A narrow tunnel passes through the centre of a planet of radius R and uniform mass density \(\rho\). A particle is released into the tunnel. What is the angular frequency of oscillation of the particle \((\omega)\) equal to? (Symbols carry their usual meaning)
\(\omega = \sqrt{\dfrac{4\pi G\rho}{3}}\)
Inside a uniform sphere of density \(\rho\), the mass enclosed within radius r is \(M(r) = \dfrac{4}{3}\pi r^3\rho\), so the gravitational field at r is \(g(r) = \dfrac{GM(r)}{r^2} = \dfrac{4}{3}\pi G\rho\, r\), directed toward the centre. The restoring force on a particle of mass m is \(F = -mg(r) = -\left(\dfrac{4}{3}\pi G\rho\right)mr\), which is linear in r and directed opposite to displacement, the signature of simple harmonic motion with \(\omega^2 = \dfrac{4}{3}\pi G\rho\). Hence \(\omega = \sqrt{\dfrac{4\pi G\rho}{3}}\).
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