The escape Velocity from the earth is `11.2 Km//s`. The escape Velocity from a planet having twice the radius and the same mean density as the earth, is :

Options

A. 11.2
B. 5.6
C. 15
D. 22.4

Answer

To find the escape velocity from a planet that has twice the radius of Earth and the same mean density, we can follow these steps:

### Step 1: Understand the formula for escape velocity
The escape velocity \( v_e \) from 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 celestial body,
- \( R \) is the radius of the celestial body.

### Step 2: Relate mass to density
The mass \( M \) of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho V \]
For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, we can write:
\[ M = \rho \left(\frac{4}{3} \pi R^3\right) \]

### Step 3: Substitute mass into the escape velocity formula
Substituting the expression for mass \( M \) into the escape velocity formula:
\[ v_e = \sqrt{\frac{2G \left(\rho \frac{4}{3} \pi R^3\right)}{R}} = \sqrt{\frac{8\pi G \rho R^2}{3}} \]
This shows that the escape velocity is proportional to the square root of the radius \( R \) when density \( \rho \) is constant.

### Step 4: Determine the escape velocity for the new planet
Given that the new planet has twice the radius of Earth, we can denote the radius of Earth as \( R \) and the radius of the new planet as \( R' = 2R \). The escape velocity for the new planet \( v_e' \) can be expressed as:
\[ v_e' = \sqrt{\frac{8\pi G \rho (2R)^2}{3}} = \sqrt{\frac{8\pi G \rho \cdot 4R^2}{3}} = \sqrt{4} \cdot \sqrt{\frac{8\pi G \rho R^2}{3}} = 2 \cdot v_e \]
where \( v_e \) is the escape velocity from Earth.

### Step 5: Calculate the escape velocity
Since the escape velocity from Earth \( v_e \) is given as \( 11.2 \, \text{km/s} \):
\[ v_e' = 2 \cdot 11.2 \, \text{km/s} = 22.4 \, \text{km/s} \]

### Conclusion
The escape velocity from the planet having twice the radius and the same mean density as Earth is \( 22.4 \, \text{km/s} \).

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