A planet has mass equal to mass of the earth but radius one fourth of radius of the earth . Then escape velocity at the surface of this planet will be

Options

A. `11.2` km/s
B. `22.4 ` km/s
C. `5.6` km/s
D. `44.8` km/s

Answer

To find the escape velocity at the surface of a planet with the same mass as Earth but with a radius one-fourth that of Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for escape velocity**:
The escape velocity (v_e) from the surface of a celestial body is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the body, and \( R \) is its radius.

2. **Identify the parameters for the new planet**:
- Mass of the planet, \( M' = M \) (mass of Earth)
- Radius of the planet, \( R' = \frac{1}{4} R \) (one-fourth the radius of Earth)

3. **Substitute the values into the escape velocity formula**:
The escape velocity for the new planet (v_e') can be expressed as:
\[ v_e' = \sqrt{\frac{2G M'}{R'}} \]
Substituting \( M' = M \) and \( R' = \frac{1}{4} R \):
\[ v_e' = \sqrt{\frac{2G M}{\frac{1}{4} R}} = \sqrt{\frac{2G M \cdot 4}{R}} = \sqrt{\frac{8GM}{R}} \]

4. **Relate this to the escape velocity of Earth**:
The escape velocity from Earth (v_e) is:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
Therefore, we can express \( v_e' \) in terms of \( v_e \):
\[ v_e' = \sqrt{4} \cdot \sqrt{\frac{2GM}{R}} = 2 \cdot v_e \]

5. **Substitute the value of Earth's escape velocity**:
The escape velocity from Earth is approximately \( 11.2 \, \text{km/s} \):
\[ v_e' = 2 \cdot 11.2 \, \text{km/s} = 22.4 \, \text{km/s} \]

### Final Answer:
The escape velocity at the surface of the planet is \( 22.4 \, \text{km/s} \).