Question

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

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

Answer 1

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

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

Potential at the centre of cavity formed
$V_P=V_1-V_2=\frac{-11GM}{8R}-\left(\frac{-3GM}{8R}\right)=\frac{-GM}{R}$