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

<p>But the moon is tidally locked to Earth. That means that all of the <a href="https://en.wikipedia.org/wiki/Lagrangian_point" rel="noreferrer">Lagrangian points</a> of the Earth-Moon system are stationary relative to the Moon surface.</p>

CC BY-SA 3.0
2017-03-22T18:24:52.643Z
That's a nice shortcut answer to this question, and it applies to all tidally locked moons or planets.
userLTK
CC BY-SA 3.0
2017-03-23T21:58:29.180Z
Earth itself is in lunarstationary, up to liberation https://en.m.wikipedia.org/wiki/Libration (Edit pending)
Grimaldi
CC BY-SA 3.0
2017-03-23T22:39:15.380Z
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.
Chappo Hasn't Forgotten
CC BY-SA 3.0
2017-03-24T13:35:14.117Z
@Chappo: I've heard that [Kordylewski cloud](https://en.wikipedia.org/wiki/Kordylewski_cloud)s are natural objects found at the Earth-Moon Lagrangians.
David Cary
CC BY-SA 3.0
2017-03-24T22:11:25.427Z
@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](https://nssdc.gsfc.nasa.gov/nmc/masterCatalog.do?sc=1990-007A), 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."
Chappo Hasn't Forgotten
CC BY-SA 4.0
2019-02-12T18:22:31.543Z
@Phillip .. Would you say then that a lunar "space elevator" pointing directly at the Earth would work?
Jack R. Woods
CC BY-SA 4.0
2019-02-12T18:35:44.347Z
@JackR.Woods Do you mean a space elevator which connects earth and moon *directly*? As in a tether from the surface of Earth to the surface of the moon?
Philipp
CC BY-SA 4.0
2019-02-13T14:22:53.207Z
@Phillip .. No, I was just thinking about a space elevator to go from lunar orbit to the surface. After thinking about it now though, I guess it would have to go all the way to the Earth to be "lunar synchronous" since the moon is tidally locked.
Jack R. Woods
CC BY-SA 4.0
2019-02-13T14:54:49.130Z
@JackR.Woods It would be possible to build a space elevator from the surface of the Moon to the [L1 point](https://en.wikipedia.org/wiki/Lagrangian_point#L1) of the Earth/Moon system (where "possible" has the usual meaning when it comes to megastructures in space: Possible in theory but impossible in practice because we neither have all the required technology nor the required economic resources).
Philipp
CC BY-SA 4.0
2019-09-19T18:57:07.683Z
@JackR.Woods a space elevator from the [moon's surface to the distance of Earth's geostationary orbit](https://science.slashdot.org/story/19/09/18/1918249/a-lunar-space-elevator-is-actually-feasible-and-inexpensive-scientists-find) is the subject of a recent paper. One catch, possibly among many: the velocities are violently mismatched between the end of the elevator and geostationary satellites.
Bob Stein
CC BY-SA 4.0
2019-09-26T20:41:00.760Z
There is - of course - a what-if from xkcd https://what-if.xkcd.com/157/
Christian
CC BY-SA 4.0
2020-12-08T15:21:42.320Z
@ChappoHasn'tForgottenMonica Their presence was confirmed since your comment ([ref](https://www.eurekalert.org/pub_releases/2018-10/ras-edc102618.php))
Asteroids With Wings
CC BY-SA 3.0
2017-03-22T15:13:34.770Z
2017-03-22T15:19:25.153Z
2017-03-22T15:19:25.153Z
Philipp
<p>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. <em>Any</em> 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.</p>
<p>As Zephyr said, the orbital distance of a selenostationary (using that term to refer to any moon, not just the Moon) orbit is:</p>
<p><span class="math-container">$$a_{\text{SL}} = \left(\frac{P^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3}$$</span></p>
<p>Where <span class="math-container">$a_{\text{SL}}$</span> is the selenostationary orbital distance, <span class="math-container">$P$</span> is the moon's orbital period, <span class="math-container">$G$</span> is the gravitational constant, and <span class="math-container">$M_{\text{Moon}}$</span> is the mass of the moon.</p>
<p>Substituting the formula for <a href="https://en.m.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body" rel="noreferrer">the orbital period <span class="math-container">$P$</span></a> into the equation yields:</p>
<p><span class="math-container">$$a_{\text{SL}} = \left(\frac{(2\pi\sqrt{a^3\over GM_{\text{Planet}} })^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3} = \left(\frac{4\pi^2a^3GM_{\text{Moon}}\over GM_{\text{Planet}}}{4\pi^2} \right)^{1/3} $$</span></p>
<p>Where <span class="math-container">$a$</span> is the moon's semi-major axis, and <span class="math-container">$M_{\text{Planet}}$</span> is the planet's mass.</p>
<p>Simplifying further results in:</p>
<p><span class="math-container">$$a_{\text{SL}} = a\sqrt[3]{\frac{M_{\text{Moon}}}{M_{\text{Planet}} }}$$</span></p>
<p>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:</p>
<p><span class="math-container">$${\frac{a_{\text{SL}}}{r_{\text{H}} }} = {\frac{a\sqrt[3]{\frac{M_{\text{Moon}}}{M_{\text{Planet}} }}} {a\sqrt[3]{\frac{M_{\text{Moon}}}{3M_{\text{Planet}} }}}} = \sqrt[3]{3}$$</span></p>
<p>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 <span class="math-container">$\sqrt[3]{3}$</span> or approximately <span class="math-container">$1.4422$</span> times it.</p>

