Timeline for Difference in gravity on both sides of the moon
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 20, 2019 at 2:04 | comment | added | David Hammen | This answer is wrong, and for many reasons. (1) The Moon is not in isostatic equilibrium. It has a frozen equatorial bulge and a frozen tidal bulge that reflect the Moon's rotation rate and closeness to the Earth four billion of years ago rather than now, along with marked disparities at the Moon's basins. (2) The Moon doesn't change shape anywhere near fast enough to counter these tidal forces. (3) This is a question about tidal forces, and this doesn't answer that question. | |
| Dec 14, 2019 at 1:40 | comment | added | user15381 | This doesn't seem right to me. The fact that the moon's surface is an equipotential is a statement about the gravitational potential $\phi$. That doesn't tell us anything about constancy of the gravitational field $\nabla\phi$. | |
| Dec 14, 2019 at 0:08 | comment | added | Ilmari Karonen | Note that the perceived gravitational force even on the surface of a body in perfect isostatic equilibrium is not necessarily constant. All we can say is that, by definition, the net force vector is everywhere normal to the surface (or else stuff on the surface would flow/roll along the tangential component). In particular, the perceived gravity on the surface of a rotating body is lower on the equator than at the poles, a fact that has been notably used in science fiction but can actually be measured even on Earth. Tidal forces have a similar effect. | |
| Dec 13, 2019 at 23:36 | comment | added | TimRias | This argument works only because the moon is tidally locked (it is also the mechanism that keeps it locked). This should probably be noted in the answer. | |
| Dec 13, 2019 at 22:36 | history | answered | Mark Olson | CC BY-SA 4.0 |