Do all celestial bodies of sufficient mass such big asteroids, moon, planets etc. have orbit such that the period of revolution is same as the sidereal period of that object (such as geostationary orbit in earth) ?

  • $\begingroup$ definitely not a "geo"stationary orbit. Not sure what the correct term is, though. $\endgroup$ Sep 25, 2013 at 8:22
  • $\begingroup$ @JanDvorak Thanks i edited that, if you still think there is a mistake you can make a suggested edit ;+) $\endgroup$
    – Hash
    Sep 25, 2013 at 8:29

1 Answer 1


This depends on a few things, in particular the Hill Sphere and the rotational rate. The hill sphere calculation can be seen below. If an object is too close to a big object, or rotates too slowly, then it won't work.

$r \approx a (1-e) \sqrt[3]{\frac{m}{3 M}}.$

If an object is too close for such an orbit to be formed, it will likely be tidally locked, such as our moon is to us. I haven't worked out all of the math yet, but I believe it's not possible to orbit in a synchronous way a tidally locked object.

The stationary orbit for the moon would be fairly well along the way to the Earth-Moon distance, although as the Moon is less massive, it will not be as far out. The only orbits which could be considered stationary are the Earth-Moon Lagrange points, of which I think all of them are to a degree for a tidally locked object. See below image.

enter image description here

In addition, a slow rotating object will just never have an altitude where it can be orbited at a stationary orbit. Venus, for instance, rotates slower than it moves around the sun, it certainly doesn't have a stationary orbit. Mercury likely doesn't, due to it's slow rotation rate. The proper way to evaluate this would be by figuring out the hill sphere, and the proper stationary altitude, and seeing if the stationary orbit is stable.

Using Wolfram Alpha, here's a few examples of the synchronous orbit / hill sphere. If less than 1, it could be stable, if more than 1, it is not stable.

  • Mercury- 1.386
  • Venus- 1.53
  • Earth- 0.02865
    • Moon- 1.522
  • Mars- 0.02079
  • Ceres- 0.0058
  • Vesta- 0.00482
  • Jupiter- 0.003164
    • Io- 1.448
    • Europa- 1.46
    • Ganymede - 1.44
    • Callisto- 1.443
  • Saturn- 0.001803
    • Titan- 1.485
    • Iapetus- 1.484
  • Uranus - 0.0012
    • Titania- 1.44
    • Oberon- 1.44
  • Neptune 0.00079
    • Triton- 1.44
  • Pluto- 0.003276
    • Charon- 1.39

I'm sure the 1.4 ish number that most objects seem to have is not a typical stationary orbit, but is one of the Lagrange points, I'm going to guess either L1 or L2. Wolfram Alpha just provides that because a normal stationary orbit isn't possible.

Bottom line is, Earth, Mars, the Gas Giants, and some of the larger asteroids have synchronous orbits, most moons will not, and neither will Venus or Mercury. Pluto actually has a satellite in synchronous orbit already, it's called Charon, but it seems unlikely it could have any others, except perhaps at Pluto-Charon Lagrange points.

  • $\begingroup$ The 1.4 ish number that most objects seem to have is because most of the objects you used are tidally locked to their primary. If you ignore the factor of $(1-e)$, the ratio of the synchronous orbit radius to the hill sphere radius for a tidally-locked object reduces to $^3\sqrt3$, or 1.442... . $\endgroup$ Dec 22, 2015 at 8:53
  • $\begingroup$ The ratio for the Earth should be 0.028 rather than 0.28. You lost a zero in translation. Your other numbers appear to be good. $\endgroup$ Dec 22, 2015 at 8:57
  • $\begingroup$ One last comment: The Hill sphere is an approximation. Since prograde equatorial orbits aren't quite as stable as retrograde orbits, it's fair to say that a synchronous orbit is definitely possible if the ratio is less than 1/3, unlikely if the ratio is more than 1/2, and not possible if the ratio is more than one. Because this ratio is about $\sqrt[3]3$ for an object that is tidally locked to its primary, a synchronous orbit about such a tidally-locked object is not possible. $\endgroup$ Dec 22, 2015 at 9:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.