Pendulum clock below a uniform-density surface ($t_{\mathrm{read}}/t=\sqrt{1-h/R}$)

Inside a spherical body of uniform density, the shell theorem gives gravitational acceleration $g(r)=g(R)r/R$. A simple pendulum at depth $h$ therefore has period $P_h=P_0/\sqrt{1-h/R}$. If its clock is calibrated at the surface, its indicated elapsed time is $t\sqrt{1-h/R}$ and it loses $t[1-\sqrt{1-h/R}]\simeq th/(2R)$. This is a mechanical clock-rate change caused by weaker local gravity, not gravitational time dilation. The uniform-density assumption is essential; Earth's actual density profile gives a different interior field.