What will be the escape speed from a planet having radius thrice that of earth and the same mean density as that of the earth ? (Take `v_(e)=11.2` km/s)

Explanation

To find the escape speed from a planet with a radius three times that of Earth and the same mean density as Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for escape velocity**:
The escape velocity (v_e) from the surface of a planet is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Relate mass to density**:
The mass \( M \) of a planet can be expressed in terms of its volume and density:
\[ M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho \]
where \( \rho \) is the mean density of the planet.

3. **Substitute mass into the escape velocity formula**:
Substituting the expression for mass into the escape velocity formula gives:
\[ v_e = \sqrt{\frac{2G \left(\frac{4}{3} \pi R^3 \rho\right)}{R}} = \sqrt{\frac{8\pi G \rho R^2}{3}} \]

4. **Determine the relationship between the new planet and Earth**:
Let the radius of Earth be \( R_e \) and its mean density be \( \rho_e \). For the new planet, the radius \( R_p \) is:
\[ R_p = 3R_e \]
Since the new planet has the same mean density as Earth, we have \( \rho_p = \rho_e \).

5. **Calculate escape velocity for the new planet**:
Substitute \( R_p \) and \( \rho_p \) into the escape velocity formula:
\[ v_{e,p} = \sqrt{\frac{8\pi G \rho_e (3R_e)^2}{3}} = \sqrt{\frac{8\pi G \rho_e \cdot 9R_e^2}{3}} = \sqrt{24\pi G \rho_e R_e^2} \]

6. **Relate the escape velocity of the new planet to Earth**:
The escape velocity from Earth is:
\[ v_{e,e} = \sqrt{\frac{8\pi G \rho_e R_e^2}{3}} \]
Therefore, we can express the escape velocity of the new planet in terms of Earth's escape velocity:
\[ v_{e,p} = \sqrt{9} \cdot v_{e,e} = 3 v_{e,e} \]

7. **Substitute Earth's escape velocity**:
Given that \( v_{e,e} = 11.2 \, \text{km/s} \):
\[ v_{e,p} = 3 \times 11.2 \, \text{km/s} = 33.6 \, \text{km/s} \]

### Final Answer:
The escape speed from the planet is \( 33.6 \, \text{km/s} \).