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I was wondering if a possible explanation is that, due to the gravity attraction of the earth, the hot, heavy, inner part of the moon, which is richer in heavy
by frankohn 4y ago
I was wondering if a possible explanation is that, due to the gravity attraction of the earth, the hot, heavy, inner part of the moon, which is richer in heavy elements, I guess, like iron is closer to the near-side surface. Since in the far-side the hot, heavy part is too far from the surface the eruption activity is small or non-existing.
Because the moon is tidally locked to the earth the pull always act in the same direction.
I was imagining the moon as a drop of liquid in the vacuum space with a denser core at the center. The heavier core would be at the center because of the moon's own gravity. In the presence of the earth's gravity the denser part of the "liquid drop" will be centered more closely to one side and so the volcanic activity would be strongly biased.
- OscarCunningham 4y agoFor the same reason there are two tides, you would expect the drop to be pulled towards Earth, but also thrown out away from Earth. https://en.wikipedia.org/wiki/Tidal_force https://en.wikipedia.org/wiki/Tidal_force
- frankohn 4y agoNo, because the moon is tidal-locked to the earth so the pull is always in the same direction. Tidal-locked means the moon always show the same face to the earth.
- CorrectHorseBat 4y agoWon't there be centrifugal pull (push?) on the other side? Just like there is tide on the moon side and on the other side
- false-mirror 4y agoYou're referring to the fictitious centrifugal force [0] which is actually a minor change in inertia—very much out shined by the centripetal force of gravity. For example, on earth if you weigh 200 lbs on the North Pole, you'd weigh about 199lbs on the equator. On the moon, this tiny effect would be wiped out by gravity. [0]:https://en.wikipedia.org/wiki/Centrifugal_force https://en.wikipedia.org/wiki/Centrifugal_force
- OscarCunningham 4y agoYour example with the North Pole refers to the centrifugal force from the Earth spinning on its axis. But the relevant centrifugal force here is not the one from the Moon spinning on its axis, but rather the one from the Moon orbiting the Earth. But in any case, you can think about the whole thing without considering centrifugal force. If the Moon wasn't orbiting but rather in freefall towards us, then the Earth would continue to have two tides right up until the moment it hit us. The point is simply that gravity weakens with distance. So the pull of the Earth on the near side of the Moon is slightly stronger than the pull of the Earth on the Moon's centre. And the Earth's pull on the far side of the Moon is slightly weaker than the pull of the Earth on the Moon's centre. So if we consider the near and far regions of the Moon relative to whatever the Moon is doing on net, it will be as though the near side is being pulled towards the Earth and the far side is being pushed away.
- Tagbert 4y agoThe effect is more of a stretch along an axis pointing toward and away from the center of mass of the Earth-Moon pair. The result is like a tide that rises in opposing directions.
- FastMonkey 4y agoThis comment is being downvoted but it's correct. The moon all moves at the same speed, which is the speed at its center of gravity. The face nearest us is moving slightly slower than it should at that distance so it is pulled towards earth. The far side is moving slightly faster, so it is pulled away from the earth. The wiki link explains it pretty well. https://en.m.wikipedia.org/wiki/Tidal_force https://en.m.wikipedia.org/wiki/Tidal_force