Question

A solid sphere of mass $M$ and radius $R$ has a spherical cavity of radius $\frac{R}{2}$ such that the centre of cavity is at a distance $\frac{R}{2}$ from the centre of the sphere. A point mass $m$ is placed inside the cavity at a distance $\frac{R}{4}$ from the centre of sphere. The gravitational force on mass $m$ is

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

Answer 1

The correct option is B $\frac{GMm}{2R^2}$

Given,

Mass of complete sphere $=M$

Let, the density of the sphere be $\rho$.

Since, the point mass (m\) is placed inside the solid sphere.

Gravitational field at the position of mass $m$ due the solid sphere is, $E_1=\frac{GM\left(\frac{R}{4}\right)}{R^3}=\frac{GM}{4R^2}$

Thus, Force acting on mass $m$ due to solid sphere is given by

$F_1=m{E}_1=\frac{GMm}{4R^2}........\left(1\right)$

Mass of cavity , $M_C=\rho \times$Volume of cavity

$\Rightarrow M_C=\frac{M}{\frac{4}{3}\pi R^3}\times \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3$

$\therefore M_C=\frac{M}{8}$

From the figure, it is evident that the mass $m$ is present at a distance of $\frac{R}{4}$ from the center of spherical cavity.

Therefore, Force acting on mass $m$ due to spherical cavity is given by

$F_2=\frac{GMm\left(\frac{R}{4}\right)}{8{\left(\frac{R}{2}\right)}^3}=\frac{GMm}{4R^2}$

By principle of superposition, force acting on the particle of mass $m$ by the sphere with cavity is

${\overrightarrow{F}}^{\prime }={\overrightarrow{F}}_1-{\overrightarrow{F}}_2$


Where,

$F_1\to$ magnitude of force due to whole solid sphere.
$F_2\to$ magnitude of force due to the mass of spherical cavity.

$\Rightarrow |F^{\prime }|=\frac{GMm}{4R^2}+\frac{GMm}{4R^2}=\frac{GMm}{2R^2}\text{units}$

Hence, option(b) is the correct answer.

Alternate approach:

We know from electrostatics, the electric field at a point inside the cavity is given by $$\overrightarrow{E}=\frac{\rho \overrightarrow{a}}{3{\epsilon }_0}$$ Since, we can replace $\frac{1}{4\pi {\epsilon }_0}$ with $G$ we can rewrite the above formula for gravitation as $${\overrightarrow{E}}_g=-\frac{4\pi G\rho \overrightarrow{a}}{3}$$ Since, $\rho =\frac{M}{\frac{4}{3}\pi R^3}$ and $a=\frac{R}{2}$ we get

${\overrightarrow{E}}_g=-\frac{2}{3}\pi G\rho R$

$\Rightarrow |E_g|=\frac{GM}{2R^2}$

$\Rightarrow |F_g|=\frac{GMm}{2R^2}$

Why this question: To make the student familiar with use of the principle of superposition when the mass of body is changed. This method can be applied for any kind of shape of a body.