Problem 83 - A Complete Resource Book in Physics for JEE Main

Question

Two particle of equal mass \(m\) go around a circle of radius \(R\) under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is (A) \(\sqrt{\frac{G m}{R}}\) (B) \(\sqrt{\frac{G m}{4 R}}\) (C) \(\sqrt{\frac{G M}{3 R}}\) (D) \(\sqrt{\frac{G m}{2 R}}\)

Step-by-Step Solution

Answer

The speed of each particle with respect to their centre of mass is (B) \(\sqrt{\frac{G m}{4 R}}\).

Step 1: Write the expressions for gravitational force and centripetal force

The expression for gravitational force between two masses \(m_1\) and \(m_2\) at a distance \(r\) is given by: \[F_g = G \frac{m_1 m_2}{r^2}\] Here, both masses are equal \(m = m_1 = m_2\) and the distance between them is twice the radius of the circle: \(r = 2R\). The expression for centripetal force acting on a mass \(m\) moving at a speed \(v\) in a circle of radius \(R\) is: \[F_c = m \frac{v^2}{R}\]

Step 2: Equate the gravitational force and centripetal force

Next, we equate the gravitational force and centripetal force: \[G \frac{m^2}{(2R)^2} = m \frac{v^2}{R}\]

Step 3: Solve for the speed v

Now, we can solve for the speed \(v\): \[\frac{G m^2}{4 R^2} = m \frac{v^2}{R}\] First, we can cancel one of the mass terms: \[\frac{G m}{4 R^2} = \frac{v^2}{R}\] And then, we can multiply both sides by \(R\) to isolate \(v^2\): \[v^2 = G \frac{m}{4R}\] Finally, we take the square root of both sides to find the speed: \[v = \sqrt{\frac{G m}{4 R}}\] So, the correct answer is (B) \(\sqrt{\frac{G m}{4 R}}\).