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Question

Two particles of equal mass (m) each move in a circle of radius (r) under the action of their mutual gravitational attraction the speed of the particle is

A
$$ \dfrac{Gm}{4r} $$
B
$$ \sqrt{\dfrac{Gm}{r}} $$
C
$$ \dfrac{4 Gm}{2r} $$
D
$$ \dfrac{1}{2} \sqrt{\dfrac{Gm}{r}} $$
Solution
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Correct option is B. $$ \dfrac{1}{2} \sqrt{\dfrac{Gm}{r}} $$
Given
Masses of the two particles are $$m_{1} = m_{2} = m $$
And radius of the two particles are $$r_{1} = r_{2} = R $$
So, the Gravitational force of attraction between the particles
$$F = \dfrac{G \times m \times m}{R^{2}} $$ .............. (i)
And centripital force $$F = \dfrac{m \times v^{2}}{R} $$ ................(ii)
From given condition (i) and (ii)
$$ \dfrac{Gmm}{(2R)^{2}} = \dfrac{mv^{2}}{R} $$,
$$v = \dfrac{Gm}{4R} = \dfrac{1}{2} \sqrt{\dfrac{Gm}{R}} $$
So the speed of the each particle is $$v = \dfrac{1}{2} \sqrt{\dfrac{Gm}{R}} $$

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