Say a spacecraft in geostationary orbit needs to be deorbited (not very common, but bear with me).

How does one decide when and how long should the retrograde burns be, in order to have started with the minimum amount of fuel?

I have searched the net but couldn't find anything useful to make these optimizations. I hope someone with expertise in orbital mechanics can help me out here.

Edit: As someone in the comments pointed out, a retrograde burn alone might be expensive. Please feel free to include any other manoeuvers, and how to calculate the timings and durations of burns required. Any hints on the computations or even softwares that could help will be greatly appreciated.

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    $\begingroup$ I think it is dangerous to assume that a retrograde burn is most efficient, rising the apoapsis and burning retrograde there or a gravity assist from the moon might be more economical. $\endgroup$
    – lijat
    Commented Nov 18, 2019 at 6:38
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    $\begingroup$ @lijat Thanks for the comment. I will edit the question to delimit. $\endgroup$ Commented Nov 18, 2019 at 7:46
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    $\begingroup$ @uhoh, strictly speaking, that's not deorbiting. You will still be orbiting the earth, but will do so as a splatter on the surface of the moon instead of being on your own. :-). So, to the OP, do we want the satellite to be burnt in earth's atmosphere or will we accept any way of safe disposal? $\endgroup$ Commented Nov 18, 2019 at 11:44
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    $\begingroup$ I don't think you'll manage to recover any significant parts of a satellite unless it's been specifically engineered for that. Your uneventful splashdown will most probably an uneventful dust shower. $\endgroup$ Commented Nov 18, 2019 at 17:19
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    $\begingroup$ Orbital mechanics 101: Anything more complex than the Hohmann solution is almost certainly going to be much slower. $\endgroup$ Commented Nov 21, 2019 at 0:41

3 Answers 3


This is not a complete answer as I won't be including the exact calculation needed to find out your burn time, but at least I will address the direct return vs bi elliptic approach.

For a return from orbit of a manned spacecraft, you want to balance two factors:

On one side you want to minimise the amount of fuel required for the operation; on the other, you want to minimise the time spent and the final speed.

The quickest and safest return would be a direct return using a Hohmann transfer orbit; burn retrograde at apogee and you will arrive to your destination orbit faster than using any other method and with the lowest reentry speed hence maximising your chance of survival.

If you use a Bi-elliptic transfer orbit you can reduce the total Delta V needed, but at the expense of more time in space for your astronauts and a higher reentry speed.

I've done a quick calculation, and a direct Hohmann transfer from a circular geostationary orbit to 100Km (and let the atmosphere do the rest) would require around 1.49Km/s (please someone confirm) and take 17 hours. A bi-elliptic going up to 380,000Km (Moon's distance to Earth, just to pick a meaningful distance for easy reference) would save you ~167m/s (11%) at the cost of 10 days in space.

Note after HopDavid's comment: Usually for this two orbits a bi-eliptic transfer should be less efficient, but as we are using the atmosphere for our final "burn", we save us the most expensive of them. The higher you go in the bi-eliptic, the more energy you have to shed in the final circularization burn, and the most fuel you save with aerobraking.

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    $\begingroup$ My numbers match yours fairly closely. Which surprised me as I had thought destination and departures orbit radii had to differ by at least a factor of 11 for bi elliptic to work. But maybe it's because we're using aerobraking to slow down at the 200 km altitude destination orbit. $\endgroup$
    – HopDavid
    Commented Nov 18, 2019 at 23:21
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    $\begingroup$ Is a plane alignment burn required for the bi-elliptic transfer? I'm not sure if the moon and a geostationary satellite would be on the same plane already. $\endgroup$ Commented Nov 19, 2019 at 3:42
  • $\begingroup$ @HopDavid, Yes, the last burn is the most expensive as you are coming from a more energetic orbit, and we are saving that fuel by aerobraking; I've added a note to the answer. Thanks for checking my numbers. $\endgroup$ Commented Nov 19, 2019 at 7:17
  • $\begingroup$ @WilliamREbenezer. No, I used "Moon's Orbit" as a generic term to say "Up to 380000Km". I've rephrased my answer to make that more clear. In this case, in which you just want to "splash down somewhere", you don't need to care about plane changes. Your geostationary orbit is an equatorial one, which is the easiest one for reentry. $\endgroup$ Commented Nov 19, 2019 at 7:19
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    $\begingroup$ Should be able to do better by using a lunar gravity assist so no second burn. $\endgroup$
    – Joshua
    Commented Jan 1, 2020 at 23:07

