Two bodies of masses 10 kg and 100 kg are seperated by a distance of 2 m. The gravitational potential at the mid-point of the line joining the two bodies is:

Options

A. `-7.3 xx 10^(-7) J//kg`
B. `-7.3 xx 10^(-8) J//kg`
C. `-7.3 xx 10^(-9) J//kg`
D. `-7.3 xx 10^(-6) J//kg`

Answer

To find the gravitational potential at the midpoint of the line joining two bodies of masses 10 kg and 100 kg separated by a distance of 2 m, we can follow these steps:

### Step 1: Understand the setup
- We have two masses: \( m_1 = 10 \, \text{kg} \) and \( m_2 = 100 \, \text{kg} \).
- The distance between the two masses is \( d = 2 \, \text{m} \).
- The midpoint (O) is \( 1 \, \text{m} \) away from each mass.

### Step 2: Write the formula for gravitational potential
The gravitational potential \( V \) at a distance \( r \) from a mass \( m \) is given by:
\[ V = -\frac{Gm}{r} \]
where \( G \) is the gravitational constant \( (6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \).

### Step 3: Calculate the potential due to each mass at the midpoint
1. **Potential due to mass \( m_1 \) (10 kg) at point O**:
\[ V_1 = -\frac{G m_1}{r_1} = -\frac{(6.67 \times 10^{-11}) \times (10)}{1} = -6.67 \times 10^{-10} \, \text{J/kg} \]

2. **Potential due to mass \( m_2 \) (100 kg) at point O**:
\[ V_2 = -\frac{G m_2}{r_2} = -\frac{(6.67 \times 10^{-11}) \times (100)}{1} = -6.67 \times 10^{-9} \, \text{J/kg} \]

### Step 4: Calculate the total gravitational potential at point O
The total gravitational potential \( V \) at point O is the sum of the potentials due to both masses:
\[ V = V_1 + V_2 \]
Substituting the values:
\[ V = -6.67 \times 10^{-10} + (-6.67 \times 10^{-9}) = -7.34 \times 10^{-9} \, \text{J/kg} \]

### Final Answer
The gravitational potential at the midpoint of the line joining the two bodies is approximately:
\[ V \approx -7.34 \times 10^{-9} \, \text{J/kg} \]