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Could there be a moon orbiting a massive planet so rapidly that the moon's own gravity is less than the сentrifugal force from the rotation?

What would be the implications in such a system? Would the shape of the moon be influenced by that (assuming it's tidally locked, for simplicity)? Could you fall "down" into the sky, if you find yourself on the wrong (outer) side of the moon?

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    $\begingroup$ This is the case for any rock out there in space which is orbiting a planet. $\endgroup$ – Steeven Jun 8 '18 at 13:02
  • $\begingroup$ @Steeven but is it the case for rocks that are big enough to be called moons? $\endgroup$ – Irdes Jun 8 '18 at 13:04
  • $\begingroup$ Read about Lagrangian Points, the L2 point in particular. $\endgroup$ – Solomon Slow Jun 8 '18 at 14:23
  • $\begingroup$ If the question was stated a little more carefully then reading about the Roche limit would be worthwhile. $\endgroup$ – dmckee --- ex-moderator kitten Jun 8 '18 at 15:45
  • $\begingroup$ @Irdes - Please wait a bit (hours) before you accept an answer. Accepting immediately discourages people from writing other, perhaps better, answers. $\endgroup$ – David Hammen Jun 8 '18 at 16:21
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In a rotating frame, a circular orbit is precisely the radius where the gravitational force is balanced by the centripetal force. Without a centripetal force to balance the attraction of gravity the objects would fall and collide into each other.

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    $\begingroup$ The gravitational force of the planet is balanced there, but I'm talking about the gravity of the moon itself. Even if a moon's orbit is circular, an astronaut on the moon's surface is still get attracted to the moon itself, right? That's the gravity I'm talking about, not the one between the moon and the planet. $\endgroup$ – Irdes Jun 8 '18 at 13:05
  • $\begingroup$ The point is that the centrifugal force of the orbit exactly balances the gravitational force of the planet at the moon's orbit. Standing on the moon, you would not be aware of the centrifugal force. The centrifugal force can be gigantic or tiny depending on the mass of the planet and the size of the orbit. The moon's surface gravity can also be tiny or huge; is independent of the planet and orbit, and depends only on the radius and mass of the moon. $\endgroup$ – S. McGrew Jun 8 '18 at 13:18
  • $\begingroup$ if we also consider the moon as extended object with its own rotation, then the same considerations apply: if it is spinning too fast to be balanced by its self-gravity, then some of it will fly out and orbit the moon in a stable orbit as explained above $\endgroup$ – lurscher Jun 8 '18 at 13:37
  • $\begingroup$ @irdes The key point is that if you're standing on the moon you have the same velocity vector as the moon, so you're also orbiting the planet; that's independent of any gravitational attraction you may have to the moon. $\endgroup$ – PM 2Ring Jun 8 '18 at 15:28

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