The minimum projection velocity of a body from the earth's surface so that it becomes the satellite of the earth `(R_(e)=6.4xx10^(6) m)`.

Options

A. `11xx10^(3) m//s`
B. `8xx10^(3) m//s`
C. `6.4xx10^(3) m//s`
D. `4xx10^(3) m//s`

Answer

To find the minimum projection velocity of a body from the Earth's surface so that it becomes a satellite of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Satellite:**
The gravitational force acting on the satellite is given by the formula:
\[ F = \frac{G M m}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the Earth to the satellite.

2. **Equate Gravitational Force to Centripetal Force:**
For a satellite in circular motion, the centripetal force required to keep it in orbit is provided by the gravitational force. Thus, we can write:
\[ \frac{m v^2}{r} = \frac{G M m}{r^2} \]
Here, \( v \) is the orbital velocity of the satellite.

3. **Cancel Out Common Terms:**
Since the mass of the satellite \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{v^2}{r} = \frac{G M}{r^2} \]

4. **Rearranging the Equation:**
Rearranging gives us:
\[ v^2 = \frac{G M}{r} \]

5. **Taking the Square Root:**
To find \( v \), we take the square root of both sides:
\[ v = \sqrt{\frac{G M}{r}} \]

6. **Substituting Values:**
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M}{R_e^2} \]
where \( R_e \) is the radius of the Earth. Therefore, we can express \( G M \) as:
\[ G M = g R_e^2 \]
Substituting this back into our equation for \( v \):
\[ v = \sqrt{\frac{g R_e^2}{R_e}} = \sqrt{g R_e} \]

7. **Using Given Values:**
We are given \( R_e = 6.4 \times 10^6 \, \text{m} \) and \( g \approx 10 \, \text{m/s}^2 \):
\[ v = \sqrt{10 \times (6.4 \times 10^6)} = \sqrt{6.4 \times 10^7} \]

8. **Calculating the Final Value:**
\[ v = \sqrt{6.4} \times 10^{3.5} \approx 8 \times 10^3 \, \text{m/s} \]

### Final Answer:
The minimum projection velocity of a body from the Earth's surface so that it becomes a satellite of the Earth is approximately:
\[ v \approx 8 \times 10^3 \, \text{m/s} \]