Years ago I was given this problem to do. I couldn't manage it at the time but was given the broad strokes of the solution. I came across it again recently and decided to have a go. Could someone please have a look and let me know if it's right? Many thanks.
Let $S_1$ be the $3$-dimensional sphere of radius $1$ centred at $(0, 0, 0)$, $S_2$ be the sphere of radius $\frac{1}{2}$ centred at $(\frac{1}{2}, 0, 0)$ and $S_3$ be the sphere of radius $\frac{1}{4}$ centred at $(−\frac{1}{4}, 0, 0)$. The eccentrically-shaped planet Zog is composed of rock of uniform density $\rho$ occupying the region within $S_1$ and outside $S_2$ and $S_3$. The regions inside $S_2$ and $S_3$ are empty.
Give an expression for Zog’s gravitational potential at a general coordinate $\textbf{x}$ that is outside $S_1$.
Is there a point in the interior of $S_3$ where a test particle would remain stably at rest?
What I've done:
For a solid sphere of density $\rho$ and radius $R$ the gravitational potential $\phi$ satisfies Poisson's equation $\nabla^2\phi=4\pi\rho G$ for $|\textbf{x}|\leq R$ and $0$ otherwise. The general solution (assuming $\phi\to0$ and $|\textbf{x}|\to\infty$) is
$$ \phi(\textbf{x},R)=\left\{\begin{array}{rcl} -\displaystyle{\frac{2\pi \rho G}{3}(3R^2-|\textbf{x}|^2)}&&\text{if }|\textbf{x}|\leq R \\ -\displaystyle{\frac{4\pi \rho G}{3}\frac{R^3}{|\textbf{x}|}}&&\text{if } |\textbf{x}|\geq R \end{array}\right. $$ Now, linear combinations of solutions are still solutions, so we can write
$$ \phi_{S_1}=\phi_{\text{Zog}}+\phi_{S_2}+\phi_{S_3}\ , $$ where $\phi_{S_1}=\phi(\textbf{x},1)$, $\phi_{S_2}=\phi(\textbf{x}-\frac{1}{2}\textbf{i}, \frac{1}{2})$ and $\phi_{S_3}=\phi(\textbf{x}+\frac{1}{4}\textbf{i},\frac{1}{4})$. Rearranging and simplifying gives
$$ \phi_{\text{Zog}}(\textbf{x})=-\frac{4\pi\rho G}{3}\Big(\frac{1}{|\textbf{x}|}-\frac{1}{8|\textbf{x}-\frac{1}{2}\textbf{i}|}-\frac{1}{64|\textbf{x}+\frac{1}{4}\textbf{i}|}\Big)\text{ , for }|\textbf{x}|\geq 1. $$
For the second part we note that for a test particle to remain stationary, we must have $\textbf{g}=-\nabla\phi=\textbf{0}$. However, within $S_3$ is also within $S_1$ but without $S_2$, so we modify our $\phi_{\text{Zog}}$ to get $$ \phi_{\text{Zog}}(\textbf{x})=-\frac{2\pi\rho G}{3}(3-|\textbf{x}|^2)+\frac{4\pi\rho G}{3}\frac{1}{8|\textbf{x}-\frac{1}{2}\textbf{i}|}+\frac{2\pi\rho G}{3}\bigg(\frac{3}{16}-|\textbf{x}+\tfrac{1}{4}\textbf{i}|^2\bigg)\text{ , for }|\textbf{x}+\tfrac{1}{4}\textbf{i}|\leq\frac{1}{4}\ . $$
Using standard facts $\nabla|\textbf{x}|^2=2\textbf{x}$ and $\nabla\big(\frac{1}{|\textbf{x}|}\big)=-\frac{\textbf{x}}{|\textbf{x}|^3}$ we get $-\nabla\phi=\textbf{0}$ is $$ -\frac{4\pi\rho G}{3}\textbf{x}+\frac{4\pi\rho G}{3}\frac{\textbf{x}-\frac{1}{2}\textbf{i}}{|\textbf{x}-\frac{1}{2}\textbf{i}|^3}+\frac{4\pi\rho G}{3}(\textbf{x}+\tfrac{1}{4}\textbf{i})=\textbf{0}. $$ Tidying it up a bit gives $$ \frac{\textbf{x}-\frac{1}{2}\textbf{i}}{|\textbf{x}-\frac{1}{2}\textbf{i}|^3}+\tfrac{1}{4}\textbf{i}=\textbf{0} $$ This implies that $\textbf{x}=(x,y,z)$ only has non-zero $x$ component. Rearranging some more leads to a cubic satisfied by $x$: $$ x^3-4x+2=0\ , $$ which has no roots for $-\frac{1}{2}\leq x\leq 0$, so there is no point within $S_3$ where there is no net gravitational force.