Hohmann transfer with a tight propellant budget
A 620 kg satellite is in a circular orbit 350 km above Earth's surface. Mission planners want it in a circular orbit at 1200 km. They propose a Hohmann transfer: a first burn at the 350 km orbit raises apogee to 1200 km, and a second burn at apogee circularizes. The engine has specific impulse 280 s, and the satellite can carry only 35 kg of usable propellant. A teammate argues that because astronauts on the ISS float, the satellite at 350 km is weightless, so the engine will need much less propellant than the rocket equation predicts. Use Earth radius 6.37×10^6 m and μ=3.986×10^14 m^3 s^-2. Ignore drag, Earth rotation, and the mass of propellant already burned when computing the first burn. The team needs a feasibility check before ordering a larger tank.
1. Calculate the total Δv for the Hohmann transfer and decide whether 35 kg of propellant is enough. Show the velocity changes at both burns and use the rocket equation.
Read the worked answer
Use r1 = 6.37×10^6 + 350×10^3 = 6.72×10^6 m and r2 = 6.37×10^6 + 1200×10^3 = 7.57×10^6 m.
Circular speeds: v1 = sqrt(μ/r1) = sqrt(3.986×10^14/6.72×10^6) = 7.70×10^3 m/s. v2 = sqrt(μ/r2) = sqrt(3.986×10^14/7.57×10^6) = 7.26×10^3 m/s.
Transfer ellipse semi-major axis: a = (r1+r2)/2 = 7.145×10^6 m. At perigee: vp = sqrt[μ(2/r1 − 1/a)] = 7.93×10^3 m/s, so Δv1 = 7.93 − 7.70 = 0.23×10^3 m/s. At apogee: va = sqrt[μ(2/r2 − 1/a)] = 7.04×10^3 m/s, so Δv2 = 7.26 − 7.04 = 0.22×10^3 m/s. Total Δv ≈ 0.45×10^3 m/s.
Rocket equation: ve = Isp g0 = 280×9.81 = 2.75×10^3 m/s. With m0 = 620+35 = 655 kg and mf = 620 kg, the maximum Δv is ve ln(655/620) = 2.75×10^3×0.0549 ≈ 151 m/s. This is far below 450 m/s, so 35 kg is not enough. The required propellant is found from m0/mf = exp(450/2747) ≈ 1.18; with mf = 620 kg, m0 ≈ 731 kg, so about 111 kg of propellant is needed. The proposal is infeasible without a larger tank or a different transfer strategy.
2. Evaluate the teammate's 'weightless' argument. Include the local gravitational field at 350 km and explain whether thrust requirements depend on it.
Read the worked answer
At 350 km, g = μ/r^2 = 3.986×10^14/(6.72×10^6)^2 ≈ 8.8 m/s^2, about 90% of the surface value. The satellite is not outside Earth's gravity. 'Weightlessness' means free fall: the satellite and everything inside it accelerate together under gravity, so there is no normal force or apparent weight. It does not mean gravity is absent.
The engine produces a non-gravitational thrust. In an inertial frame, the burn changes the satellite's velocity, and the rocket equation still applies. Gravity acts on the satellite during the burn, but for an impulsive maneuver the required Δv is set by the before-and-after orbital velocities, not by the local g. A burn at perigee is actually more efficient in energy terms because the satellite is moving faster (Oberth effect), but that does not make the propellant requirement smaller than the rocket equation predicts for the same Δv. The teammate is confusing free fall with zero gravity.
3. If only the first burn is performed and the second is cancelled, describe the resulting orbit and calculate its period. Where does the satellite move fastest?
Read the worked answer
After the first burn, the orbit is an ellipse with perigee radius rp = 6.72×10^6 m (350 km altitude) and apogee radius ra = 7.57×10^6 m (1200 km altitude). Its semi-major axis is a = (rp+ra)/2 = 7.145×10^6 m.
Kepler's third law gives T = 2π sqrt(a^3/μ). Here a^3 ≈ 3.65×10^20 m^3, so a^3/μ ≈ 9.16×10^5 s^2, sqrt ≈ 957 s, and T ≈ 2π×957 ≈ 6.01×10^3 s ≈ 100 min.
The satellite reaches 1200 km at apogee, but it is not circular there: its apogee speed is 7.04 km/s, less than the local circular speed 7.26 km/s, so it will fall back toward perigee. It moves fastest at perigee (350 km) and slowest at apogee. The second burn is needed to raise the apogee speed to the circular value and keep the satellite at 1200 km.