JEE Physics 2023 Question with Solution

Given: Two identical particles each have mass $m$ and each moves in a circular path of radius $a$ about the common centre of mass.

Find: The angular speed $\omega$ of each particle.

For equal masses, the common centre of mass lies midway between them, so the separation between the particles is

$$2a$$

The gravitational force between the two particles is

$$F = \frac{Gm^2}{(2a)^2} = \frac{Gm^2}{4a^2}$$

This same force provides the centripetal force for each particle. For one particle,

$$F_c = m\omega^2 a$$

Equating gravitational force and centripetal force,

$$m\omega^2 a = \frac{Gm^2}{4a^2}$$

Cancelling $m$,

$$\omega^2 = \frac{Gm}{4a^3}$$

Therefore,

$$\omega = \sqrt{\frac{Gm}{4a^3}}$$

So, the correct option is B.

Given: Each mass $m$ moves in a circle of radius $a$, so the two masses are separated by $2a$.

Find: The angular speed $\omega$.

The mutual gravitational force is

$$\frac{Gm^2}{(2a)^2}$$

and this must equal the centripetal force on one mass,

$$m\omega^2 a$$

So,

$$m\omega^2 a = \frac{Gm^2}{4a^2}$$

which gives

$$\omega = \sqrt{\frac{Gm}{4a^3}}$$

The correct option is B. The key shortcut is to notice immediately that the separation is $2a$, not $a$.