`v_(e)` and `v_(p)` denotes the escape velocity from the earth and another planet having twice the radius and the same mean density as the earth. Then

Options

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

Answer

To solve the problem of comparing the escape velocities \( v_e \) (from Earth) and \( v_p \) (from another planet with twice the radius and the same mean density as Earth), we can follow these steps:

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

### Step 2: Calculate the escape velocity from Earth
Let the radius of Earth be \( r \) and its mean density be \( \rho \). The mass \( M \) of Earth can be expressed as:
\[ M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi r^3 \rho \]
Substituting this into the escape velocity formula for Earth, we have:
\[ v_e = \sqrt{\frac{2G \left(\frac{4}{3} \pi r^3 \rho\right)}{r}} = \sqrt{\frac{8\pi G \rho r^2}{3}} \]

### Step 3: Calculate the escape velocity from the other planet
The radius of the other planet is \( 2r \), and since it has the same mean density \( \rho \), its mass \( M_p \) is:
\[ M_p = \frac{4}{3} \pi (2r)^3 \rho = \frac{4}{3} \pi (8r^3) \rho = \frac{32}{3} \pi r^3 \rho \]
Now, substituting this mass into the escape velocity formula for the planet:
\[ v_p = \sqrt{\frac{2G M_p}{2r}} = \sqrt{\frac{2G \left(\frac{32}{3} \pi r^3 \rho\right)}{2r}} = \sqrt{\frac{32\pi G \rho r^2}{3}} \]

### Step 4: Compare \( v_e \) and \( v_p \)
Now we have:
- \( v_e = \sqrt{\frac{8\pi G \rho r^2}{3}} \)
- \( v_p = \sqrt{\frac{32\pi G \rho r^2}{3}} \)

To compare these, we can see that:
\[ v_p = \sqrt{4} \cdot v_e = 2v_e \]

### Conclusion
Thus, the escape velocity from the planet \( v_p \) is twice that of the escape velocity from Earth \( v_e \):
\[ v_p = 2v_e \]

### Final Answer
The answer to the question is that the escape velocity from the planet is twice the escape velocity from Earth.