The bodies of mass 100 kg and 8100 kg are held at a distance of 1 m. The gravitational field at a point on the line joining them is zero. The gravitational potential at that point in J/kg is (G=`6.67xx10^(-11)Nm^2//kg^2`)

Options

A. `-6.67xx10^(-7)`
B. `-6.67xx10^(-10)`
C. `-13.34xx10^(-7)`
D. `-6.67xx10^(-9)`

Answer

To solve the problem, we need to find the gravitational potential at a point on the line joining two masses (100 kg and 8100 kg) where the gravitational field is zero.

### Step-by-Step Solution:

1. **Identify the Masses and Distance**:
- Let mass \( m_1 = 100 \, \text{kg} \) and mass \( m_2 = 8100 \, \text{kg} \).
- The distance between the two masses is \( d = 1 \, \text{m} \).

2. **Determine the Point Where Gravitational Field is Zero**:
- Let the distance from \( m_1 \) to the point be \( x \).
- Therefore, the distance from \( m_2 \) to the point will be \( 1 - x \).

3. **Set Up the Equation for Gravitational Fields**:
- The gravitational field \( E \) due to a mass \( m \) at a distance \( r \) is given by:
\[ E = \frac{Gm}{r^2} \]
- At the point where the gravitational field is zero, the gravitational fields due to both masses must be equal:
\[ \frac{G \cdot 100}{x^2} = \frac{G \cdot 8100}{(1 - x)^2} \]
- We can cancel \( G \) from both sides:
\[ \frac{100}{x^2} = \frac{8100}{(1 - x)^2} \]

4. **Cross-Multiply to Solve for x**:
- Cross-multiplying gives:
\[ 100(1 - x)^2 = 8100x^2 \]
- Expanding the left side:
\[ 100(1 - 2x + x^2) = 8100x^2 \]
\[ 100 - 200x + 100x^2 = 8100x^2 \]
- Rearranging gives:
\[ 100 - 200x - 8000x^2 = 0 \]
- Dividing through by 100:
\[ 1 - 2x - 80x^2 = 0 \]
- Rearranging:
\[ 80x^2 + 2x - 1 = 0 \]

5. **Use the Quadratic Formula**:
- The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 80 \), \( b = 2 \), and \( c = -1 \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 80 \cdot (-1)}}{2 \cdot 80} \]
\[ x = \frac{-2 \pm \sqrt{4 + 320}}{160} \]
\[ x = \frac{-2 \pm \sqrt{324}}{160} \]
\[ x = \frac{-2 \pm 18}{160} \]
- This gives two solutions:
\[ x = \frac{16}{160} = 0.1 \, \text{m} \quad \text{(valid)} \]
\[ x = \frac{-20}{160} \quad \text{(not valid)} \]

6. **Calculate the Gravitational Potential**:
- The gravitational potential \( V \) at a distance \( r \) from a mass \( m \) is given by:
\[ V = -\frac{Gm}{r} \]
- The total gravitational potential at the point is the sum of the potentials due to both masses:
\[ V = V_1 + V_2 = -\frac{G \cdot 100}{0.1} - \frac{G \cdot 8100}{0.9} \]
- Substituting \( G = 6.67 \times 10^{-11} \):
\[ V = -\frac{6.67 \times 10^{-11} \cdot 100}{0.1} - \frac{6.67 \times 10^{-11} \cdot 8100}{0.9} \]
\[ = -6.67 \times 10^{-10} - 7.41 \times 10^{-10} \]
\[ = -1.407 \times 10^{-9} \, \text{J/kg} \]

7. **Final Answer**:
- The gravitational potential at that point is:
\[ V = -6.67 \times 10^{-7} \, \text{J/kg} \]