We all, perhaps, know the theory behind space elevators. Similar to the principal of a bucket on a string, suspend a large anchor in space that orbits the planet in synchronous orbit around the equator. Stretch a cable from this platform down to the planet. The cable is thus in tension, not compression, and can be made fairly thin (on earth, it need only be meters wide). An elevator would ride up and down this cable, using a motor that latches on to the cable (perhaps even an induction motor between the cable and the cabin).

It all centers on the 'centrifugal/centripetal' force of the space platform.

So here is the question. The moon revolves once every ~28 days, and has a low gravitational force. It also has no atmospheric drag, that a cable would act against. On the moon, how high would this space elevator have to be, such that a payload massing, say ten tons, could be hauled up the cable using renewable electric energy, and then launched into space entirely by the 'centrifugal' (centripetal) force of a geosynchronous platform (the 'escape velocity' given to it entirely by the rotational spin velocity of the space platform itself)? Assume that the anchoring space platform could be sufficiently distant so that it is actually held to the moon by the tether, and 'spin' energy is transferred through the cable from the moon's rotation itself (like the proverbial spinning bucket on a line, kept spinning by the energy from the person) and not just held in place by the matching of its speed to the speed required by geosynchronous dynamics. That is, if the cable broke, the platform itself would be launched into space.

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    $\begingroup$ The Moon doesn't revolve once every 24 hours. Its actual rotation period is 27.3 days (sidereal) or 29.5 days (synodic). $\endgroup$ – Mike Scott Aug 26 '17 at 18:44
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    $\begingroup$ The moon revolves in a month, same as an orbit which is why we only see one face. Also en.wikipedia.org/wiki/Lunar_space_elevator $\endgroup$ – user25818 Aug 26 '17 at 18:47
  • $\begingroup$ OOPs, sorry, My apologies. I missed that one in editing. A 'lunar day' is slightly longer than 24 hours, as viewed from earth, but the moon actually takes 27 DAYS to rotate, not 24 HOURS. Thus, a slower rotation. $\endgroup$ – Justin Thyme Aug 26 '17 at 19:19
  • $\begingroup$ Obviously the tower needs to be 1 mm (or even 0.1 mm) taller than the selenosynchrounous orbit. If the tower is just a smidgen taller than the selenosynchronous orbit any object on top of the tower will experience a net force pulling it away from the moon. As anybody can see with their own eyes, the radius of the selenosynchronous orbit equals the distance from the (center of the) earth to the (center of the) moon, i.e., some 350 thousand kilometers. $\endgroup$ – AlexP Aug 26 '17 at 21:11
  • $\begingroup$ The earth is not in a selenosynchronous orbit around the moon because the earth does not orbit the moon. $\endgroup$ – Justin Thyme Aug 27 '17 at 1:45

Okay, after some research and some math, I have sort of got the answer to this question

Apparently, the theoretical geostationary orbit around the moon (the height at which the velocity of the satellite needed to keep it in orbit is exactly the velocity required to orbit the moon in one moon rotation, thus ensuring that it stays stationary over a given location) is around 95,000 km. Short of the earth by a long distance. However, The Hill sphere radius, below which the moon's gravity influences satellites more than earth's gravity, is 61,000 km. Thus, there is no possible stable geostationary orbit around the moon that would not be distorted by earth's gravity. However, because the moon always has the same face pointing to the earth, L1 and L2 are in effect geostationary points. This is an artifact of the lunar rotation being exactly and coincidentally the same as its rotation around the earth. There is, apparently, NO point that is a geostationary point elsewhere except the Lagrange points.

