Is there a stable geostationary orbit around the Moon?

My feeling is, that the orbit would collide with Earth, because of the Moon's slow rotation.

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    $\begingroup$ because the math I didn't check out. I'll try again later $\endgroup$
    – Christian
    Commented Mar 22, 2017 at 12:59
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    $\begingroup$ Here's Adamo giving a pretty practical and accessible talk on the stability of Lunar orbits. There does not appear to exist any stable Lunar centric orbit. The Moon is pretty choosy. She prefers taking another hit rather than hanging out regularly with anyone but Earth. $\endgroup$
    – LocalFluff
    Commented Mar 22, 2017 at 13:05
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    $\begingroup$ Just to remark on your Wolfram Alpha result, you didn't tell it the units for the mass of the Moon. You just left it a raw number so of course Alpha didn't know to cancel the $kg$ in $G$. If you throw in these units, you get a number with the correct unit output. $\endgroup$
    – zephyr
    Commented Mar 22, 2017 at 15:54
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    $\begingroup$ Yes thanks. I feel kinda stupid, but the answers still told me new things about Hill spheres and that the moon does not have a stable orbit at all. So the question was worth asking $\endgroup$
    – Christian
    Commented Mar 22, 2017 at 16:56
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    $\begingroup$ Well, obviously the earth is in a lunastationary orbit, since it is always in line with a point in the centre of the "visible side" of the moon. So any object orbiting above the equator of the moon at the same distance as the earth would also be lunastationary, if it weren't for the presence of the earth. The problem becomes dealing with the earth's pull on such an object, in addition to the moon's pull. It's not a two-body problem any more. $\endgroup$ Commented Mar 22, 2017 at 23:36

3 Answers 3


First off, such an orbit wouldn't be a geostationary orbit since geo- refers to the Earth. A more appropriate name would be lunarstationary or selenostationary. I'm not sure if there is an officially accepted term since you rarely hear people talk about such an orbit.

You can calculate the orbital distance of a selenostationary orbit using Kepler's law:

$$a = \left(\frac{P^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3}$$

In this case, $a$ is your orbital distance of interest, $P$ is the orbital period (which we know to be 27.321 days or 2360534 seconds), $G$ is just the gravitational constant, and hopefully it is obvious that $M_{\text{Moon}}$ is the mass of the Moon. All we have to do is plug in numbers. I find that

$$a = 88,417\:\mathrm{km}=0.23\:\mathrm{Earth\mathit{-}Moon\:Distance}$$

So I at least match your calculation pretty well. I think you were just relying on Wolfram Alpha a bit too much to get the units right. The units do work out right though.

If you want to determine if this orbit can exist however, you need to do a bit more work. As a first step, calculate the Moon's Hill Sphere. This is the radius at which the Moon still maintains control over it's satellite, without the Earth causing problems. The equation for this radius is given by

$$r \approx a_{\text{Moon}}(1-e_{\text{Moon}})\sqrt[3]{\frac{M_{\text{Moon}}}{3M_{\text{Earth}}}}$$

In this equation, $a_{\text{Moon}} = 348,399\:\mathrm{km}$ is the Moon's semi-major axis around the Earth and $e_{\text{Moon}} = 0.0549$ is the Moon's orbital eccentricity. I'm sure you can figure out that the $M$'s are the masses of the respective bodies. Just plug and chug and you get

$$r \approx 52,700\:\mathrm{km}$$

A more careful calculation, including the effects of the Sun is slightly more optimistic and provides a Hill radius of $r = 58,050\:\mathrm{km}$. In either case though, hopefully you can see that the radius for a selenostationary orbit is much farther than the Hill radius, meaning that no stable orbit can be achieved as it would be too much perturbed by the Earth and/or the Sun.

One final, semi-related point. It turns out almost no orbits around the Moon are stable, even if they're inside the Hill radius. This is primarily to do with mass concentrations (or mascons) in the Moon's crust and mantle which make the gravitational field non-uniform and act to degrade orbits. There are only a handful of "stable" orbits and these are only achieved by orbiting in such a way as to miss passing over these mascons.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – called2voyage
    Commented Mar 26, 2017 at 2:26
  • $\begingroup$ They will eventually need to place some satellites at this distance so as to serve as a communications relay for future Lunar exploration on the 'dark side' of the Moon...eventually they could also serve to provide Location data for Drones or Astronauts during activities on the Lunar surface. $\endgroup$
    – Nicolas
    Commented Nov 22, 2022 at 6:00

As the answer by zephyr describes very well, there are very few stable orbits around the moon, and none of them are stationary.

But the moon is tidally locked to Earth. That means that all of the Lagrangian points of the Earth-Moon system are stationary relative to the Moon surface.

