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Question

Physics Question on Gravitation

Two identical particles each of mass m, move in circular path due to their own mutual gravitational force. Find the velocity of the particle if the radius of the circular path is a.

A

4Gma\frac{\sqrt{4Gm}}{a}

B

Gm2a\frac{\sqrt{Gm}}{2a}

C

2Gma\frac{\sqrt{2Gm}}{a}

D

Gm4a\frac{\sqrt{Gm}}{4a}

Answer

Gm4a\frac{\sqrt{Gm}}{4a}

Explanation

Solution

Two identical particles each of mass m,

mv2a=Gm24a2\frac{mv^2}{a}=\frac{Gm^2}{4a^2}

⇒v=Gm4a\Rightarrow v=\sqrt{\frac{Gm}{4a}}

Hence the correct option is (D):\text{Hence the correct option is (D):} Gm4a\frac{\sqrt{Gm}}{4a}