Question

The gravitational potential at the centre \( O \) of a uniform solid hemisphere of mass \( M \) and radius \( R \) is Zero \( -\frac{G M}{2 R} \) \( -\frac{3 G M}{2 R} \) \( -\frac{3 G M}{4 R} \)

Answer 1

Let the elemental disc or radius Rsinθ and thickness Rdθ. it mass

dM = \(\frac{M}{\frac23\pi R^3}\pi\)(Rsinθ)²(Rdθsinθ)

dM = \(\frac{3M}{2R^3}\)R²sin²θ Rsinθ dθ

dM = \(\frac{3M}2\)sin³θ dθ

Net potential due to a element under consideration at the centre at the hemisphere

dV = \(-\frac{2GdM}{r(cosec\theta-cot\theta)}\) (potential due to circular phase)

dV = \(\frac{3GMsin^3\theta(cosec\theta-cot\theta)}{(Rsin\theta)}d\theta\) 

dV = \(-\frac{+3GM}{R}\)(sin²θ cosec θ - sin²θ cot θ)

dV = \(-\frac{3GM}R\)(sin θ - cos θ) dθ

there

V = \(-\frac{3GM}R\)\(\int\limits_0^{\pi/2}(sin\theta-cos\theta.sin\theta)d\theta\) 

V = \(-\frac{3GM}R\)  \([-cos\theta+\frac{cos^2\theta}2]_0^{\pi/2}\)

V =  \(-\frac{3GM}R\)