Find gravitational potential inside uniform ball

  • #1

Homework Statement

Find the potential and force on a mass m outside and inside the Earth in terms of g, the acceleration due to gravity, assuming Earth has uniform density and radius R.

Homework Equations

For a mass m, the potential energy of it in the gravitational field of a spherical shell of radius r' and uniform mass distribution m' is given by
γ=const. if m is inside the shell
γ=-Cm'/r if m is outside the shell, where r is the distance from the center of the sphere to the mass and C is a constant

It has the answers provided,
F=mgR^2/r^2, γ=-mgR^2/r
F=mgr/R, γ=mg/2R*(r^2-3R^2).

As a hint, to find the constants, when r=R, F=mg.

The Attempt at a Solution

I managed to get the first part outside the earth, though whether or not the method I used is legitimate I don't know.

I integrated from 0 to R [itex]\int-Cm'/rdr'[/itex] to find the total potential β=-CMR/r, where M is the mass of the Earth (integrating from 0 to R, the sum of the masses of the shells will total M). Taking the negative gradient to find the force, I find F=-CMR/r^2er. At r=R, F=mg, so C=mgR/M. β=-mgR^2/r and F=-mgR^2/r^2er

Getting the inside potential is a little more difficult. I know that it involves integrating from 0 to r [itex]\int-Cm'/rdr'[/itex] and maybe r to R [itex]\int Adr'[/itex]. I've tried two separate methods, where the second term was the integral as above and the second as just a constant in itself. I do know that at r=R, both inside/outside forces and inside/outside potentials should be equal, but neither attempt yielded anything remotely similar to the answer provided in the problem.

Since the method I used in part 1 didn't work for this, I believe it is at least partly a fluke.

Answers and Replies

  • #2
I assume that by inside the Earth they want you to assume that the Earth is a hollow sphere, in that case the force inside is 0, since the gravitational potential is equal to the work done in moving something from infinity to that point and there is no net force inside the Earth what does that make the potential?
  • #3
The potential would be constant in that case. This isn't consistent with the problem, though, since the potential I am supposed to be getting is mg/2R*(r^2-3R^2) which isn't constant.

I just erased some previous work since I thought I discovered an error but it doesn't change anything as far as I can tell.

What I just tried doing was 0 to r [itex]\int-Cm'/rdr'[/itex] = -CM(r^3/R^3) In this case I am not integrating over the whole ball so I can't put down M as the mass, r^3/R^3 is the ratio of volumes, so times M would give the mass of the portion I integrated over. This part of the potential has an extra r in it. I am not certain how to approach integrating over the constant potentials provided by the shells from r to R.

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