The ratio of escape velocity at earth `(v_(e))` to the escape velocity at a planet `(v_(p))` whose radius and mean density in twice as that of earth is

Options

A. `1 : 2`
B. `1 : 2 sqrt(2)`
C. ` 1 : 4 `
D. `1 : 2 `

Explanation

To find the ratio of escape velocity at Earth `(v_e)` to the escape velocity at a planet `(v_p)` whose radius and mean density are twice that of Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Define 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.

2. **Escape Velocity for Earth**:
For Earth, we denote the radius as \( R_e \) and the mass as \( M_e \). Thus, the escape velocity at Earth \( v_e \) is:
\[ v_e = \sqrt{\frac{2GM_e}{R_e}} \]

3. **Mass of Earth in terms of Density**:
The mass of Earth can be expressed in terms of its density \( \rho \):
\[ M_e = \rho \cdot V_e = \rho \cdot \left(\frac{4}{3} \pi R_e^3\right) \]
Therefore, substituting this into the escape velocity formula gives:
\[ v_e = \sqrt{\frac{2G \left(\rho \cdot \frac{4}{3} \pi R_e^3\right)}{R_e}} = \sqrt{\frac{8\pi G \rho R_e^2}{3}} \]

4. **Escape Velocity for the Planet**:
For the planet, its radius \( R_p \) is twice that of Earth, so:
\[ R_p = 2R_e \]
The density of the planet is also twice that of Earth:
\[ \rho_p = 2\rho \]
The mass of the planet \( M_p \) can be calculated as:
\[ M_p = \rho_p \cdot V_p = 2\rho \cdot \left(\frac{4}{3} \pi (2R_e)^3\right) = 2\rho \cdot \left(\frac{4}{3} \pi \cdot 8R_e^3\right) = \frac{64}{3} \pi \rho R_e^3 \]
Now, substituting this into the escape velocity formula for the planet gives:
\[ v_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2G \left(\frac{64}{3} \pi \rho R_e^3\right)}{2R_e}} = \sqrt{\frac{64\pi G \rho R_e^2}{3}} \]

5. **Finding the Ratio**:
Now we can find the ratio of the escape velocities:
\[ \frac{v_e}{v_p} = \frac{\sqrt{\frac{8\pi G \rho R_e^2}{3}}}{\sqrt{\frac{64\pi G \rho R_e^2}{3}}} \]
Simplifying this gives:
\[ \frac{v_e}{v_p} = \frac{\sqrt{8}}{\sqrt{64}} = \frac{2\sqrt{2}}{8} = \frac{1}{2\sqrt{2}} \]

### Final Answer:
The ratio of escape velocity at Earth to the escape velocity at the planet is:
\[ \frac{v_e}{v_p} = \frac{1}{2\sqrt{2}} \]