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 with respect to their center of

Answer 1

Step 1: Given

1. The equal mass of the particles is $m$
2. The radius of the circle is $R$.

Step 2: Formula used

1. Gravitational force $=\frac{GMm}{R^2}$
2. Centrifugal force $=m{\omega }^{2}R$

Step 3: Solution

Let the speed of rotation of each particle be $v$.

By equating gravitational force with centrifugal force

$\frac{G{m}^2}{{\left(2R\right)}^2}=m{\omega }^{2}R$

$G=$ Gravitational constant

$2R=$The distance between the center of the two particles.

$\omega =$ Angular velocity of the particles

Now,

$$\begin{gathered} \Rightarrow \frac{G{m}^2}{4R^2}=m{\omega }^{2}R \\ \Rightarrow \frac{G{m}^2}{4R^2·mR}={\omega }^2 \\ \Rightarrow \frac{Gm}{4R^3}={\omega }^2 \\ \Rightarrow \omega =\sqrt{\frac{Gm}{4R^3}}......\left(1\right) \end{gathered}$$

From the relation between angular velocity and linear velocity

$v=\omega R$

$$\begin{gathered} \Rightarrow v=\sqrt{\frac{Gm}{4R^3}}\times R\left[Puttingvalueof\omega from\left(1\right)\right] \\ \Rightarrow v=\sqrt{\frac{Gm{R}^2}{4R^3}} \\ \Rightarrow v=\sqrt{\frac{Gm}{4R}} \end{gathered}$$

Hence the required answer is $v=\sqrt{\frac{Gm}{4R}}$.