CC BY-SA 4.0
2023-11-30T21:41:17.693Z
I honestly was not expecting to get such an interesting answer 6 years later. Thank you :)
Christian
CC BY-SA 4.0
2023-12-01T00:21:36.720Z
@Christian No problem! I only learned this very recently and just had to share it.
user267545
CC BY-SA 4.0
2023-12-11T19:07:16.877Z
@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?
Michael_1812
CC BY-SA 4.0
2023-12-12T06:40:37.843Z
@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.
user267545
CC BY-SA 4.0
2023-12-13T03:10:42.483Z
@user267545 Good observation. What makes you think that the planets of red dwarfs are tidally locked? I am asking because I am aware of one yellow dwarf, WASP-12, and one subgiant, Kepler-1658, whose inner planets are almost certainly nonsynchronised.
Michael_1812
CC BY-SA 4.0
2023-12-13T07:24:49.223Z
@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 (https://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.
user267545
CC BY-SA 4.0
2023-12-13T13:10:15.920Z
@user267545  Thank you for the link. I agree that the stronger the tidal force the more likely synchronisation -- but only under the condition that no external perturbation sustains a finite eccentricity value (lest the planet shares our Mercury's lot).
Michael_1812
CC BY-SA 4.0
2023-11-30T05:57:42.670Z
2023-12-01T00:23:06.737Z
2023-12-01T00:23:06.737Z
user267545
<p>First off, such an orbit wouldn't be a <em>geostationary</em> orbit since <em>geo-</em> refers to the Earth. A more appropriate name would be <em>lunarstationary</em> or <em>selenostationary</em>. I'm not sure if there is an officially accepted term since you rarely hear people talk about such an orbit.</p>

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

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

<p>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</p>

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

<p>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.</p>

<p>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 <a href="https://en.wikipedia.org/wiki/Hill_sphere" rel="noreferrer">Hill Sphere</a>. 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</p>

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

<p>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</p>

<p>$$r \approx 52,700\:\mathrm{km}$$</p>

<p>A <a href="http://articles.adsabs.harvard.edu//full/1964SvA.....7..618C/0000622.000.html" rel="noreferrer">more careful calculation</a>, 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.</p>

<p>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 <a href="https://en.wikipedia.org/wiki/Mass_concentration_(astronomy)" rel="noreferrer">mass concentrations</a> (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.</p>

CC BY-SA 3.0
2017-03-26T02:26:25.367Z
Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/56029/discussion-on-answer-by-zephyr-is-it-possible-to-achieve-a-stable-lunarstationa).
called2voyage
CC BY-SA 4.0
2022-11-22T06:00:55.510Z
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.
Nicolas
CC BY-SA 3.0
2017-03-22T15:14:32.860Z
2017-03-22T15:56:03.350Z
2017-03-22T15:56:03.350Z
zephyr
<p>Is there a stable geostationary orbit around the Moon?</p>
<p>My feeling is, that the orbit would collide with Earth, because of the Moon's slow rotation.</p>

