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1$\begingroup$ Whats a tidal leash? The same thing as a tide? $\endgroup$– Confuse-ray30Commented Jun 20 at 13:35
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$\begingroup$ The tides are caused by the moon's gravity pulling on the sea causing it to bulge slightly at the part of the Earth facing the moon* As the Earth rotates this bulge moves over the surface of the Earth. Equal and opposite, if the mopon pulls this bulge, then this bulge pulls the moon. That's waht I mean by a tidal leash * Gross simplification. To be technical there is a bulge facing the moon as a result of stronger lunar gravity, and away from the moon as a result of string centrifugal force, but this is not relevant to the question $\endgroup$– Francis CagneyCommented Jun 20 at 14:34
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$\begingroup$ Then your argument is flawed since gravitational pull is independent of what it pulls on. So even if the tidal effect (in amplitude) on mountains and other solid matter is smaller, the force acting on the moon is the same. (Barring some effects from the fact matter moves a tiny bit nearer). So no, it shouldnt change of any significance $\endgroup$– Confuse-ray30Commented Jun 20 at 14:45
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1$\begingroup$ Work cannot be done without force, force can act without work. Also, the force acting on stuff really only depends on distance and mass, so since frozen water is as "heavy" as liquid and neglecting the slight increase in height, the force is the same. Even including friction and whatnot. $\endgroup$– Confuse-ray30Commented Jun 20 at 14:53
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1$\begingroup$ Two additions: The first is to clarify my first statement: If you hold something up (i.e. you is not necessarily you as a person) you exert force on an object without doing any work. The second concerns the "no work: statement: You do increase some energy, imagine the ice as being springs on molecules. Then the slight stretching of ice is work, given by (a toy model) $1/2 kx^2$. Where $k$ is some spring constant, x the displacement. $\endgroup$– Confuse-ray30Commented Jun 20 at 14:57
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