= Solution
For uniform Earth density, the <shell theorem> says that only <mass> within radius $r$ contributes to the <gravitational field> there. Since enclosed <mass> is proportional to $r^3$,
$$
g(r)=\frac{GM(r)}{r^2}=g(R)\frac rR.
$$
At depth $h$, $g_h/g_0=1-h/R$. A <simple pendulum> has small-<amplitude> period $P=2\pi\sqrt{l/g}$, so the <pendulum clock below a uniform-density surface> runs at the fraction $P_0/P_h=\sqrt{1-h/R}$ of its surface-calibrated rate. After a surface-clock hour, its lag is
$$
\boxed{\Delta t=3600\left[1-\sqrt{1-\frac{1000}{6.371\times10^6}}\right]\,\mathrm s\simeq0.283\,\mathrm s.}
$$
The mine clock is slower because gravity is weaker under the stated uniform-density assumption. The expansion $\Delta t\simeq3600h/(2R)$ gives the same result. This is a change in the mechanical pendulum period, not a relativistic clock effect; a realistic Earth density profile would require a different interior-gravity calculation.
Back to article page
