Two escape speed from the surface of earth is `V_(e)`. The escape speed from the surface of a planet whose mass and radius are double that of earth will be.

Options

A. `V_(e)`
B. `2V_(e)`
C. `4V_(e)`
D. `2sqrt(2)V_(e)`

Answer

To find the escape speed from the surface of a planet whose mass and radius are double that of Earth, we can follow these steps:

### Step 1: Understand the formula for escape speed
The escape speed \( V_e \) 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,
- \( R \) is the radius of the body.

### Step 2: Define the parameters for Earth
Let:
- The mass of Earth be \( M_e \),
- The radius of Earth be \( R_e \).

Thus, the escape speed from Earth is:
\[ V_e = \sqrt{\frac{2GM_e}{R_e}} \]

### Step 3: Define the parameters for the new planet
According to the problem, the mass and radius of the new planet are double that of Earth:
- Mass of the planet \( M_p = 2M_e \),
- Radius of the planet \( R_p = 2R_e \).

### Step 4: Substitute the parameters into the escape speed formula for the new planet
Using the escape speed formula for the new planet, we have:
\[ V_{ep} = \sqrt{\frac{2G(2M_e)}{2R_e}} \]

### Step 5: Simplify the expression
Now, simplify the expression:
\[ V_{ep} = \sqrt{\frac{2G(2M_e)}{2R_e}} = \sqrt{\frac{2GM_e}{R_e}} = V_e \]

### Conclusion
Thus, the escape speed from the surface of the planet whose mass and radius are double that of Earth is equal to the escape speed from Earth:
\[ V_{ep} = V_e \]

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

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