CC BY-SA 3.0
2017-03-22T12:55:15.230Z
Instead of "feeling,'  why not calculate the orbital radius for a body with the mass of the Moon and a rotational period of 28 days -- i.e. pretend Earth doesn't exist?
Carl Witthoft
CC BY-SA 3.0
2017-03-22T12:59:31.027Z
because the math I didn't check out. I'll try again later
Christian
CC BY-SA 3.0
2017-03-22T13:05:20.707Z
[Here's Adamo](http://spirit.as.utexas.edu/~fiso/telecon13-15/Adamo_11-6-13/) 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.
LocalFluff
CC BY-SA 3.0
2017-03-22T14:46:52.477Z
Your result with cube root kg does not make sense because you forgot the units on mass. If you [do this right](https://www.wolframalpha.com/input/?i=%28%28%2827.322+days%29%2F%282+pi%29%29%5E2*%284.9048695e12+m%5E3%2Fs%5E2%29%29%5E1%2F3) you will have obtained a value of 88470 km, which is outside of the Moon's sphere of influence.
David Hammen
CC BY-SA 3.0
2017-03-22T15:54:29.817Z
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](https://www.wolframalpha.com/input/?i=((27day)%5E2*G*(7.349+%C2%B7+10%5E(22)+kg)+%2F+(4pi%5E2))%5E1%2F3).
zephyr
CC BY-SA 3.0
2017-03-22T16:56:06.833Z
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
Christian
CC BY-SA 3.0
2017-03-22T20:18:59.970Z
A geostationary orbit means that the object always remains above the same point on the earth, in other words it appears to be stationary. This is useful for communication satellites because the antenna does not need to track the satellite. The antenna is just pointed at the satellite and since the satellite does not move, the antenna does not need to move. The moon moves with respect to the earth, therefore it is not in a geostationary orbit.
Tyler Durden
CC BY-SA 3.0
2017-03-22T20:41:32.780Z
@TylerDurden that answers a different question
Christian
CC BY-SA 3.0
2017-03-22T22:22:54.327Z
@Christian: No, that answers the question that was ASKED - whether the moon is in a geostationary orbit.  Obviously it's not.  Now maybe you really intended to ask whether there are any stable SELENOSTATIONARY orbits, in which case you need to re-write the question and title.
jamesqf
CC BY-SA 3.0
2017-03-22T23:36:10.703Z
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.
Dawood ibn Kareem
CC BY-SA 3.0
2017-03-23T02:08:49.250Z
@TylerDurden - when reading the title on the HNQ I too took it to mean "is the moon in a geostationary orbit."  And the answer was obviously no so I clicked to see how folks would answer.  I was surprised to see what the question was really getting at.
Tracy Cramer
CC BY-SA 3.0
2017-03-23T11:34:20.147Z
Can someone think of an improved wording for the title?  I had the same response as Tracy - "No, of course the moon isn't in a geostationary orbit".
Toby Speight
CC BY-SA 3.0
2017-03-23T12:01:24.063Z
Well if I put lunar stationary. Does everyone know what that is?
Christian
CC BY-SA 3.0
2017-03-23T12:03:47.740Z
I removed the math from the question, as the answer already covers it.
Christian
CC BY-SA 3.0
2017-03-23T20:41:15.077Z
@Christian "selenostationary" or such would be more appropriate (since geo- is from the greek ge) but even less intuitive. I think the title should use "luna-stationary" in quotes. Ah, I didn't even look at the first answer -- zephyr is right.
Peter - Reinstate Monica
CC BY-SA 3.0
2017-03-24T03:55:19.110Z
The correct term for an arbitrary body seems to be "synchronous". If you never talk about it, play KSP more.
Joshua
CC BY-SA 3.0
2017-03-24T13:21:10.540Z
So if the earth is in selenostationary orbit around the moon, it would seem like you could place an object in selenostationary orbit on the other side of the moon from the earth.  It would then technically be orbiting the moon and the earth like a double pendulum which isn't hinging.
Joshua
CC BY-SA 3.0
2017-03-27T23:50:05.240Z
Okay .. here's a related thought. When I saw this question I thought, "Lunar Space Elevators". I see that even though L1 and L2 are unstable, that spacecraft can do small "halo" orbits around these Lagrangian points. I wonder if a space elevator "quasi-selenostationary" station would work in these "halo" orbits.
Jack R. Woods
CC BY-SA 3.0
2017-03-28T07:23:40.877Z
@JackR.Woods That's how we got to the question over lunch
Christian
CC BY-SA 4.0
2021-05-23T20:55:52.953Z
The Earth's Moon is truly a massive Body unto itself so of course moving beyond being a purely natural phenomenon the answer to this question is *yes with good station keeping.* The question that would need answering is "does the Moon have any Trojans?" and proceed from an evidence based approach to getting a reality based answer.
Doctor Zhivago
<orbit><the-moon>
Is it possible to achieve a stable "selenostationary" orbit around the Moon?
Christian
# Is it possible to achieve a stable "selenostationary" orbit around the Moon?

