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, we need to find the point where the gravitational forces exerted by the Earth and the Moon on a particle are equal in magnitude but opposite in direction.

### Given:
- Let the mass of the Moon be \( M_m \).
- Then, the mass of the Earth \( M_e = 81 M_m \).
- The distance between the centers of the Earth and the Moon is \( D \).

### Step 1: Define the distances
Let \( x \) be the distance from the center of the Earth to the point where the gravitational force is zero. Consequently, the distance from the center of the Moon to this point will be \( D - x \).

### Step 2: Write the gravitational force equations
The gravitational force exerted by the Earth on the particle at distance \( x \) is given by:
\[ F_e = \frac{G M_e m}{x^2} = \frac{G (81 M_m) m}{x^2} \]
where \( G \) is the gravitational constant and \( m \) is the mass of the particle.

The gravitational force exerted by the Moon on the particle at distance \( D - x \) is:
\[ F_m = \frac{G M_m m}{(D - x)^2} \]

### Step 3: Set the forces equal
For the gravitational forces to balance, we set \( F_e = F_m \):
\[ \frac{G (81 M_m) m}{x^2} = \frac{G M_m m}{(D - x)^2} \]

### Step 4: Simplify the equation
We can cancel \( G \) and \( m \) from both sides:
\[ \frac{81 M_m}{x^2} = \frac{M_m}{(D - x)^2} \]

Now, we can cancel \( M_m \) (assuming \( M_m \neq 0 \)):
\[ \frac{81}{x^2} = \frac{1}{(D - x)^2} \]

### Step 5: Cross-multiply to solve for \( x \)
Cross-multiplying gives us:
\[ 81 (D - x)^2 = x^2 \]

### Step 6: Expand and rearrange
Expanding the left side:
\[ 81 (D^2 - 2Dx + x^2) = x^2 \]
\[ 81D^2 - 162Dx + 81x^2 = x^2 \]

Rearranging the equation:
\[ 80x^2 - 162Dx + 81D^2 = 0 \]

### Step 7: Solve the quadratic equation
This is a standard quadratic equation of the form \( ax^2 + bx + c = 0 \) where:
- \( a = 80 \)
- \( b = -162D \)
- \( c = 81D^2 \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{162D \pm \sqrt{(-162D)^2 - 4 \cdot 80 \cdot 81D^2}}{2 \cdot 80} \]
\[ x = \frac{162D \pm \sqrt{26244D^2 - 25920D^2}}{160} \]
\[ x = \frac{162D \pm \sqrt{324D^2}}{160} \]
\[ x = \frac{162D \pm 18D}{160} \]

### Step 8: Calculate the two possible solutions
1. \( x = \frac{180D}{160} = \frac{9D}{8} \)
2. \( x = \frac{144D}{160} = \frac{9D}{10} \)

Since \( x \) must be less than \( D \), we take:
\[ x = \frac{9D}{10} \]

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