An asteroid is moving directly towards the centre of the earth. When at a distance of 4R (R is the radius of the earth) from the earths centre, it has a speed of `2sqrt2` km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km/s)? Give your answer to the nearest integer in kilometer/s _________.

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy (kinetic energy + potential energy) of the asteroid remains constant as it moves towards the Earth.

### Step-by-Step Solution:

1. **Identify Initial and Final States:**
- Initial distance from the center of the Earth, \( r_i = 4R \)
- Initial speed of the asteroid, \( v_i = 2\sqrt{2} \) km/s
- Final distance from the center of the Earth, \( r_f = R \)
- Final speed of the asteroid, \( v_f \) (this is what we want to find)

2. **Write the Conservation of Mechanical Energy Equation:**
The total mechanical energy at the initial position (4R) is equal to the total mechanical energy at the final position (R):
\[ KE_i + PE_i = KE_f + PE_f \]
Where:
- \( KE = \frac{1}{2} mv^2 \) (Kinetic Energy)
- \( PE = -\frac{G M m}{r} \) (Potential Energy)

3. **Substituting the Values:**
- Initial Kinetic Energy:
\[ KE_i = \frac{1}{2} m (2\sqrt{2})^2 = \frac{1}{2} m \cdot 8 = 4m \]
- Initial Potential Energy:
\[ PE_i = -\frac{G M m}{4R} \]
- Final Kinetic Energy:
\[ KE_f = \frac{1}{2} m v_f^2 \]
- Final Potential Energy:
\[ PE_f = -\frac{G M m}{R} \]

4. **Set Up the Equation:**
\[ 4m - \frac{G M m}{4R} = \frac{1}{2} m v_f^2 - \frac{G M m}{R} \]

5. **Cancel Out the Mass (m):**
Since mass \( m \) is common in all terms, we can cancel it out:
\[ 4 - \frac{G M}{4R} = \frac{1}{2} v_f^2 - \frac{G M}{R} \]

6. **Rearranging the Equation:**
Multiply the entire equation by 2 to eliminate the fraction:
\[ 8 - \frac{G M}{2R} = v_f^2 - \frac{2G M}{R} \]
Rearranging gives:
\[ v_f^2 = 8 + \frac{G M}{2R} \]

7. **Substituting \( \frac{G M}{R} \):**
We know that the escape velocity \( v_e \) from the surface of the Earth is given by:
\[ v_e = \sqrt{\frac{2G M}{R}} = 11.2 \text{ km/s} \]
Therefore:
\[ \frac{G M}{R} = \frac{v_e^2}{2} = \frac{(11.2)^2}{2} = 62.72 \text{ km}^2/\text{s}^2 \]
Hence:
\[ \frac{G M}{2R} = \frac{62.72}{2} = 31.36 \text{ km}^2/\text{s}^2 \]

8. **Final Calculation:**
Substitute back into the equation for \( v_f^2 \):
\[ v_f^2 = 8 + 31.36 = 39.36 \]
Taking the square root:
\[ v_f = \sqrt{39.36} \approx 6.27 \text{ km/s} \]

9. **Rounding to the Nearest Integer:**
The final speed of the asteroid when it hits the surface of the Earth is approximately:
\[ v_f \approx 6 \text{ km/s} \]

### Final Answer:
The speed of the asteroid when it hits the surface of the Earth is **6 km/s**.