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    $\begingroup$ That's a nice shortcut answer to this question, and it applies to all tidally locked moons or planets. $\endgroup$
    – userLTK
    Commented Mar 22, 2017 at 18:24
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    $\begingroup$ Earth itself is in lunarstationary, up to liberation en.m.wikipedia.org/wiki/Libration (Edit pending) $\endgroup$
    – Grimaldi
    Commented Mar 23, 2017 at 21:58
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    $\begingroup$ The Lagrangian points are stationary, as they're defined geometrically (or should that be geo-selenometrically?), but they are not stable due to the perturbative effect of the Sun's gravity, and an object at such a point would require an occasional boost to maintain its position. Hence no natural objects found at the Earth-Moon Lagrangians. $\endgroup$ Commented Mar 23, 2017 at 22:39
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    $\begingroup$ @Chappo: I've heard that Kordylewski clouds are natural objects found at the Earth-Moon Lagrangians. $\endgroup$
    – David Cary
    Commented Mar 24, 2017 at 13:35
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    $\begingroup$ @DavidCary: the existence of the Kordylewski clouds, at the L4 and L5 Lagrangian points, is disputed. One of the objectives of Japan's Hiten space probe was to find evidence for the clouds. To quote NASA, Hiten was "put into a looping orbit which passed through the L4 and L5 stable libration points to look for trapped dust particles. No obvious increase was found." $\endgroup$ Commented Mar 24, 2017 at 22:11

All of the other answers in this thread are completely correct, however I think there's a point missed by focusing only on the Moon-Earth system. Any tidally locked moon will always have its synchronous orbital radius outside of its Hill Sphere. That is to say, no tidally locked moon can have a satellite with the same orbital period as its rotational period.

As Zephyr said, the orbital distance of a selenostationary (using that term to refer to any moon, not just the Moon) orbit is:

$$a_{\text{SL}} = \left(\frac{P^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3}$$

Where $a_{\text{SL}}$ is the selenostationary orbital distance, $P$ is the moon's orbital period, $G$ is the gravitational constant, and $M_{\text{Moon}}$ is the mass of the moon.

Substituting the formula for the orbital period $P$ into the equation yields:

$$a_{\text{SL}} = \left(\frac{(2\pi\sqrt{a^3\over GM_{\text{Planet}} })^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3} = \left(\frac{4\pi^2a^3GM_{\text{Moon}}\over GM_{\text{Planet}}}{4\pi^2} \right)^{1/3} $$

Where $a$ is the moon's semi-major axis, and $M_{\text{Planet}}$ is the planet's mass.

Simplifying further results in:

$$a_{\text{SL}} = a\sqrt[3]{\frac{M_{\text{Moon}}}{M_{\text{Planet}} }}$$

Looks familiar? This is the exact same as the formula for the moon's Hill Sphere, with the only difference being the 3 in the denominator. This means that the ratio of any tidally locked moon's selenostationary orbital radius to its Hill Sphere is:

$${\frac{a_{\text{SL}}}{r_{\text{H}} }} = {\frac{a\sqrt[3]{\frac{M_{\text{Moon}}}{M_{\text{Planet}} }}} {a\sqrt[3]{\frac{M_{\text{Moon}}}{3M_{\text{Planet}} }}}} = \sqrt[3]{3}$$

In other words, not only will the selenostationary orbital distance for any tidally locked moon always be outside its Hill Sphere, it will be exactly $\sqrt[3]{3}$ or approximately $1.4422$ times it.

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    $\begingroup$ I honestly was not expecting to get such an interesting answer 6 years later. Thank you :) $\endgroup$
    – Christian
    Commented Nov 30, 2023 at 21:41
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    $\begingroup$ @Christian No problem! I only learned this very recently and just had to share it. $\endgroup$
    – user267545
    Commented Dec 1, 2023 at 0:21
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    $\begingroup$ @user267545 Thank you for this interesting observation. If I got it right, we cannot have a stable setting with a planet synchronised both with its star and its moon. If a moon synchronises its orbital motion with the planet's rotation, then such a moon will be outside the planet's Hill sphere. Correct? $\endgroup$ Commented Dec 11, 2023 at 19:07
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    $\begingroup$ @Michael_1812 Correct. In any 3 body system where an intermediate body (i.e. a planet) orbits around a massive body (i.e. a star) and is tidally locked to it, then all satellites of the intermediate body will necessarily orbit faster than the body rotates around its axis. In other words, any moons orbiting planets of red dwarfs (which are highly likely to be tidally locked) will necessarily orbit faster than their planet rotates. This would lead to tidal deceleration and on long timscales, cause the satellites to crash into the planet or break up once they enter the Roche limit. $\endgroup$
    – user267545
    Commented Dec 12, 2023 at 6:40
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    $\begingroup$ @Michael_1812 I'm a bit confused as to the examples you presented. Not only are both of those not red dwarfs, they're even larger and more massive than the Sun. However the reason planets orbiting around red dwarfs are more likely to be tidally locked is that the smaller, less massive, and cooler a star is, the more tidal forces a planet experiences. And it scales exponentially. Check out the answer to this (astronomy.stackexchange.com/questions/40746) question for further info. Habitable planets around red dwarfs (class M) experience >100x the amount of tidal forces as Earth does. $\endgroup$
    – user267545
    Commented Dec 13, 2023 at 7:24

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