Question

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Q. two particles of equal mass go round a circle of radius 'R 'under the action of their mutual gravitational attraction. find the speed of each particle. ?

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Answer 1

Solutions:-

The particles will always remain by metrically opposite so that the force on each particle will be directed along the radius. Consider the motion of one of the particles. the the force on the particle is

$F = \ \frac{G {m}^{2} }{4 {R}^{2} }$

if if the speed is v ,

its acceleration is

$\\ \frac{ {v}^{2} }{R}$

Thus, by Newton's law,

$\frac{g {m}^{2} }{4 {R}^{2} } = \frac{m {v}^{2} }{R} \\ \: \: \: \: \: \: \: \: \: or \: \\ = > v = \sqrt{ \frac{Gm}{4R} }$

Hope it will be helpful to you

Answer 2

Explanation:

Hope u get it.........