The escape velocity from the surface of the earth is `V_(e)`. The escape velcotiy from the surface of a planet whose mass and radius are three times those of the earth, will be

Options

A. `v_(e)`
B. `3v_(e)`
C. `9v_(e)`
D. `(1)/(3v_(e))`

Answer

To find the escape velocity from the surface of a planet whose mass and radius are three times those of the 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 universal gravitational constant, \( M \) is the mass of the body, and \( R \) is the radius of the body.

2. **Identify the parameters for Earth**:
Let the mass of the Earth be \( M_E \) and the radius of the Earth be \( R_E \). The escape velocity from the surface of the Earth is given as \( V_e \).

3. **Parameters for the new planet**:
For the new planet, we are given:
- Mass \( M_P = 3M_E \) (three times the mass of the Earth)
- Radius \( R_P = 3R_E \) (three times the radius of the Earth)

4. **Substitute the parameters into the escape velocity formula for the new planet**:
Using the escape velocity formula for the new planet:
\[ V_P = \sqrt{\frac{2G \cdot M_P}{R_P}} \]
Substituting \( M_P \) and \( R_P \):
\[ V_P = \sqrt{\frac{2G \cdot (3M_E)}{3R_E}} \]

5. **Simplify the expression**:
We can simplify the expression:
\[ V_P = \sqrt{\frac{2G \cdot 3M_E}{3R_E}} = \sqrt{\frac{2GM_E}{R_E}} \]
Since \( V_e = \sqrt{\frac{2GM_E}{R_E}} \) (the escape velocity from Earth), we can substitute:
\[ V_P = V_e \]

6. **Conclusion**:
Thus, the escape velocity from the surface of the planet is equal to the escape velocity from the surface of the Earth:
\[ V_P = V_e \]

### Final Answer:
The escape velocity from the surface of the planet is \( V_e \).