Question

A man of mass $m$ starts falling towards a planet of mass $M$ and radius $R.$ As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces, a spherical shell of negligible thickness of mass $2M/3$ and a point mass $M/3$ at the centre. Change in the force of gravity experienced by the man is

• A.
  $\frac{2}{3}\frac{GMm}{R^2}$
• B. $0$
• C.
  $\frac{1}{3}\frac{GMm}{R^2}$
• D.
  $\frac{4}{3}\frac{GMm}{R^2}$

Answer 1

The correct option is A

$\frac{2}{3}\frac{GMm}{R^2}$

At the surface of the planet, just outside it, the graviational field will be due to the both spherical shell of mass $2M/3$ and point mass $M/3$ and is calculated as,

$E_1=\frac{G\left(2M/3\right)}{R^2}+\frac{G\left(M/3\right)}{R^2}$

$\Rightarrow E_1=\frac{GM}{R^2}$

Just below the surface of the spherical shell the field will be only due to point mass because the field due to spherical shell inside is zero, hence

$E_2=\frac{G\left(M/3\right)}{R^2}=\frac{GM}{3R^2}$

So, change in gravitational field strength will be

$\Delta E=E_1-E_2=\frac{2Gm}{3R^2}$

So, force of gravity experienced by man of mass $m$ is given by

$F_g=m\Delta E=\frac{2GMm}{3R^2}$

Hence, option $\left(a\right)$ is the correct answer.

Why this question? It is absolutely important to be well versed with the field due to a solid sphere and a hollow sphere both inside and outside. This concept is extremely important form a JEE standpoint. $\text{Key Concept:}$ Field due to a hollow sphere of radius $R$ ouside is $-GM/R^2$ and inside is zero.