= Solution
Choose zero <gravitational potential> at infinity. Apply <superposition> to a uniform sphere of positive density and the two voids represented by spheres of negative density. The <shell theorem> makes every component act as a point mass outside the large sphere. With $c_2=(1/2,0,0)$ and $c_3=(-1/4,0,0)$,
$$
\boxed{\Phi(x)=-\frac{4\pi G\rho}{3}\left[\frac1{|x|}-\frac{1/8}{|x-c_2|}-\frac{1/64}{|x-c_3|}\right],\qquad |x|>1.}
$$
The volume factors are the cubes of the three radii.
Inside the smaller void, the large sphere and the removed small sphere have interior fields $-Cx$ and $+C(x-c_3)$, where $C=4\pi G\rho/3$. They combine into the <uniform gravitational field in an off-center spherical cavity>, $-Cc_3=C(1/4,0,0)$. The point is outside the other void's sphere, whose removed mass contributes a repulsive field. Thus
$$
\mathbf g(x)=C\left[\frac14e_1+\frac18\frac{x-c_2}{|x-c_2|^3}\right].
$$
An equilibrium must lie on the line through the two centers. Write $x=qe_1$ with $-1/2<q<0$. Its equation is $1/4-1/[8(1/2-q)^2]=0$, giving
$$
\boxed{x_*=(1/2-1/\sqrt2,0,0).}
$$
This point is inside the smaller void, but it is not stable. Let $d=|x_*-c_2|=1/\sqrt2$. The field's <Jacobian matrix> has eigenvalues
$$
-\frac{C}{4d^3},\qquad\frac{C}{8d^3},\qquad\frac{C}{8d^3}.
$$
The axial displacement is restoring, while transverse displacements accelerate away from equilibrium. Therefore \b[there is no point in that void where a test particle remains stably at rest]. The <cavity equilibrium destabilized by a second void> is a saddle; existence of a zero force is insufficient for stability.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-3-cavity-equilibrium.png]
{title=The two spherical voids and the unstable equilibrium inside the smaller void}
Back to article page
