Question

Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

• A. $\frac{1}{2}\sqrt{\frac{Gm}{R}}$
• B. $\sqrt{\frac{4Gm}{R}}$
• C. $\sqrt{\frac{Gm}{2R}}$
• D. $\frac{1}{2R}\sqrt{\frac{1}{Gm}}$

Answer 1

Answer: (A)

The correct option is A $\frac{1}{2}\sqrt{\frac{Gm}{R}}$

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 $F=\frac{G\times m\times m}{R^2}--------\left(1\right)$

Ans, centripital force $F=\frac{m\times v^2}{R}------\left(2\right)$

From given condition (1) and (2)
$\frac{Gmm}{(2R{)}^2}=\frac{m{v}^2}{R}$,

$v=\frac{Gm}{4R}=\frac{1}{2}\sqrt{\frac{Gm}{R}}$

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