How high does a tower have to be, so that centrifugal force launches a payload into space?

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.

posted over 7 years ago

Justin Thyme‭

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