I know that the rotation period of the moon equals its revolution period. It's just so astonishing that these 2 values have such a small difference. I mean, what is the probability of these 2 values to be practically the same? I don't believe this to be a coincidence. It's just too much for a coincidence. What could have caused this?
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1$\begingroup$ Just the fact that a question concerns astronomical bodies doesn't necessarily make it inappropriate for a physics site ;-) This is actually a physics (or astrophysics) question. $\endgroup$– David ZCommented Jan 29, 2011 at 0:23
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$\begingroup$ @David: Yep, I said the same thing in a comment on my post. Can't fault the questioner for being deferent though. :) $\endgroup$– NoldorinCommented Jan 29, 2011 at 0:26
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2$\begingroup$ Same question on Astro.SE: astronomy.stackexchange.com/q/16/476 $\endgroup$– Qmechanic ♦Commented Nov 23, 2013 at 19:18
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$\begingroup$ digipro.com/Trials/moon.html $\endgroup$– Dawood ibn KareemCommented Aug 20, 2016 at 11:18
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3 Answers
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This is a gravitational phenomenon known as tidal lock. It is closely related to the phenomenon of tides on Earth, hence the name.
Tidal locking is an effect caused by the gravitational gradient from the near side to the far side of the moon. (That is, the continuous variation of the gravitational field strength across the Moon.) The end result is that the Moon rotates around its own axis with the same period as which it rotates around the Earth, causing the face of one hemisphere always to point towards the Earth.
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2$\begingroup$ wow, didn't expect such a clear answer :) Thanks for clearing this. $\endgroup$– AlexCommented Jan 29, 2011 at 0:07
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$\begingroup$ No problem. It's an good question if you thought of it yourself, as you seem to have done! $\endgroup$– NoldorinCommented Jan 29, 2011 at 0:21
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$\begingroup$ Also, just for reference, this really falls under the category of astrophysics more than astronomy, so you're quite welcome to ask such questions here. In any case, the categories are pretty vague. $\endgroup$– NoldorinCommented Jan 29, 2011 at 0:23
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2$\begingroup$ Remember, that the moon was once much closer to the earth, and the field strength of the tidal compnent scales like the derivitive of the inverse R squared gravitational field (or like inverse R cubed), so the tidal forces on the moon were orders of magnitude greater in the distant past. All the modern tides have to do is maintain the lock, that was easier established in an earlier era. $\endgroup$ Commented Jan 29, 2011 at 5:27
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$\begingroup$ @Omega: This is true. Though to be fair, I suspect that even if the Moon had begun its life at this radius from Earth, it's had billions of years to get into tidal lock anyway... $\endgroup$– NoldorinCommented Jan 29, 2011 at 16:43
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Begin by imagining that the moon isn't quite a perfect sphere. One side is just a little bigger than the other. As the moon rotates, the heavier face will swing around towards the earth a little faster, and it will swing away from the earth a little slower, since it feels a stronger gravitational attraction via its larger mass.
Since gravity is a conservative force, you might think that this continues forever - but it doesn't! The moon isn't totally rigid; rocks can slide around both on the surface and even inside the moon. The heavier lump actually slides through the moon to try to stay facing the earth, which causes friction inside the moon. That friction heats up the rocks, but the heat is slowly lost into space.
Now we have a conservation of energy problem: the rotational energy of the moon is being converted to thermal energy in the rocks, and that thermal energy is being slowly leaked out of the system. The only resolution is that the rotation slowly stops over many millions of years.
Finally, let's return to our assumption of a small mass imbalance. Is this true? Almost certainly - all we need is the fact that the moon isn't a truly perfect sphere.
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4$\begingroup$ Not so great an explanation: a rigid unbalanced sphere won't stop rotating by this mechanism because gravity is conservative. You can't explain it without converting the energy of rotation to some other form (i.e. heat or gravitational potential as the whole body moves away from the primary). $\endgroup$ Commented Jan 29, 2011 at 0:51
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3$\begingroup$ dmckee - quite true, but the moon isn't truly rigid, just very nearly so. The bulging heavier-mass side actually slides through the moon just a tad, and the heat loss from this kinetic process is where the energy goes. $\endgroup$ Commented Jan 29, 2011 at 0:58
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1$\begingroup$ I have given the mod up, but I also think that the effect of rotational energy loss should be made a little clearer, not just as an edit at the end. The problem with the answer as it stands, is that, until the last sentence, you are forced to think of a pendulum, and then to ask "Where is the braking force coming from?" That braking force, aka, energy loss, is the real answer. And while there is friction in space, I suspect, but cannot prove, it would not be enough to slow down the moons rotation significantly, thus this other force should really be expanded upon. $\endgroup$ Commented Jan 29, 2011 at 17:35
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$\begingroup$ @Paul Wagland - Yeah, I was in a rush when I saw dmckee's comment. Answer is now more thoroughly edited. $\endgroup$ Commented Jan 29, 2011 at 18:54
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$\begingroup$ @dmckee: I would think a rigid unbalanced planet would "rotate" much as would e.g. a lopsided cylinder rolling on a plane; if it has enough rotational energy, it will speed up and slow down while making complete rotations, but otherwise it would rock back and forth. Would my intuition be correct? I wonder if there are estimates of the moon's asymmetry which are sufficiently accurate to allow an estimation of what its period would be? $\endgroup$– supercatCommented Jun 9, 2014 at 16:25
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Comment to Spencer Nelson's answer: Consider a spherically symmetric satellite. Let us, for simplicity, consider the idealized situation where all atoms of the satellite are organized into one big crystal held together by covalent bonds, so that all the individual atoms are only allowed to make movements that doesn't break any of their covalent bonds. Even such an idealized satellite can get tidally locked with Earth, because it would still get elastically stretched into an oval shape by tidal forces in the direction of Earth.
answered Mar 10, 2011 at 15:29
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$\begingroup$ ""..where all atoms of the satellite are organized into one big crystal held together by covalent bonds, so that all the individual atoms are only allowed to make movements that doesn't break any of their covalent bonds."" Please explain, how You will achieve the dissipation of that ideal moons rotational energy. $\endgroup$– GeorgCommented Mar 10, 2011 at 15:35
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$\begingroup$ Partial answer @ Georg: Let us, for simplicity, assume a circular orbit and ignore the Sun and other masses (although it is not essential). If the satellite has residual spin (as viewed from the rotating reference frame where the orbit plane and the center-of-masses of satellite and Earth are all fixed), then the oval shape will be out of sync with the direction to Earth (because it takes time to elastically re-shape). The tidal forces will create a net torque on the tilted oval shape, which will slow the spinning. $\endgroup$– Qmechanic ♦Commented Mar 10, 2011 at 19:10
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$\begingroup$ Comment to the answer(v1): Spencer Nelson nowadays calls himself kharybdis. $\endgroup$– Qmechanic ♦Commented Dec 23, 2012 at 22:09
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$\begingroup$ @Qmechanic And is now "spencer nelson". $\endgroup$– JABCommented Nov 9, 2017 at 5:16