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)))`

Options

A. `2pisqrt((a^(1))/(3Gm))`
B. `2pisqrt((a^(2))/(3Gm))`
C. `2pisqrt((a^(3))/(2Gm))`
D. `2pisqrt((a^(3))/(3Gm))`

Explanation

To solve the problem, we need to find the initial velocity that should be given to each particle so that they move in a circular path while maintaining the original mutual separation `a`. We will also find the time period of this circular motion.

### Step 1: Understanding the Configuration
We have three particles of mass `m` located at the vertices of an equilateral triangle with side length `a`. Each particle experiences gravitational forces due to the other two particles.

### Step 2: Gravitational Force Calculation
The gravitational force \( F \) between any two particles is given by the formula:
\[ F = \frac{G m^2}{a^2} \]
where \( G \) is the gravitational constant.

### Step 3: Net Force Calculation
Since each particle is acted upon by the gravitational forces from the other two particles, we need to find the net force acting on one particle. The angle between the forces due to the two other particles is \( 60^\circ \).

Using the law of cosines, the net force \( F_{\text{net}} \) can be calculated as:
\[ F_{\text{net}} = \sqrt{F^2 + F^2 + 2F \cdot F \cdot \cos(60^\circ)} = \sqrt{F^2 + F^2 + F^2} = \sqrt{3F^2} = \sqrt{3}F \]
Substituting the value of \( F \):
\[ F_{\text{net}} = \sqrt{3} \cdot \frac{G m^2}{a^2} = \frac{\sqrt{3} G m^2}{a^2} \]

### Step 4: Centripetal Force Requirement
For the particles to move in a circular path, the net gravitational force must provide the necessary centripetal force. The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the velocity of the particles and \( r \) is the radius of the circular path.

### Step 5: Finding the Radius
The radius \( r \) of the circular motion can be determined from the geometry of the equilateral triangle. The distance from the center of the triangle to a vertex is given by:
\[ r = \frac{a}{\sqrt{3}} \]

### Step 6: Equating Forces
Setting the net gravitational force equal to the centripetal force:
\[ \frac{\sqrt{3} G m^2}{a^2} = \frac{m v^2}{\frac{a}{\sqrt{3}}} \]
Simplifying this equation:
\[ \frac{\sqrt{3} G m^2}{a^2} = \frac{m v^2 \sqrt{3}}{a} \]
Cancelling \( m \) from both sides and rearranging gives:
\[ v^2 = \frac{G m}{a} \]
Thus, the initial velocity \( v \) is:
\[ v = \sqrt{\frac{G m}{a}} \]

### Step 7: Time Period Calculation
The time period \( T \) of the circular motion can be calculated using the formula:
\[ T = \frac{\text{Circumference}}{\text{Velocity}} = \frac{2 \pi r}{v} \]
Substituting the values of \( r \) and \( v \):
\[ T = \frac{2 \pi \left(\frac{a}{\sqrt{3}}\right)}{\sqrt{\frac{G m}{a}}} \]
This simplifies to:
\[ T = 2 \pi \frac{a}{\sqrt{3}} \cdot \sqrt{\frac{a}{G m}} = 2 \pi \sqrt{\frac{a^3}{3 G m}} \]

### Final Answers
- 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:
\[ T = 2 \pi \sqrt{\frac{a^3}{3 G m}} \]