The gravitational potential energy of an isolated system of three particles, each of mass `m`, at the three corners of an equilateral triangle of side `l` is

Options

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

Explanation

To find the gravitational potential energy of an isolated system of three particles, each of mass \( m \), located at the corners of an equilateral triangle with side length \( l \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Configuration**:
- We have three particles, each of mass \( m \), placed at the corners of an equilateral triangle.
- Let's denote the corners of the triangle as \( A \), \( B \), and \( C \).

2. **Calculate the Gravitational Potential Energy Between Each Pair of Masses**:
- The gravitational potential energy \( U \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by the formula:
\[ U = -\frac{G m_1 m_2}{r} \]
- In our case, since all masses are equal (\( m_1 = m_2 = m = m_3 \)), and the distance between any two masses is \( l \):
\[ U_{AB} = -\frac{G m m}{l} = -\frac{G m^2}{l} \]
\[ U_{BC} = -\frac{G m m}{l} = -\frac{G m^2}{l} \]
\[ U_{CA} = -\frac{G m m}{l} = -\frac{G m^2}{l} \]

3. **Sum the Gravitational Potential Energies**:
- The total gravitational potential energy \( U_{total} \) of the system is the sum of the potential energies of all pairs:
\[ U_{total} = U_{AB} + U_{BC} + U_{CA} \]
- Substituting the values we calculated:
\[ U_{total} = -\frac{G m^2}{l} - \frac{G m^2}{l} - \frac{G m^2}{l} \]
\[ U_{total} = -3 \frac{G m^2}{l} \]

4. **Final Result**:
- Therefore, the gravitational potential energy of the isolated system of three particles is:
\[ U_{total} = -\frac{3 G m^2}{l} \]

### Conclusion:
The gravitational potential energy of the system is \( -\frac{3 G m^2}{l} \).