A particle is projected from earth's surface with velocity such that is rises to a height equal to earth's radius. Find the approximate value of initial velocity of projection in km/s. Assuming R = 6400 km, `M = 6xx10^(24)` kg.

Answer

To solve the problem of finding the initial velocity of a particle projected from the Earth's surface to a height equal to the Earth's radius, we can use the principle of conservation of energy. Here’s the step-by-step solution:

### Step 1: Understand the Problem
We need to find the initial velocity \( v \) required for a particle to reach a height \( h \) equal to the Earth's radius \( R \). Given:
- \( R = 6400 \) km
- \( M = 6 \times 10^{24} \) kg

### Step 2: Set Up the Conservation of Energy Equation
At the Earth's surface (initial position), the total energy \( E_i \) is the sum of kinetic energy \( KE \) and gravitational potential energy \( PE \):
\[ E_i = KE + PE = \frac{1}{2} m v^2 - \frac{GMm}{R} \]
At the maximum height \( h = R \) (final position), the total energy \( E_f \) is:
\[ E_f = PE = -\frac{GMm}{2R} \]
where \( G \) is the gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

### Step 3: Apply Conservation of Energy
According to the conservation of energy:
\[ E_i = E_f \]
Substituting the expressions for \( E_i \) and \( E_f \):
\[ \frac{1}{2} m v^2 - \frac{GMm}{R} = -\frac{GMm}{2R} \]

### Step 4: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 - \frac{GM}{R} = -\frac{GM}{2R} \]
Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{GM}{R} - \frac{GM}{2R} \]
\[ \frac{1}{2} v^2 = \frac{GM}{2R} \]

### Step 5: Solve for \( v \)
Multiplying both sides by 2:
\[ v^2 = \frac{GM}{R} \]
Taking the square root:
\[ v = \sqrt{\frac{GM}{R}} \]

### Step 6: Substitute Values
Now, we substitute the values of \( G \), \( M \), and \( R \):
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M = 6 \times 10^{24} \, \text{kg} \)
- \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)

Calculating \( v \):
\[ v = \sqrt{\frac{(6.67 \times 10^{-11})(6 \times 10^{24})}{6400 \times 10^3}} \]

### Step 7: Calculate the Result
Calculating the numerator:
\[ 6.67 \times 10^{-11} \times 6 \times 10^{24} = 4.002 \times 10^{14} \]
Calculating the denominator:
\[ 6400 \times 10^3 = 6.4 \times 10^6 \]
Thus,
\[ v = \sqrt{\frac{4.002 \times 10^{14}}{6.4 \times 10^6}} = \sqrt{6.25 \times 10^7} \approx 7900 \, \text{m/s} \approx 8 \, \text{km/s} \]

### Final Answer
The approximate value of the initial velocity of projection is:
\[ \boxed{8 \, \text{km/s}} \]