D "slightly greater than 2" at orbital energieshttps://ntrs.nasa.gov/citations/19730010970Cited in: Chapters 6, 19; assumptions
R16	NASA Small Spacecraft Systems Virtual Institute, State-of-the-Art of Small Spacecraft Technology, Ch. 4 "In-Space Propulsion", 2026 edition.Where: §4.6.1: hydrazine thrusters typically 200-235 s (1 N class and up); Table 4-1https://www.nasa.gov/smallsat-institute/sst-soa/in-space_propulsion/Cited in: Chapter 9; assumptions
R17	ArianeGroup GmbH, "400N bipropellant apogee motors" product page (S400-12, S400-15), © 2002-2024.Where: Isp at nominal point 318 s (S400-12), 321 s (S400-15), MMH/MONhttps://space-propulsion.com/spacecraft-propulsion/apogee-motors/index.htmlCited in: Chapters 17, 19; assumptions
R18	Lovell, R.R. and O'Malley, T.A., Station Keeping of High Power Communication Satellites, NASA TM X-2136, Lewis Research Center, Nov 1970.Where: Summary (pdf p. 4): inclination grows 0.85 deg/yr; triaxiality E-W Δv < 1.75 m/s/yr; Introduction (pdf p. 5): 46 m/s N-S and 2 m/s E-W per year "documented in a number of reports"https://ntrs.nasa.gov/citations/19710001770Cited in: Chapters 17, 19; assumptions
R19	Decoto, J. and Loerch, P. (Orbital ATK), "Technique for GEO RSO Station Keeping Characterization and Maneuver Detection", AMOS Technical Conference, 2015.Where: §2: about 0.85 deg/yr for a beginning-of-life GEO satellite (citing Soop 1994), about 46 m/s/yr N-Shttps://amostech.com/TechnicalPapers/2015/SSA/Decoto.pdfCited in: Chapter 17; assumptions
R20	Inter-Agency Space Debris Coordination Committee, IADC Space Debris Mitigation Guidelines, IADC-02-01 Rev. 2, Mar 2020.Where: §5.3.1, p. 11: ΔH_min = 235 km + 1000 C_R A/m, e ≤ 0.003https://orbitaldebris.jsc.nasa.gov/library/iadc-space-debris-guidelines-revision-2.pdfCited in: Chapters 17, 19; assumptions
R21	NASA JSC Flight Operations Directorate, TOPO office, ISS public ephemeris ISS.OEM_J2K_EPH.txt (CCSDS OEM, created 2026-11-02).Where: header: MASS = 473406.50 kg, DRAG_AREA = 1511.80 m², DRAG_COEFF = 1.10https://nasa-public-data.s3.amazonaws.com/iss-coords/current/ISS_OEM/ISS.OEM_J2K_EPH.txtCited in: Chapter 6; assumptions
R22	ESA, "Types of orbits", 30 Mar 2020.Where: GEO altitude 35,786 km; LEO below 2000 km; SSO 600-800 km; L1/L2 about 1.5 million kmhttps://www.esa.int/Enabling_Support/Space_Transportation/Types_of_orbitsCited in: Chapters 7, 16, 17
R23	Cornish, N.J., "What is a Lagrange Point?", NASA Science, 27 Mar 2018, updated 1 Nov 2024.Where: L1-L3 unstable (about 23 days), L4/L5 stable if mass ratio > 24.96https://science.nasa.gov/resource/what-is-a-lagrange-point/Cited in: Chapter 16
R24	NASA Science, "Webb Orbit".Where: L2 about 1.5 million km from Earth; orbit about L2 takes about 6 months; small periodic thrusthttps://science.nasa.gov/mission/webb/orbit/Cited in: Chapter 16
R25	ESA, "Europe's Spaceport".Where: Kourou at latitude 5°3′ N; Earth rotation adds 460 m/shttps://www.esa.int/Enabling_Support/Space_Transportation/Europe_s_Spaceport/Europe_s_Spaceport2Cited in: Chapters 10, 18, 19
R26	NIST Time and Frequency Division, "Leap second and UT1-UTC information" (created 6 Aug 2009, updated 28 Oct 2026).Where: TAI − UTC = 37 s; leap seconds keep UT1 − UTC within ±0.9 s; DUT1 broadcasthttps://www.nist.gov/pml/time-and-frequency-division/time-realization/leap-secondsCited in: Chapter 1
R27	U.S. Naval Observatory, Astronomical Applications Dept., "Computing Approximate Sidereal Time".Where: GMST formula in hours from days since J2000.0; LST = GMST + east longitude/15; sidereal time = hour angle of the vernal equinoxhttps://aa.usno.navy.mil/faq/GASTCited in: Chapter 2
R28	Petit, G. and Luzum, B. (eds.), IERS Conventions (2010), IERS TN 36, Ch. 5 (10 Aug 2012 update).Where: eq. (5.43), argument F14: general precession in longitude p_A = 0.02438175 t rad (t in Julian centuries) → 50.29″/yr, about 25,770 yr per turnhttps://iers-conventions.obspm.fr/content/chapter5/icc5.pdfCited in: Chapter 2
R29	NASA Glenn Research Center, "Ideal Rocket Equation", Beginner's Guide to Aeronautics, updated 21 Nov 2023.Where: Δu = V_eq ln(m_f/m_e) = Isp g₀ ln MR; with gravity Δu = V_eq ln MR − g₀ t_bhttps://www1.grc.nasa.gov/beginners-guide-to-aeronautics/ideal-rocket-equation/Cited in: Chapter 9

