Three identical bodies (each mass M) are placed at vertices of an equilateral triangle of arm L, keeping the triangle as such by which angular speed the bodies should rotated in their gravitional fields so that the triangle moves along circumference of circular orbit :

Options

A. `sqrt((3GM)/L^(3))`
B. `sqrt((GM)/L^(3))`
C. `sqrt((GM)/(3L^(3)))`
D. `3sqrt((GM)/L^(3))`

Answer

To solve the problem, we need to find the angular speed (ω) at which three identical bodies (each of mass M) placed at the vertices of an equilateral triangle of side length L should rotate in their gravitational fields so that the triangle moves along the circumference of a circular orbit.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- Each body experiences a gravitational force due to the other two bodies. This gravitational force (F_g) is attractive.
- The bodies also require a centripetal force (F_c) to maintain circular motion.

2. **Gravitational Force Calculation**:
- The gravitational force between two masses M separated by a distance L is given by:
\[ F_g = \frac{G M^2}{L^2} \]
- Since each mass experiences the gravitational pull from the other two masses, the resultant gravitational force acting on one mass due to the other two can be calculated.

3. **Resultant Gravitational Force**:
- The angle between the lines connecting the masses is 60 degrees (since it’s an equilateral triangle).
- The resultant force (F) from the two gravitational forces can be calculated using vector addition:
\[ F = 2F_g \cos(30^\circ) = 2 \left(\frac{G M^2}{L^2}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3} G M^2}{L^2} \]

4. **Centripetal Force Calculation**:
- The centripetal force required to keep one mass in circular motion is given by:
\[ F_c = M \omega^2 r \]
- The radius (r) of the circular path for the triangle can be derived as:
\[ r = \frac{L}{\sqrt{3}} \]
- Therefore, the centripetal force becomes:
\[ F_c = M \omega^2 \left(\frac{L}{\sqrt{3}}\right) \]

5. **Setting Forces Equal**:
- For the triangle to maintain its circular motion, the resultant gravitational force must equal the centripetal force:
\[ \frac{\sqrt{3} G M^2}{L^2} = M \omega^2 \left(\frac{L}{\sqrt{3}}\right) \]

6. **Solving for Angular Speed (ω)**:
- Cancel M from both sides:
\[ \frac{\sqrt{3} G M}{L^2} = \omega^2 \left(\frac{L}{\sqrt{3}}\right) \]
- Rearranging gives:
\[ \omega^2 = \frac{\sqrt{3} G M}{L^2} \cdot \frac{\sqrt{3}}{L} = \frac{3 G M}{L^3} \]
- Taking the square root:
\[ \omega = \sqrt{\frac{3 G M}{L^3}} \]

### Final Answer:
The angular speed (ω) at which the bodies should rotate is:
\[ \omega = \sqrt{\frac{3 G M}{L^3}} \]