### Is it possible to achieve a stable "lunarstationary" orbit around the moon?

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.

because the math I didn't check out. I'll try again later

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.

Your result with cube root kg does not make sense because you forgot the units on mass. If you do this right you will have obtained a value of 88470 km, which is outside of the Moon's sphere of influence.

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%5E2*G*(7.349+%C2%B7+10%5E(22)+kg)+%2F+(4pi%5E2))%5E1%2F3).

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

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.

@TylerDurden that answers a different question

@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.

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.

@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.

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".

Well if I put lunar stationary. Does everyone know what that is?

I removed the math from the question, as the answer already covers it.

@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.

The correct term for an arbitrary body seems to be "synchronous". If you never talk about it, play KSP more.

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.

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.

@JackR.Woods That's how we got to the question over lunch

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.

zephyr Correct answer

5 years agoFirst 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.

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Carl Witthoft 5 years ago

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?