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).

Question

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Answer 1

By superposition principle, $v_{1} = \frac{- G M}{2 R ^{3}} [3 R^{2} - (\frac{R}{2})^{2}]$
$= - \frac{1 1 G M}{8 R ^{3}}$
Also, $v_{2} = - \frac{3}{2} \frac{G ( M / 8 )}{( R / 2 )} = \frac{- 3 G M}{8 R}$
The required potential is, $v = v_{1} - v_{2}$
$= - \frac{1 1 G M}{8 R} - (- \frac{3 G M}{8 R})$
$V = - \frac{G M}{R}$

Answer 2

By superposition principle,

v_{1} = \frac{- G M}{2 R ^{3}} [3 R^{2} - (\frac{R}{2})^{2}]

= - \frac{1 1 G M}{8 R ^{3}}

Also,

v_{2} = - \frac{3}{2} \frac{G ( M / 8 )}{( R / 2 )} = \frac{- 3 G M}{8 R}

The required potential is,

v = v_{1} - v_{2}

= - \frac{1 1 G M}{8 R} - (- \frac{3 G M}{8 R})

V = - \frac{G M}{R}