Question

From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $\left(\frac{R}{2}\right)$ 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. $-\frac{GM}{2R}$
• C. $-\frac{-2GM}{3R}$
• D. $-\frac{-2GM}{R}$

Answer 1

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



Solid sphere is of mass $M$, radius $R$.
Spherical portion removed has radius $\frac{R}{2}$,
Therefore, the volume of the portion removed is 1/8 of the volume of the sphere.
Therefore, its mass is $\frac{M}{8}$.
C is a point at the centre of the cavity at a distance of $\frac{R}{2}$ from the centre of the sphere.
Potential at the centre of cavity
$=V_{\text{solid sphere}}-V_{\text{removed part}}$
$=\frac{-GM[3R^2-(R/2{)}^2]}{2R^3}-\frac{-GM/8\left[3\right(R/2{)}^2-0^2]}{2(R/2{)}^3}$
$=-\frac{GM}{R}$