From a solid sphere of mass M and radius R, a spherical portion of...

Central idea is, consider the cavity as negative a mass and apply the superposition of gravitational potential.

Consider the cavity formed in a solid sphere as shown in the figure.

The gravitational potential at infinite distance,

$V = \dfrac{- GM}{r}$

$⇒ V\left(\infty\right) = 0$

According to the question, we can write potential at an internal point $P$ due to complete solid sphere,

${V}_{s} = - \dfrac{GM}{2{R}^{3}}\left[3{R}^{2} - {\left(\dfrac{R}{2}\right)}^{2}\right]$

$= \dfrac{- GM}{2{R}^{3}}\left[3{R}^{2} - \dfrac{{R}^{2}}{4}\right]$

$= \dfrac{- GM}{2{R}^{3}}\left[\dfrac{11{R}^{2}}{4}\right] = \dfrac{- 11GM}{8R}$

Mass of removed part

$= \dfrac{M}{\dfrac{4}{3} \times \pi{R}^{3}} \times \dfrac{4}{3}\pi{\left(\dfrac{R}{2}\right)}^{3} = \dfrac{M}{8}$

The potential at a point $P$ due to removed part

${V}_{C} = \dfrac{- 3}{2} \times \dfrac{\left(\dfrac{GM}{8}\right)}{R} = \dfrac{- 3GM}{8R}$

Thus, potential due to the remaining part at the point $P\text{,}$

${V}_{P} = {V}_{s} - {V}_{c} = \dfrac{- 11GM}{8R} - \left(- \dfrac{3GM}{8R}\right)$

$\dfrac{\left(- 11 + 3\right)GM}{8R} = \dfrac{- GM}{R}$