Question

Two particles of equal mass $\mathrm{2m}$ go around a circle of radius $\mathrm{R/2}$ under the action of their mutual gravitational attraction. The speed of each particle is
A. $\text{ 4}\sqrt{\dfrac{Gmm}{R}}$
B. $\sqrt{\dfrac{4Gm}{R}}$
C. $\sqrt{\dfrac{G m}{2 R}}$
D. $\dfrac{1}{2R}\sqrt{\dfrac{1}{Gm}}$

Answer 1

The gravitational attraction between the particles will contribute to the circular motion of the centrifugal force required. Find the gravitational attraction force that is due to each other. To derive the expression of speed, equate the value with the centrifugal force required.

Formula used:
Newton's law of gravitation gives us the following equation,
$F=G \dfrac{m_{1} m_{2}}{r^{2}}$
Where
$F$ is the gravitational force between the two particles,
$G$ is the universal gravitational constant
$m_{1}$ is the mass of the first particle
$m_{2}$ is the mass of the second particle
$r$ is the distance between two particles.

Complete answer:
With the force of gravitational attraction, ALL objects attract each other. Gravity has become universal. This force of gravitational attraction depends directly on both objects' masses and is inversely proportional to the square of the distance separating their centers.
Given,
Masses of the two particles are $\mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{2m}$
And radius of the two particles are ${r_1} = {r_2} = \dfrac{R}{2}$
So, The Gravitational force of attraction between the particles
$\text{F = }\dfrac{G\times 2m\times 2m}{{{(\dfrac{R}{2})}^{2}}}$...(1)
And, centripetal force $\text{F}=\dfrac{\text{m}\times {{\text{v}}^{2}}}{\text{R}}$- (2)
From given condition (1) and (2)
$\Rightarrow$ $\text{F = }\dfrac{G\times 2m\times 2m}{{{(\dfrac{R}{2})}^{2}}}$=$\text{F}=\dfrac{\text{m}\times {{\text{v}}^{2}}}{\text{R}}$
$\begin{aligned} & \Rightarrow {{\text{V}}^{2}}\text{ = }\frac{16Gm}{R} \\\ & \therefore \text{V}=\text{ 4}\sqrt{\dfrac{Gmm}{R}} \\\ \end{aligned}$
So, the speed of each particle is V=$\text{ 4}\sqrt{\dfrac{Gmm}{R}}$

Note:
Without drawing the free body diagram, this problem was easy enough to complete (FBD). But drawing the FBD is a great practice as we can integrate all the forces in the same diagram. If the effects of any other planets are not assumed, the planetary motions are guided in a similar way. The foundation of planetary motion is this problem.