An asteroid of mass `m (m lt lt m_E)` is approaching with a velocity 12 km/s when it is at distance of 10 R from the centre of earth (where R is radius of earth). When it reaches at the surface of Earth, its velocity is (Nearest Integer) in km/s

Explanation

To solve the problem of the asteroid approaching Earth, we will use the principle of conservation of mechanical energy. The total mechanical energy (kinetic + potential) at the initial position (10R from the center of Earth) will be equal to the total mechanical energy at the final position (on the surface of Earth).

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial velocity of the asteroid, \( u = 12 \, \text{km/s} = 12000 \, \text{m/s} \)
- Initial distance from the center of Earth, \( r_i = 10R \)
- Mass of the asteroid, \( m \) (where \( m \ll m_E \))
- Gravitational potential energy at distance \( r_i \):
\[ U_i = -\frac{G m_E m}{r_i} = -\frac{G m_E m}{10R} \]

2. **Calculate Initial Kinetic Energy:**
- Initial kinetic energy, \( K_i = \frac{1}{2} m u^2 \)

3. **Identify Final Conditions:**
- Final distance from the center of Earth, \( r_f = R \)
- Final velocity of the asteroid, \( v \) (to be determined)
- Gravitational potential energy at distance \( r_f \):
\[ U_f = -\frac{G m_E m}{R} \]
- Final kinetic energy, \( K_f = \frac{1}{2} m v^2 \)

4. **Apply Conservation of Energy:**
- According to the conservation of energy:
\[ K_i + U_i = K_f + U_f \]
- Substituting the energies:
\[ \frac{1}{2} m u^2 - \frac{G m_E m}{10R} = \frac{1}{2} m v^2 - \frac{G m_E m}{R} \]

5. **Simplify the Equation:**
- Cancel \( m \) from all terms (since \( m \) is common and non-zero):
\[ \frac{1}{2} u^2 - \frac{G m_E}{10R} = \frac{1}{2} v^2 - \frac{G m_E}{R} \]
- Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{1}{2} u^2 + \frac{G m_E}{R} - \frac{G m_E}{10R} \]
- Combine the potential energy terms:
\[ \frac{1}{2} v^2 = \frac{1}{2} u^2 + \frac{G m_E}{R} \left(1 - \frac{1}{10}\right) = \frac{1}{2} u^2 + \frac{9G m_E}{10R} \]

6. **Substitute Known Values:**
- Substitute \( u = 12000 \, \text{m/s} \):
\[ v^2 = u^2 + \frac{18G m_E}{10R} \]
- Using \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \), \( m_E = 6 \times 10^{24} \, \text{kg} \), and \( R = 6.4 \times 10^6 \, \text{m} \):
\[ v^2 = (12000)^2 + \frac{18 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{10 \times 6.4 \times 10^6} \]

7. **Calculate the Value:**
- Calculate \( \frac{18 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{10 \times 6.4 \times 10^6} \):
\[ = \frac{18 \times 6.67 \times 6}{10 \times 6.4} \times 10^{7} \approx 9.81 \times 10^7 \, \text{m}^2/\text{s}^2 \]
- Now calculate \( v^2 \):
\[ v^2 \approx 144000000 + 9.81 \times 10^7 = 256000000 \]
- Taking the square root gives:
\[ v \approx \sqrt{256000000} \approx 16000 \, \text{m/s} = 16 \, \text{km/s} \]

### Final Answer:
The velocity of the asteroid when it reaches the surface of the Earth is approximately **16 km/s**.