# Why is the gravitational potential inside a hollow sphere same as that of the gravitational potential on the surface of the hollow sphere? [duplicate]

Gravitational potential inside a hollow sphere is given by $$V(r)=\frac{-Gm}{R}$$ Why is it the same as the gravitational potential on the surface of the hollow sphere, which is given by $$\frac{-Gm}{R}$$ too? Where as in the case of a solid sphere the gravitational potential is not constant inside the solid sphere, Why is it so?

• By "hollow sphere" do you mean an infinitely thin spherical shell, or a spherical shell with a non-zero thickness? The two have different internal potentials. Commented May 8, 2022 at 6:34
• Infinitely thin spherical shell. Commented May 8, 2022 at 6:52
• g = 0 inside the sphere, thus no extra work is needed to move inside the sphere. So the potential is the same as on the surface. Commented May 8, 2022 at 6:55

The 'Shell theorem' states that inside a hollow sphere there is no net gravitational pull. This is because the pull of all the parts of the surface cancel each other out perfectly. This is not the case for the solid sphere, because all the masses between a test object and the centre don't cancel eachother out and there is a net gravitational pull towards the centre.

You can derive this but first of all we can note three possible potential difference possibilities. It is instructive to think about what this would mean for the gravitational force inside the hollow sphere

Three possibilities for potential

The potential inside the hollow sphere can either be:

1. Lower than the surface: $$\Delta V< 0$$

This would mean there would be a potential difference between the inside and the surface. This would result a mass to get pulled towards the surface, since $$F = - \Delta V/\Delta r$$. Thi is not entirely unintuitive, however because of the shell theorem this will be not true

1. Equal to the surface $$\Delta V= 0$$

This would mean there would be no potential difference, and thus no force. The mass would stay where it is, since $$F = - 0/\Delta r = 0$$. This corresponds to the shell theorem, and this is why there is no potential difference

1. Higher than surface $$\Delta V > 0$$

This would mean the mass gets repelled by the surface and driven to centre point. This is not the case.

Graphs of the gravitational potential

Graphically the potential of a hollow sphere looks like this:

For a solid sphere it is:

Gauss' law for gravitational field

You can use Gauss' law for gravity to quickly show that there is no net gravitational pull. It is more commonly used for the electric field, however it is also valid for the gravitational field. It basically says that if you have a closed bag in space, the net gravitational field per unit area is proportional to the mass enclosed:

$$\int \vec{g(r)} \cdot dA = 4 \pi G M_{\text{enclosed}}$$ If we choose a spherical surface it simplifies to

$$\vec{g(r)} 4 \pi r^2 = 4 \pi G M_{\text{enclosed}}$$

For a hollow sphere $$M_{\text{enclosed}}=0$$, for a solid sphere $$M_{\text{enclosed}} \neq 0 \implies \vec{g(r)} \neq 0$$

The gravitational potential from the gravitational field is given by:

$$V(r) = \int \vec{g(r)} dr$$

• Yeah this clarified my doubt, Thanks a lot. Commented May 8, 2022 at 6:50

Suppose one point on the shell is of potential V0. Due to spherical symmetry(with constant mass density throughtout the shell), every other point on the shell will be of that same potential. Now, the gravitation potential obeys laplace's equation inside the spherical shell due to absence of mass, thus it cannot have amaximum or a minimum inside. But if it's different at one point within, it's going to have either a maximum or a minimum. So the potential inside the hollow spherical shell has to be the same as that on ita surface.

• got it, thanks. Commented May 8, 2022 at 6:51