Question

Four identical particles of equal masses $1kg$ made to move along the circumference of a circle of radius $1m$ under the action of their own mutual gravitational attraction. The speed of each particle will be

• A. $\frac{1}{2}\sqrt{\left(1+2\sqrt{2}\right)G}$
• B. $\sqrt{\left(1+2\sqrt{2}\right)G}$
• C. $\sqrt{\frac{G}{2}\left(2\sqrt{2}-1\right)}$
• D. $\sqrt{\frac{G}{2}\left(1+2\sqrt{2}\right)}$

Answer 1

The correct option is A

$\frac{1}{2}\sqrt{\left(1+2\sqrt{2}\right)G}$

Step 1. Given Data,

Four identical particles of equal masses $M=$$1kg$,

Circle of radius $R=1m$

Step 2. Calculate the speed of each particle,

From the above diagram the net force on mass $M$,

Net Force $F_c=F_1+F_2\cos \left(45°\right)+F_2\cos \left(45°\right)$

$$\begin{gathered} F_1+F_2\cos 45°+F_2\cos 45°=F_c \\ \Rightarrow F_1+2F_2\cos 45°=F_c \end{gathered}$$

Since the four masses are moving in a circle, hence the net force acting on them will be centripetal force.

$F_c=$Centripetal force $=\frac{m{v}^2}{r}$ (where, $m$ is mass, $v$ is velocity and $r$ is radius).

We know that, Gravitational Force $F=\frac{GMm}{R^2}$

(where, $G$ is universal gravitational constant.)

Here, $m=M$

$$\begin{gathered} \Rightarrow F_1+2F_2\cos 45°=F_c \\ \Rightarrow \frac{G{M}^2}{\left(2R^2\right)}+2\left[\frac{G{M}^2}{\left(2R^2\right)}\right]\cos 45°=\frac{M{V}^2}{R} \\ \Rightarrow \frac{G{M}^2}{4R^2}+\frac{G{M}^2}{2\sqrt{2}{R}^2}=\frac{M{V}^2}{R} \\ \Rightarrow \frac{GM}{4R}+\frac{GM}{2\sqrt{2}R}=V^2 \\ \Rightarrow \sqrt{\frac{GM}{4R}+\frac{GM}{2\sqrt{2}R}}=V \\ \Rightarrow \sqrt{\frac{GM}{4R}\left(1+2\sqrt{2}\right)}=V \\ \Rightarrow \frac{1}{2}\sqrt{\frac{GM}{R}\left(1+2\sqrt{2}\right)}=V \end{gathered}$$

Putting values of mass $M=1kg$ and radius $R=1m$,

$\Rightarrow V=\frac{1}{2}\sqrt{G\left(1+2\sqrt{2}\right)}$

Hence, the correct option is $A$.