I understand that more power is required to launch a satellite into retrograde orbit, but once it's up there how much does the orbit differ from the more usual West to East? Would you still use $v\approx\sqrt{\frac{GM}{r}}$ to find the speed or are there more things to consider, such as the drag in LEO.

I'm assuming a simplified circular orbit.

  • $\begingroup$ This is a good question, but you didn't mention to what precision you'd like any effects discussed? Since you mention atmospheric drag in LEO, I imagine you'd want answers to be fairly precise? I calculate for ISS altitude change in speed of atmosphere of ± 0.27 km/s which is at ISS speed (7.71 km/s) ± 0.355% change in its decay rate, so ~ + 0.7% faster orbital decay (from about 2 km/month) if it was orbiting retrograde. But there might be other things to consider, such as tidal perturbations that I don't have any numbers for. $\endgroup$
    – TildalWave
    Mar 31, 2014 at 15:53
  • $\begingroup$ BTW speed itself doesn't change prograde to retrograde, since it defines your orbital altitude, but the decay rate (and with it precession, or simplifying - eccentricity) would, if your orbit has something more to work against than merely orbit around a single, perfectly homogeneous gravitational field alone (and no system is perfect, so there's always something, like tidal perturbations, gravity anomalies, atmospheric/exospheric drag, insolation intervals, magnetosphere, solar winds,... at play). $\endgroup$
    – TildalWave
    Mar 31, 2014 at 16:04

1 Answer 1


To first order, no. Considering only Newtonian gravitational forces and a spherical gravitational field, the satellite has no way to tell if the body beneath it is spinning or not.

The exceptions would be a) if you are close enough to a body with an atmosphere to measure its drag. Since the atmosphere rotates with the body, the drag will be higher retrograde than prograde. b) If the gravitational field is not spherical (and they never are), bumps in the gravity field will go by faster in the retrograde than the prograde orbits. And c) if you can measure accurately enough to see General Relativity frame dragging.

  • $\begingroup$ I'm not sure I agree with point 'b'; bumps in the gravity field should be conservative irrespective of the direction you fly over them. Having a zero net effect on the orbit. $\endgroup$
    – ThePlanMan
    Apr 16, 2014 at 11:44
  • 2
    $\begingroup$ As a mass concentration approaches from ahead you are sped up a smidge. As it retreats behind, you are slowed down a smidge. The duration of those accelerations are shorter if you are retrograde vs. prograde. You can use these accelerations to map the gravity field of the body measure its spin axis and spin rate. You would need to track your orbit to measure the accelerations, or you could have a detector inside your vehicle to measure tidal forces using the relative motion of two free masses. $\endgroup$
    – Mark Adler
    Apr 16, 2014 at 16:20
  • $\begingroup$ I agree that the second by second speeds would change due to this, but the question asks about a simplified spherical orbit. I assumed the conservative effect of zonal harmonics would not come under 'simplified'. Then again, when did anything ever work as assumed! :) $\endgroup$
    – ThePlanMan
    Apr 16, 2014 at 16:23
  • $\begingroup$ Since gravity is a conserving force, would the bumps going by faster in retrograde have any effect? (non-relativistic) Of course in a lumpy gravity field I'm not sure if the semi-major axis or period are even well defined anymore at this level. $\endgroup$
    – uhoh
    Jul 2, 2017 at 3:26

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