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

The correct option is v = (1/2)√(Gm/R).

To find the speed of each particle, let us analyze the forces acting on the system step-by-step:

Step 1: Understand the geometry of the motion
Two particles, each of mass $m$, are moving in a circular path of radius $R$. Since they move under their mutual gravitational attraction, they must always be diametrically opposite to each other to maintain a stable circular orbit. Therefore, the distance $d$ between the two particles is equal to the diameter of the circle:
$d=2R$

Step 2: Determine the gravitational force
According to Newton's law of universal gravitation, the attractive force $F_g$ between the two particles is given by:
$F_g=\frac{G\cdot m\cdot m}{d^2}$
Substituting $d=2R$ into the equation:
$F_g=\frac{G{m}^2}{{\left(2R\right)}^2}=\frac{G{m}^2}{4R^2}$

Step 3: Relate gravitational force to centripetal force
For a particle to move in a circle of radius $R$ with a constant speed $v$, it requires a centripetal force $F_c$ directed towards the center of the circle:
$F_c=\frac{m{v}^2}{R}$
Here, the mutual gravitational force provides the necessary centripetal force. Therefore, we can set them equal:
$F_c=F_g$
$\frac{m{v}^2}{R}=\frac{G{m}^2}{4R^2}$

Step 4: Solve for the speed v
We can simplify the equation by dividing both sides by $m$ and multiplying both sides by $R$:
$v^2=\frac{Gm}{4R}$
Taking the square root of both sides gives:
$v=\sqrt{\frac{Gm}{4R}}$
$v=\frac{1}{2}\sqrt{\frac{Gm}{R}}$