Escape velocity on the surface of earth is 11.2 km/s . Escape velocity from a planet whose mass is the same as that of earth and radius 1/4 that of earth is

Options

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

Explanation

To solve the problem of finding the escape velocity from a planet whose mass is the same as that of Earth but with a radius of 1/4 that of Earth, we can use the formula for escape velocity:

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

Where:
- \( v_e \) is the escape velocity,
- \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \text{m}^3/\text{kg s}^2 \)),
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step 1: Identify the known values
- The escape velocity on the surface of Earth \( v_{e, \text{Earth}} = 11.2 \, \text{km/s} = 11200 \, \text{m/s} \).
- The mass of the planet \( M \) is the same as that of Earth.
- The radius of the planet \( R = \frac{1}{4} R_{\text{Earth}} \).

### Step 2: Write the escape velocity for Earth
For Earth, we can express the escape velocity as:
\[ v_{e, \text{Earth}} = \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} \]

### Step 3: Write the escape velocity for the new planet
For the planet with radius \( \frac{1}{4} R_{\text{Earth}} \):
\[ v_{e, \text{planet}} = \sqrt{\frac{2GM_{\text{Earth}}}{\frac{1}{4} R_{\text{Earth}}}} = \sqrt{\frac{2GM_{\text{Earth}} \cdot 4}{R_{\text{Earth}}}} = \sqrt{4} \cdot \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} = 2 \cdot v_{e, \text{Earth}} \]

### Step 4: Calculate the escape velocity for the new planet
Substituting the escape velocity of Earth:
\[ v_{e, \text{planet}} = 2 \cdot 11.2 \, \text{km/s} = 22.4 \, \text{km/s} \]

### Final Answer
The escape velocity from the planet whose mass is the same as that of Earth and radius is \( \frac{1}{4} \) that of Earth is \( 22.4 \, \text{km/s} \).

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