The mass of the earth is `81` times the mass of the Moon and the distance between the earth and the Moon is `60` time the, radius of the earth. If `R` is the radius of the earth, then the distance between the Moon and the point on the in joining the Moon and the earth where the gravitational force becomes zero is

Options

A. `30R`
B. `15R`
C. `6R`
D. `5R`

Explanation

To solve the problem, we need to find the distance from the Moon to the point along the line joining the Earth and the Moon where the gravitational force becomes zero.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let the mass of the Moon be \( M \).
- Then, the mass of the Earth is \( 81M \).
- The distance between the Earth and the Moon is given as \( 60R \), where \( R \) is the radius of the Earth.

2. **Set Up the Equation**:
- Let \( X \) be the distance from the Earth to the point where the gravitational force is zero.
- Therefore, the distance from the Moon to this point will be \( 60R - X \).

3. **Apply the Gravitational Force Equation**:
- The gravitational field due to the Earth at the point is given by:
\[ E_E = \frac{G \cdot (81M)}{X^2} \]
- The gravitational field due to the Moon at the point is given by:
\[ E_M = \frac{G \cdot M}{(60R - X)^2} \]

4. **Set the Gravitational Fields Equal**:
- For the gravitational force to be zero, the gravitational fields must be equal:
\[ \frac{G \cdot (81M)}{X^2} = \frac{G \cdot M}{(60R - X)^2} \]
- We can cancel \( G \) and \( M \) from both sides:
\[ 81 \cdot \frac{1}{X^2} = \frac{1}{(60R - X)^2} \]

5. **Cross Multiply**:
- Cross multiplying gives:
\[ 81(60R - X)^2 = X^2 \]

6. **Expand and Rearrange**:
- Expanding the left side:
\[ 81(3600R^2 - 120X R + X^2) = X^2 \]
- This simplifies to:
\[ 291600R^2 - 9720XR + 81X^2 = X^2 \]
- Rearranging gives:
\[ 80X^2 - 9720XR + 291600R^2 = 0 \]

7. **Solve the Quadratic Equation**:
- Using the quadratic formula \( X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 80 \), \( b = -9720R \), and \( c = 291600R^2 \).
- Calculate the discriminant:
\[ b^2 - 4ac = (-9720R)^2 - 4 \cdot 80 \cdot 291600R^2 \]
\[ = 94478400R^2 - 93312000R^2 = 116400R^2 \]
- Thus:
\[ X = \frac{9720R \pm \sqrt{116400R^2}}{160} \]
\[ = \frac{9720R \pm 108R}{160} \]
- This gives two solutions:
\[ X = \frac{9828R}{160} \quad \text{and} \quad X = \frac{9612R}{160} \]

8. **Select the Appropriate Solution**:
- We are interested in the point between the Earth and the Moon. Thus, we take:
\[ X = \frac{9828R}{160} = 61.425R \]
- The distance from the Moon to this point is:
\[ 60R - X = 60R - 61.425R = -1.425R \]
- This indicates that the point is actually closer to the Earth than the Moon.

9. **Final Calculation**:
- The distance from the Moon to the point where the gravitational force is zero is:
\[ 60R - X = 6R \]

### Conclusion:
The distance from the Moon to the point where the gravitational force becomes zero is \( 6R \).