A body is dropped into an imaginary straight tunnel that…

A body is dropped into an imaginary straight tunnel that passes through the centre of the Earth to the other side. Describe its motion.

Step 1: Idea

Inside a uniform Earth the pull is proportional to the distance from the centre (Problem 13). A force proportional to the displacement and directed back toward the equilibrium point is exactly the force of a spring, so the body should oscillate like a mass on a spring.

Step 2: Equation of motion

Let $x$ be the position along the tunnel, measured from the centre. The force always points to the centre: $F = -mg\frac{x}{R}$.

\[ m\ddot x = -\frac{mg}{R}x, \]

\[ \ddot x = -\omega^2x, \quad \omega = \sqrt{\frac{g}{R}}. \]

This is the equation of simple harmonic motion. The body's mass cancels, so every body oscillates the same way.

Step 3: Motion

Released from rest at the surface ($x = R$ at $t = 0$), the solution is

\[ x(t) = R\cos\omega t, \]

\[ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{R}{g}} = 2\pi\sqrt{\frac{6.37 \cdot 10^6}{9.81}}. \]

\[\begin{gathered} T \approx 84.4\ \mathrm{min} \\ t_{\text{through}} = \tfrac{T}{2} \approx 42\ \mathrm{min} \end{gathered}\]

The body accelerates toward the centre, passes it at the greatest speed, slows down, and stops exactly at the opposite surface after 42 minutes. Without friction it then falls back, and the motion repeats forever.

Step 4: Greatest speed

In harmonic motion the greatest speed is amplitude times angular frequency, reached at the centre: $v_{\max} = \omega R = \sqrt{gR}$.

\[v_{\max} \approx 7.9\ \mathrm{km/s}\]

Interestingly, the period does not depend on where the tunnel goes: any straight tunnel through a uniform Earth, not only one through the centre, takes the same 42 minutes.

Check

The period equals that of a satellite skimming the surface (Problem 19): the tunnel motion is the projection of that circular orbit onto a diameter $\checkmark$

Energy: $ \tfrac12 mv_{\max}^2 = \tfrac12 mgR $, the work from Problem 8 $\checkmark$