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. `(3Gm^(2))/(a)`
B. `(3Gm^(2))/(2a)`
C. `(Gm^(2))/(2a)`
D. `(Gm^(2))/(a)`

Answer

To find the work done on the system to increase the sides of the triangle from \( a \) to \( 2a \), we can follow these steps:

### Step 1: Calculate the initial gravitational potential energy (U_initial) of the triangle.
For three particles of mass \( m \) at the corners of an equilateral triangle of side \( a \), the gravitational potential energy \( U \) is given by the formula:

\[ U = -\frac{G m_1 m_2}{r_{12}} - \frac{G m_1 m_3}{r_{13}} - \frac{G m_2 m_3}{r_{23}} \]

In our case, since all masses are equal and the distances between each pair of masses are equal to \( a \):

\[ U_{\text{initial}} = -\frac{G m^2}{a} - \frac{G m^2}{a} - \frac{G m^2}{a} = -\frac{3G m^2}{a} \]

### Step 2: Calculate the final gravitational potential energy (U_final) of the triangle.
When the sides of the triangle are increased to \( 2a \), the potential energy becomes:

\[ U_{\text{final}} = -\frac{G m^2}{2a} - \frac{G m^2}{2a} - \frac{G m^2}{2a} = -\frac{3G m^2}{2a} \]

### Step 3: Calculate the change in gravitational potential energy (ΔU).
The work done on the system is equal to the change in gravitational potential energy:

\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]

Substituting the values we calculated:

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

### Step 4: Simplify the expression for ΔU.
\[ \Delta U = -\frac{3G m^2}{2a} + \frac{3G m^2}{a} \]

To combine these fractions, we can express \( \frac{3G m^2}{a} \) with a common denominator:

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

### Step 5: Conclusion
The work done on the system to increase the sides of the triangle to \( 2a \) is:

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