Three masses of `1kg`, `2kg`, and `3kg`, are placed at the vertices of an equilateral triangle of side `1m`. Find the gravitational potential energy of this system.
Take `G= 6.67xx10^(-11) N-m^(2)//kg^(2)`.

Explanation

To find the gravitational potential energy of the system of three masses placed at the vertices of an equilateral triangle, we can follow these steps:

### Step 1: Identify the masses and distances
We have three masses:
- \( m_1 = 1 \, \text{kg} \)
- \( m_2 = 2 \, \text{kg} \)
- \( m_3 = 3 \, \text{kg} \)

The distance between each pair of masses (the side of the equilateral triangle) is:
- \( r_{12} = r_{23} = r_{31} = 1 \, \text{m} \)

### Step 2: Use the formula for gravitational potential energy
The gravitational potential energy \( U \) of a system of point masses is given by the formula:
\[ U = -G \left( \frac{m_1 m_2}{r_{12}} + \frac{m_2 m_3}{r_{23}} + \frac{m_3 m_1}{r_{31}} \right) \]
Where \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

### Step 3: Calculate each term in the formula
1. For the pair \( m_1 \) and \( m_2 \):
\[ U_{12} = -G \frac{m_1 m_2}{r_{12}} = -6.67 \times 10^{-11} \frac{1 \times 2}{1} = -6.67 \times 10^{-11} \times 2 = -1.334 \times 10^{-10} \, \text{J} \]

2. For the pair \( m_2 \) and \( m_3 \):
\[ U_{23} = -G \frac{m_2 m_3}{r_{23}} = -6.67 \times 10^{-11} \frac{2 \times 3}{1} = -6.67 \times 10^{-11} \times 6 = -4.002 \times 10^{-10} \, \text{J} \]

3. For the pair \( m_3 \) and \( m_1 \):
\[ U_{31} = -G \frac{m_3 m_1}{r_{31}} = -6.67 \times 10^{-11} \frac{3 \times 1}{1} = -6.67 \times 10^{-11} \times 3 = -2.001 \times 10^{-10} \, \text{J} \]

### Step 4: Sum the potential energies
Now we sum all the potential energies:
\[ U = U_{12} + U_{23} + U_{31} \]
\[ U = -1.334 \times 10^{-10} + (-4.002 \times 10^{-10}) + (-2.001 \times 10^{-10}) \]
\[ U = -7.337 \times 10^{-10} \, \text{J} \]

### Final Answer
The total gravitational potential energy of the system is:
\[ U = -7.337 \times 10^{-10} \, \text{J} \]

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