Question

Two particles of equal mass '\(m\)' move in a circle of radius '\(r\)' under the action of their mutual gravitational attraction. The speed of each particle will be ( \(\text{G} =\) Universal gravitational constant)

• A. \(\sqrt{\frac{\text{Gm}}{4r}}\)
• B. \(\sqrt{\frac{\text{Gm}}{r}}\)
• C. \(\sqrt{\frac{\text{Gm}}{2r}}\)
• D. \(\sqrt{\frac{4\text{Gm}}{r}}\)

Hint:

In mutual rotation, distance between masses is \(2r\), but circular radius is \(r\).

Answer 1

The Correct Option is A

Concept:
For two equal masses rotating due to mutual gravitation:

• Both revolve about their common centre of mass
• Distance between them = \(2r\)
• Gravitational force provides centripetal force

Step 1: Write gravitational force. \[ F = \frac{G m^2}{(2r)^2} = \frac{G m^2}{4r^2} \]
Step 2: Write centripetal force. Each mass moves in circle of radius \(r\): \[ F = \frac{mv^2}{r} \]
Step 3: Equate forces. \[ \frac{mv^2}{r} = \frac{G m^2}{4r^2} \] Cancel \(m\): \[ v^2 = \frac{G m}{4r} \]
Step 4: Conclusion. \[ v = \sqrt{\frac{Gm}{4r}} \]