A particle is fired vertically upward fom earth's surface and it goes up to a maximum height of 6400 km. find the initial speed of particle.

Explanation

To find the initial speed of a particle fired vertically upward from the Earth's surface that reaches a maximum height of 6400 km, we can use the principle of conservation of mechanical energy.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Maximum height \( h = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)
- Radius of the Earth \( R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)
- Gravitational constant \( G \) and mass of the Earth \( M \) will be used in calculations.

2. **Understand the Energy Conservation Principle:**
The total mechanical energy (potential energy + kinetic energy) at the Earth's surface must equal the total mechanical energy at the maximum height. At the maximum height, the kinetic energy will be zero because the particle comes to a stop.

3. **Write the Energy Conservation Equation:**
\[ \text{Initial Potential Energy} + \text{Initial Kinetic Energy} = \text{Final Potential Energy} + \text{Final Kinetic Energy} \]
\[ -\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{2R} + 0 \]
Here, \( m \) is the mass of the particle, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

4. **Simplify the Equation:**
The mass \( m \) cancels out from both sides:
\[ -\frac{GM}{R} + \frac{1}{2}v^2 = -\frac{GM}{2R} \]
Rearranging gives:
\[ \frac{1}{2}v^2 = -\frac{GM}{2R} + \frac{GM}{R} \]
\[ \frac{1}{2}v^2 = \frac{GM}{2R} \]

5. **Solve for Initial Speed \( v \):**
Multiply both sides by 2:
\[ v^2 = \frac{GM}{R} \]
Taking the square root:
\[ v = \sqrt{\frac{GM}{R}} \]

6. **Substitute Known Values:**
- Use \( G = 6.67 \times 10^{-11} \, \text{m}^3/\text{kg s}^2 \)
- Mass of the Earth \( M = 6 \times 10^{24} \, \text{kg} \)
- Radius of the Earth \( R = 6.4 \times 10^6 \, \text{m} \)

Now, substituting these values:
\[ v = \sqrt{\frac{(6.67 \times 10^{-11})(6 \times 10^{24})}{6.4 \times 10^6}} \]

7. **Calculate \( v \):**
\[ v = \sqrt{\frac{4.002 \times 10^{14}}{6.4 \times 10^6}} = \sqrt{6.25 \times 10^7} \approx 7900 \, \text{m/s} \]

8. **Convert to km/s:**
\[ v \approx 7.9 \, \text{km/s} \]

### Final Answer:
The initial speed of the particle is approximately \( 7.9 \, \text{km/s} \).