Question

From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $R/2$ 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. $\frac{2GM}{R}$
• D. $\frac{-GM}{2R}$

Answer 1

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

As we know $v\left(\infty \right)=0$

$v_s=\frac{GM}{2R^3}[3R^2-(\frac{R}{2}{)}^2]=\frac{-11GM}{8R}$

Mass of the removed$=\frac{M}{\frac{4}{3}\pi R^3}\times \frac{4}{3}(\frac{R}{2}{)}^3=\frac{M}{8}$

$r_0=-\frac{3}{2}\times \frac{GM/8}{R}=\frac{11GM}{8R}-\left[\frac{-3GM}{8R}\right]$

Also $\frac{GM}{2R^3}[3R^2-(\frac{R}{2}{)}^2]=\frac{11GM}{8R}$

$v_c=-\frac{3}{2}\times \frac{GM/8}{R}=\frac{11GM}{8R}-\left[\frac{-3GM}{8R}\right]$

$v_p$(final$=V_s-v_c)$$=\frac{11GM}{8R}-\left[\frac{-3GM}{8R}\right]=\frac{-GM}{R}$