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- v=\[\dfrac{1}{2R}\sqrt{\dfrac{1}{GM}}\]
B- \[v=\sqrt{\dfrac{GM}{2R}}\]
C- \[v=\dfrac{1}{2}\sqrt{\dfrac{GM}{R}}\]
D- \[v=\sqrt{\dfrac{4GM}{R}}\]

Answer 1

Hint: Whenever a particle moves in a circular motion then the force which is responsible to bring it in this motion must be equal to the centripetal force. Centripetal force always acts towards the centre of the circle. Here, both the particles having the same mass move in a circle under the influence of mutual gravitational force, so it must be equal to the centripetal force.

Complete step by step answer:
Mass of each particle=M
Gravitational force is given by: \[F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{x}^{2}}}\], where x is the distance between the two bodies. Here it will be equal to the diameter of the circle, 2R. Putting the values we get, \[F=\dfrac{G{{M}^{2}}}{4{{R}^{2}}}\]
But this must be equal to the centripetal force which is: \[F=\dfrac{M{{v}^{2}}}{R}\]
Thus, putting them equal we get,
$ \dfrac{G{{M}^{2}}}{4{{R}^{2}}}=\dfrac{M{{v}^{2}}}{R} \\ $
$\Rightarrow {{v}^{2}}=\dfrac{GM}{4R} \\ $
$\therefore v=\dfrac{1}{2}\sqrt{\dfrac{GM}{R}} \\ $

So, the correct answer is “Option C”.

Note:
The gravitational force is always an attractive force. Here the bodies were moving under the influence of their own gravitational attraction. And the centripetal force comes into play whenever the body moves in a circular motion. Here the distance between the two will be 2R because they are always diametrically opposite, if they are not then it will be impossible for them to undergo circular motion and that too under the influence of gravitational force. Also, when we calculate the centripetal force, the radius of the circle is to be taken.