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. (a)`D/2`
B. (b)`(2D)/3`
C. (c)`(4D)/3`
D. (d)`(9D)/10`

Explanation

To find the distance from the center of the Earth where the gravitational force is zero, we can follow these steps:

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

Given that the mass of the Earth is 81 times the mass of the Moon, we can express this as:
\[ M = 81m \]

### Step 2: Set Up the Gravitational Forces
The gravitational force exerted by the Earth on an object at distance \( x \) from its center is given by:
\[ F_E = \frac{GMm}{x^2} \]
where \( G \) is the gravitational constant.

The gravitational force exerted by the Moon on the same object at distance \( D - x \) from the center of the Moon is given by:
\[ F_M = \frac{Gm^2}{(D - x)^2} \]

### Step 3: Set the Forces Equal
For the gravitational force to be zero, these two forces must be equal:
\[ \frac{GMm}{x^2} = \frac{Gm^2}{(D - x)^2} \]

### Step 4: Simplify the Equation
We can cancel \( G \) and \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{M}{x^2} = \frac{m}{(D - x)^2} \]
Substituting \( M = 81m \):
\[ \frac{81m}{x^2} = \frac{m}{(D - x)^2} \]

### Step 5: Cancel \( m \) and Rearrange
Cancel \( m \):
\[ \frac{81}{x^2} = \frac{1}{(D - x)^2} \]
Cross-multiplying gives:
\[ 81(D - x)^2 = x^2 \]

### Step 6: Expand and Rearrange
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: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 80 \), \( b = -162D \), and \( c = 81D^2 \):
\[ x = \frac{162D \pm \sqrt{(-162D)^2 - 4 \cdot 80 \cdot 81D^2}}{2 \cdot 80} \]

Calculating the discriminant:
\[ (-162D)^2 - 4 \cdot 80 \cdot 81D^2 = 26244D^2 - 25920D^2 = 324D^2 \]

Thus, we have:
\[ x = \frac{162D \pm 18D}{160} \]

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

### Step 9: Choose the Valid Solution
Since \( x \) must be less than \( D \) (the distance from the center of the Earth to the point where the gravitational force is zero), the valid solution is:
\[ x = \frac{9D}{10} \]

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

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