Past exam of the physics 1 paper of the University of Oxford / 2014 / 2

Let $x$ be the signed distance from the Earth's centre along the polar tunnel, with $x(0)=r_0$ and $\dot{x}(0)=0$. The shell theorem leaves only the enclosed mass $4\pi \rho |x{|}^3/3$ contributing to the gravitational field. ConsequentlyThis is simple harmonic motion with $\omega =\sqrt{g/r_0}$ and $x=r_0\cos (\omega t)$. Crossing the centre takes a quarter-period; reaching the opposite surface takes a half-period:The body's mass cancels. The polar diameter lies on the rotation axis, so the tunnel motion has no additional centrifugal force or Coriolis force from the Earth's rotation. A realistic radial density profile would change the harmonic equation.

A straight diameter at nonzero angle $\chi$ to the rotation axis moves the body toward that axis during the inward half of its trip. With no external axial torque, the total axial angular momentum of the Earth and body is conserved. The body's contribution to the moment of inertia decreases toward the centre, so the Earth spins slightly faster; the reverse happens on the outward half, restoring the original rotation rate at the far surface.

For a body of mass $m$ constrained to follow a tunnel fixed in the Earth, an illustrative angular-momentum balance isHere $I_E$ is the Earth's axial moment of inertia, treated as constant, and $x$ is the signed distance along the diameter. A frictionless guide can still exert normal forces: these supply the transverse acceleration and exchange axial angular momentum with the rotating Earth. Simply allowing an unconstrained body to fall would not keep it inside a narrow nonpolar straight tunnel. The effect vanishes for $\chi =0$ and is largest for an equatorial diameter.