The gravitational potential energy of a system of three particles of mass m each kept at the vertices of equilateral triangle of side x will be

Options

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

Answer

To find the gravitational potential energy of a system of three particles of mass \( m \) each, placed at the vertices of an equilateral triangle with side length \( x \), we can follow these steps:

### Step 1: Understand the System
We have three particles, each of mass \( m \), located at the vertices of an equilateral triangle. Let's label the vertices as \( A \), \( B \), and \( C \).

### Step 2: Write the Formula for Gravitational Potential Energy
The gravitational potential energy \( U \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by the formula:
\[ U = -\frac{G m_1 m_2}{r} \]
where \( G \) is the gravitational constant.

### Step 3: Calculate the Potential Energy Between Each Pair of Masses
1. **Potential Energy between Masses at A and B**:
\[ U_{AB} = -\frac{G m m}{x} = -\frac{G m^2}{x} \]

2. **Potential Energy between Masses at B and C**:
\[ U_{BC} = -\frac{G m m}{x} = -\frac{G m^2}{x} \]

3. **Potential Energy between Masses at C and A**:
\[ U_{CA} = -\frac{G m m}{x} = -\frac{G m^2}{x} \]

### Step 4: Sum the Potential Energies
The total gravitational potential energy \( U \) of the system is the sum of the potential energies of each pair:
\[ U = U_{AB} + U_{BC} + U_{CA} \]
Substituting the values we calculated:
\[ U = -\frac{G m^2}{x} - \frac{G m^2}{x} - \frac{G m^2}{x} \]
\[ U = -\frac{3G m^2}{x} \]

### Final Answer
Thus, the gravitational potential energy of the system of three particles is:
\[ U = -\frac{3G m^2}{x} \]