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It makes sense that tidal forces lead to tidal locking. Celestial bodies have varying densities and shapes, so some orientations have a lower gravitational potential, and eventually the tendency will be for the body to settle into the orientation of lowest potential, similar to how dice with imperfections, when buoyant in water, will roll until the denser side faces down.

For the Earth and the Moon, it seems like the story is that the denser side faces earth:

The mass of the Moon is not evenly distributed; mass concentrations, called Mascons, lie beneath many of the lunar basins, and the center of mass of the Moon is displaced several kilometers towards the Earth.

However, unlike the die floating in water, in a system of two orbiting celestial bodies (like the Earth and Moon) each body has two gravitational wells. These two areas of lower gravitational potential cause, for instance, the two high tides on earth. The explanation straightforward: Bodies oribiting around their mutual barycenter experience a centripetal acceleration which is equal to the gravitational force only at the center of mass. There is a net inward force on the near side of each body (where the acceleration is less than the gravitational force), and a net outward force on the outside (where the acceleration is greater than the gravitational force).

Since there are two areas of locally lower gravitational potential, it seems equally likely (or at least quite possible) that the mass concentrations of the Moon would end up on the far side of the Earth. Is this reasoning correct? Are there examples of such systems?

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  • $\begingroup$ Follow-up: The causality also partially goes in the reverse direction, where the tidal locking caused a shift that promoted more volcanic activity on the near side [0] "As the Moon became tidally locked to the much bigger Earth, the entire core of the Moon shifted slightly closer to the Earth, and consequently closer to the near side’s crust." My above question predates this effect, but it raises a natural follow-up: If tidal forces go in both directions, why would the moon's core move toward earth and not away? Was that also chance? [0]: futurism.com/the-moons-fixed-face $\endgroup$
    – kotoole
    May 20 at 13:23
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    $\begingroup$ I don't think it's equally likely for mascons to end up on the far side. If you look at my diagram of the combined gravitational and centrifugal pseudo-potential in the co-rotating frame here: space.stackexchange.com/a/57679/38535 you can see that the pseudo-potential is slightly lower on the L1 (near) side than on the L2 (far) side. $\endgroup$
    – PM 2Ring
    May 20 at 13:44
  • $\begingroup$ BTW, you can easily hack that script to print potentials. Eg, insert for u in (0.9, 1.1): print(-func(x=u, y=0)) anywhere after func is defined. $\endgroup$
    – PM 2Ring
    May 20 at 14:06
  • $\begingroup$ I suspect the case of "dense side farther away" is one of unstable equilibrium. Any tiny perturbation will lead to rotation & rocking until the dense side is near. $\endgroup$ May 23 at 12:33
  • $\begingroup$ @CarlWitthoft I don't think that's true. It would be for dice in water but for bodies in free fall, having the densest part maximally far from earth should be the locally lowest potential configuration. pm2ring points out that the potential on the near side is actually globally lower, but both should be local minima. $\endgroup$
    – kotoole
    May 23 at 12:41

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a system of two orbiting celestial bodies (like the earth and moon) has two gravitational wells: One with a greater inward force on the inside (where the rotational acceleration is less than the gravitational force), and one with greater outward force on the outside (where the rotational acceleration is greater than the gravitational force). This is what causes two high tides on earth.

I don't think this is correct.

The tidal forces exerted on Earth by the Sun are caused by the gradient of the Sun's gravitational field. The Sun's gravity is slightly greater closer to the Sun and slightly less on the far side. These tidal forces have nothing to do with rotational accelerations. A non-rotating Earth-like planet falling directly into the Sun would have the same tidal forces at 1 AU that we experience on our rotating, orbiting Earth.

Suppose there was a tidally locked moon that was more dense on the far side. A gravitational asymmetry between two hemispheres of any plane passing through the moon's center and planet's center would result in gravitational torque, gradually spinning the denser side of the moon towards the planet. Perhaps unintuitively, tides exert no torques on tidally locked bodies, since there is no tidal bulge lag. So tidal torques might slow the spin, but not stop it. Eventually the moon's center of gravity would be closer to the planet than the moon's geometric center, like our Moon with respect to Earth.

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  • $\begingroup$ In that quote I'm talking about the revolutions of the earth and moon around their mutual barycenter. Their self-rotations of course don't matter, I will edit to clarify. This revolution, at the center of mass, is equal to the force of gravity, but on the far side the centrifugal acceleration is greater than gravity, just as it's less than gravity on the near side. Both sides are lower-potential than the center. On earth this causes two tidal bulges, see e.g. noc.ac.uk/files/documents/business/Double-Bulge-Explanation.pdf $\endgroup$
    – kotoole
    May 20 at 21:57
  • $\begingroup$ The revolutions of the Earth and the Moon around their mutual barycenter also don't matter. If the Earth and Moon were in free fall towards one another, the same tidal forces would be in effect at the same distances. $\endgroup$
    – Connor Garcia
    May 21 at 14:29
  • $\begingroup$ Sure, but in reality the acceleration is in a circular path so I used that description. The difference shouldn't impact the question. $\endgroup$
    – kotoole
    May 22 at 1:53
  • $\begingroup$ To clarify, if you were in free fall holding two masses on springs above and below you, you would see each spring stretch from tidal forces: the one below you because gravity is greater below, and the one above you because gravity is less above. The weights would tend to each sit stable in place, along the axis of gravitational pull, but in opposite directions. Your 100% right these are the same forces, and this supports my point. $\endgroup$
    – kotoole
    May 23 at 12:47

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