Question

A uniform ring of mass $M$ and radius $R$ is placed directly above a uniform sphere of mass $8M$ and of radius $R$. The center of the ring is at a distance of $d=\sqrt{3}R$ from the center of the sphere. The gravitational attraction between the sphere and ring is

Answer 1

Gravitational field due to ring at a distance

$d=\sqrt{3}R$ on its axis is

$E=\frac{GMd}{{(R^2+d^2)}^{\frac{3}{2}}}$

$=\frac{GM\sqrt{3}R}{{(R^2+{\left(\sqrt{3}R\right)}^2)}^{\frac{3}{2}}}$

$=\frac{GM\sqrt{3}R}{{(R^2+{3R}^2)}^{\frac{3}{2}}}$

$=\frac{GM\sqrt{3}R}{{\left({4R}^2\right)}^{\frac{3}{2}}}$

$=\frac{\sqrt{3}GMR}{8R^3}$

$=\frac{\sqrt{3}GM}{8R^2}$

Force on the sphere $=$ mass of sphere $\times E$

$=8ME$

$=8M\times \frac{\sqrt{3}GM}{8R^2}$

$=\frac{\sqrt{3}G{M}^2}{R^2}$