(Asked by: username_0 on 2017-03-22)

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.

**Tags:** orbit, the-moon
> **Comments:**
> **username_1**: Instead of "feeling,'  why not calculate the orbital radius for a body with the mass of the Moon and a rotational period of 28 days -- i.e. pretend Earth doesn't exist?  
> **username_0**: because the math I didn't check out. I'll try again later  
> **username_2**: [Here's Adamo](http://spirit.as.utexas.edu/~fiso/telecon13-15/Adamo_11-6-13/) 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.  
> **username_3**: Your result with cube root kg does not make sense because you forgot the units on mass. If you [do this right](https://www.wolframalpha.com/input/?i=%28%28%2827.322+days%29%2F%282+pi%29%29%5E2*%284.9048695e12+m%5E3%2Fs%5E2%29%29%5E1%2F3) you will have obtained a value of 88470 km, which is outside of the Moon's sphere of influence.  
> **username_4**: 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](https://www.wolframalpha.com/input/?i=((27day)%5E2*G*(7.349+%C2%B7+10%5E(22)+kg)+%2F+(4pi%5E2))%5E1%2F3).  
> **username_0**: 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  
> **username_5**: A geostationary orbit means that the object always remains above the same point on the earth, in other words it appears to be stationary. This is useful for communication satellites because the antenna does not need to track the satellite. The antenna is just pointed at the satellite and since the satellite does not move, the antenna does not need to move. The moon moves with respect to the earth, therefore it is not in a geostationary orbit.  
> **username_0**: @TylerDurden that answers a different question  
> **username_6**: @Christian: No, that answers the question that was ASKED - whether the moon is in a geostationary orbit.  Obviously it's not.  Now maybe you really intended to ask whether there are any stable SELENOSTATIONARY orbits, in which case you need to re-write the question and title.  
> **username_7**: 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.  
> **username_8**: @TylerDurden - when reading the title on the HNQ I too took it to mean "is the moon in a geostationary orbit."  And the answer was obviously no so I clicked to see how folks would answer.  I was surprised to see what the question was really getting at.  
> **username_9**: Can someone think of an improved wording for the title?  I had the same response as Tracy - "No, of course the moon isn't in a geostationary orbit".  
> **username_0**: Well if I put lunar stationary. Does everyone know what that is?  
> **username_0**: I removed the math from the question, as the answer already covers it.  
> **username_10**: @Christian "selenostationary" or such would be more appropriate (since geo- is from the greek ge) but even less intuitive. I think the title should use "luna-stationary" in quotes. Ah, I didn't even look at the first answer -- zephyr is right.  
> **username_11**: The correct term for an arbitrary body seems to be "synchronous". If you never talk about it, play KSP more.  
> **username_11**: So if the earth is in selenostationary orbit around the moon, it would seem like you could place an object in selenostationary orbit on the other side of the moon from the earth.  It would then technically be orbiting the moon and the earth like a double pendulum which isn't hinging.  
> **username_12**: Okay .. here's a related thought. When I saw this question I thought, "Lunar Space Elevators". I see that even though L1 and L2 are unstable, that spacecraft can do small "halo" orbits around these Lagrangian points. I wonder if a space elevator "quasi-selenostationary" station would work in these "halo" orbits.  
> **username_0**: @JackR.Woods That's how we got to the question over lunch  
> **username_13**: The Earth's Moon is truly a massive Body unto itself so of course moving beyond being a purely natural phenomenon the answer to this question is *yes with good station keeping.* The question that would need answering is "does the Moon have any Trojans?" and proceed from an evidence based approach to getting a reality based answer.  

