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)

$$V_{\text{resultant}} = V_{\text{original sphere}} - V_{\text{removed sphere}}$$

The mass of the removed part is $$m = M \times \left( \frac{R/2}{R} \right)^3 = \frac{M}{8}$$

$$V_{\text{in}} = -\frac{GM}{2R^3} (3R^2 - x^2)$$

Substituting $$x = \frac{R}{2}$$:

$$V_{\text{orig}} = -\frac{GM}{2R^3} \left[ 3R^2 - \left(\frac{R}{2}\right)^2 \right] = -\frac{GM}{2R^3} \left( 3R^2 - \frac{R^2}{4} \right)$$

$$V_{\text{orig}} = -\frac{GM}{2R^3} \left( \frac{11R^2}{4} \right) = -\frac{11GM}{8R}$$

$$V_{\text{center}} = -\frac{3Gm}{2r}$$

Substituting $$m = \frac{M}{8}$$ and $$r = \frac{R}{2}$$:

$$V_{\text{rem}} = -\frac{3G(M/8)}{2(R/2)} = -\frac{3GM/8}{R} = -\frac{3GM}{8R}$$

$$V_P = V_{\text{orig}} - V_{\text{rem}}$$

$$V_P = \left( -\frac{11GM}{8R} \right) - \left( -\frac{3GM}{8R} \right)$$

$$V_P = -\frac{GM}{R}$$