Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V / 3

For any closed surface enclosing a region $V$, Gauss's law for gravity states that the outward flux of the gravitational field isHere $G$ is the gravitational constant and $M_V$ is the enclosed mass; the negative sign expresses attractive gravity.

Use the gravitational potential sign conventions specified by this paper: $\mathbf{g}=\nabla \phi$. Thus $\phi$ is the negative of the usual Newtonian gravitational potential $\Phi$ defined by $\mathbf{g}=-\nabla \Phi$. For example, setting the value at infinity to zero gives $\phi =GM/r$ for a point mass. The divergence theorem now givesSince $V$ is arbitrary, continuity of the integrands implies the local Poisson's equationConversely, integrating this Poisson's equation over any $V$ and applying the divergence theorem recovers Gauss's law for gravity. At a discontinuity of the mass density, the local equation is interpreted piecewise, or globally in the sense of distributions.

The mass density depends only on $r$, so it has spherical symmetry. For this isolated planet, choose the Newtonian gravitational potential to vanish at infinity, or equivalently choose $\phi =-\Phi$ to do so. There is no imposed external gravitational field. Rotating the mass distribution therefore leaves $\phi$ unchanged: $\phi =\phi (r)$.

The gradient in spherical coordinates consequently has no angular components, and the paper's gravitational potential sign conventions giveAt the centre there is no distinguished radial direction, and spherical symmetry makes $\mathbf{g}(0)=0$.

On a sphere of radius $r$, the radial gravitational field has constant magnitude. Thus Gauss's law for gravity becomeswhere $M(r)$ is the enclosed mass. In the hollow core $M(r)=0$. In the material shell, put $u=r/r_0$ and use the spherical coordinates volume element:Outside the shell the enclosed mass is the totalConsequently,At $r=0$ the displayed value refers to the radial derivative of the constant core value. The expressions match at both shell boundaries. There is no surface sheet of mass, so a jump in the volume mass density does not produce a jump in the gravitational field. The core has zero gravitational field, and the exterior gravitational field equals that of a central point mass $M_{\ast }$, as in the shell theorem.

The core contributes no change in $\phi$, since ${\phi }^{\prime }=0$ there. Integrating the radial gravitational field across the shell, with $u=r/r_0$, givesUsing the supplied antiderivative,HenceThe negative sign agrees with ${\phi }^{\prime }(r)<0$ in the material shell under the paper's gravitational potential sign conventions.