Understanding gravitational potential due to spherical shell

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SUMMARY

The gravitational potential due to a uniform spherical shell at a point outside the shell is equivalent to the potential created by a particle of the same mass located at the center of the shell. This principle is supported by Gauss's Law, which states that the gravitational flux integral over a surface depends solely on the enclosed mass. For spherically symmetric mass distributions, the calculations simplify significantly, allowing for quick derivation of gravitational potential values.

PREREQUISITES
  • Understanding of gravitational potential and its mathematical representation
  • Familiarity with Gauss's Law for gravity
  • Basic knowledge of spherical symmetry in mass distributions
  • Ability to perform integrals in physics contexts
NEXT STEPS
  • Study Gauss's Law for gravity in detail
  • Explore gravitational potential calculations for various mass distributions
  • Learn about the implications of spherical symmetry in gravitational fields
  • Investigate advanced topics in gravitational physics, such as potential energy in multi-body systems
USEFUL FOR

Students of physics, educators teaching gravitational concepts, and anyone interested in understanding gravitational potential and its derivations through Gauss's Law.

ThinkerCorny
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I know that gravitational potential due to uniform sherical shell at a point outside the shell is equivalent to the potential due to particle of same mass situated at the centre and got proof here http://m.sparknotes.com/physics/gravitation/potential/section3.rhtml. But I was looking for more intuitive proof. Can anyone help me? It really bother me.
 
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For some values of "intuitive", Gauss' Law might do.

https://en.m.wikipedia.org/wiki/Gauss's_law_for_gravity

Short version is that the integral of the gravitational flux over the surface of some volume depends only on the enclosed mass. If you have a spherically symmetric mass distribution and consider the surface of a concentric spherical volume then the answer drops out in a couple of lines (hint: in this case, ##\vec g.d\vec A=gdA##).
 
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