---

## 3 answers

### Answer by username_4 on 2017-03-22. Score: 79 | Accepted answer

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](https://en.wikipedia.org/wiki/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](http://articles.adsabs.harvard.edu//full/1964SvA.....7..618C/0000622.000.html), 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](https://en.wikipedia.org/wiki/Mass_concentration_(astronomy)) (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.

> **Comments:**
> **username_23**: Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/56029/discussion-on-answer-by-zephyr-is-it-possible-to-achieve-a-stable-lunarstationa).  
> **username_24**: 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.  

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### Answer by username_14 on 2017-03-22. Score: 67

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](https://en.wikipedia.org/wiki/Lagrangian_point) of the Earth-Moon system are stationary relative to the Moon surface.

> **Comments:**
> **username_15**: That's a nice shortcut answer to this question, and it applies to all tidally locked moons or planets.  
> **username_16**: Earth itself is in lunarstationary, up to liberation https://en.m.wikipedia.org/wiki/Libration (Edit pending)  
> **username_17**: 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.  
> **username_18**: @Chappo: I've heard that [Kordylewski cloud](https://en.wikipedia.org/wiki/Kordylewski_cloud)s are natural objects found at the Earth-Moon Lagrangians.  
> **username_17**: @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](https://nssdc.gsfc.nasa.gov/nmc/masterCatalog.do?sc=1990-007A), 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."  
> **username_12**: @Phillip .. Would you say then that a lunar "space elevator" pointing directly at the Earth would work?  
> **username_14**: @JackR.Woods Do you mean a space elevator which connects earth and moon *directly*? As in a tether from the surface of Earth to the surface of the moon?  
> **username_12**: @Phillip .. No, I was just thinking about a space elevator to go from lunar orbit to the surface. After thinking about it now though, I guess it would have to go all the way to the Earth to be "lunar synchronous" since the moon is tidally locked.  
> **username_14**: @JackR.Woods It would be possible to build a space elevator from the surface of the Moon to the [L1 point](https://en.wikipedia.org/wiki/Lagrangian_point#L1) of the Earth/Moon system (where "possible" has the usual meaning when it comes to megastructures in space: Possible in theory but impossible in practice because we neither have all the required technology nor the required economic resources).  
> **username_19**: @JackR.Woods a space elevator from the [moon's surface to the distance of Earth's geostationary orbit](https://science.slashdot.org/story/19/09/18/1918249/a-lunar-space-elevator-is-actually-feasible-and-inexpensive-scientists-find) is the subject of a recent paper. One catch, possibly among many: the velocities are violently mismatched between the end of the elevator and geostationary satellites.  
> **username_0**: There is - of course - a what-if from xkcd https://what-if.xkcd.com/157/  
> **username_20**: @ChappoHasn'tForgottenMonica Their presence was confirmed since your comment ([ref](https://www.eurekalert.org/pub_releases/2018-10/ras-edc102618.php))  

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### Answer by username_21 on 2023-11-30. Score: 10

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 $P$](https://en.m.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body) 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.

> **Comments:**
> **username_0**: I honestly was not expecting to get such an interesting answer 6 years later. Thank you :)  
> **username_21**: @Christian No problem! I only learned this very recently and just had to share it.  
> **username_22**: @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?  
> **username_21**: @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.  
> **username_22**: @user267545 Good observation. What makes you think that the planets of red dwarfs are tidally locked? I am asking because I am aware of one yellow dwarf, WASP-12, and one subgiant, Kepler-1658, whose inner planets are almost certainly nonsynchronised.  
> **username_21**: @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 (https://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.  
> **username_22**: @user267545  Thank you for the link. I agree that the stronger the tidal force the more likely synchronisation -- but only under the condition that no external perturbation sustains a finite eccentricity value (lest the planet shares our Mercury's lot).  

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