Escape velocity on a planet is e v . If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes

Options

A. `4 v_(e)`
B. `2 v_(e)`
C. `v_(e)`
D. `1/2 v_(e)`

Answer

To determine the new escape velocity when the mass of the planet is increased to four times its original value while the radius remains constant, we can follow these steps:

### Step 1: Understand the formula for escape velocity
The escape velocity (v_e) from the surface of 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,
- \( R \) is the radius of the planet.

### Step 2: Identify the initial conditions
Let:
- The initial mass of the planet be \( M \).
- The initial radius of the planet be \( R \).
- The initial escape velocity be \( v_e \).

From the formula, we have:
\[ v_e = \sqrt{\frac{2GM}{R}} \]

### Step 3: Modify the mass of the planet
According to the problem, the mass of the planet becomes four times its original mass:
\[ M' = 4M \]

### Step 4: Calculate the new escape velocity
Since the radius remains the same, we can substitute \( M' \) into the escape velocity formula:
\[ v_e' = \sqrt{\frac{2G(4M)}{R}} \]

### Step 5: Simplify the expression
This simplifies to:
\[ v_e' = \sqrt{\frac{8GM}{R}} = \sqrt{4 \cdot \frac{2GM}{R}} = 2\sqrt{\frac{2GM}{R}} \]

### Step 6: Relate the new escape velocity to the initial escape velocity
Since \( v_e = \sqrt{\frac{2GM}{R}} \), we can express the new escape velocity as:
\[ v_e' = 2v_e \]

### Conclusion
Thus, the new escape velocity when the mass of the planet is increased to four times its original value while keeping the radius constant is:
\[ v_e' = 2v_e \]