Question

Two identical particles of mass 1 kg each go round a circle of radius R under the action of their mutual gravitational attraction. The angular speed of each particle is :

• A. $\sqrt{\frac{G}{2R^3}}$
• B. $\frac{1}{2}\sqrt{\frac{G}{R^3}}$
• C. $\frac{1}{2R}\sqrt{\frac{1}{G}}$
• D. $\sqrt{\frac{2G}{R^3}}$

Hint:

In a two-body gravitational problem where particles orbit a common center, remember that the distance in the gravitational force formula ($F=Gm_1m_2/d^2$) is the distance between the two bodies, while the radius in the centripetal force formula ($F=m\omega^2r$) is the distance of one body from the center of rotation.

Answer 1

The Correct Option is B

Let the two identical particles have mass $m=1$ kg.
They move in a circle of radius R, so they are always diametrically opposite each other.
The center of the circle is the center of mass of the two-particle system.
The distance between the two particles is constant and equal to the diameter of the circle, $d = 2R$.
Consider one of the particles. The only force acting on it is the gravitational attraction from the other particle. This force is directed towards the center of the circle and provides the necessary centripetal force.
Gravitational force: $F_g = \frac{G m_1 m_2}{d^2} = \frac{G m^2}{(2R)^2} = \frac{G m^2}{4R^2}$.
Centripetal force: $F_c = m a_c = m \omega^2 R$, where $\omega$ is the angular speed.
Equating the two forces:
$m \omega^2 R = \frac{G m^2}{4R^2}$.
We can cancel one factor of 'm' from both sides.
$\omega^2 R = \frac{G m}{4R^2}$.
Now, solve for $\omega^2$:
$\omega^2 = \frac{G m}{4R^3}$.
Take the square root to find the angular speed $\omega$:
$\omega = \sqrt{\frac{G m}{4R^3}}$.
Given that the mass $m = 1$ kg:
$\omega = \sqrt{\frac{G(1)}{4R^3}} = \sqrt{\frac{G}{4R^3}} = \frac{\sqrt{G}}{2\sqrt{R^3}} = \frac{1}{2}\sqrt{\frac{G}{R^3}}$.