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 determine the angular speed (ω) at which three identical bodies, each of mass M, must rotate around the center of an equilateral triangle formed by their positions, such that the triangle moves along the circumference of a circular orbit.

### Step-by-Step Solution:

1. **Understanding the System**:
We have three identical bodies placed at the vertices of an equilateral triangle with side length L. Each body experiences gravitational attraction due to the other two bodies.

2. **Centripetal Force Requirement**:
For the triangle to move in a circular path, each body must experience a centripetal force that keeps it in circular motion. The centripetal force (F_c) required for a body of mass M moving in a circle of radius r with angular speed ω is given by:
\[ F_c = M \omega^2 r \]

3. **Finding the Radius**:
The distance from the center of the triangle to any vertex (the radius r) can be calculated as:
\[ r = \frac{L}{\sqrt{3}} \]
This is derived from the geometry of the equilateral triangle.

4. **Gravitational Force Between the Bodies**:
The gravitational force (F_g) between any two bodies of mass M separated by a distance L is given by Newton's law of gravitation:
\[ F_g = \frac{G M^2}{L^2} \]
where G is the gravitational constant.

5. **Net Gravitational Force on One Body**:
Each body experiences gravitational attraction from the other two bodies. The net gravitational force acting on one body due to the other two can be calculated. The effective force can be found using vector addition, resulting in:
\[ F_{net} = \sqrt{F_g^2 + F_g^2 - 2F_gF_g\cos(120^\circ)} = \sqrt{F_g^2 + F_g^2 + F_g^2} = \sqrt{3} F_g \]
Thus,
\[ F_{net} = \sqrt{3} \cdot \frac{G M^2}{L^2} \]

6. **Setting the Forces Equal**:
For the triangle to maintain its shape and rotate, the net gravitational force must equal the required centripetal force:
\[ M \omega^2 \left(\frac{L}{\sqrt{3}}\right) = \sqrt{3} \cdot \frac{G M^2}{L^2} \]

7. **Simplifying the Equation**:
Cancel M from both sides (assuming M ≠ 0):
\[ \omega^2 \frac{L}{\sqrt{3}} = \sqrt{3} \cdot \frac{G M}{L^2} \]
Rearranging gives:
\[ \omega^2 = \frac{3 G M}{L^3} \]

8. **Finding Angular Speed**:
Taking the square root to find ω:
\[ \omega = \sqrt{\frac{3 G M}{L^3}} \]

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