Choosing a Transfer Strategy from LEO to Geostationary Orbit
A 4,500 kg communications satellite, including 220 kg of hydrazine reserved for station-keeping, is injected into a 300 km circular orbit at 28.5° inclination. The apogee engine has 420 N thrust and Isp 310 s (effective exhaust velocity 3.04 km/s); its propellant is carried separately from the hydrazine. Mission design is comparing three routes to geostationary orbit: (i) a 300 km × 35,786 km Hohmann transfer with a separate inclination change performed at apogee; (ii) a single combined apogee burn that both raises perigee and removes the inclination; (iii) a supersynchronous transfer with 45,000 km apogee, followed by a combined burn at that apogee. Use μ = 398,600 km³/s² and R_E = 6,378 km. The programme wants the cheapest propellant option; structures is worried about tank volume and the attitude control team about attitude authority during long burns. Propellant margin is 2%.
1. Compute the total Δv for routes (i) and (ii) and the propellant mass each needs. Which do you recommend and by how much?
Read the worked answer
Circular speed at 300 km: v₁ = √(398,600/6,678) = 7.726 km/s.
Transfer ellipse, a = (6,678 + 42,164)/2 = 24,421 km:
- Perigee speed v_p = √[398,600 × (2/6,678 − 1/24,421)] = 10.152 km/s → Δv₁ = 2.426 km/s
- Apogee speed v_a = √[398,600 × (2/42,164 − 1/24,421)] = 1.608 km/s
- GEO speed v_G = √(398,600/42,164) = 3.075 km/s → apogee raise Δv₂ = 1.467 km/s
Route (i), separate plane change at apogee: Δv_plane = 2 v_a sin(Δi/2) = 2 × 1.608 × sin 14.25° = 0.792 km/s. Total = 2.426 + 1.467 + 0.792 = 4.685 km/s.
Route (ii), combined burn: the burn must rotate the velocity vector through 28.5° while raising the speed from 1.608 to 3.075 km/s:
Δv = √(1.608² + 3.075² − 2 × 1.608 × 3.075 × cos 28.5°) = √(2.586 + 9.454 − 8.692) = 1.830 km/s
Total = 2.426 + 1.830 = 4.256 km/s, a saving of 0.429 km/s over the separate plane change.
Propellant (Tsiolkovsky, m₀ = 4,500 kg, v_e = 3.04 km/s):
- Route (i): m_p/m₀ = 1 − exp(−4.685/3.04) = 0.786 → 3,536 kg
- Route (ii): m_p/m₀ = 1 − exp(−4.256/3.04) = 0.754 → 3,391 kg
Route (ii) saves about 145 kg of propellant, which is either 145 kg more payload or a smaller tank. It is the clear choice, subject to the spacecraft being able to hold attitude while thrusting off the velocity vector through a large angle.
2. Evaluate the supersynchronous option (iii). Compute its Δv and decide whether the additional complexity is justified.
Read the worked answer
Perigee burn to a 6,678 × 45,000 km ellipse (a = 25,839 km):
v_p = √[398,600 × (2/6,678 − 1/25,839)] = 10.196 km/s → Δv₁ = 2.470 km/s (0.044 more than the standard Hohmann)
At the 45,000 km apogee, v_a = √[398,600 × (2/45,000 − 1/25,839)] = 1.513 km/s.
To raise perigee to 42,164 km, the new ellipse has a = (45,000 + 42,164)/2 = 43,582 km, giving an apogee speed of √[398,600 × (2/45,000 − 1/43,582)] = 2.927 km/s.
Combining this raise with the 28.5° plane change:
Δv = √(1.513² + 2.927² − 2 × 1.513 × 2.927 × cos 28.5°) = √(2.289 + 8.567 − 7.786) = 1.752 km/s
Total = 2.470 + 1.752 = 4.222 km/s, against 4.256 km/s for route (ii).
Propellant: m_p/m₀ = 1 − exp(−4.222/3.04) = 0.751 → 3,378 kg, i.e. a saving of only about 13 kg, under 0.5% of the propellant load.
Verdict: not justified. The saving comes from performing the plane change where the velocity is lowest (1.513 km/s against 1.608 km/s), but it is largely cancelled by the higher perigee burn needed to reach a 45,000 km apogee. Against that you pay: a larger apogee motor burn at an altitude where the orbit period is longer, a longer time to station, more thermal and power stress during the extended transfer, a re-designed apogee engine operating point, and a Δv that is sensitive to the exact apogee altitude (the optimum is narrow and shifts with Isp and thrust). A 2% propellant margin fits either option, so the decision is not propellant-limited - it should be made on schedule, risk and cost, where route (ii) wins.
3. Check whether the impulsive Δv assumption is valid for a 420 N engine, and state what this implies for the manoeuvre design and for the choice between the routes.
Read the worked answer
Mass flow at 420 N and v_e = 3,040 m/s: ṁ = 420/3,040 = 0.138 kg/s.
For route (ii) the burn consumes about 3,391 kg, so the total burn time is 3,391/0.138 ≈ 24,600 s ≈ 6.8 hours. The GTO period is 2π√(a³/μ) with a = 24,421 km: a³/μ = 3.65×10⁷ s², so P = 2π × 6,045 = 10.6 hours.
A 6.8-hour burn is comparable to the orbital period, so the impulsive assumption fails outright: the spacecraft spends most of the burn far from apogee, the thrust direction must be steered continuously, and the losses appear as gravity and steering losses (hundreds of metres per second, not tens) plus a large plane-change penalty because the plane change is no longer done at a single low-velocity point.
Even the perigee burn alone is not impulsive: Δv₁ = 2.426 km/s from 4,500 kg leaves about 2,026 kg, so 2,474 kg is burned in 2,474/0.138 ≈ 17,900 s ≈ 5 hours at a point where the spacecraft is near perigee only briefly.
Implications.
- The manoeuvre must be split. In practice: several perigee burns over successive revolutions to raise apogee progressively, then a final apogee burn that raises perigee and removes inclination together. This recovers most of the impulsive Δv and adds only tens of metres per second.
- Each burn must be short compared with the period in which the spacecraft is near the burn point, so the burn arc is limited and the number of revolutions is set by that limit.
- The comparison between routes (ii) and (iii) narrows further: the supersynchronous option's 13 kg saving was computed impulsively and would be swamped by the steering losses of an extra long burn.
- The attitude control team's concern is real: a combined burn holding a fixed inertial thrust direction through 28.5° requires wheel or thruster authority that does not saturate over tens of minutes, and the tank must be able to feed the engine in the resulting attitude over many restarts. Verify that the apogee engine can be restarted the required number of times and that the pressurant budget covers it.