Gravitation Inside Earth — Question 8

Question 8

An ideal diameter tunnel turns interior gravity into harmonic motion

A narrow frictionless evacuated diameter tunnel is drilled through a uniform, nonrotating planet of radius $R$ and surface gravity $g_s$. Neglect removed mass. A particle is released from rest at one surface. You may use the solution $x(t)=x_0\cos(\omega t)$ for $\ddot x=-\omega^2x$.

1. Find the motion and the time to reach the opposite surface.

2. Find the speed at the center and explain its mass independence.

Question 8 - Solution

An ideal diameter tunnel turns interior gravity into harmonic motion

(a) Derivation and calculation

Inside a uniform planet, signed displacement $x$ obeys $$\ddot x=-\frac{g_s}{R}x=-\omega^2x,\qquad \omega=\sqrt{g_s/R}.$$ With $x(0)=R$ and $\dot x(0)=0$, $$\boxed{x(t)=R\cos(\omega t)},\qquad \boxed{t_{\mathrm{opposite}}=\frac\pi\omega=\pi\sqrt{R/g_s}}.$$

(b) Result and interpretation

Differentiating gives $$v(t)=-R\omega\sin(\omega t).$$ At the first center crossing, $\omega t=\pi/2$, so $$\boxed{|v_{\mathrm{center}}|=R\omega=\sqrt{g_sR}}.$$ The particle mass cancels from $m\ddot x=-mg_sx/R$, giving the same motion for every test mass.

Physical check

The speed vanishes at both surfaces and is greatest at the center; these conclusions require the ideal uniform, nonrotating planet.