Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 40 / 3 / i

Use gravitational acceleration $\mathbf{g}=\nabla \psi$. Through the plane with upward unit normal, the signed gravitational flux per unit area of the point mass isThus its downward flux magnitude is $Gmb/(R^2+b^2{)}^{3/2}$. Subtracting a constant from the gravitational potential does not change this flux.

The Kuzmin reflection method gives the even relative potential $\psi =Gm/\sqrt{R^2+(|z|+b{)}^2}$. It is harmonic above and below the plane because the corresponding point masses lie outside the respective half-spaces. At the plane the derivative jump isIntegrating the Poisson equation for Newtonian gravity ${\nabla }^2\psi =-4\pi G\Sigma (R)\delta (z)$ through a thin layer, with $\delta$ the Dirac delta, equates this jump to $-4\pi G\Sigma$. ThereforeThis is a Kuzmin disk. As a check, its total mass is $2\pi {\int }_0^{\infty }R\Sigma (R)dR=m$.

Replace the point mass by $dm=\mu (b)db$ and superpose the Kuzmin disks. Linearity of the Poisson equation for Newtonian gravity gives, whenever the force integral converges,Both choices of potential reference give this same surface density. If the unreferenced potential diverges, perform the reference subtraction in each component before taking the integral limit; the force and the surface density may still be finite.