Q 17. Consider a ‘liquid giant’. A planet made completely of a liquid of density \(\rho\), and of radius \(R\). Consider the density of planet to remain constant with depth. A small spherical object if density \(\rho_s\) is gently dropped from the surface of the planet. Find its position as a function of time.

Idea:

This question is simple, and all we must do is write the force equation at a general point, as shown in the diagram.

\[F = \rho_s V g - \rho V g\]

\[\implies a = g_s \left(1 - {x \over R}\right) \left( 1 - {\rho \over \rho_s }\right)\]

This is the equation for Simple Harmonic Motion.

\[\implies v = \sqrt{2 g_s \left( x - \frac{x^2}{2R} \right)\left( 1 - {\rho \over \rho_s }\right)}\]

\[\implies x = R \left[ 1 - \cos \left( t \sqrt{ {4 \over 3} \pi G \rho \left( 1 - {\rho \over \rho_s }\right)} \right) \right]\]

Further, we can find its time period:

\[T = \sqrt{ { 3 \pi \rho_s \over G \rho(\rho_s - \rho) } }\]