Assume the moon was flying freely in space, and not rotating at all. If if then happened to pass earth at just the right distance and velocity to fall into orbit, would its absolute rotation be altered in any way, or would its angular momentum (relative to the rest of the universe) remain at 0?
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$\begingroup$ Well, the Moon is tidally locked to Earth, so eventually there will be one rotation per orbit. $\endgroup$– Qmechanic ♦Commented Feb 20, 2016 at 17:49
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$\begingroup$ Well I know the moon is now, but did that have to happen? Would it always happen, automatically? $\endgroup$– LudwikCommented Feb 20, 2016 at 18:00
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1$\begingroup$ It, for all practical purposes, always happens because there's always some degree of tidal bulge and the tidal bulge always tends towards tidal locking, but, it can take a very long time. Even with the young Earth-Moon system, it (probably) took millions of years for the Moon to lose it's rotation (or gain rotation to correspond to the orbit is also possible) to tidally lock with the Earth. It can take a very long time. Billions or trillions of years or more, depending on the tidal forces and how much angular momentum there is to start with. $\endgroup$– userLTKCommented Feb 20, 2016 at 19:07
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2 Answers
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In theory, Yes, when it passes earth at just the right velocity and angle -
It would get some rotational motion due to the difference of earth's gravity between the far side of the moon and the near side. And due to the angle between the lines that join two sides to the CG of earth. That is what you need to rotate something. Far and near side means at different places on its circumference and diagonally opposite points.
If it was a point particle, then it would not gain any rotation because the angle between lines joining far side and near side to CG of earth would be same line, i.e. 0 angle.
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$\begingroup$ Wouldn't that not be "in theory" at all, but very real, though quite small. I agree with you on the point particle, though the idea of a point particle having angular momentum is in itself, a bit of a conundrum. :-) $\endgroup$– userLTKCommented Feb 20, 2016 at 18:53
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$\begingroup$ I agree, I said in theory, because not sure about its magnitude, or it would be too complex to calculate it. $\endgroup$– kpvCommented Feb 20, 2016 at 19:00
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$\begingroup$ On the point particle: Can you really speak of the rotation of a point? I thank you for the rest of your answer by the way, clared things up for me. $\endgroup$– LudwikCommented Feb 20, 2016 at 19:37
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1$\begingroup$ No, I can not, because, it does not make any sense. Just included it to make the other part to be contrasted. $\endgroup$– kpvCommented Feb 20, 2016 at 19:39
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$\begingroup$ The point particle example is actually important, tides are a product, not just of the gravitational strength of the planet but also the size, density and density variation and, for lack of a better word, stretchy-ness and/or liquidity of the moon. Size of the moon in this example, matters. $\endgroup$– userLTKCommented Feb 20, 2016 at 19:52
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A couple of points, too long for comment, but not meant as a complete answer.
Tidal locking doesn't mean zero angular momentum. The Moon has angular momentum. It rotates at nearly exactly the same speed as it's sidereal orbit, every 27.3 days. The Moon is also unusually lopsided, it's near side being denser than it's far side, but I think we can ignore that aspect for your question. Tidal locking requires rotation,otherwise, we'd see different faces of the moon as it went about it's orbit. Tidal locking appears to be zero rotation, but it's not.
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$\begingroup$ Hm, I'm trying to comprehend what exactly you're saying here. I know tidal lock is not zero rotation, it's "rotating as fast as you're orbiting". I was asking whether it would always happen or if a body could retain "zero rotation" while being in orbit, and apparently, it cannot. Anyway, you seem to be speaking about some tidal-caused reverse rotation force. Could you summarize that to make it easier to understand? My education level as to physics is high school completed, but not more (I think; I'm not in the US.) $\endgroup$– LudwikCommented Feb 20, 2016 at 21:05
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$\begingroup$ The effect is tiny, but think of two objects flying in parallel at the same velocity, as in your example, passing by the Earth. The nearer of the two objects would be drawn towards the earth a little faster and curve around the earth a little more. Now imagine they are tied together, the nearer object moving a little faster creates a reverse rotation between the two objects provided they hold together. This works with a sphere as well (er, I think), because the tidal forces on the sphere aren't uniform. I'm going to give this some thought and research though, just to be sure. $\endgroup$– userLTKCommented Feb 20, 2016 at 21:40
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$\begingroup$ @Ludwik giving it some thought, I think I'm wrong on that point. It would work with a linear object not a spherical one. I'm going to delete that part of the answer. $\endgroup$– userLTKCommented Feb 21, 2016 at 0:24