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 provides the centripetal force for circular motion, meaning \(F_{\text{gravity}} = F_{\text{centripetal}}\). The gravitational force between two particles is given by \(F = \frac{G M_1 M_2}{d^2}\), where \( G \) is the gravitational constant, \( M_1 = M_2 = m \), and \( d = 2a \) is the distance between them. Substituting these values yields \(F = \frac{G m^2}{(2a)^2} = \frac{G m^2}{4a^2}\). For circular motion, the centripetal force is \(F = m \omega^2 r\), with \( r = a \). Equating the two expressions for force, we get \(\frac{G m^2}{4a^2} = m \omega^2 a\). Simplifying this equation leads to \(\omega^2 = \frac{G m}{4a^3}\). Taking the square root gives the angular velocity: \(\omega = \sqrt{\frac{G m}{4a^3}}\). Final Answer: \[\boxed{\sqrt{\frac{G m}{4a^3}}}\]