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  • 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 :<div c

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 :

Option: 1

\sqrt{\frac{4Gm}{r}}


Option: 2

\sqrt{\frac{Gm}{4r}}


Option: 3

\sqrt{\frac{Gm}{r}}


Option: 4

\sqrt{\frac{Gm}{2r}}


Answers (1)

\begin{aligned} & \frac{m v^2}{r}=\frac{G m \cdot m}{(2 r)^2} \\ & \frac{v^2}{r}=\frac{G m}{4 r^2} \\ & V=\sqrt{\frac{G m}{4 r}} \end{aligned}

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Kshitij

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