Two bodies of mass `10^(2)` kg and `10^(3)` kg are lying 1 m apart .
The gravitational potential at the mid-point of the line joining them is

Answer

To find the gravitational potential at the midpoint of the line joining two bodies of mass \(10^2 \, \text{kg}\) and \(10^3 \, \text{kg}\) that are 1 meter apart, we can follow these steps:

### Step 1: Identify the masses and distance
Let:
- Mass \(m_1 = 10^2 \, \text{kg} = 100 \, \text{kg}\)
- Mass \(m_2 = 10^3 \, \text{kg} = 1000 \, \text{kg}\)
- Distance between the two masses \(d = 1 \, \text{m}\)

The midpoint between the two masses is at a distance of \(0.5 \, \text{m}\) from each mass.

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

### Step 3: Calculate the gravitational potential due to each mass at the midpoint
1. **Gravitational potential due to mass \(m_1\)** at the midpoint:
\[ U_1 = -\frac{G m_1}{r_1} = -\frac{G \cdot 100}{0.5} \]

2. **Gravitational potential due to mass \(m_2\)** at the midpoint:
\[ U_2 = -\frac{G m_2}{r_2} = -\frac{G \cdot 1000}{0.5} \]

### Step 4: Combine the potentials
The total gravitational potential \(U_{\text{total}}\) at the midpoint is the sum of the potentials due to both masses:
\[ U_{\text{total}} = U_1 + U_2 \]
Substituting the expressions for \(U_1\) and \(U_2\):
\[ U_{\text{total}} = -\frac{G \cdot 100}{0.5} - \frac{G \cdot 1000}{0.5} \]
Factoring out the common terms:
\[ U_{\text{total}} = -\frac{G}{0.5} (100 + 1000) = -\frac{G}{0.5} \cdot 1100 \]

### Step 5: Substitute the value of \(G\)
Now substituting \(G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\):
\[ U_{\text{total}} = -\frac{6.67 \times 10^{-11}}{0.5} \cdot 1100 \]

### Step 6: Calculate the final value
Calculating the above expression:
\[ U_{\text{total}} = -\frac{6.67 \times 10^{-11} \cdot 1100}{0.5} = -6.67 \times 10^{-11} \cdot 2200 \]
\[ U_{\text{total}} = -1.4674 \times 10^{-7} \, \text{J} \]

### Final Answer
Thus, the gravitational potential at the midpoint of the line joining the two bodies is approximately:
\[ U_{\text{total}} \approx -1.47 \times 10^{-7} \, \text{J} \]