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 mass equal to that of Earth and radius one-fourth of Earth's radius, we can use the formula for escape velocity:

\[ v = \sqrt{\frac{2GM}{R}} \]

Where:
- \( v \) = escape velocity
- \( G \) = universal gravitational constant (\(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\))
- \( M \) = mass of the planet
- \( R \) = radius of the planet

### Step-by-Step Solution:

1. **Identify the mass and radius of the planet**:
- Given that the mass of the planet \( M \) is equal to the mass of Earth \( M_{E} \).
- The radius of the planet \( R \) is one-fourth of the radius of Earth \( R_{E} \):
\[ R = \frac{1}{4} R_{E} \]

2. **Substituting the values into the escape velocity formula**:
- We can substitute \( M \) and \( R \) into the escape velocity formula:
\[ v = \sqrt{\frac{2G M_{E}}{R}} \]
- Replacing \( R \) with \( \frac{1}{4} R_{E} \):
\[ v = \sqrt{\frac{2G M_{E}}{\frac{1}{4} R_{E}}} \]

3. **Simplifying the expression**:
- The expression simplifies as follows:
\[ v = \sqrt{\frac{2G M_{E}}{\frac{1}{4} R_{E}}} = \sqrt{8 \frac{GM_{E}}{R_{E}}} \]

4. **Using the escape velocity of Earth**:
- The escape velocity from the surface of Earth is given by:
\[ v_{E} = \sqrt{\frac{2GM_{E}}{R_{E}}} \]
- Thus, we can express the escape velocity for the planet as:
\[ v = \sqrt{8} \cdot v_{E} = 2\sqrt{2} \cdot v_{E} \]

5. **Calculating the escape velocity**:
- The escape velocity from Earth is approximately \( 11.2 \, \text{km/s} \) or \( 11200 \, \text{m/s} \).
- Therefore, the escape velocity for the planet becomes:
\[ v = 2\sqrt{2} \cdot 11200 \, \text{m/s} \]
- Calculating \( 2\sqrt{2} \):
\[ 2\sqrt{2} \approx 2.828 \]
- Thus:
\[ v \approx 2.828 \cdot 11200 \approx 31600 \, \text{m/s} \]

### Final Answer:
The escape velocity at the surface of the planet is approximately \( 31600 \, \text{m/s} \).