Question 9 of 223

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 :

A$\sqrt{\dfrac{Gm}{2r}}$B$\sqrt{\dfrac{4Gm}{r}}$C$\sqrt{\dfrac{Gm}{r}}$D$\sqrt{\dfrac{Gm}{4r}}$

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Answer: (D) $\sqrt{\dfrac{Gm}{4r}}$

Let speed of both the particles be $v$.
Then, gravitational force between the particles will provide the required centripetal force.
Therefore,
$\dfrac{G\left(m\right)\left(m\right)}{{\left(2r\right)}^{2}}=\dfrac{m{v}^{2}}{r}$
$\Rightarrow v=\sqrt{\dfrac{Gm}{4r}}$

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