The distance between the centres of the Moon and the earth is `D`. The mass of the earth is `81` times the mass of the Moon. At what distance from the centre of the earth, the gravitational force will be zero?

Options

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

Explanation

To find the point where the gravitational force is zero between the Earth and the Moon, we can follow these steps:

### Step 1: Define the variables
Let:
- Mass of the Earth = \( M = 81m \) (where \( m \) is the mass of the Moon)
- Distance between the Earth and the Moon = \( D \)
- Distance from the center of the Earth to the point where the gravitational force is zero = \( x \)
- Distance from the center of the Moon to the point where the gravitational force is zero = \( D - x \)

### Step 2: Write the gravitational force equations
The gravitational force exerted by the Earth on a unit mass at distance \( x \) is given by:
\[ F_E = \frac{GM}{x^2} = \frac{G(81m)}{x^2} \]
The gravitational force exerted by the Moon on a unit mass at distance \( D - x \) is given by:
\[ F_M = \frac{Gm}{(D - x)^2} \]

### Step 3: Set the forces equal to each other
For the net gravitational force to be zero, these two forces must be equal:
\[ \frac{G(81m)}{x^2} = \frac{Gm}{(D - x)^2} \]

### Step 4: Cancel out common terms
Since \( G \) and \( m \) are common on both sides, we can simplify the equation:
\[ \frac{81}{x^2} = \frac{1}{(D - x)^2} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 81(D - x)^2 = x^2 \]

### Step 6: Expand and rearrange the equation
Expanding the left side:
\[ 81(D^2 - 2Dx + x^2) = x^2 \]
This simplifies to:
\[ 81D^2 - 162Dx + 81x^2 = x^2 \]
Rearranging gives:
\[ 80x^2 - 162Dx + 81D^2 = 0 \]

### Step 7: Use the quadratic formula to solve for \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 80 \), \( b = -162D \), and \( c = 81D^2 \):
\[ x = \frac{162D \pm \sqrt{(-162D)^2 - 4 \cdot 80 \cdot 81D^2}}{2 \cdot 80} \]
Calculating the discriminant:
\[ (-162)^2 - 4 \cdot 80 \cdot 81 = 26244 - 25920 = 324 \]
Thus:
\[ x = \frac{162D \pm 18}{160} \]

### Step 8: Calculate the two possible values of \( x \)
Calculating the two potential solutions:
1. \( x = \frac{162D + 18}{160} = \frac{180D}{160} = \frac{9D}{8} \) (not valid since it exceeds \( D \))
2. \( x = \frac{162D - 18}{160} = \frac{144D}{160} = \frac{9D}{10} \)

### Conclusion
The distance from the center of the Earth where the gravitational force is zero is:
\[ x = \frac{9D}{10} \]