Finding gravitational potential inside solid sphere

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SUMMARY

The discussion focuses on calculating the gravitational potential and force inside and outside a solid sphere, specifically Earth, using the concept of spherical shells. The gravitational potential outside the sphere is derived as -CM/r, while the potential inside is expressed as D - CMr²/R³, where D is a constant related to the mass of the shells enclosing the mass m. The participant initially arrives at two different values for the constant CM, which is resolved by correctly calculating the potential for the case when mass m is inside the Earth. Ultimately, both calculations yield the same value for CM, confirming the consistency of the gravitational potential equations.

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  • Understanding of gravitational potential and force concepts
  • Familiarity with integration techniques in physics
  • Knowledge of spherical symmetry in mass distributions
  • Basic principles of calculus and vector analysis
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  • Explore the concept of gravitational fields in non-uniform density spheres
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demonelite123
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So I am given that the gravitational potential of a mass m a distance r away from the center of a spherical shell with mass m' is -Cm'/r for m outside the shell and constant for m inside ths shell.

I am to find the potentials inside and outside a solid sphere (the earth) of radius R as well as the gravitational force inside and outside on a mass m.

I thought of the Earth as a lot of spherical shells of mass dm so if the mass of the solid sphere is M, i integrated for example -Cdm/r from m = 0 to m = M to get -CM/r outside the sphere. Then taking the negative gradient, i find F = (-CM/r2) er. Then since the gravitational force on the surface of the Earth is -mg, i see that -CM/R2) = -mg or CM = mgR2.

now for inside the sphere, i have the potential to be D - CM'/r where D is a constant (due to the shells that enclose the mass m) and M' is the total mass of the shells that do not enclose the mass m. since the sphere has uniform density, we have M'/M = r3/R3 so the potential is D - CMr2/R3. Taking the negative gradient once again, i get 2CMr/R3 and since the force at the surface is -mg, i get 2CM/R2 = -mg or CM = (-1/2)mgR2.

but earlier, i got that the constant CM = mgR2.

why does my constant CM have 2 different values? have i done something wrong?
 
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i think i figured it out. i didn't calculate the potential correctly for the case that m was inside the earth. while my -CM'/r was correct, the term D was incorrect. after correctly calculating the potential due to all the shells that enclose the mass m, i do indeed get the same value of CM as i did in the other case.
 

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