Earth's geostationary orbit has many human-made satellites which have helped vastly with communications and research. Aside from this artificial collection, are there any other objects orbiting a massive body at synchronous or stationary orbits? Generally, since masses are considered to be point masses, an orbit of any radius can be constructed, including ones that would require crossing through an atmosphere or the surface. Are there any such orbits that can't exist because of a combination of body radius, strong gravitational pull, and fast rotation? You can also consider cases were the rotation is so slow that the orbital radius would require the object to exit the massive body's gravitational sphere-of-influence (Venus).

My next question is how do you calculate the orbital parameters of geostationary orbits on other bodies (specifically orbital velocity and radius)?

My thought is that we are very lucky on Earth to have such a geostationary orbit that resides inside our Hill Sphere, outside of our atmosphere, and at a distance that serves well for communication. These sorts of systems around other bodies could prove very useful in the future.

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    $\begingroup$ I suppose one could say that Charon and Pluto are "geo"synchronous. In multi-moon systems like Jupiter, there might not be any stable "geo"synchronous orbits because of the gravitational disturbance of moons passing by. $\endgroup$
    – LocalFluff
    Commented Oct 1, 2014 at 6:31
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    $\begingroup$ Certainly not for Mercury or Venus. The orbital radius required for 'once per rotation' would be far outside each planet's Hill Sphere. $\endgroup$ Commented Oct 1, 2014 at 8:39
  • $\begingroup$ @Andrew Thompson Venus would indeed not have a GEO (or VEO?)! The very definition of a geosynchronous orbit seems to be quite arbitrary or at least just a coincidentally practical thing for our communication satellites. It depends on how the motherbody rotates. And what part of it: its core, its mantle, its surface, its atmosphere? Besides from tidal locking, there's nothing special with geosynchronous orbits. Television on Venus will never be as cheap as on Earth. $\endgroup$
    – LocalFluff
    Commented Oct 1, 2014 at 13:42
  • $\begingroup$ @LoclaFluff You are both right that Venus, with a rotation every 243 Earth days, would not work. The very definition of a geosynchronous orbit is that the orbiting body's orbital period matches the rotation of the mother-body. Such an orbit (theoretically) exists for every body. I want to know which ones are "valid" in the sense that you could place an object there for a somewhat stable orbit. $\endgroup$
    – Stu
    Commented Oct 1, 2014 at 13:45
  • $\begingroup$ @LocalFluff Each synchronous orbit's radius and speed would definitely depend on the rotation of the mother body but it also depends on the mass. $\endgroup$
    – Stu
    Commented Oct 1, 2014 at 13:51

1 Answer 1


Geosynchronous orbit radius can be calculated imposing the orbital period to be equal to the Earth's rotational period, which results in: $R_{GEO}=\sqrt[3]{\frac{GM_ET_{rot}^2}{4\pi^2}}$

where $G = 6.673 \cdot 10^{-11} \frac{Nm^2}{kg^2}$ is the universal gravitational constant, $M_E = 5.97\cdot 10^{24} kg$ is the Earth Mass and $T_{rot} = 86164 s$ is the Earth rotation period. A circular orbit having the resulting radius ($46164 km$ for Earth) is called Geosynchronous; if it also have 0 inclination it is a Geostationary orbit, since a spacecraft put in such an orbit will always be over the same point on the Earth.

The orbital speed on any circular orbit can be calculated with the following formula:

$ V_c=\sqrt{\frac{GM}{R}}$

In the Geosynchronous case this results in about $3.07 km/s$

The same computations can be performed for any celestial body using the appropriate values: $R_{synch}=\sqrt[3]{\frac{GM_{planet}T_{rot,planet}^2}{4\pi^2}}$

After the computation of the radius, you could compare it with the planet's radius and the radius of the Hill Sphere (Sphere of influence of the planet).

In the following, I report the results of an approximate computation for each Solar System planet (plus the Moon), considering the Hill sphere relative to the Sun (to the Earth for the Moon):

Mercury: $R_{synch}=242843 km$ , $R_H=220594 km$ , $R_{planet}=2440 km$ , $V_c=0.3 km/s$

Venus: $R_{synch}=1535681 km$ , $R_H=1010369 km$ , $R_{planet}=6052 km$ , $V_c=0.46 km/s$

Moon: $R_{synch}=88463 km$ , $R_H=129417 km$ , $R_{planet}=1737 km$ , $V_c=0.24 km/s$

Mars: $R_{synch}=20429 km$ , $R_H=1083941 km$ , $R_{planet}=3390 km$ , $V_c=1.45 km/s$

Jupiter: $R_{synch}=160052 km$ , $R_H=53155071 km$ , $R_{planet}=69911 km$ , $V_c=28.14 km/s$

Saturn: $R_{synch}=111606 km$ , $R_H=65439558 km$ , $R_{planet}=58232 km$ , $V_c=18.43 km/s$

Uranus: $R_{synch}=82674 km$ , $R_H=70064595 km$ , $R_{planet}=25362 km$ , $V_c=8.37 km/s$

Neptune: $R_{synch}=83395 km$ , $R_H=115863626 km$ , $R_{planet}=24622 km$ , $V_c=9.03 km/s$

Pluto: $R_{synch}=18892 km$ , $R_H=7633076 km$ , $R_{planet}=1184 km$ , $V_c=0.22 km/s$

As you can see from these numbers, the orbits relative to Mercury and Venus would be outside the Hill sphere. Every orbit, instead, is well above the relative planet surface.

These computations only determine the radius of the orbits, then, their stability has to be considered, which is a much more complicated matter. Even the Geostationary orbit is not stable and satellites spend fuel to maintain this orbit. In general each orbit will be influenced by the actual shape of the planet (flatness, asymmetries...), by the gravitational attraction exerted by other nearby bodies like moons (and even by more distant but bigger ones like the Sun or Jupiter) and many other factors. A precise computation of these effects requires an accurate knowledge of the Solar System dynamics.

Once one of these orbit has been proved to exist, nothing would prevent a natural body to follow it. Charon and Pluto are mutually tidally locked which means that Charon is essentially in one of these orbits.

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    $\begingroup$ Yes, I am aware of many different perturbations that can happen to an orbiting body. For stability I am just looking for cheap approximations. Thank you for the answer, though, it is very useful! $\endgroup$
    – Stu
    Commented Oct 1, 2014 at 15:24
  • $\begingroup$ You messed up the numbers for the moon, it should be 61,500km for the hill sphere. $\endgroup$ Commented May 10, 2023 at 8:09

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