How long would it take for a spacecraft with a mass of 30,000 kg, to go from LEO to reach and match the speed of a space station in geostationary orbit so it could dock with the station, using a Hohmann transfer orbit?

LEO speed 7.8 km/s, 160 km above the Earth GEO speed 3 km/s, height of 36,000 km ESA Space Transportation page

Thrust of the craft's engine 45 kN, modeled on the CSM. Wikipedia-Apollo CSM

Are there any other parameters I should include? If any of the information doesn't make sense please let me know.

  • 2
    $\begingroup$ For a rough approximation: There's a Hohmann transfer calculator here that will give you the required delta-v (which affects the burn time); the transfer trajectory is one-half of an orbit with perigee in LEO and apogee at GEO, so compute the orbital period and divide by two to get the transfer time. Since the thrust is pretty low, the burn time will be significant; the idealized instant-impulse Hohmann transfer doesn't apply, but it'll be close to correct. $\endgroup$ May 24, 2019 at 20:42

1 Answer 1


The time in transfer will be a good approximation - replacing the long burns with impulsive.

LEO altitude 160 km + 6,371km Earth average radius, so periapsis of the transfer orbit is 6531km from Earth center.

Geostationary orbit radius: 42,164 km

Semi-major axis is the average between the two. a=24347km

$T=2 \pi \sqrt{a^3 \over GM}$

The period of that orbit is 37810 seconds so the time to pass from periapsis to apoapsis is half of that - 18904 seconds or 5 hours 15 minutes and 4 seconds.

You may add about 1650s departure burn and a 933s arrival (circularization) burn, depending on how you measure the time. In reality likely less because the craft will have spent a good portion of fuel by then, so the mass will be lower and so the acceleration higher, but that's not possible to calculate without specific impulse.

ps. what sort of sadist with bottomless pockets put a space station at GEO? Not only it's obscenely expensive to reach, it's also in the middle of the outer Van Allen belt!


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