Gravitational potential at the centre of curv | vedclass

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Question

(D) The gravitational potential $V$ due to a mass element $dm$ at a distance $r$ is given by $V = -\int \frac{G dm}{r}.$ For a hemispherical bowl of r

Vedclass · Accepted Answer

$(A)$ and $(C)$ both

Vedclass · Answer

(D) The gravitational potential $V$ due to a mass element $dm$ at a distance $r$ is given by $V = -\int \frac{G dm}{r}.$ For a hemispherical bowl of radius $R$ and mass $M,$ every point on the surface is at a distance $R$ from the centre. Thus,$V = -\int \frac{G dm}{R} = -\frac{G}{R} \int dm = -\frac{GM}{R}.$ For a thin uniform wire of mass $M$ bent into a semicircle of radius $R,$ every point on the wire is at a distance $R$ from the centre of curvature. Thus,$V = -\int \frac{G dm}{R} = -\frac{GM}{R}.$ Since the distance $R$ is constant for all mass elements in both cases,the potential remains $-GM/R.$ This holds true regardless of the distribution of mass along the wire,as long as every element is at distance $R.$ Therefore,both $(A)$ and $(C)$ are correct.

Gravitational potential at the centre of curvature of a hemispherical bowl of radius $R$ and mass $M$ is $V.$

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