Escape velocity from a planet is `v_(e)`. If its mass is increased to 16 times and its radius is increased to 4 times, then the new escape velocity would be

Options

A. `v_(e)`
B. `sqrt(2)v_(e)`
C. `2v_(e)`
D. `2 sqrt(2)v_(e)`

Explanation

To find the new escape velocity after the mass and radius of the planet are changed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for escape velocity**:
The escape velocity \( v_e \) from a planet is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Identify the changes in mass and radius**:
According to the problem:
- The mass of the planet is increased to 16 times its original mass:
\[ M' = 16M \]
- The radius of the planet is increased to 4 times its original radius:
\[ R' = 4R \]

3. **Substitute the new mass and radius into the escape velocity formula**:
The new escape velocity \( v_e' \) can be calculated as follows:
\[ v_e' = \sqrt{\frac{2G M'}{R'}} \]
Substituting \( M' \) and \( R' \):
\[ v_e' = \sqrt{\frac{2G (16M)}{4R}} \]

4. **Simplify the expression**:
\[ v_e' = \sqrt{\frac{16 \cdot 2GM}{4R}} = \sqrt{4 \cdot \frac{2GM}{R}} \]
Since \( \sqrt{4} = 2 \):
\[ v_e' = 2 \sqrt{\frac{2GM}{R}} \]

5. **Relate it back to the original escape velocity**:
The original escape velocity \( v_e \) is:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
Therefore, we can express the new escape velocity in terms of the original escape velocity:
\[ v_e' = 2 v_e \]

### Final Answer:
The new escape velocity is \( 2 v_e \).