Is it possible to achieve a stable "selenostationary" orbit around the Moon?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.