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 three particles each of mass `m` situated at the vertices of an equilateral triangle of side `a`, we need to find the initial velocity that should be given to each particle and the time period of their circular motion.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have three particles (m1, m2, m3) each of mass `m` located at the vertices of an equilateral triangle with side length `a`. The angle between any two sides of the triangle is 60 degrees.

2. **Finding the Radius of Circular Motion**:
- The center of the circle in which the particles will move lies at a distance from each vertex. This distance can be calculated as follows:
- The distance from a vertex to the center of the triangle (circumcenter) is given by \( r = \frac{a}{\sqrt{3}} \).

3. **Calculating the Gravitational Force**:
- The gravitational force between any two particles is given by the formula:
\[ F = \frac{G m^2}{a^2} \]
- Since there are two forces acting on each particle due to the other two particles, we need to find the net force acting on one particle.

4. **Finding the Net Force**:
- The two gravitational forces acting on a particle from the other two particles can be resolved into components. The net force \( F_{net} \) can be calculated using the formula for the resultant of two forces at an angle:
\[ F_{net} = \sqrt{F^2 + F^2 + 2F \cdot F \cdot \cos(60^\circ)} = \sqrt{2F^2 + 2F^2 \cdot \frac{1}{2}} = \sqrt{3F^2} = \sqrt{3}F \]
- Substituting for \( F \):
\[ F_{net} = \sqrt{3} \cdot \frac{G m^2}{a^2} \]

5. **Equating to Centripetal Force**:
- The net gravitational force provides the centripetal force required for circular motion:
\[ F_{net} = \frac{m v^2}{r} \]
- Substituting for \( r \):
\[ \sqrt{3} \cdot \frac{G m^2}{a^2} = \frac{m v^2}{\frac{a}{\sqrt{3}}} \]
- Rearranging gives:
\[ v^2 = \frac{\sqrt{3} G m^2}{a^2} \cdot \frac{a}{\sqrt{3}} = \frac{G m}{a} \]
- Therefore, the initial velocity \( v \) is:
\[ v = \sqrt{\frac{G m}{a}} \]

6. **Finding the Time Period**:
- The time period \( T \) of circular motion can be calculated using the formula:
\[ T = \frac{2 \pi r}{v} \]
- Substituting for \( r \) and \( v \):
\[ T = \frac{2 \pi \cdot \frac{a}{\sqrt{3}}}{\sqrt{\frac{G m}{a}}} \]
- Simplifying gives:
\[ T = \frac{2 \pi a}{\sqrt{3}} \cdot \sqrt{\frac{a}{G m}} = \frac{2 \pi a^{3/2}}{\sqrt{3 G m}} \]

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