The escape velocity for the earth is `V_(e)` . The escape velocity for a planet whose radius `(1)/(4)` th radius of earth and mass haif that of earth is

Options

A. `(v_(a))/(sqrt(2))`
B. `sqrt(2) v_(a)`
C. `2 v_(a)`
D. `(v_(a))/(2)`

Explanation

To find the escape velocity for a planet whose radius is one-fourth that of Earth and mass is half that of Earth, we can follow these steps:

### Step 1: Understand the formula for escape velocity
The escape velocity \( V \) from the surface of a celestial body is given by the formula:
\[ V = \sqrt{\frac{2GM}{R}} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the celestial body,
- \( R \) is the radius of the celestial body.

### Step 2: Define the parameters for Earth
Let:
- Mass of Earth \( M_E = M \)
- Radius of Earth \( R_E = R \)
- Escape velocity from Earth \( V_E = \sqrt{\frac{2GM_E}{R_E}} \)

### Step 3: Define the parameters for the new planet
For the new planet:
- Mass \( M_P = \frac{1}{2} M_E = \frac{M}{2} \)
- Radius \( R_P = \frac{1}{4} R_E = \frac{R}{4} \)

### Step 4: Write the escape velocity formula for the new planet
Using the escape velocity formula for the new planet:
\[ V_P = \sqrt{\frac{2GM_P}{R_P}} \]
Substituting the values of \( M_P \) and \( R_P \):
\[ V_P = \sqrt{\frac{2G \left(\frac{M}{2}\right)}{\frac{R}{4}}} \]

### Step 5: Simplify the expression
Now simplify the expression:
\[ V_P = \sqrt{\frac{2G \cdot \frac{M}{2}}{\frac{R}{4}}} = \sqrt{\frac{GM}{R} \cdot 4} = \sqrt{4 \cdot \frac{GM}{R}} = 2\sqrt{\frac{GM}{R}} \]

### Step 6: Relate it to the escape velocity of Earth
We know that:
\[ V_E = \sqrt{\frac{2GM}{R}} \]
Thus, we can express \( V_P \) in terms of \( V_E \):
\[ V_P = 2 \cdot \sqrt{\frac{GM}{R}} = 2 \cdot \frac{V_E}{\sqrt{2}} = \sqrt{2} \cdot V_E \]

### Final Answer
The escape velocity for the planet is:
\[ V_P = \sqrt{2} V_E \]