Star Tracker

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Orbital mechanics sample pack

Six worked problems. Every number is computed by a script, not typed by hand. Constants: mu = 398600.4418 km^3/s^2, R_E = 6378.137 km, J2 = 1.08262668e-3, sidereal day 86164.0905 s (Vallado, 4th ed.). Try each problem first, then open the working.

1. How fast is the ISS going?

Given. Circular orbit at h = 420 km altitude. R_E = 6378.137 km, mu = 398600.4418 km^3/s^2.

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  1. r = R_E + h = 6378.137 + 420 = 6798.137 km
  2. v = sqrt(mu / r) = sqrt(398600.4418 / 6798.137) = 7.657 km/s
  3. T = 2*pi*sqrt(r^3 / mu) = 5578.2 s = 92.97 min

Answer. v = 7.66 km/s, T = 93.0 min. About 15.5 orbits per day.

2. Escape from low Earth orbit

Given. Same orbit as problem 1 (r = 6798.137 km). What extra speed makes the craft escape Earth?

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  1. v_esc = sqrt(2*mu / r) = sqrt(2 * 398600.4418 / 6798.137) = 10.829 km/s
  2. v_esc = sqrt(2) * v_circ: check 10.829 km/s
  3. delta-v = v_esc - v_circ = 10.829 - 7.657 = 3.172 km/s

Answer. delta-v = 3.17 km/s. Escape costs only 41% more speed than you already have.

3. Hohmann transfer from LEO to GEO

Given. Start in the circular orbit of problem 1 (r1 = 6798.137 km). Reach a circular geostationary orbit (period = one sidereal day, 86164.0905 s). Both burns impulsive, same plane.

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  1. GEO radius: r2 = (mu * (T/2pi)^2)^(1/3) = 42164.2 km (altitude 35786 km)
  2. Transfer ellipse: a_t = (r1 + r2)/2 = 24481.2 km
  3. v1 (circular, LEO) = 7.657 km/s; v_perigee on transfer = sqrt(mu(2/r1 - 1/a_t)) = 10.049 km/s
  4. delta-v_1 = 10.049 - 7.657 = 2.392 km/s
  5. v_apogee on transfer = 1.620 km/s; v2 (circular, GEO) = 3.075 km/s
  6. delta-v_2 = 3.075 - 1.620 = 1.454 km/s
  7. Total = 3.846 km/s; time of flight = half the transfer period = 5.29 h

Answer. delta-v total = 3.85 km/s, flight time 5.3 h. The second burn is the smaller one.

4. A Molniya orbit: fast at perigee, slow at apogee

Given. Perigee altitude 600 km, apogee altitude 39750 km.

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  1. r_p = 6978.137 km, r_a = 46128.137 km, a = 26553.137 km, e = (r_a - r_p)/(r_a + r_p) = 0.7372
  2. v_p = sqrt(mu(2/r_p - 1/a)) = 9.961 km/s
  3. v_a = sqrt(mu(2/r_a - 1/a)) = 1.507 km/s
  4. Check with angular momentum: r_p v_p = 69512.6, r_a v_a = 69512.6 km^2/s (equal)
  5. T = 11.96 h, close to half a sidereal day (11.97 h)

Answer. v_p = 9.96 km/s, v_a = 1.51 km/s, T = 12.0 h. The satellite spends most of its time near apogee, which is the whole point.

5. Sun-synchronous inclination at 700 km

Given. Circular orbit, h = 700 km. Find the inclination whose node drifts 360 deg per year (eastward, 1.991e-07 rad/s). J2 = 0.00108262668.

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  1. r = 7078.137 km, mean motion n = sqrt(mu/r^3) = 1.060206e-03 rad/s
  2. Node rate (J2 only): dOmega/dt = -1.5 n J2 (R_E/r)^2 cos i
  3. cos i = -dOmega/dt / (1.5 n J2 (R_E/r)^2) = -0.14242
  4. i = acos(-0.14242) = 98.19 deg

Answer. i = 98.2 deg. Slightly retrograde, which is why every Earth-observation satellite at this height sits near 98 deg.

6. Why you change planes at apogee

Given. A 28.5 deg plane change (Cape Canaveral latitude to equatorial). Compare doing it at LEO speed (7.657 km/s) and at GEO speed (3.075 km/s).

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  1. delta-v = 2 v sin(di/2)
  2. At LEO: 2 * 7.657 * sin(14.25 deg) = 3.770 km/s
  3. At GEO: 2 * 3.075 * sin(14.25 deg) = 1.514 km/s

Answer. 3.77 km/s at LEO versus 1.51 km/s at GEO. Same angle, 2.5 times the cost. Combine the plane change with the GEO circularisation burn.

Original work, not course material. A study aid, not assignment answers. More packs (thermal, power, ADCS) go to the email list first. Join the list.