The mass of a planet is `1/25`th of the mass of the earth. The distance between the planet and the earth is `6xx 10^8` m. At what distance from the earth, the gravity becomes zero

Options

A. `1xx10^8` m
B. `0.32xx10^8` m
C. `5xx10^8` m
D. `0.032xx10^8` m

Answer

To solve the problem, we need to find the distance from the Earth where the gravitational forces from the Earth and the planet cancel each other out, resulting in zero net gravitational force.

### Step-by-Step Solution:

1. **Identify the Masses and Distance:**
- Let the mass of the Earth be \( M \).
- The mass of the planet is \( m = \frac{1}{25} M \).
- The distance between the Earth and the planet is given as \( r = 6 \times 10^8 \) m.

2. **Set Up the Equation for Gravitational Forces:**
- Let \( x \) be the distance from the Earth to the point where the gravitational force is zero.
- The distance from the planet to this point will then be \( r - x \).
- The gravitational force exerted by the Earth on a mass at distance \( x \) is given by:
\[ F_E = \frac{G M}{x^2} \]
- The gravitational force exerted by the planet on the same mass at distance \( r - x \) is given by:
\[ F_P = \frac{G m}{(r - x)^2} \]

3. **Set the Forces Equal:**
- For the net gravitational force to be zero, we set \( F_E = F_P \):
\[ \frac{G M}{x^2} = \frac{G m}{(r - x)^2} \]
- The \( G \) cancels out:
\[ \frac{M}{x^2} = \frac{m}{(r - x)^2} \]

4. **Substitute the Mass of the Planet:**
- Substitute \( m = \frac{1}{25} M \):
\[ \frac{M}{x^2} = \frac{\frac{1}{25} M}{(r - x)^2} \]
- Cancel \( M \) from both sides:
\[ \frac{1}{x^2} = \frac{\frac{1}{25}}{(r - x)^2} \]

5. **Cross Multiply:**
- Cross multiplying gives:
\[ (r - x)^2 = \frac{1}{25} x^2 \]

6. **Take Square Roots:**
- Taking the square root of both sides:
\[ r - x = \frac{1}{5} x \]

7. **Rearranging the Equation:**
- Rearranging gives:
\[ r = x + \frac{1}{5} x = \frac{6}{5} x \]
- Thus:
\[ x = \frac{5}{6} r \]

8. **Substituting the Value of \( r \):**
- Now substituting \( r = 6 \times 10^8 \) m:
\[ x = \frac{5}{6} \times 6 \times 10^8 = 5 \times 10^8 \text{ m} \]

### Final Answer:
The distance from the Earth where the gravitational force becomes zero is \( 5 \times 10^8 \) m.