Question

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

Answer 1

Answer:

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

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

From given condition (1) and (2)

$$\dfrac{Gmm}{(2R)^2}=\dfrac{mv^2}{R}$$,

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

So the speed of the each particle is $$v=\dfrac{1}

Answer 2

Answer:

=Mv^2/R considering the the The mass of each particle as M and the rotation speed V mutual force of gravitational attraction is equal to GM^2(2r) ^2