Four particles of equal mass are moving round a circle of radius r due to their mutual gravitational attraction . Find the angular velocity of each particle .

Answer

To find the angular velocity of each particle in a system of four equal mass particles moving in a circle due to their mutual gravitational attraction, we can follow these steps:

### Step 1: Understand the System
We have four particles of equal mass \( m \) positioned at the corners of a square inscribed in a circle of radius \( r \). Each particle experiences gravitational attraction from the other three particles.

### Step 2: Calculate the Gravitational Force Between Particles
The gravitational force between any two particles is given by Newton's law of gravitation:
\[ F = \frac{G m^2}{d^2} \]
where \( G \) is the gravitational constant and \( d \) is the distance between the two particles.

For two particles at the corners of the square, the distance \( d \) can be calculated using the Pythagorean theorem:
\[ d = \sqrt{(r)^2 + (r)^2} = r\sqrt{2} \]
Thus, the gravitational force between any two particles is:
\[ F = \frac{G m^2}{(r\sqrt{2})^2} = \frac{G m^2}{2r^2} \]

### Step 3: Resolve Forces Acting on Each Particle
Each particle experiences a net gravitational force due to the other three particles. The forces can be resolved into components. The angle between the line connecting the center of the circle to the particles and the line connecting two particles is \( 45^\circ \).

The net force acting towards the center of the circle (centripetal force) can be expressed as:
\[ F_{\text{net}} = 2F \cos(45^\circ) + F \]
Substituting the value of \( F \):
\[ F_{\text{net}} = 2 \left(\frac{G m^2}{2r^2}\right) \frac{1}{\sqrt{2}} + \frac{G m^2}{2r^2} \]
This simplifies to:
\[ F_{\text{net}} = \frac{G m^2}{r^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{2} \right) \]

### Step 4: Apply Centripetal Force Condition
For circular motion, the net gravitational force must equal the centripetal force required to keep the particle in circular motion:
\[ F_{\text{net}} = \frac{m v^2}{r} \]
Setting the two expressions equal gives:
\[ \frac{G m^2}{r^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{2} \right) = \frac{m v^2}{r} \]

### Step 5: Solve for Velocity \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{G m}{r} \left( \frac{1}{\sqrt{2}} + \frac{1}{2} \right) \]

### Step 6: Calculate Angular Velocity \( \omega \)
The angular velocity \( \omega \) is related to linear velocity \( v \) by:
\[ \omega = \frac{v}{r} \]
Substituting for \( v \):
\[ \omega = \frac{1}{r} \sqrt{\frac{G m}{r} \left( \frac{1}{\sqrt{2}} + \frac{1}{2} \right)} \]

### Final Expression
Thus, the angular velocity \( \omega \) of each particle is:
\[ \omega = \sqrt{\frac{G m}{r^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{2} \right)} \]