Question

A solid sphere of mass $m$ and radius $r$ is placed inside a hollow thin spherical shell of mass $M$ and radius $R$ as shown in figure. A particle of mass $m^{\prime }$ is placed on the line joining the two centres at a distance $x$ from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if $x>2R$.

• A. $\frac{2Gm{m}^{\prime }}{(x-r{)}^2}+\frac{GM{m}^{\prime }}{(x-R{)}^2}$
• B. $\frac{GM{m}^{\prime }}{(x-R{)}^2}+\frac{2Gm{m}^{\prime }}{(x-r{)}^2}$
• C. $\frac{GM{m}^{\prime }}{(x-R{)}^2}+\frac{GM{m}^{\prime }}{(x-r{)}^2}$
• D. $\frac{GM{m}^{\prime }}{(x-R{)}^2}+\frac{Gm{m}^{\prime }}{(x-r{)}^2}$

Answer 1

The correct option is D $\frac{GM{m}^{\prime }}{(x-R{)}^2}+\frac{Gm{m}^{\prime }}{(x-r{)}^2}$

The particle of mass $m^{\prime }$ is placed at $x>2R$, shown in the figure.


The point is outside of the solid sphere at a distance $x-r$ from its centre and also outside of the spherical shell at a distance $x-R$ from the center of shell.

So, the net gravitational field at the point where $m^{\prime }$ is placed will be due the sum of the fields due to solid sphere and shell.

$E_{net}=E_{sphere}+E_{shell}$

Substituting the values,

$\Rightarrow E_{net}=\frac{Gm}{(x-r{)}^2}+\frac{GM}{(x-R{)}^2}$

So, the net force on $m^{{}^{\prime }}$ due to this field is,

$F=m^{{}^{\prime }}{E}_{net}$

$\Rightarrow F=\frac{Gm{m}^{\prime }}{(x-r{)}^2}+\frac{GM{m}^{\prime }}{(x-R{)}^2}$

Hence, option (d) is the correct answer.

Key Concept: Field due to solid sphere: Outside: $\frac{Gm}{r^2}$ Inside: $\frac{Gm(R-r)}{r^3}$ Field due to a hollow sphere: Outside: $\frac{Gm}{r^2}$ Inside: 0 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 from JEE standpoint.