An asteroid of mass `m` is approaching earth, initially at a distance `10R_(E)` with speed `v_(i)`. It hits earth with a speed `v_(f)` (`R_(E)` and `M_(E)` are radius and mass of earth),. Then

Options

A. `v_(f)^(2)=v_(i)^(2)+(2Gm)/(R_(E))(1+(1)/(10))`
B. `v_(f)^(2)=v_(i)^(2)+(2GM_(E))/(R_(E))(1+(1)/(10))`
C. `v_(f)^(2)=v_(i)^(2)+(2GM_(E))/(R_(E))(1-(1)/(10))`
D. `v_(f)^(2)=v_(i)^(2)+(2Gm)/(R_(E))(1-(1)/(10))`

Explanation

To solve the problem of an asteroid approaching Earth and colliding with it, we will use the principle of conservation of mechanical energy. The total mechanical energy (kinetic + potential) at the initial position must equal the total mechanical energy at the final position.

### Step-by-Step Solution:

1. **Identify Initial and Final States:**
- Initial distance from Earth: \(10R_E\)
- Initial speed of the asteroid: \(v_i\)
- Final distance from Earth (when it hits): \(R_E\)
- Final speed of the asteroid: \(v_f\)
- Mass of the asteroid: \(m\)
- Mass of the Earth: \(M_E\)
- Gravitational constant: \(G\)

2. **Write the Initial Energy:**
- The initial kinetic energy (KE_initial) is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m v_i^2 \]
- The initial potential energy (PE_initial) is given by:
\[ PE_{\text{initial}} = -\frac{G M_E m}{10 R_E} \]
- Therefore, the total initial energy (E_initial) is:
\[ E_{\text{initial}} = KE_{\text{initial}} + PE_{\text{initial}} = \frac{1}{2} m v_i^2 - \frac{G M_E m}{10 R_E} \]

3. **Write the Final Energy:**
- The final kinetic energy (KE_final) is given by:
\[ KE_{\text{final}} = \frac{1}{2} m v_f^2 \]
- The final potential energy (PE_final) is given by:
\[ PE_{\text{final}} = -\frac{G M_E m}{R_E} \]
- Therefore, the total final energy (E_final) is:
\[ E_{\text{final}} = KE_{\text{final}} + PE_{\text{final}} = \frac{1}{2} m v_f^2 - \frac{G M_E m}{R_E} \]

4. **Set Initial Energy Equal to Final Energy:**
\[ \frac{1}{2} m v_i^2 - \frac{G M_E m}{10 R_E} = \frac{1}{2} m v_f^2 - \frac{G M_E m}{R_E} \]

5. **Simplify the Equation:**
- Cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ \frac{1}{2} v_i^2 - \frac{G M_E}{10 R_E} = \frac{1}{2} v_f^2 - \frac{G M_E}{R_E} \]
- Rearranging gives:
\[ \frac{1}{2} v_f^2 = \frac{1}{2} v_i^2 + \frac{G M_E}{R_E} - \frac{G M_E}{10 R_E} \]
- Combine the potential energy terms:
\[ \frac{1}{2} v_f^2 = \frac{1}{2} v_i^2 + \frac{G M_E}{R_E} \left(1 - \frac{1}{10}\right) \]
- This simplifies to:
\[ \frac{1}{2} v_f^2 = \frac{1}{2} v_i^2 + \frac{9 G M_E}{10 R_E} \]

6. **Multiply by 2 to Solve for \(v_f^2\):**
\[ v_f^2 = v_i^2 + \frac{18 G M_E}{10 R_E} \]
- This can be rewritten as:
\[ v_f^2 = v_i^2 + \frac{9 G M_E}{5 R_E} \]

### Final Result:
The final velocity \(v_f\) of the asteroid when it hits the Earth is given by:
\[ v_f = \sqrt{v_i^2 + \frac{9 G M_E}{5 R_E}} \]