If the mass of moon is `(M)/(81)`, where M is the mass of earth, find the distance of the point where gravitational field due to earth and moon cancel each other, from the centre of moon. Given the distance between centres of earth and moon is 60 R where R is the radius of earth

Options

A. 4 R
B. 8 R
C. 12 R
D. 6 R

Explanation

To find the distance from the center of the moon where the gravitational fields due to the Earth and the Moon cancel each other, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( M \) = mass of the Earth
- \( m = \frac{M}{81} \) = mass of the Moon
- \( R \) = radius of the Earth
- The distance between the centers of the Earth and the Moon = \( 60R \)

### Step 2: Set Up the Gravitational Field Equations
The gravitational field \( E \) due to a mass \( M \) at a distance \( r \) is given by the formula:
\[ E = \frac{GM}{r^2} \]
Where \( G \) is the gravitational constant.

### Step 3: Determine the Distance from the Center of the Moon
Let \( x \) be the distance from the center of the Moon to the point where the gravitational fields cancel each other. Therefore, the distance from the center of the Earth to this point will be:
\[ (60R - x) \]

### Step 4: Set the Gravitational Fields Equal
At the point where the gravitational fields cancel, we have:
\[ \text{Gravitational field due to Earth} = \text{Gravitational field due to Moon} \]
This can be expressed as:
\[ \frac{GM}{(60R - x)^2} = \frac{G \left(\frac{M}{81}\right)}{x^2} \]

### Step 5: Simplify the Equation
We can cancel \( G \) from both sides:
\[ \frac{M}{(60R - x)^2} = \frac{M/81}{x^2} \]
Now, cancel \( M \):
\[ \frac{1}{(60R - x)^2} = \frac{1/81}{x^2} \]

### Step 6: Cross Multiply
Cross multiplying gives:
\[ x^2 = \frac{(60R - x)^2}{81} \]

### Step 7: Expand and Rearrange
Expanding the right side:
\[ x^2 = \frac{(3600R^2 - 120R x + x^2)}{81} \]
Multiplying through by 81 to eliminate the fraction:
\[ 81x^2 = 3600R^2 - 120R x + x^2 \]
Rearranging gives:
\[ 80x^2 + 120R x - 3600R^2 = 0 \]

### Step 8: Solve the Quadratic Equation
This is a standard quadratic equation in the form \( ax^2 + bx + c = 0 \), where:
- \( a = 80 \)
- \( b = 120R \)
- \( c = -3600R^2 \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-120R \pm \sqrt{(120R)^2 - 4 \cdot 80 \cdot (-3600R^2)}}{2 \cdot 80} \]
Calculating the discriminant:
\[ = \frac{-120R \pm \sqrt{14400R^2 + 1152000R^2}}{160} = \frac{-120R \pm \sqrt{1166400R^2}}{160} = \frac{-120R \pm 1080R}{160} \]

### Step 9: Calculate the Positive Root
Taking the positive root:
\[ x = \frac{960R}{160} = 6R \]

### Conclusion
The distance from the center of the Moon where the gravitational fields cancel each other is \( 6R \).

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