A uniform ring of mass m and radius a is placed directly above a uniform sphere of mass m and of equal to radius. The centre of the ring is at a distance $\sqrt{3}$ a from the centre of the sphere. The gravitational force (Fl exerted by the sphere on the ring is

$\frac{\sqrt{3} G Mm}{8 a^{2}}$

$\frac{2 G M m}{3 a^{2}}$

$\frac{7 G Mm}{\sqrt{2} a^{2}}$

$\frac{3 G Mm}{a^{2}}$

Solution

$\frac{\sqrt{3} G Mm}{8 a^{2}}$

Now, the gravitational field due to the ring at a distance d = $\sqrt{3}$ a on its axis is

E = $\frac{G md}{{(a^{2} + d^{2})}^{\frac{3}{2}}}$

E = $\frac{\sqrt{3} G m a}{{(a^{2} + {(\sqrt{3} a)}^{2})}^{\frac{3}{2}}}$

E = $\frac{\sqrt{3} Gm}{8 a^{2}}$

The force in a particle of mass M placed here is

Force on the sphere = Mass of sphere × gravitational field

F = M E

F = $\frac{\sqrt{3} GMm}{8 a^{2}}$

This is also the force due to the sphere in the ring.

Some More Questions From Gravitation Chapter

For Daily Free Study Material Join wiredfaculty