Past exam of the astrophysics course of the University of Oxford / 2021 / 6 / d / Solution

Let the uniform thin galactic disk have outer radius $R_d$ and surface mass density $\Sigma =M/(\pi R_d^2)$, where $M$ is the mass calculated in a. If the redistribution retains the stated outer radius, set $R_d=16\mathrm{kpc}$; otherwise $M$ alone does not determine both $R_d$ and $\Sigma$.

A ring of radius $s$ and width $ds$ has mass $dM=2\pi \Sigma sds$. By axial symmetry its transverse gravitational force components cancel. At signed height $z$, its vertical force on a star of mass $m$ isIntegrating the rings gives, for $z\ne 0$,ThusFor a positive distance above the plane, this is $\mathbf{F}=-2\pi Gm\Sigma [1-z/\sqrt{z^2+R_d^2}]\hat{\mathbf{z}}$, directed toward the disk. With $R_d=16\mathrm{kpc}$, the inferred $\Sigma \simeq 178.3M_{\odot }\mathrm{p}{\mathrm{c}}^{-2}$ sets its normalization.

The far-field expansion gives $F_z\simeq -GMm\mathrm{sgn}(z)/z^2$, as required for a point-like total mass. Close to either face, $F_z\to \mp 2\pi Gm\Sigma$, the infinite-sheet limit. The ideal zero-thickness model has a jump in gravitational field at $z=0$; the formula is understood through its one-sided limits. Its symmetric mean is zero on the plane, and a disk of finite thickness resolves the discontinuity.