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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)

Answers (1)

@Mannika

Outside the surface

V=-\frac{GM}{r}

on the surface

V_{surface}=-\frac{GM}{R}

Inside the surface 

r<R

V=-\frac{GM}{2R}\left[3-\left ( \frac{r}{R} \right )^{2}\right]

\left (3R^{2}-\left ( \frac{R}{2} \right )^{2} \right )

=\frac{-11GM}{8R}

Mass of spherical portion to be removed

=\frac{M}{8}

Potential at point P due to spehrical portion to be removed =\frac{-3GM}{8R}

=\frac{-11GM}{8R}-(\frac{-3GM}{8R})=\frac{-GM}{R}

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