The escape velocity from the earth is about 11 km/s. The escape velocity from a planet having twice the radius and the twice mean density as the earth, is

Options

A. `22 kms^(-1)`
B. `11 km s^(-1)`
C. `5.5 km s^(-1)`
D. `15.5 km s^(-1)`

Answer

To find the escape velocity from a planet that has twice the radius and twice the mean density of the 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,
- \( R \) is the radius of the body.

### Step 2: Express mass in terms of density and volume
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, the mass can be rewritten as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 3: Substitute mass into the escape velocity formula
Substituting the expression for mass into the escape velocity formula gives:
\[ v = \sqrt{\frac{2G(\rho \cdot \frac{4}{3} \pi R^3)}{R}} = \sqrt{\frac{8\pi G \rho R^2}{3}} \]

### Step 4: Compare the escape velocity of the planet with that of the Earth
Let the escape velocity from Earth be \( v_e \) and the radius and density of Earth be \( R_e \) and \( \rho_e \) respectively. Then:
\[ v_e = \sqrt{\frac{8\pi G \rho_e R_e^2}{3}} \]
For the new planet, we have:
- Radius \( R_p = 2R_e \)
- Density \( \rho_p = 2\rho_e \)

Substituting these values into the escape velocity formula for the planet:
\[ v_p = \sqrt{\frac{8\pi G (2\rho_e) (2R_e)^2}{3}} = \sqrt{\frac{8\pi G (2\rho_e) (4R_e^2)}{3}} = \sqrt{\frac{32\pi G \rho_e R_e^2}{3}} \]

### Step 5: Relate the escape velocities
Now we can relate \( v_p \) to \( v_e \):
\[ v_p = \sqrt{4} \cdot \sqrt{\frac{8\pi G \rho_e R_e^2}{3}} = 2v_e \]
Given that \( v_e \approx 11 \, \text{km/s} \):
\[ v_p = 2 \times 11 \, \text{km/s} = 22 \, \text{km/s} \]

### Step 6: Final calculation
However, we need to consider the factor of \( \sqrt{2} \) from the density and radius relationship:
\[ v_p = v_e \cdot \sqrt{8} = 11 \cdot \sqrt{8} = 11 \cdot 2\sqrt{2} \approx 11 \cdot 2.828 \approx 31.1 \, \text{km/s} \]

### Conclusion
Thus, the escape velocity from the planet is approximately:
\[ \boxed{31.1 \, \text{km/s}} \]