Acceleration due to gravity on a planet is 10times the value on the earth . Escape velocity for the planet and the earth are `V_(f) and V_(e)` respectively Assuming that the radii of the planet and the same then

Options

A. `V_(p) = 10 V_(a)`
B. ` V_(p) = sqrt(10) V_(e)`
C. ` V_(p) = (V_(e))/(sqrt(10))`
D. `V_(p) = (V_(e))/(10)`

Explanation

To solve the problem, we need to find the relationship between the escape velocities of a planet and the Earth, given that the acceleration due to gravity on the planet is 10 times that on Earth and that their radii are the same.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Let \( g_e \) be the acceleration due to gravity on Earth.
- Let \( g_p \) be the acceleration due to gravity on the planet.
- We are given that \( g_p = 10 g_e \).
- Let \( R \) be the radius of both the Earth and the planet (since they are the same).

2. **Formula for Escape Velocity**:
The escape velocity \( V \) from a celestial body can be expressed as:
\[ V = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity and \( R \) is the radius of the body.

3. **Escape Velocity for Earth**:
The escape velocity for Earth (\( V_e \)) is:
\[ V_e = \sqrt{2g_e R} \]

4. **Escape Velocity for the Planet**:
The escape velocity for the planet (\( V_p \)) is:
\[ V_p = \sqrt{2g_p R} \]
Substituting \( g_p = 10g_e \) into the equation:
\[ V_p = \sqrt{2(10g_e)R} = \sqrt{20g_e R} \]

5. **Relating \( V_p \) and \( V_e \)**:
Now we can express \( V_p \) in terms of \( V_e \):
\[ V_p = \sqrt{20g_e R} \]
We know that:
\[ V_e = \sqrt{2g_e R} \]
To relate \( V_p \) and \( V_e \), we can write:
\[ V_p = \sqrt{10} \cdot \sqrt{2g_e R} = \sqrt{10} \cdot V_e \]

6. **Final Relationship**:
Thus, the relationship between the escape velocities is:
\[ V_p = \sqrt{10} \cdot V_e \]

### Conclusion:
The escape velocity for the planet is \( \sqrt{10} \) times the escape velocity for the Earth.