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I was reading The First Three Minutes and in the book it talked about how if a stone fell in a cave deep in the Earth we could calculate it's motion as if the surface of the Earth is the surface of the cave and only consider the mass closer to the centre of the Earth than the stone. I'm struggling to see how the mass above the stone pulling it up can be ignored.

Qmechanic
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user1070280
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1 Answers1

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This is an application of the shell theorem. The theorem holds that a) a spherically symmetric shell has the same external gravitational field as an equivalent point mass located at its center and b) the shell has zero internal gravitational field. The latter is the consequence of cancellation of gravitational fields. Consider a point off center within the shell. While the shell is closer on one side to the point, there's more mass on the other side. It turns out these two effects cancel. The shell theorem can be derived formally either from Gauss's law for gravity or integration of Newton's law of universal gravitation across the shell.

Therefore, if we assume the mass distribution of Earth is spherically symmetric (true to a good approximation), we can decompose it into a set of spherically symmetric shells. The shells with a radius smaller than the distance from the Earth's center contribute to the gravitational field in the cave. Those with larger radius do not. Hence, the gravitational field in the cave is equivalent to that of the part of the Earth closer to its center than the cave.

FTT
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  • Ah I see. I was struggling to derive if the effect of the shorter distance between the shell and the asymmetrical point and the extra mass on the opposing side really do cancel out. Thanks! – user1070280 Dec 18 '22 at 12:02