Question

Two particles of equal mass m go round 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. $\frac{1}{2R}\sqrt{\frac{1}{Gm}}$

Hint:

The force that brings an article in circular motion is equal to the centripetal force. The centripetal force acts towards the circle's centre.

Answer 1

The Correct Option is A

The force that brings an article in circular motion is equal to the centripetal force. The centripetal force acts towards the circle's centre. In the question, both particles with the same mass moves in a circle, hence the gravitational force must be equal to the centripetal force.

Complete step-by-step answer:

Mass of each particle = M

Gravitational force is given by \(F=\frac{Gm_{1}m_{2}}{x^{2}}\), 

where x refers to the distance between two bodies. 

In the question, the distance between two bodies is equal to the diameter of the circle, i.e. 2R. 

Putting the values in the equation, we get, \(F=\frac{GM^{2}}{4R^{2}}\) 

Now, this must be equal to the centripetal force which is: \(F=\frac{Mv^{2}}{R}\)

Putting them equally we get,

\(\frac{GM^{2}}{4R^{2}}=\frac{Mv^2}{R}\)

\(⇒v^2=\frac{GM}{4R}\)

\(\therefore v=\frac{1}{2}\sqrt\frac{GM}{R}\)

So, the correct answer is Option A.

Note: The gravitational force is an attractive force. The centripetal force is generated whenever the body moves in a circular motion. The distance between the two bodies will be 2R as they are diametrically opposite at each point in time, otherwise, it will be impossible for them to undergo circular motion under the influence of gravitational force.

Answer 2

Both particles diametrically move in opposite positions along the circular path with a radius of R. The gravitational force provides the required centripetal force.

From the given condition 
\(\frac{Gmm}{\left(2R\right)^{2}}=\frac{mv^{2}}{R}, v=\frac{Gm}{4R}\)

v = \(\frac{1}{2}\sqrt{\frac{Gm}{R}}\)