Question

A thin spherical shell of total mass $M$ and radius $R$ is held fixed. There is a small hole in the shell. A mass $m$ is released from rest a distance $R$ from the hole along a line that passes through the hole and also through the centre of the shell. This mass subsequently moves under the gravitational force of the shell. The mass travels in time $x\sqrt{\frac{R^3}{GM}}$ from the hole to the point diametrically opposite. Find $x$.

Answer 1

Lets divide the scenario into two steps:
1) Movement till the hole
2) Movement inside the spherical shell through hole

For step 1:

As energy is conserved, change in kinetic energy + change in potential energy $=0$
$\Rightarrow (K_f-K_i)+(P_f-P_i)=0$

$\Rightarrow \frac{m{v}^2}{2}=\frac{GMm}{R}-\frac{GMm}{2R}$

Solving this we get $v=(\frac{GM}{R}{)}^{0.5}$

Now, for step 2:

Inside the shell, potential is constant and net force is zero.

So, it moves with constant velocity of $v={\left(\frac{GM}{R}\right)}^{0.5}$

$\Rightarrow$ Time taken $=\frac{2R}{v}=2(\frac{R^3}{GM}{)}^{0.5}$

Hence, $x=2$