A planet has a mass 120 times that of earth and radius is 5 times the radius of earth. What is the escape velocity at this planet if the escape velocity at earth's surface is `11.2 km s^(-1)`.

Answer

To find the escape velocity on the surface of a planet with a mass 120 times that of Earth and a radius 5 times that of Earth, we can use the formula for escape velocity:

\[ V_e = \sqrt{\frac{2GM}{R}} \]

where:
- \( V_e \) is the escape velocity,
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step-by-step Solution:

1. **Identify the mass and radius of the planet**:
- Given that the mass of the planet \( M_p = 120 \times M_e \) (where \( M_e \) is the mass of Earth).
- Given that the radius of the planet \( R_p = 5 \times R_e \) (where \( R_e \) is the radius of Earth).

2. **Substitute the values into the escape velocity formula**:
- The escape velocity on the planet can be expressed as:
\[ V_p = \sqrt{\frac{2G \cdot M_p}{R_p}} = \sqrt{\frac{2G \cdot (120 \cdot M_e)}{5 \cdot R_e}} \]

3. **Simplify the expression**:
- This can be rewritten as:
\[ V_p = \sqrt{\frac{120 \cdot 2G \cdot M_e}{5 \cdot R_e}} = \sqrt{24 \cdot \frac{2G \cdot M_e}{R_e}} \]

4. **Recognize the escape velocity at Earth's surface**:
- The escape velocity at Earth's surface is given as \( V_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \, \text{km/s} \).

5. **Substitute the escape velocity of Earth into the equation**:
- Therefore, we can express \( V_p \) as:
\[ V_p = \sqrt{24} \cdot V_e = \sqrt{24} \cdot 11.2 \, \text{km/s} \]

6. **Calculate the escape velocity**:
- Calculate \( \sqrt{24} \):
\[ \sqrt{24} \approx 4.899 \]
- Now, multiply by \( 11.2 \):
\[ V_p \approx 4.899 \cdot 11.2 \approx 54.88 \, \text{km/s} \]

### Final Answer:
The escape velocity at the planet is approximately \( 54.88 \, \text{km/s} \).