Question

From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $\frac{R}{2}$ is removed as shown in figure. Taking the gravitational potential $V=0$ at $\infty$, the potential at the centre of the cavity thus formed is
($G=$ Gravitational constant)

• A. $-\frac{GM}{2R}$
• B. $-\frac{GM}{R}$
• C. $-\frac{2GM}{3R}$
• D. $-\frac{2GM}{R}$

Answer 1

The correct option is B $-\frac{GM}{R}$

Given that,
Gravitational potential at infinity $=0$
Mass of complete solid sphere $=M$
Radius of complete solid sphere $=R$
Radius of the removed solid sphere$=\frac{R}{2}$

Density of the solid sphere, $\rho =\frac{M}{\frac{4}{3}\pi R^3}$

Mass of the small sphere, $M_s=\frac{M}{\frac{4}{3}\pi R^3}\times \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3$
$\Rightarrow M_s=\frac{M}{8}$



Considering the formation of cavity as a negative mas of radius $\frac{R}{2}$ placed on the solid sphere of mass $M$ and radius $R$.

The distance between centre of bigger sphere and small sphere = $R-\frac{R}{2}=\frac{R}{2}$

The gravitational potential at the point $P$ due to solid sphere is given by
$V_1=-\frac{GM}{2R^3}(3R^2-{\left(\frac{R}{2}\right)}^2)$
$\Rightarrow V_1=-\frac{11GM}{8R}$

The gravitation potential at the centre of the removed sphere is
$V_2=-\frac{3G{M}_s}{2\times \frac{R}{2}}$
$\Rightarrow V_2=-\frac{3\times \frac{GM}{8}}{R}$
$\Rightarrow V_2=-\frac{3GM}{8R}$

But we have assumed that the cavity is formed by negative mass.

Hence,
$V_2=+\frac{3GM}{8R}$

Now applying superposition of potential to get the net potential at the centre of the cavity.

$V_{net}=V_1+V_2$
$\Rightarrow V_{net}=-\frac{11GM}{8R}+\frac{3GM}{8R}$
$V_{net}=-\frac{GM}{R}$

Hence, option (b) is correct.

Tip: In problem involving cavities, always apply the principle of superposition to obtain the potential at the required position. The cavity can be thought as a negative mass is placed there.