The distance of the centres of moon the earth is D. The mass of earth is 81 times the mass of the moon. At what distance from the centre of the earth, the gravitational force on a particle will be zero.

Options

A. `(D)/(2)`
B. `(2D)/(3)`
C. `(4D)/(3)`
D. `(9D)/(10)`

Explanation

To solve the problem of finding the distance from the center of the Earth where the gravitational force on a particle will be zero, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Variables:**
- Let the mass of the Earth be \( M_E \).
- Let the mass of the Moon be \( M_M \).
- Given that \( M_E = 81 M_M \).
- The distance between the centers of the Earth and the Moon is \( D \).

2. **Identify the Point of Interest:**
- Let \( x \) be the distance from the center of the Earth to the point where the gravitational force is zero.
- Therefore, the distance from the center of the Moon to this point will be \( D - x \).

3. **Set Up the Gravitational Forces:**
- The gravitational force exerted by the Earth on the particle at distance \( x \) is given by:
\[ F_E = \frac{G M_E m}{x^2} \]
- The gravitational force exerted by the Moon on the particle at distance \( D - x \) is given by:
\[ F_M = \frac{G M_M m}{(D - x)^2} \]
- Here, \( G \) is the gravitational constant and \( m \) is the mass of the particle.

4. **Set the Forces Equal:**
- For the net gravitational force to be zero, these two forces must be equal:
\[ F_E = F_M \]
- Substituting the expressions for the forces, we get:
\[ \frac{G M_E m}{x^2} = \frac{G M_M m}{(D - x)^2} \]
- We can cancel \( G \) and \( m \) from both sides:
\[ \frac{M_E}{x^2} = \frac{M_M}{(D - x)^2} \]

5. **Substitute the Mass Relationship:**
- Substitute \( M_E = 81 M_M \):
\[ \frac{81 M_M}{x^2} = \frac{M_M}{(D - x)^2} \]
- Cancel \( M_M \) from both sides:
\[ \frac{81}{x^2} = \frac{1}{(D - x)^2} \]

6. **Cross-Multiply:**
- Cross-multiplying gives:
\[ 81 (D - x)^2 = x^2 \]

7. **Expand and Rearrange:**
- Expanding the left side:
\[ 81 (D^2 - 2Dx + x^2) = x^2 \]
- Rearranging gives:
\[ 81D^2 - 162Dx + 81x^2 - x^2 = 0 \]
- This simplifies to:
\[ 80x^2 - 162Dx + 81D^2 = 0 \]

8. **Solve the Quadratic Equation:**
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 80 \), \( b = -162D \), and \( c = 81D^2 \).
\[ x = \frac{162D \pm \sqrt{(-162D)^2 - 4 \cdot 80 \cdot 81D^2}}{2 \cdot 80} \]
- Calculate the discriminant:
\[ = 26244D^2 - 25920D^2 = 324D^2 \]
- Thus, the square root is:
\[ \sqrt{324D^2} = 18D \]
- Therefore:
\[ x = \frac{162D \pm 18D}{160} \]

9. **Find the Valid Solution:**
- This gives two possible solutions:
\[ x = \frac{180D}{160} = \frac{9D}{8} \quad \text{(not valid, as it exceeds D)} \]
\[ x = \frac{144D}{160} = \frac{9D}{10} \quad \text{(valid)} \]

### Final Answer:
The distance from the center of the Earth where the gravitational force on a particle will be zero is \( \frac{9D}{10} \).