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 apply the principle of conservation of mechanical energy. The total mechanical energy (kinetic energy + gravitational potential energy) of the asteroid at a distance of 4R from the center of the Earth will be equal to its total mechanical energy just before it hits the surface of the Earth.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Initial distance from the center of the Earth, \( r_i = 4R \)
- Initial speed of the asteroid, \( v_i = 2\sqrt{2} \) km/s
- Escape velocity from the Earth, \( v_e = 11.2 \) km/s

2. **Calculate the initial kinetic energy (KE) and gravitational potential energy (PE):**
- Kinetic Energy at \( r_i \):
\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (2\sqrt{2})^2 = \frac{1}{2} m \cdot 8 = 4m \text{ (in km}^2\text{/s}^2\text{)} \]
- Gravitational Potential Energy at \( r_i \):
\[ PE_i = -\frac{GMm}{r_i} = -\frac{GMm}{4R} \]

3. **Identify the final conditions:**
- Final distance from the center of the Earth, \( r_f = R \)
- Final speed of the asteroid, \( v_f \) (unknown)

4. **Calculate the final kinetic energy (KE) and gravitational potential energy (PE):**
- Kinetic Energy just before hitting the Earth:
\[ KE_f = \frac{1}{2} m v_f^2 \]
- Gravitational Potential Energy at the surface of the Earth:
\[ PE_f = -\frac{GMm}{R} \]

5. **Apply conservation of mechanical energy:**
\[ KE_i + PE_i = KE_f + PE_f \]
Substituting the expressions we have:
\[ 4m - \frac{GMm}{4R} = \frac{1}{2} m v_f^2 - \frac{GMm}{R} \]

6. **Cancel the mass \( m \) from the equation:**
\[ 4 - \frac{GM}{4R} = \frac{1}{2} v_f^2 - \frac{GM}{R} \]

7. **Rearranging the equation:**
\[ \frac{1}{2} v_f^2 = 4 - \frac{GM}{4R} + \frac{GM}{R} \]
\[ \frac{1}{2} v_f^2 = 4 + \frac{3GM}{4R} \]

8. **Substituting \( \frac{GM}{R} \) using escape velocity:**
\[ v_e^2 = \frac{2GM}{R} \Rightarrow GM = \frac{v_e^2 R}{2} \]
Substituting \( v_e = 11.2 \) km/s:
\[ GM = \frac{(11.2)^2 R}{2} \]

9. **Substituting back into the equation:**
\[ \frac{1}{2} v_f^2 = 4 + \frac{3 \cdot \frac{(11.2)^2 R}{2}}{4R} \]
\[ \frac{1}{2} v_f^2 = 4 + \frac{3 \cdot 11.2^2}{8} \]

10. **Calculating \( 11.2^2 \):**
\[ 11.2^2 = 125.44 \]
\[ \frac{3 \cdot 125.44}{8} = 47.04 \]
\[ \frac{1}{2} v_f^2 = 4 + 47.04 = 51.04 \]

11. **Solving for \( v_f^2 \):**
\[ v_f^2 = 2 \cdot 51.04 = 102.08 \]

12. **Taking the square root to find \( v_f \):**
\[ v_f = \sqrt{102.08} \approx 10.1 \text{ km/s} \]

13. **Rounding to the nearest integer:**
\[ v_f \approx 10 \text{ km/s} \]

### Final Answer:
The speed of the asteroid when it hits the surface of the Earth is approximately **10 km/s**.