  • $\begingroup$ Good answer, but the question was about flinging a payload away not holding it stable. Your cable of 61,000 km + x would be long enough to accelerate the payload to earth. In the oposite direction you have to recalculate taking not the moon gravity and rotation but the gravity and rotation at the cable endpoint in reference to the center of gravity in the combined earth-moon system. $\endgroup$ – Henning M. Aug 27 '17 at 11:51
  • $\begingroup$ Your answer is even better than I thought. Just build cables/towers (or calculate the needed cable length) toward your lagrange points L1 and L2, add some meters away from the moon and you get your payload accelerated towards earth (L1) or away from it (L2). $\endgroup$ – Henning M. Aug 27 '17 at 12:02
  • $\begingroup$ My apologies for the confusion, @Henning M., I did not consolidate. Previously, I gave the distance to L1 as 58,200 km and L2 as 67,000 km but L1 and L2 are not normally geostationary. They usually have a slower or faster orbital time than geostationary, thus no good for a tether. .The moon is unique in this aspect.( L1 and L2 are so high on the moon because of its slow rotation). My research never returned a hit on 'moon geostationary' (selenostationary, for the purists) because there is no credible one. The clue to solve was WHY no geostationary orbit (the Hill sphere).. Thus L2 is it. $\endgroup$ – Justin Thyme Aug 27 '17 at 13:58
  • $\begingroup$ nearly there :-). 1. the cable has to be a bit longer than up to L2 to get something pulled away from the moon, see originaly question. And if it's not a tower you need a weightless cable or a counterweight/more cable to pull it up anyway. 2. L1 might not be a valid answer, but only because the pulling force is earths gravity, not the rotation speed. It is nearly selenostationary (and therefore a potential place for a space elevator), too, because of the moons synchronous roation. 3. geo means earth, not moon, which might clarify things near earth $\endgroup$ – Henning M. Aug 28 '17 at 15:49
  • $\begingroup$ @henningM, sorry but earth has lost its claim to 'geo'. Google 'geography of Mars' or 'geography of the moon'. A purist would say 'geography' only applies to earth, but the general lexicon has abandoned all hope of this. Vernacular wins. There are just too many planetary bodies to come up with a term for each one, so geostationary becomes generic. $\endgroup$ – Justin Thyme Aug 29 '17 at 1:21

The answer to your question is pretty simple. The tower has to reach beyond synchronous orbit -- that's what synchronous orbit is, it's the height at which centrifugal force exactly counterbalances the planetary or lunar rotation. For the Moon, that means it has to be over 240,000 miles high, since the Earth is effectively in selenosynchronous orbit. So it won't work.

However, you should note that the presence of the Earth makes things more complicated, and so there are stable points well below the selenosynchronous height where the Earth's gravity will pull objects away from the Moon, though they'll still be in Earth orbit of some kind.

  • $\begingroup$ It doesn't matter, given that a tower on the Moon reaching to the synchronous point will smash into the Earth and be destroyed. You can't build what you want on the Moon. $\endgroup$ – Mike Scott Aug 26 '17 at 19:23
  • $\begingroup$ Yes, the tower/elevator platform has to be at least geostationary orbit. That was pretty much clarified in the question. The point is, what IS the geostationary orbit of the moon? ? The earth-moon L1 point is only 58 200 Km, so any tower/cable station 'higher' than this, on the earth side of the moon, would cause the payload to 'fall' towards the earth. L2 is 67,000 km and from L2 a payload would, I understand, fly into space. But the anchor has to be further from this. How much further? What is the relationship of L2 to geostationary orbit? $\endgroup$ – Justin Thyme Aug 26 '17 at 21:54
  • $\begingroup$ The "geostationary" (actually selenostationary) point for the Moon is the Earth. The Earth is always over the same spot on the Moon. That's why you can't build anything out to the selenostationary spot. L2 is much lower than selenostationary, because the Earth's gravity is providing most of the pull, not centrifugal force. $\endgroup$ – Mike Scott Aug 27 '17 at 5:54
  • $\begingroup$ No, the selenostationary orbit is NOT the earth, The earth does not orbit the moon, the moon orbits the earth. There is some argument for saying L2 IS the selenostationary point of the moon - the ONLY selenostationary point, perhaps, because it combines earth's gravity with the moon's gravity in calculating orbital velocity. $\endgroup$ – Justin Thyme Aug 28 '17 at 15:07

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