Question

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

• A. \( \sqrt{\frac{Gm}{2a^3}} \)
• B. \( \sqrt{\frac{Gm}{8a^3}} \)
• C. \( \sqrt{\frac{Gm}{4a^3}} \)
• D. \( \sqrt{\frac{Gm}{a^3}} \)

Hint:

When solving problems involving gravitationally bound circular motion, simplify by expressing the gravitational force in terms of the centripetal force. Keep track of distances, as they directly impact force calculations.

Answer 1

The Correct Option is C

The gravitational force acts as the centripetal force for circular motion: \[ F_{\text{gravity}} = F_{\text{centripetal}}. \] The gravitational force between two particles is expressed as: \[ F = \frac{G M_1 M_2}{d^2}, \] where: - \( G \) is the gravitational constant, - \( M_1 = M_2 = m \), - \( d = 2a \), the distance between the two particles. Substitute \( M_1 = M_2 = m \) and \( d = 2a \): \[ F = \frac{G m^2}{(2a)^2} = \frac{G m^2}{4a^2}. \] For circular motion: \[ F = m \omega^2 r, \] where \( r = a \). Equating \( F \): \[ \frac{G m^2}{4a^2} = m \omega^2 a. \] Simplify the equation: \[ \omega^2 = \frac{G m}{4a^3}. \] Take the square root to find \( \omega \): \[ \omega = \sqrt{\frac{G m}{4a^3}}. \] Final Answer: \[ \boxed{\sqrt{\frac{G m}{4a^3}}} \]