= Hollow-sphere gravitational transit
{title2=$t_{\rm cavity}=\sqrt{3/(2\pi G\rho)}$}
For a spherical shell of uniform <mass density> $\rho$, inner radius $a$ and outer radius $2a$, suppose a negligible-width radial tunnel permits a small <mass> $m$ to fall from rest at the outer surface. The enclosed <mass> at radius $r$ within the material is $4\pi\rho(r^3-a^3)/3$. The <spherical shell theorem> gives inward <gravitational acceleration> $g(r)=4\pi G\rho(r-a^3/r^2)/3$. Integrating from $a$ to $2a$ gives released <gravitational potential energy> $4\pi G\rho ma^2/3$, hence speed $a\sqrt{8\pi G\rho/3}$ at the cavity. The cavity has zero <gravitational field>, so the diameter transit time is the title expression, independent of $a$ and $m$.
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