Gravitational force from spherical shell

Question

Say we have a point mass $\mu$ located at $(0,0,R)$ and a spherical shell (not considering it's volume!) of radius $R$ located in the origin. So we have a particle standing right on top of the sphere and we want to determine the total gravitational force exerted on the particle by the spherical shell.

Since the force is directed along the z-axis, we only have to consider the z-component of the force. Moreover, suppose the shell has homogeneous density $\sigma$. We can then write $dM = \sigma dA = \sigma R^2 \sin \theta d\theta d\phi$. So we have $$ dF_z = \frac {(z-R)G\mu dM}{\sqrt{x^2 + y^2 + (z-R)^2}^{\ 3}} = \sigma \mu G R^2 \frac { (z-R)\sin \theta d\theta d\phi }{ \sqrt{x^2 + y^2 + (z-R)^2}^{\ 3} } $$ We then integrate over the entire sphere: $$ F_z = \iint \sigma \mu G R^2 \frac { (z-R)\sin \theta d\theta d\phi }{ \sqrt{x^2 + y^2 + (z-R)^2}^{\ 3} } $$ This turns out to equal $-2\pi G \sigma \mu$ and $\sigma = \frac M{4\pi R^2}$, where $M$ is the total mass of the spherical shell, which yields $-\frac {GM\mu}{2R^2}$.

What I do not understand why there is a factor 2 in the denominator, should it really be there? Just for the protocol, we get the classical formula if the particle is outside the shell and zero if it is inside the shell.

Answer 1

The factor of two is correct as far as the integral goes; it comes from the unphysical situation of having your test mass exactly on the thin shell. Intuitively, you get the average of the "just outside" result (as if mass is concentrated at the centre) and the "just inside" result of zero.

A more physical thing to do would be to `regulate' the calculation somehow, to find something more meaningful. For example, you could give the shell a small but finite thickness, in which case the force interpolates linearly between the inside and outside results as you pass through it.

Alternatively, you could cut a very small circular hole in the shell where you're passing through. Then the force would smoothly change from zero on the inside to the expected outside result in a distance of order the size of the hole. (If the hole has radius $\epsilon<<R$, at a height $h<<R$ above the surface the force will be modified by a factor $\frac{h}{\sqrt{h^2+\epsilon ^2}}+1$).