Question 98: Gravitation

Question

Two identical particles each of mass ' $m$ ' go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be :

Accepted Answer

The gravitational force between two particles of mass $m$ separated by a distance $r$ is given by:
$$
F = \frac{Gm^2}{r^2}
$$
where $G$ is the gravitational constant. In this problem, the two particles are moving in a circular orbit of radius $a$ under the influence of their mutual gravitational attraction. Therefore, the gravitational force between the two particles provides the necessary centripetal force to keep them in circular motion.
The centripetal force required for a particle of mass $m$ moving in a circle of radius $a$ with angular speed $\omega$ is given by:
$$
F_{	ext{centripetal}} = m\omega^2a
$$
Setting the gravitational force equal to the centripetal force, we get:
$$
\frac{Gm^2}{r^2} = m\omega^2a
$$
Substituting $r = 2a$ (since the two particles are separated by a distance equal to twice the radius of the circle), we get:
$$
\frac{Gm^2}{(2a)^2} = m\omega^2a
$$
Simplifying, we get:
$$
\omega^2 = \frac{Gm}{4a^3}
$$
Taking the square root of both sides, we get:
$$
\omega = \sqrt{\frac{Gm}{4a^3}}
$$
Therefore, the angular speed of each particle is $\sqrt{\frac{Gm}{4a^3}}$.

Answer

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

Answer

$\sqrt{\frac{G m}{a^{3}}}$

Answer

$\sqrt{\frac{G m}{8 a^{3}}}$

Answer

$\sqrt{\frac{G m}{4 a^{3}}}$

Gravitation

Two identical particles each of mass ' m ' go round a circle of radius a under the action of their mutual gravitational attraction. The angular speed of each particle will be :

Choose the correct answer: