The boides of masses 100 kg and 8100 kg are held at a distance of 1 m. The gravitational field at a point on the line joining at them is zero. The gravitational potential at the point in J/kg is `(G = 6.67 xx 10^(-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 step by step, we will first identify the relevant concepts and then calculate the gravitational potential at the point where the gravitational field is zero.

### Step 1: Determine the position of the null point
The gravitational field is zero at a point on the line joining two masses when the gravitational forces exerted by both masses on a small test mass are equal.

Let:
- \( m_1 = 100 \, \text{kg} \)
- \( m_2 = 8100 \, \text{kg} \)
- The distance between the two masses \( d = 1 \, \text{m} \)

Let \( x \) be the distance from the mass \( m_1 \) (100 kg) to the null point. Therefore, the distance from \( m_2 \) (8100 kg) to the null point will be \( (1 - x) \).

Using the condition for the gravitational field to be zero:
\[ \frac{G m_1}{x^2} = \frac{G m_2}{(1 - x)^2} \]

Since \( G \) is a constant, we can cancel it out:
\[ \frac{100}{x^2} = \frac{8100}{(1 - x)^2} \]

Cross-multiplying gives:
\[ 100(1 - x)^2 = 8100x^2 \]

### Step 2: Simplify the equation
Expanding the left side:
\[ 100(1 - 2x + x^2) = 8100x^2 \]
\[ 100 - 200x + 100x^2 = 8100x^2 \]

Rearranging the equation:
\[ 100 - 200x + 100x^2 - 8100x^2 = 0 \]
\[ 100 - 200x - 8000x^2 = 0 \]

### Step 3: Solve the quadratic equation
Rearranging gives:
\[ 8000x^2 + 200x - 100 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 8000, b = 200, c = -100 \):
\[ b^2 - 4ac = 200^2 - 4 \cdot 8000 \cdot (-100) = 40000 + 3200000 = 3240000 \]
\[ x = \frac{-200 \pm \sqrt{3240000}}{2 \cdot 8000} \]
Calculating the square root:
\[ \sqrt{3240000} = 1800 \]
Thus,
\[ x = \frac{-200 \pm 1800}{16000} \]

Taking the positive root:
\[ x = \frac{1600}{16000} = 0.1 \, \text{m} \]

### Step 4: Calculate the gravitational potential at the null point
The gravitational potential \( V \) at a point due to a mass \( m \) is given by:
\[ V = -\frac{Gm}{r} \]

At the null point (0.1 m from \( m_1 \) and 0.9 m from \( m_2 \)):
\[ V = -\frac{G m_1}{0.1} - \frac{G m_2}{0.9} \]
Substituting the values:
\[ V = -\frac{6.67 \times 10^{-11} \times 100}{0.1} - \frac{6.67 \times 10^{-11} \times 8100}{0.9} \]
Calculating each term:
\[ V = -\left(6.67 \times 10^{-11} \times 1000 + 6.67 \times 10^{-11} \times 9000\right) \]
\[ = -\left(6.67 \times 10^{-8} + 6.67 \times 10^{-7}\right) \]
\[ = -6.67 \times 10^{-7} \, \text{J/kg} \]

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