Question

Three particles, each of mass $M$, are situated at the vertices of an equilateral triangle of side $a$. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining their original separation $a$. The initial velocity that should be given to each particle and the time period of circular motion is, respectively.

Answer 1

Step 1: Given data

$\text{Mass}=M$

$\text{side}=a$

$\theta =60°$

Step 2:To Find:

The initial velocity that should be given to each particle and the time period of circular motion.

Step 3: Calculate initial velocity

Gravitational force can be calculated using the formula given as,

$F=G\frac{m_1{m}_2}{r^2}$

Where, $G$ is the gravitational force, $m_1$and $m_2$ are the masses, and $r$ is the distance between the center of the masses.

The resultant Gravitational force acting on each particle is:

$$\begin{gathered} F=\sqrt{{\left(\frac{G{m}^2}{a^2}\right)}^2+{\left(\frac{G{m}^2}{a^2}\right)}^2+2\left(\frac{G{m}^2}{a^2}\right)\left(\frac{G{m}^2}{a^2}\right)\cos 60°} \\ =\sqrt{3}\frac{G{m}^2}{a^2} \end{gathered}$$

As per the figure on the right side, as the particles are expected to move in a circle of radius $r$ while maintaining original separation $a$,

$$\begin{gathered} \frac{{\displaystyle \frac{a}{2}}}{r}=\cos 30° \\ r=\frac{a}{\sqrt{3}} \end{gathered}$$

Centripetal force is balanced by resultant gravitational force.

$$\begin{gathered} \frac{m{v}^2}{r}=\sqrt{3}\frac{G{m}^2}{a^2} \\ \frac{\sqrt{3}m{v}^2}{a}=\sqrt{3}\frac{G{m}^2}{a^2} \\ v=\sqrt{\frac{Gm}{a}} \end{gathered}$$

Step 4: Calculate the time period

Calculate the time period as follows:

$$\begin{gathered} \mathrm{T}=\frac{\mathrm{Distance}}{\mathrm{Velocity}} \\ T=\frac{2\pi r}{v}...\left(1\right) \end{gathered}$$

Substitute the value of $v$ and $r$ in equation (1).

$$\begin{gathered} T=\frac{2\pi r}{v} \\ =\frac{2\pi a\sqrt{a}}{\sqrt{3}\times \sqrt{Gm}} \\ =2\pi \sqrt{\frac{a^3}{3Gm}} \end{gathered}$$

Therefore, the initial velocity and the time period will be $v=\sqrt{\frac{Gm}{a}}$ and $T=2\pi \sqrt{\frac{a^3}{3Gm}}$.