The following table provides some useful velocities

Description Velocity (m/s) Difference from Circular GEO (m/s)
Circular GEO Orbit 3075 0
Elliptical (100km/GEO) Apogee 1586 -1489
Elliptical (GEO/Moon Intercept) Perigee 4112 1037
Escape from GEO 4348 1274

The most efficient way to deorbit would probably be to wait for the moon to be at the perigee of its orbit and then time a 1037 m/s prograde burn to raise the satellite's orbit just enough to make the satellite crash into the moon.

It might be possible to do a similar prograde burn but instead, lead the moon by just a bit. Then the moon's gravity could be used to alter the satellite's orbit so that it falls back down to Earth.

With a prograde burn of 1274 m/s, you can deorbit the satellite by getting it up to Earth escape velocity. You could reduce the delta-v required for this by using a lunar gravity assist.

Of course, the satellite will still be orbiting the sun. If that's a problem, then another tiny burn can send it from "infinity" back down to Earth, essentially performing a bi-elliptic transfer of sorts.

If you want to go directly from GEO to 100 km so that you can aerobrake and reenter quickly, this will require a retrograde burn of 1489 m/s (452 m/s more delta-v than the most efficient option).

Since you asked "how", most of these calculations can be done by using the Vis-viva equation.

  • $\begingroup$ According to a comment the goal is to safely land a manned spacecraft back on Earth. OTOH, you're certainly entitled to ignore that info and just answer the question body. ;) $\endgroup$
    – PM 2Ring
    Commented Feb 25 at 19:32
  • $\begingroup$ It sounds like the two competitors for lowest delta-v to 100km are both at the lunar apogee, either 1) helping to get it directly to Earth, or 2) to Earth-escape where a small burn can then send it to Earth; is that a reasonable interpretation of your answer? $\endgroup$
    – uhoh
    Commented Feb 28 at 0:43
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    $\begingroup$ The first is raising the satellite's apogee to the lunar perigee. The second, which takes more delta-v, but doesn't rely on the moon, is Earth-escape. In both cases, I described an option - you can also turn the satellite around and send it back to Earth. It doesn't take much additional delta-v to do that in either case. $\endgroup$
    – phil1008
    Commented Feb 28 at 0:57
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    $\begingroup$ Basically you want to do the reverse of what HGS-1/PAS-22 did. $\endgroup$
    – user71659
    Commented Feb 29 at 23:36
  • $\begingroup$ @user71659 good point! I always get the recoveries of PAS-22 and AMC-14 confused. See also this answer and Would lunar flyby be less costly in delta-v than direct change from ISS's to Hubble's orbit? and Spacecraft Maneuvers as Intellectual Property? Wow! $\endgroup$
    – uhoh
    Commented Mar 1 at 2:29

It doesn't matter when, as the system is rotationally symmetric. A retrograde burn is almost certainly the most efficient method (excluding any very long time period wait-for-perturbations-to-become-significant ones).

It is uncommon to de-orbit geostationary object so you might not find much about it directly (the common approach is to increase the orbital radius to a "graveyard" orbit). However, it's the opposite maneuver to an apogee-kick, which you will find plenty of info on. The summary is to get back to a geostationary transfer orbit, you need to reduce orbital velocity to 1.64 km/s (Geostationary orbit speed is 3.07 km/s). Hence you need to burn retrograde by 1.43 km/s. How long this will take will depend on the thrust to weight ratio.

  • $\begingroup$ For a deorbit you burn at apoapsis. Unless your orbit is perfectly circular the time matters. $\endgroup$ Commented Nov 21, 2019 at 0:40
  • $\begingroup$ @LorenPechtel as its geostationary, it must be circular. otherwise you are correct. $\endgroup$
    – drjpizzle
    Commented Nov 23, 2019 at 1:25
  • $\begingroup$ Yes and no. It depends on your engines. If they are high thrust the burn will be over fast enough it doesn't matter but if they're low thrust and you want absolutely the minimum fuel consumption you'll want to split your burn up, doing a bit each time you get back up to geosync. I've plowed too many NERVA rockets into Kerbin from ignoring this issue! $\endgroup$ Commented Nov 23, 2019 at 5:15

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