Orbital mechanics

How to use this pack

Chapters

You will be able to

From two bodies to one

Kepler's three laws

Circular orbits

Which day?

Which clock?

Key equations

Worked examples

How long is a lap of the Earth?[R1]

Why the solar day is four minutes longer[R2, R5]

Practice

An Earth-observation orbit

Choose the altitude from the period[R2]

Kepler's third law from the second

Common mistakes

Check yourself

Sources for this chapter

You will be able to

The orbit equation

The conic family

Six numbers place an orbit

Two frames

Where it sits in the sky

Frames that drift

Key equations

Worked examples

Shape of a transfer ellipse

Where is it, in Earth-centred coordinates?[R1]

Practice

From altitudes to a and e

A highly elliptical orbit

An escaping hyperbola

Common mistakes

Check yourself

Recap from earlier chapters

Sources for this chapter

You will be able to

Specific energy

Vis-viva

Angular momentum and flight-path angle

Escape

Key equations

Worked examples

Speed and climb angle partway up

Watch a computer conserve energy

Practice

Escape from an 800 km orbit[R1]

Orbit from one radar fix

A probe on its way out

Common mistakes

Check yourself

Recap from earlier chapters

Sources for this chapter

You will be able to

Three anomalies

Kepler's equation

The hyperbolic version

Checking your answer

Key equations

Worked examples

Solve Kepler's equation by Newton's method

How long to climb to ν = 150°?

Practice

A quarter-turn on a transfer orbit

Where will it be in 40 minutes?

Leaving on a hyperbola

Common mistakes

Check yourself

Recap from earlier chapters

Sources for this chapter

You will be able to

Three helper vectors

Quadrant checks

Back to r and v

When elements break

Key equations

Worked examples

How long is a lap of the Earth?^[R1]

Why the solar day is four minutes longer^{[R2, R5]}

Choose the altitude from the period^[R2]

Where is it, in Earth-centred coordinates?^[R1]

Escape from an 800 km orbit^[R1]

How fast does the orbit plane turn?^[R1]

How fast does drag pull a small satellite down?^{[R14, R15]}

Node drift of a mid-inclination orbit^[R1]

Drag or sunlight: which pushes harder at 700 km?^{[R8, R14]}

Reboosting a space station^{[R14, R21]}

A sun-synchronous orbit at 600 km^[R5]

Sun-synchronous at 900 km^[R5]