The escape velocity on a planet is v. If the radius of the planet contracts to `(1/4)^(th)` the present value without any change in its mass, then the escape velocity becomes

Options

A. v
B. v/2
C. 2v
D. `sqrt2v`

Answer

To solve the problem, we need to determine how the escape velocity changes when the radius of the planet contracts to one-fourth of its original value while keeping the mass constant.

### Step-by-Step Solution:

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

2. **Identify the New Radius**:
The problem states that the radius of the planet contracts to one-fourth of its original value. If the original radius is \( R \), the new radius \( R' \) is:
\[ R' = \frac{R}{4} \]

3. **Substitute the New Radius into the Escape Velocity Formula**:
We need to find the new escape velocity \( v' \) using the new radius \( R' \):
\[ v' = \sqrt{\frac{2GM}{R'}} \]
Substituting \( R' \) into the equation gives:
\[ v' = \sqrt{\frac{2GM}{\frac{R}{4}}} \]

4. **Simplify the Expression**:
Simplifying the expression for \( v' \):
\[ v' = \sqrt{\frac{2GM \cdot 4}{R}} = \sqrt{\frac{8GM}{R}} \]

5. **Relate the New Escape Velocity to the Original Escape Velocity**:
We know that the original escape velocity \( v \) is:
\[ v = \sqrt{\frac{2GM}{R}} \]
Therefore, we can express \( v' \) in terms of \( v \):
\[ v' = \sqrt{4} \cdot \sqrt{\frac{2GM}{R}} = 2v \]

6. **Conclusion**:
Thus, the new escape velocity \( v' \) when the radius contracts to one-fourth is:
\[ v' = 2v \]

### Final Answer:
The escape velocity becomes \( 2v \).