Gravitational potential of a point within a hollow sphere

[eep]

You may be interested in not the potential, but the gravitational self-energy aka gravitational binding energy of the shell. That would be the amount of energy needed to disassemble the shell.

(If you're not, just skip this part below).

If you imagine pulling the shell apart, you can see that you would need to apply a force (GMdm/r^2) to every mass element dm, from r to infinity.

As the shell expands, the force equation won't change - to calculatse the required force, you can still use the simplification that the force is the same as if all the mass were at the center of the shell.

This intergal gives a binding energy of GM^2/r for a uniform shell.

Contrast this to a sphere, where the binding energy is (3/5)GM^2/r

http://en.wikipedia.org/wiki/Gravitational_binding_energy

The binding energy of a sphere is not just the sum of the binding energies of its shells, because the shells interact with each other.

This is what I've been trying to calculate, because the derivation I saw didn't make any sense to me. Here's what I saw:

http://scienceworld.wolfram.com/physics/SphereGravitationalPotentialEnergy.html

What they're doing is taking a spherical shell of radius r, thickness r+dr, and calculating the potential between that spherical shell and all the mass containing within the shell. What *I* don't get is why can they ignore all the mass outside the shell, and why can they ignore the potential energy due to the interactions of the mass that lies *on* the shell.

I've been trying to do it without somebody just giving me the answer, which is why I didn't ask what the self-binding energy was outright :)

EDIT: Also, in pulling the shell apart, how can it be GMdm/r^2, as all the little bits that make up M are not equi-distance from dm. I also don't see how you can simplify things by saying you can concentrate all the mass at the center of the shell, since we're trying to calculate the force *on* the shell, due to the other parts of the shell, not inside or outside of it.

 

[pervect]

This is what I've been trying to calculate, because the derivation I saw didn't make any sense to me. Here's what I saw:

http://scienceworld.wolfram.com/physics/SphereGravitationalPotentialEnergy.html

What they're doing is taking a spherical shell of radius r, thickness r+dr, and calculating the potential between that spherical shell and all the mass containing within the shell. What *I* don't get is why can they ignore all the mass outside the shell, and why can they ignore the potential energy due to the interactions of the mass that lies *on* the shell.

I've been trying to do it without somebody just giving me the answer, which is why I didn't ask what the self-binding energy was outright :)

EDIT: Also, in pulling the shell apart, how can it be GMdm/r^2, as all the little bits that make up M are not equi-distance from dm. I also don't see how you can simplify things by saying you can concentrate all the mass at the center of the shell, since we're trying to calculate the force *on* the shell, due to the other parts of the shell, not inside or outside of it.

Take a spherical mass. The amount of energy required to remove any small piece of the mass from its surface to infinity will be

E = G*M*dm/r

where r is the distance of the piece of the mass element dm from the center.

This comes from the gravitational potential of a sphere.

Now, to remove a thin spherical shell, the energy required is just

E = G*Mencl*(rho*4*Pi*r^2*dr)/r

from the above formula. This works because the shell is so thin that it doesn't have any significant self energy. (Can you see why the self energy of just the shell is on the order dm^2 when the shell is small, while the energy due to the shell coupling to the rest of the mass is of the order M*dm?)

Mencl is the mass of the enclosed sphere, which is just

Mencl = rho*(4/3)*Pi*r^3

We make r vary downard from R to 0, R being the radius of the mass. As we peel away each layer of the sphere, Mencl drops.Put this all together, mash it around, do the intergal, and you should get the desired result, after back-substituing for the total mass M in terms of density, rho, and radius, R.

 

[eep]

Ah, now I understand. A much better explanation. I thought that perhaps you could say the that due to the thinness of the shell it doesn't have any significant self-energy, but it makes much more sense because the self energy of the shell will be on the order of dm^2. Thank you.