Question

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

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

Answer 1

Answer: (D)

Potential at point O (Centre of cavity) before removing the spherical portion,

$V_1=\frac{-GM}{2R^3}\left(3R^2-{\left(\frac{R}{2}\right)}^2\right)=\frac{-GM}{2R^3}\left(3R^2-{\left(\frac{R}{4}\right)}^2\right)=\frac{-11GM}{8R}$
Mass of spherical portion to be removed,
$M^{\prime }=\frac{M{V}^{\prime }}{V}=\frac{m\frac{4\pi }{3}{\left(\frac{R}{2}\right)}^3}{\frac{4\pi }{3}{R}^3}=\frac{M}{8}$
Potential at point P due to spherical portion to be removed
$V_2=\frac{-3G{M}^{\prime }}{2R^{\prime }}=\frac{-3G(M/8)}{2(R/2)}=\frac{-3GM}{8R}$
Potential at the centre of cavity formed
$V_P=V_1-V_2=\frac{-11GM}{8R}-\left(\frac{-3GM}{8R}\right)=\frac{-GM}{R}$