If the radius of a planet is four times that of earth and the value of g is same for both, the escape velocity on the planet will be

Options

A. 11.2 km/s
B. 5 . 6 km/s
C. 22.4 km/s
D. None

Explanation

To solve the problem, we need to find the escape velocity of a planet whose radius is four times that of Earth, given that the value of gravitational acceleration (g) is the same for both.

### Step-by-Step Solution:

1. **Understand the Variables**:
- Let the radius of Earth be \( R_e \).
- The radius of the planet is \( R_1 = 4R_e \).
- The gravitational acceleration on both Earth and the planet is \( g \).

2. **Use the Formula for Escape Velocity**:
The escape velocity \( V_e \) is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

3. **Relate Mass and Radius Using Gravitational Acceleration**:
The gravitational acceleration \( g \) is defined as:
\[ g = \frac{GM}{R^2} \]
For Earth:
\[ g = \frac{GM_e}{R_e^2} \]
For the planet:
\[ g = \frac{GM_1}{R_1^2} \]
Since \( R_1 = 4R_e \), we can substitute:
\[ g = \frac{GM_1}{(4R_e)^2} = \frac{GM_1}{16R_e^2} \]

4. **Set the Gravitational Accelerations Equal**:
Since \( g \) is the same for both Earth and the planet:
\[ \frac{GM_1}{16R_e^2} = \frac{GM_e}{R_e^2} \]
Simplifying gives:
\[ M_1 = 16M_e \]

5. **Substitute Mass and Radius into the Escape Velocity Formula**:
Now, substituting \( M_1 \) and \( R_1 \) into the escape velocity formula for the planet:
\[ V_{e1} = \sqrt{\frac{2G(16M_e)}{4R_e}} = \sqrt{\frac{32GM_e}{4R_e}} = \sqrt{\frac{8GM_e}{R_e}} \]

6. **Relate to Escape Velocity on Earth**:
The escape velocity from Earth \( V_{e} \) is:
\[ V_e = \sqrt{\frac{2GM_e}{R_e}} \]
Therefore, we can express \( V_{e1} \) as:
\[ V_{e1} = \sqrt{4} \cdot V_e = 2V_e \]

7. **Calculate the Escape Velocity**:
Given that the escape velocity from Earth \( V_e \) is approximately \( 11.2 \, \text{km/s} \):
\[ V_{e1} = 2 \times 11.2 \, \text{km/s} = 22.4 \, \text{km/s} \]

### Conclusion:
The escape velocity on the planet is \( 22.4 \, \text{km/s} \).