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In Newtons's shell theorem, the net gravitational force is zero inside a hollow sphere, if the gravitational force is proporional to $1/r^2$.

In 2D, the net force is zero inside if the force is proportional to $1/r$.

Is there a general result in $d$ spatial dimensions showing that net force is zero inside if the force is proportional to $1/r^{d-1}$?

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  • $\begingroup$ Getting the logic right is nontrivial. The only reason for taking the force to vary like $1/r^{d-1}$ is because you want Gauss's law to hold. Depending on what you think of as a starting set of assumptions, this doesn't even need to be true. If it is true, it can be considered either trivial or nontrivial. $\endgroup$ – Ben Crowell May 22 '18 at 19:11
  • $\begingroup$ Thanks, I'll read about Gauss law in more details. $\endgroup$ – SrjD May 23 '18 at 12:05
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Yes; this follows from Gauss's law for gravity. Over a closed surface enclosing a mass $M$ we have $\int \vec{g}\cdot d\vec{S}\propto -GM$, where the required proportionality constant is the surface of a unit sphere. If the chosen surface is a sphere containing a spherically symmetric density, such as a uniform density or a point mass, the result follows.

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  • $\begingroup$ Thanks ! It's somehow quite hard to show in n-D by integration and geometrical arguments only, even recursively. $\endgroup$ – SrjD May 22 '18 at 16:26
  • $\begingroup$ @J.G. It’s not that he chose not to, it’s just that new members don’t have enough rep to upvote. $\endgroup$ – knzhou May 22 '18 at 16:48

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