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. 6R
B. 8R
C. 2R
D. 4R

Answer

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: Understand the setup
We have two celestial bodies: the Earth and the Moon. The mass of the Moon is given as \( \frac{M}{81} \), where \( M \) is the mass of the Earth. The distance between the centers of the Earth and the Moon is \( 60R \), where \( R \) is the radius of the Earth.

### Step 2: Define the variables
Let:
- \( m_E = M \) (mass of the Earth)
- \( m_M = \frac{M}{81} \) (mass of the Moon)
- \( d = 60R \) (distance between the Earth and Moon)
- \( x \) = distance from the center of the Moon to the point where gravitational fields cancel each other.

### Step 3: Set up the gravitational field equations
The gravitational field \( E \) due to a mass \( m \) at a distance \( r \) is given by:
\[ E = \frac{Gm}{r^2} \]
where \( G \) is the gravitational constant.

At point \( P \) (where the fields cancel), the gravitational field due to the Earth \( E_E \) and the gravitational field due to the Moon \( E_M \) must be equal in magnitude but opposite in direction:
\[ E_E = E_M \]

### Step 4: Write the expressions for gravitational fields
1. The gravitational field due to the Earth at point \( P \):
\[ E_E = \frac{GM}{(60R - x)^2} \]

2. The gravitational field due to the Moon at point \( P \):
\[ E_M = \frac{G \left(\frac{M}{81}\right)}{x^2} \]

### Step 5: Set the fields equal
Setting \( E_E \) equal to \( E_M \):
\[ \frac{GM}{(60R - x)^2} = \frac{G \left(\frac{M}{81}\right)}{x^2} \]

### Step 6: Cancel out common terms
Since \( G \) and \( M \) are common on both sides, we can cancel them:
\[ \frac{1}{(60R - x)^2} = \frac{1}{81 x^2} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ 81 x^2 = (60R - x)^2 \]

### Step 8: Expand and rearrange
Expanding the right side:
\[ 81 x^2 = 3600R^2 - 120xR + x^2 \]
Rearranging gives:
\[ 80 x^2 + 120 x R - 3600 R^2 = 0 \]

### Step 9: Solve the quadratic equation
This is a standard quadratic equation of the form \( ax^2 + bx + c = 0 \):
- \( a = 80 \)
- \( b = 120R \)
- \( c = -3600R^2 \)

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:
\[ (120R)^2 + 4 \cdot 80 \cdot 3600R^2 = 14400R^2 + 1152000R^2 = 1166400R^2 \]
Thus,
\[ x = \frac{-120R \pm \sqrt{1166400}R}{160} \]
Calculating \( \sqrt{1166400} = 1080 \):
\[ x = \frac{-120R \pm 1080R}{160} \]
Calculating the two possible values:
1. \( x = \frac{960R}{160} = 6R \)
2. \( x = \frac{-120R - 1080R}{160} \) (not physically meaningful as it would give a negative distance)

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