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Why is the gravitational potential at the surface of a hollow sphere equal to the gravitational potential inside the sphere which is $-\frac {Gm}{r}$ ? Does this mean that the potential is the same at every place inside the sphere?

N.B > The hollow sphere generates the Gravitational field

Qmechanic
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user70421
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  • how lengthy and detailed answer are you looking for? this shell theorem can be found in textbooks and wikis, so probably that is a good place to look for. in case you look for something else, let me know. – Zoltan Zimboras Mar 17 '18 at 22:00
  • If you help me with both figure and details it would be helpful . Yeah I saw some books but some critical theories they mentioned that I am not familiar with . – user70421 Mar 17 '18 at 22:05
  • In my answer, I gave you this link: https://en.wikipedia.org/wiki/Shell_theorem#Inside_a_shell , where you can find a a nice figure that explains it. Would be harder for me to draw something better than that. – Zoltan Zimboras Mar 17 '18 at 22:07

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Yes, inside the hollow shell the potential is the same anywhere, thus the net gravitational force on a test mass inside the shell is zero. This the famous shell theorem of classical gravitation theory, which is explained in many textbooks (see for a description at this link). The reason for this is basically Gauss's flux theorem for gravity.

Zoltan Zimboras
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