Question

A uniform ring of mass $m$ and radius $a$ is placed directly above a uniform sphere of mass $M$ and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance $a\sqrt{3}$ as shows in the figure. The gravitational force exerted by the sphere on the ring will be

• A. $\frac{\sqrt{3}GMm}{4a^2}$
• B. $\frac{\sqrt{3}GMm}{8a^2}$
• C. $\frac{\sqrt{5}GMm}{4a^2}$
• D. $\frac{2GMm}{\sqrt{3}{a}^2}$

Answer 1

The correct option is B $\frac{\sqrt{3}GMm}{8a^2}$

For the given problem, the uniform spherical mass can be considered to be a point mass placed at its center.


For the situation shown in above figure, the gravitational field at any point on the ring of mass $m$ will be due to to a single mass $M$ placed at the centre of sphere.

Thus, the force on ring is due to the gravitational force by a point mass $M$ placed at the centre of sphere.

Now, the gravitational field due to ring at the centre of sphere is,

$E_g=\frac{Gmd}{{[a^2+d^2]}^{3/2}}$

$\Rightarrow E_g=\frac{Gm\left(\sqrt{3}a\right)}{{[a^2+(\sqrt{3}a{)}^2]}^{3/2}}$

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

According to the Newton's third law,

$|\text{Force exerted by ring=Force exerted by sphere}|$

Thus, force experienced by sphere of mass $M$ due to ring,

$F_g=M{E}_g$

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

Thus, magnitude of gravitational force exerted by sphere on ring will be $\frac{\sqrt{3}GMm}{8a^2}$

$\text{Alternative Method:}$

Consider a small mass of $dm$ at the periphery of the ring.

The uniform spherical mass can be considered to be a point mass placed at its center


Force on $dm$ due to sphere,

$dF=\frac{GMdm}{(2a{)}^2}$ = $\frac{GMdm}{4a^2}$

Consider a symmetric element $dm$ at the diametrically opposite side as shown in figure.

The component $2F\cos \theta$ will get cancelled and only the component $dF\sin \theta$ will exist.

This will happen for all diametrically opposite elements.

Thus net force
$F=\int dF\sin \theta$

From figure, $\sin \theta =\frac{\sqrt{3}}{2}$

$\Rightarrow F=\int \frac{GM}{4a^2}dm\frac{\sqrt{3}}{2}$

$\Rightarrow F=\frac{\sqrt{3}GM}{8a^2}\int dm$

As, $\int dm=m$

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

Why this question? In such problems by taking advantage of symmetry of sphere, we can asssume it as a point mass at its centre. Then the gravitational field due to ring can be obtained using formula for an axial point.