Question

A particle of mass m is placed at a distance of $4R$ from the centre of a huge uniform sphere of mass $M$ and radius $R$. A spherical cavity of diameter $R$ is made in the sphere as shown in the figure. If the gravitational interaction potential energy of the system of mass $m$ and the remaining sphere after making the cavity is $-\frac{\lambda GMm}{28R}$ Find the value of $\lambda$

Answer 1

Since the radius of the cavity is half the radius of the sphere it was made in, the mass removed to make it is one-eighth of the mass of the sphere.

Gravitational Potential Energy $=P{E}_{sys}$

$\Rightarrow {PE}_{sys}={PE}_{duetosphere}-{PE}_{duetoremovedmass}$

${PE}_{duetosphere}=-\frac{GMm}{4R}{PE}_{duetoremovedmass}=-\frac{Gm\left(\frac{M}{8}\right)}{4R-\frac{R}{2}}=-\frac{GMm}{28R}$.

So, ${PE}_{sys}=-\frac{GMm}{4R}+\frac{GMm}{28R}=-\frac{6GMm}{28R}$

Therefore, $\lambda =6$