Three particles, each of the 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 the original mutual separation `a`. Find the initial velocity that should be given to each particle and also the time period of the circular motion. `(F=(Gm_(1)m_(2))/(r^(2)))`

Explanation

To solve the problem of finding the initial velocity and time period of the circular motion of three particles situated at the vertices of an equilateral triangle, we will follow these steps:

### Step 1: Understand the Forces Acting on the Particles
Each particle experiences gravitational forces due to the other two particles. The gravitational force between any two particles is given by:

\[ F = \frac{G m^2}{a^2} \]

where \( G \) is the gravitational constant, \( m \) is the mass of each particle, and \( a \) is the distance between the particles.

### Step 2: Calculate the Resultant Gravitational Force
Since the particles are at the vertices of an equilateral triangle, the angle between the forces acting on one particle due to the other two is \( 60^\circ \). Therefore, we can calculate the resultant force \( F_r \) acting on one particle using the law of cosines:

\[ F_r = \sqrt{F^2 + F^2 + 2F^2 \cos(60^\circ)} = \sqrt{2F^2 + 2F^2 \cdot \frac{1}{2}} = \sqrt{3F^2} = F \sqrt{3} \]

### Step 3: Relate the Resultant Force to Centripetal Force
The resultant gravitational force \( F_r \) provides the necessary centripetal force for circular motion. The centripetal force \( F_c \) required for circular motion is given by:

\[ F_c = \frac{mv^2}{R} \]

where \( v \) is the velocity of the particle and \( R \) is the radius of the circular path.

### Step 4: Set the Forces Equal
Since the gravitational force provides the centripetal force, we can set them equal:

\[ F \sqrt{3} = \frac{mv^2}{R} \]

Substituting \( F = \frac{G m^2}{a^2} \):

\[ \frac{G m^2}{a^2} \sqrt{3} = \frac{mv^2}{R} \]

### Step 5: Simplify the Equation
We can cancel one \( m \) from both sides:

\[ \frac{G m \sqrt{3}}{a^2} = \frac{v^2}{R} \]

Rearranging gives:

\[ v^2 = \frac{G m \sqrt{3} R}{a^2} \]

### Step 6: Find the Relation Between \( R \) and \( a \)
To find \( R \), we can drop a perpendicular from the center of the triangle to one side. The height of the triangle is given by:

\[ h = \frac{\sqrt{3}}{2} a \]

The radius \( R \) of the circular path is:

\[ R = \frac{a}{\sqrt{3}} \]

### Step 7: Substitute \( R \) into the Velocity Equation
Substituting \( R \) into the velocity equation:

\[ v^2 = \frac{G m \sqrt{3} \left(\frac{a}{\sqrt{3}}\right)}{a^2} = \frac{G m}{a} \]

Thus, the initial velocity \( v \) is:

\[ v = \sqrt{\frac{G m}{a}} \]

### Step 8: Calculate the Time Period of Circular Motion
The time period \( T \) of circular motion is given by:

\[ T = \frac{2 \pi R}{v} \]

Substituting \( R \) and \( v \):

\[ T = \frac{2 \pi \left(\frac{a}{\sqrt{3}}\right)}{\sqrt{\frac{G m}{a}}} \]

Simplifying gives:

\[ T = \frac{2 \pi a^{3/2}}{\sqrt{3 G m}} \]

### Final Results
- The initial velocity \( v \) that should be given to each particle is:

\[ v = \sqrt{\frac{G m}{a}} \]

- The time period \( T \) of the circular motion is:

\[ T = \frac{2 \pi a^{3/2}}{\sqrt{3 G m}} \]