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 between two masses, we can follow these steps:

### Step 1: Identify the masses and their positions
- Mass \( m_1 = 10 \, \text{kg} \) (located at one end)
- Mass \( m_2 = 100 \, \text{kg} \) (located at the other end)
- Distance between the two masses \( d = 2 \, \text{m} \)

### Step 2: Determine the midpoint distance
Since the two masses are separated by 2 meters, the midpoint is located 1 meter from each mass.

### Step 3: Use the formula for gravitational potential
The gravitational potential \( V \) at a point due to a mass \( m \) is given by:
\[ V = -\frac{Gm}{r} \]
where:
- \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \)
- \( m \) is the mass
- \( r \) is the distance from the mass to the point where the potential is being calculated

### Step 4: Calculate the gravitational potential due to each mass at the midpoint
1. **For mass \( m_1 = 10 \, \text{kg} \)**:
\[ V_1 = -\frac{G \cdot m_1}{r_1} = -\frac{(6.674 \times 10^{-11}) \cdot 10}{1} = -6.674 \times 10^{-10} \, \text{J/kg} \]

2. **For mass \( m_2 = 100 \, \text{kg} \)**:
\[ V_2 = -\frac{G \cdot m_2}{r_2} = -\frac{(6.674 \times 10^{-11}) \cdot 100}{1} = -6.674 \times 10^{-9} \, \text{J/kg} \]

### Step 5: Sum the potentials
The total gravitational potential \( V \) at the midpoint is the sum of the potentials due to both masses:
\[ V = V_1 + V_2 = -6.674 \times 10^{-10} + (-6.674 \times 10^{-9}) = -7.3414 \times 10^{-9} \, \text{J/kg} \]

### Step 6: Final result
Thus, the gravitational potential at the midpoint is approximately:
\[ V \approx -7.3414 \times 10^{-9} \, \text{J/kg} \]