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A geostationary equatorial orbit (GEO) is a circular geosynchronous orbit in the plane of the Earth's equator with a radius of approximately 42,164 km (26,199 mi) (measured from the center of the Earth). A satellite in such an orbit is at an altitude of approximately 35,786 km (22,236 mi) above mean sea level.

And then what is the altitude of a satellite orbiting the Moon and synchronous with the surface i.e. remaining approximately over one area of the Moon?

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    $\begingroup$ astronomy.stackexchange.com/questions/20499/… $\endgroup$
    – Organic Marble
    Dec 7, 2020 at 16:44
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    $\begingroup$ @uhoh OP didn't mention the sun at all. What's confusing? $\endgroup$
    – Russell Borogove
    Dec 7, 2020 at 17:58
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    $\begingroup$ @RussellBorogove The last two words of the body are "... Moon orbit?" and that conflicts with the last three words of the title "... Moon synchronous orbit?" One sounds like an orbit around the Moon synchronous with something else (potentially the Sun) because it sounds just like "Sun synchronous orbit" with "Sun" replaced by "Moon", and the other sounds like an orbit the Earth with a period synchronous with the Moon. These are not confusing, they are contradictory! so a clarification is in order. $\endgroup$
    – uhoh
    Dec 7, 2020 at 23:42
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    $\begingroup$ Upvoted and edited to make OP's obvious-to-me intent more obvious to you. $\endgroup$
    – Russell Borogove
    Dec 7, 2020 at 23:56
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    $\begingroup$ Further edited to make OP's obvious-to-a-few intent more obvious to future readers in general. Leaving the down vote because of OP's hesitance to clarify this and other questions. $\endgroup$
    – uhoh
    Dec 8, 2020 at 0:06

2 Answers 2

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As computed in this answer on the Astronomy SE (h/t Organic Marble), the altitude would be 88,417km if the Earth weren't present, but an orbit of that height is outside of the moon's sphere of influence. There is no stable orbit around the moon with a 28-day period.

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  • $\begingroup$ Why is it higher than GEO altitude? Because the Moon is smaller or because its rotation is slower than Earth? $\endgroup$
    – Joe Jobs
    Dec 7, 2020 at 19:22
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    $\begingroup$ Because its rotation is much, much slower. If you look at the equation in the linked answer, you'll see the variable terms are the orbital period and the mass of the primary. The lower mass of the moon would require a lower altitude for a given period, not higher. $\endgroup$
    – Russell Borogove
    Dec 7, 2020 at 20:04
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    $\begingroup$ Yep. For any tide-locked body B, the Hill Sphere is always smaller than the synchronous orbit radius. The period of the largest orbit that can fit inside the Hill Sphere of B is always about 58% of the orbital period of B. $\endgroup$
    – notovny
    Dec 8, 2020 at 1:06
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    $\begingroup$ @notovny Oh, neat. Is that independent of density? $\endgroup$
    – Russell Borogove
    Dec 8, 2020 at 1:07
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    $\begingroup$ @RussellBorogove As long as body B is much less massive than body it orbits, as that allows a simpler formula for Hill Sphere size. The ratio of periods appears to be exactly $\frac{1}{\sqrt{3}}$ $\endgroup$
    – notovny
    Dec 8, 2020 at 2:17
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While the OP never clarified the question themselves, edits and comments suggest that what is wanted is an orbit that leaves an artificial satellite roughly above a specific site on the Moon's surface and visible from that point, analogous to how a geosynchronous satellite remains roughly above a specific site on the Earth's surface.

@Russell Borogove's answer which invokes @zephyr's answer in Astronomy SE is correct if one a priori requires the central body being orbited to be the Moon.

However:

tl;dr: There are two potential solutions to this problem but neither is an orbit around the Moon per se.:

  1. Exploit three-body orbits
  2. Attempt to use a Moon-synchronous Earth orbit.

1. Three-body orbit around an Earth-Moon Lagrange point.

EM L1: A Halo orbit associated with Earth-Moon L1 would keep a spacecraft between the Earth and the Sun and fairly close to the Moon. L1 (and L2 below) fall roughly at the Hill sphere radius $R(M_{Moon}/3 M_{Earth})^{1/3}$. With a separation $R$ of 385,000 km and a mass ratio of about 1/81 that would be about 60,000 km above the Moon's surface and meander around that area by several thousand km.

Long term it wouldn't be stable and requires some station keeping, but of course geosynchronous orbits around the Earth also require station-keeping to keep from drifting east-west to be over a substantially different point on the Earth, as well as to maintain low inclination.

EM L2: Similar to EM L1 above except visible to the far hemisphere of the Moon.

This orbit is currently being used by the Queqiao communications satellite, see below.

EM L4 & L5: similar. Since the Earth/Moon mass ratio of about 81 is greater than the critical value of $(25 + \sqrt{621})/2 \approx 24.96$ one can put spacecraft in orbits associated with these points, which will remain roughly above points 60 degrees east and west of the average sub-Earth point on the Moon's surface.

All station-keeping caveats apply here as well.

2. Moon-synchronous Earth orbit

The title of the question was originally What is the altitude of a Moon synchronous orbit? so I'd started down this path before clarifications were made for the OP by third parties.

Results: nice try but doesn't work.

While a high inclination Sun-synchronous orbit around the Earth can always keep the Sun in view by slowly precessing around the Earth (once per year), there is no solution that could precess once per lunar month of about 27.3 days.

The ISS precesses around with a period of the order of sixty days for example, but at it's inclination it is usually (but maybe not always) hidden from the Moon once per orbit.

Queqiao remained visible continuously both to Chang'e 4 and Yutu 2 on the far side of the Moon and to Earth

enter image description here

Source: Wikimedia Commons by Loren Roberts for The Planetary Society https://www.planetary.org/space-images/change-4-mission-profile

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    $\begingroup$ It might be worth mentioning that L4 and L5 are more stable than L1 and L2, so less station-keeping would be required when using them. $\endgroup$
    – Pitto
    Dec 8, 2020 at 2:22
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    $\begingroup$ It's a general fact of Lagrange points: en.wikipedia.org/wiki/Lagrange_point#L4_and_L5_points $\endgroup$
    – Pitto
    Dec 8, 2020 at 2:32
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    $\begingroup$ @Pitto with constraints on system mass ratios - "The triangular points (L4 and L5) are stable equilibria, provided that the ratio of M1/M2 is greater than 24.96 This is the case for the Sun–Earth system, the Sun–Jupiter system, and, by a smaller margin, the Earth–Moon system." $\endgroup$
    – crasic
    Dec 8, 2020 at 2:44
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    $\begingroup$ @JoeJobs that would be the Moon itself, so of course it's possible. That reminds me that I'd meant to use the real orbital period of 27.3 days, not the synodic period of 29.5, I'll update the number. $\endgroup$
    – uhoh
    Dec 8, 2020 at 12:08
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    $\begingroup$ With all the nit-picking of the OP, I hope you won't mind my asking how a halo at E-M L1 "would keep a spacecraft between the Earth and the Sun" $\endgroup$
    – Jeff
    Dec 14, 2022 at 5:06

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