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

Answer

To solve the problem of finding 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. **Understand the Mass Relationship**:
- Given that the mass of the Earth (M) is 81 times that of the Moon (m):
\[ M = 81m \]

2. **Set Up the Problem**:
- 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. Let this point be at a distance \(x\) from the center of the Earth.

3. **Gravitational Field Equations**:
- The gravitational field \(E\) due to a mass \(M\) at a distance \(r\) is given by:
\[ E = \frac{GM}{r^2} \]
- At point \(P\), the gravitational field due to the Earth is:
\[ E_E = \frac{GM}{x^2} \]
- The gravitational field due to the Moon at point \(P\) (which is at a distance \(d - x\) from the Moon) is:
\[ E_m = \frac{Gm}{(d - x)^2} \]

4. **Setting the Gravitational Fields Equal**:
- For the gravitational field to be zero at point \(P\), the fields due to the Earth and the Moon must be equal in magnitude:
\[ \frac{GM}{x^2} = \frac{Gm}{(d - x)^2} \]
- Cancel \(G\) from both sides:
\[ \frac{M}{x^2} = \frac{m}{(d - x)^2} \]

5. **Substituting the Mass Relationship**:
- Substitute \(M = 81m\) into the equation:
\[ \frac{81m}{x^2} = \frac{m}{(d - x)^2} \]
- Cancel \(m\) from both sides:
\[ \frac{81}{x^2} = \frac{1}{(d - x)^2} \]

6. **Cross-Multiplying**:
- Cross-multiply to eliminate the fractions:
\[ 81(d - x)^2 = x^2 \]

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

8. **Using the Quadratic Formula**:
- This is a quadratic equation in the form \(ax^2 + bx + c = 0\):
- \(a = 80\)
- \(b = -162d\)
- \(c = 81d^2\)
- The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Substitute the values:
\[ x = \frac{162d \pm \sqrt{(-162d)^2 - 4 \cdot 80 \cdot 81d^2}}{2 \cdot 80} \]

9. **Calculating the Discriminant**:
- Calculate the discriminant:
\[ (-162)^2 - 4 \cdot 80 \cdot 81 = 26244 - 25920 = 324 \]
- Thus:
\[ x = \frac{162d \pm 18}{160} \]

10. **Finding the Positive Root**:
- The two possible solutions for \(x\) are:
\[ x = \frac{180d}{160} = \frac{9d}{8} \quad \text{(not valid, as it exceeds d)} \]
\[ x = \frac{144d}{160} = \frac{9d}{10} \]

11. **Conclusion**:
- The point where the gravitational field is zero is at a distance \(x = \frac{9d}{10}\) from the center of the Earth, which means it is \(d - x = \frac{d}{10}\) from the center of the Moon.

### Final Answer:
The point on the line joining their centers at which the gravitational field due to them is zero is at a distance of \(\frac{d}{10}\) from the center of the Moon.