Three point masses each of mass `m` rotate in a circle of radius `r` with constant angular velocity `omega` due to their mutual gravitational attraction. If at any instant, the masses are on the vertices of an equilateral triangle of side `a`, then the value of `omega` is

Options

A. `sqrt((Gm)/(a^(3)))`
B. `sqrt((3Gm)/(a^(3)))`
C. `sqrt((Gm)/(3a^(3)))`
D. None of these

Explanation

To solve the problem, we need to find the angular velocity \( \omega \) of three point masses each of mass \( m \) that are rotating in a circle of radius \( r \) and are positioned at the vertices of an equilateral triangle with side length \( a \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
Each mass is positioned at the vertices of an equilateral triangle. The distance from the center of the triangle to any vertex (which is the radius \( r \)) can be derived from the properties of the triangle. For an equilateral triangle with side length \( a \), the radius \( r \) can be expressed as:
\[ r = \frac{a}{\sqrt{3}} \]

2. **Gravitational Force Between the Masses**:
The gravitational force \( F \) between any two masses \( m \) separated by a distance \( a \) is given by Newton's law of gravitation:
\[ F = \frac{G m^2}{a^2} \]
where \( G \) is the gravitational constant.

3. **Centripetal Force Requirement**:
Each mass experiences a centripetal force due to the gravitational attraction from the other two masses. The net gravitational force acting on one mass can be calculated by considering the contributions from the other two masses. The resultant gravitational force \( F_{net} \) acting on one mass due to the other two can be derived as:
\[ F_{net} = 2 \cdot \frac{G m^2}{a^2} \cdot \cos(30^\circ) = 2 \cdot \frac{G m^2}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G m^2}{a^2} \]

4. **Setting the Forces Equal**:
The gravitational force must equal the required centripetal force for circular motion, which is given by:
\[ F_{centripetal} = m \omega^2 r \]
Substituting \( r = \frac{a}{\sqrt{3}} \):
\[ F_{centripetal} = m \omega^2 \left(\frac{a}{\sqrt{3}}\right) \]

5. **Equating the Forces**:
Setting the net gravitational force equal to the centripetal force:
\[ \frac{\sqrt{3} G m^2}{a^2} = m \omega^2 \left(\frac{a}{\sqrt{3}}\right) \]

6. **Solving for Angular Velocity \( \omega \)**:
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{\sqrt{3} G m}{a^2} = \omega^2 \left(\frac{a}{\sqrt{3}}\right) \]
Rearranging gives:
\[ \omega^2 = \frac{\sqrt{3} G m}{a^2} \cdot \frac{\sqrt{3}}{a} = \frac{3 G m}{a^3} \]
Taking the square root:
\[ \omega = \sqrt{\frac{3 G m}{a^3}} \]

### Final Answer:
The value of \( \omega \) is:
\[ \omega = \sqrt{\frac{3 G m}{a^3}} \]