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

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Answer 1

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
$F = \frac{G \times m \times m}{R ^{2}}$ .............. (i)

And centripital force $F = \frac{m \times v ^{2}}{R}$ ................(ii)

From given condition (i) and (ii)
$\frac{G m m}{( 2 R ) ^{2}} = \frac{m v ^{2}}{R}$ ,

$v = \frac{G m}{4 R} = \frac{1}{2} \frac{G m}{R}$

So the speed of the each particle is $v = \frac{1}{2} \frac{G m}{R}$

Answer 2

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 = \frac{G \times m \times m}{R ^{2}}

.............. (i)

And centripital force

F = \frac{m \times v ^{2}}{R}

................(ii)

From given condition (i) and (ii)

\frac{G m m}{( 2 R ) ^{2}} = \frac{m v ^{2}}{R}

,

v = \frac{G m}{4 R} = \frac{1}{2} \frac{G m}{R}

So the speed of the each particle is

v = \frac{1}{2} \frac{G m}{R}