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 centre of the cavity thus formed is :
($G=$ gravitational constant)

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

Answer 1

Answer: (A)

Gravitational potential at any inside point is given as

$V=-\frac{GM}{2R^3}(3R^2-r^2)$....(i)

for $r=\frac{R}{2}V=-\frac{11GM}{8R}$

Subtracting potential due to cavity of mass $M_c=\frac{M}{8}$ and $R_c=\frac{R}{2}$

Gravitational potential at center is obtained by substituting r=0 in equation (i) $=-\frac{3G{M}_c}{2R_c}$

$V=-\frac{11GM}{8R}-(-\frac{3G{M}_c}{2R_c})=-\frac{11GM}{8R}+\frac{3G\frac{M}{8}}{2\frac{R}{2}}\Rightarrow V=-\frac{GM}{R}$