Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V resit / Question 7 / a

Use the convention that the Newtonian gravitational potential tends to zero at infinity and the gravitational field is $-\nabla \varphi$. A source element of mass $dm$ at position vector ${\mathbf{r}}^{\prime }$ contributes $-Gdm/|\mathbf{r}-{\mathbf{r}}^{\prime }|$, where $G$ is the gravitational constant. For the curve, $dm=\gamma d{s}^{\prime }$, soHere $d{s}^{\prime }$ is arc length along the source curve; the expression is finite away from it. This normalization fixes the otherwise arbitrary additive constant.

For the surface, a source element has mass $dm=\sigma d{S}^{\prime }$. Superposing its Newtonian gravitational potential gives the surface integralHere $d{S}^{\prime }$ is the source surface area element, ${\mathbf{r}}^{\prime }$ the source position vector, and $G$ the gravitational constant. The same zero-at-infinity convention applies.

For a volume source of constant mass density $\rho$, use $dm=\rho d{V}^{\prime }$. Its Newtonian gravitational potential is the volume integralThe source and observation position vectors are ${\mathbf{r}}^{\prime }$ and $\mathbf{r}$, respectively; $G$ is the gravitational constant. With this sign convention positive source mass gives negative Newtonian gravitational potential.