= Solution
The tunnel is taken thin enough that removal of its material does not significantly disturb spherical <symmetry>. For $a\le r\le2a$, the enclosed <mass> is
$$
M(r)=\frac{4\pi\rho}{3}(r^3-a^3).
$$
By the <spherical shell theorem>, the inward <gravitational acceleration> has magnitude
$$
g(r)=\frac{GM(r)}{r^2}=\frac{4\pi G\rho}{3}\left(r-\frac{a^3}{r^2}\right).
$$
The released <gravitational potential energy> is the positive <work> done by gravity while the <mass> falls from radius $2a$ to $a$:
$$
\begin{aligned}
U_A-U_B&=m\int_a^{2a}g(r)\,dr\\
&=\frac{4\pi G\rho m}{3}\left[\frac{r^2}{2}+\frac{a^3}{r}\right]_a^{2a}\\
&=\boxed{\frac{4\pi G\rho ma^2}{3}.}
\end{aligned}
$$
By <conservation of mechanical energy>,
$$
\frac12mv_B^2=U_A-U_B,\qquad\boxed{v_B=a\sqrt{\frac{8\pi G\rho}{3}}.}
$$
Inside the cavity, the <spherical shell theorem> gives zero <gravitational field>, so the <velocity> stays constant. The original figure places B and C at opposite ends of a cavity diameter, a distance $2a$ apart. Therefore
$$
\boxed{t_{BC}=\frac{2a}{v_B}=\sqrt{\frac3{2\pi G\rho}}.}
$$
The quantity $G\rho$ has units $\mathrm{s^{-2}}$, so this is a <time>. There are no numerical values for $a$ or $\rho$ in the paper. The transit <time> is independent of $a$ and the falling <mass> $m$. The cavity's <gravitational potential> is constant, not necessarily zero; only its <gradient> vanishes. The calculation is a <hollow-sphere gravitational transit>.
Back to article page
