Two point masses 100 kg and 25 kg are situated at two points 2 m apart, then the gravitational potential midway between them will be

Options

A. `-228xx10^(-11)J//kg`
B. `-25xx10^(-11)J//kg`
C. `-8xx10^(-10J//kg`
D. `-833xx10^(-11)J//kg`

Explanation

To find the gravitational potential midway between two point masses, we can follow these steps:

### Step 1: Identify the masses and distance
We have two point masses:
- \( m_1 = 100 \, \text{kg} \)
- \( m_2 = 25 \, \text{kg} \)

The distance between them is \( d = 2 \, \text{m} \).

### Step 2: Determine the position of the midpoint
The midpoint \( O \) is located at a distance of \( 1 \, \text{m} \) from each mass since the total distance is \( 2 \, \text{m} \).

### Step 3: Calculate the gravitational potential at point O due to each mass
The gravitational potential \( V \) at a distance \( r \) from a point mass \( m \) is given by the formula:
\[ V = -\frac{Gm}{r} \]
where \( G \) is the universal gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

#### For mass \( m_1 \):
- Distance from \( m_1 \) to point \( O \) is \( r_1 = 1 \, \text{m} \).
\[ V_1 = -\frac{G m_1}{r_1} = -\frac{6.67 \times 10^{-11} \times 100}{1} = -6.67 \times 10^{-9} \, \text{J/kg} \]

#### For mass \( m_2 \):
- Distance from \( m_2 \) to point \( O \) is \( r_2 = 1 \, \text{m} \).
\[ V_2 = -\frac{G m_2}{r_2} = -\frac{6.67 \times 10^{-11} \times 25}{1} = -1.6675 \times 10^{-9} \, \text{J/kg} \]

### Step 4: Calculate the net gravitational potential at point O
The net gravitational potential \( V_{\text{net}} \) at point \( O \) is the sum of the potentials due to both masses:
\[ V_{\text{net}} = V_1 + V_2 \]
Substituting the values:
\[ V_{\text{net}} = -6.67 \times 10^{-9} + (-1.6675 \times 10^{-9}) = -8.3375 \times 10^{-9} \, \text{J/kg} \]

### Step 5: Express the final answer
The gravitational potential midway between the two masses is:
\[ V_{\text{net}} = -8.34 \times 10^{-9} \, \text{J/kg} \]

### Summary
Thus, the gravitational potential midway between the two point masses is approximately:
\[ \boxed{-8.34 \times 10^{-9} \, \text{J/kg}} \]