Find the PE of three objects of masses 1 kg, 2 kg and 3 kg placed at the three vertices of an equilateral triangle of side 20 cm.

Options

A. 50 G J
B. `-55 G` J
C. 5 G J
D. 60 G J

Answer

To find the gravitational potential energy (PE) of three objects of masses 1 kg, 2 kg, and 3 kg placed at the vertices of an equilateral triangle with a side length of 20 cm, we can follow these steps:

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

These masses are located at the vertices of an equilateral triangle with a side length of 20 cm, which is equivalent to 0.2 m.

### Step 2: Use the Formula for Gravitational Potential Energy
The gravitational potential energy between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by the formula:
\[ PE = -\frac{G m_1 m_2}{r} \]
where \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

### Step 3: Calculate the Potential Energy Between Each Pair of Masses
1. **Between \( m_1 \) and \( m_2 \)**:
\[ PE_{12} = -\frac{G \cdot 1 \cdot 2}{0.2} = -\frac{2G}{0.2} = -10G \]

2. **Between \( m_2 \) and \( m_3 \)**:
\[ PE_{23} = -\frac{G \cdot 2 \cdot 3}{0.2} = -\frac{6G}{0.2} = -30G \]

3. **Between \( m_3 \) and \( m_1 \)**:
\[ PE_{31} = -\frac{G \cdot 3 \cdot 1}{0.2} = -\frac{3G}{0.2} = -15G \]

### Step 4: Sum the Potential Energies
Now, we can find the total gravitational potential energy of the system by summing the potential energies calculated above:
\[ PE_{\text{total}} = PE_{12} + PE_{23} + PE_{31} \]
Substituting the values:
\[ PE_{\text{total}} = -10G - 30G - 15G = -55G \]

### Step 5: Final Result
The total gravitational potential energy of the system is:
\[ PE_{\text{total}} = -55G \, \text{Joules} \]