Three particle of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.

Options

A. `-(3)/(2) (GM^(2))/(a^2)`
B. `(-3 GM^(2))/(a^2)`
C. `(3)/(2) (GM^(2))/(a^2)`
D. `(3 GM^(2))/(l)`

Explanation

To find the work done on the system of three particles of mass \( m \) each, placed at the corners of an equilateral triangle with side length \( a \), when the side length is increased to \( 2a \), we will follow these steps:

### Step 1: Calculate the Initial Potential Energy
The gravitational potential energy \( U \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:

\[ U = -\frac{G m_1 m_2}{r} \]

For our system, we have three particles (let's call them \( P_1, P_2, P_3 \)) at the corners of the triangle. The initial potential energy \( U_i \) when the side length is \( a \) can be calculated as follows:

- The potential energy between \( P_1 \) and \( P_2 \) is:

\[ U_{12} = -\frac{G m^2}{a} \]

- The potential energy between \( P_1 \) and \( P_3 \) is:

\[ U_{13} = -\frac{G m^2}{a} \]

- The potential energy between \( P_2 \) and \( P_3 \) is:

\[ U_{23} = -\frac{G m^2}{a} \]

Thus, the total initial potential energy \( U_i \) is:

\[ U_i = U_{12} + U_{13} + U_{23} = -\frac{G m^2}{a} - \frac{G m^2}{a} - \frac{G m^2}{a} = -\frac{3G m^2}{a} \]

### Step 2: Calculate the Final Potential Energy
Now, we need to calculate the final potential energy \( U_f \) when the side length is increased to \( 2a \):

- The potential energy between \( P_1 \) and \( P_2 \) is:

\[ U_{12}' = -\frac{G m^2}{2a} \]

- The potential energy between \( P_1 \) and \( P_3 \) is:

\[ U_{13}' = -\frac{G m^2}{2a} \]

- The potential energy between \( P_2 \) and \( P_3 \) is:

\[ U_{23}' = -\frac{G m^2}{2a} \]

Thus, the total final potential energy \( U_f \) is:

\[ U_f = U_{12}' + U_{13}' + U_{23}' = -\frac{G m^2}{2a} - \frac{G m^2}{2a} - \frac{G m^2}{2a} = -\frac{3G m^2}{2a} \]

### Step 3: Calculate the Work Done
The work done \( W \) on the system is equal to the change in potential energy, which can be calculated as:

\[ W = U_f - U_i \]

Substituting the values we found:

\[ W = \left(-\frac{3G m^2}{2a}\right) - \left(-\frac{3G m^2}{a}\right) \]

This simplifies to:

\[ W = -\frac{3G m^2}{2a} + \frac{3G m^2}{a} = -\frac{3G m^2}{2a} + \frac{6G m^2}{2a} = \frac{3G m^2}{2a} \]

Thus, the work done on the system to increase the sides of the triangle to \( 2a \) is:

\[ W = \frac{3G m^2}{2a} \]

### Final Answer
The work done is:

\[ \boxed{\frac{3G m^2}{2a}} \]