Mass of the earth is 81 times that of the moon . If the distance between their centers is d, then the point on the line joining their centers at which the gravitational field due to them is zero is

Options

A. `d/10` from the center of moon
B. `d/10` from the center of earth
C. `(9d)/10` from the center of moon
D. `d/2` from each other

Answer

To find the point on the line joining the centers of the Earth and the Moon where the gravitational field is zero, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Masses**:
- Let the mass of the Moon be \( M \).
- The mass of the Earth is \( 81M \) (since it is given that the mass of the Earth is 81 times that of the Moon).

2. **Define the Distance**:
- Let the distance between the centers of the Earth and the Moon be \( d \).
- We need to find a point \( P \) on the line joining the centers of the Earth and the Moon where the gravitational field is zero.

3. **Set Up the Distances**:
- Let \( r \) be the distance from the Earth to the point \( P \).
- Therefore, the distance from the Moon to the point \( P \) will be \( d - r \).

4. **Gravitational Field Equations**:
- The gravitational field due to the Earth at point \( P \) is given by:
\[ E_E = \frac{G \cdot (81M)}{r^2} \]
- The gravitational field due to the Moon at point \( P \) is given by:
\[ E_M = \frac{G \cdot M}{(d - r)^2} \]

5. **Set the Gravitational Fields Equal**:
- Since we want the net gravitational field to be zero, we set the magnitudes of the gravitational fields equal:
\[ \frac{G \cdot (81M)}{r^2} = \frac{G \cdot M}{(d - r)^2} \]

6. **Cancel Common Terms**:
- We can cancel \( G \) and \( M \) from both sides:
\[ \frac{81}{r^2} = \frac{1}{(d - r)^2} \]

7. **Cross Multiply**:
- Cross multiplying gives:
\[ 81(d - r)^2 = r^2 \]

8. **Expand and Rearrange**:
- Expanding the left side:
\[ 81(d^2 - 2dr + r^2) = r^2 \]
- Rearranging gives:
\[ 81d^2 - 162dr + 81r^2 = r^2 \]
- Combine like terms:
\[ 80r^2 - 162dr + 81d^2 = 0 \]

9. **Use the Quadratic Formula**:
- This is a quadratic equation in \( r \). We can use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:
- \( a = 80 \)
- \( b = -162d \)
- \( c = 81d^2 \)
- Plugging in these values:
\[ r = \frac{162d \pm \sqrt{(-162d)^2 - 4 \cdot 80 \cdot 81d^2}}{2 \cdot 80} \]
- Simplifying the discriminant:
\[ = \frac{162d \pm \sqrt{26244d^2 - 25920d^2}}{160} \]
\[ = \frac{162d \pm \sqrt{324d^2}}{160} \]
\[ = \frac{162d \pm 18d}{160} \]

10. **Calculate the Possible Values for \( r \)**:
- This gives us two possible solutions:
- \( r = \frac{180d}{160} = \frac{9d}{8} \) (not valid, as it exceeds \( d \))
- \( r = \frac{144d}{160} = \frac{9d}{10} \) (valid)

11. **Conclusion**:
- Thus, the point on the line joining the centers of the Earth and the Moon where the gravitational field is zero is at a distance \( r = \frac{9d}{10} \) from the Earth.

### Final Answer:
The point at which the gravitational field due to the Earth and the Moon is zero is located at a distance of \( \frac{9d}{10} \) from the Earth.