Find the orbital velocity of an artifical satellite of the earth in an orbital close to the earth?

Explanation

To find the orbital velocity of an artificial satellite in an orbit close to the Earth, we can use the following steps:

### Step 1: Understand the Formula for Orbital Velocity
The formula for the orbital velocity \( v \) of a satellite is given by:
\[ v = \sqrt{\frac{GM}{r}} \]
where:
- \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \text{m}^3/\text{kg s}^2 \)),
- \( M \) is the mass of the Earth (\( 5.972 \times 10^{24} \, \text{kg} \)),
- \( r \) is the distance from the center of the Earth to the satellite.

### Step 2: Determine the Value of \( r \)
Since the satellite is in an orbit close to the Earth, we can approximate \( r \) as the radius of the Earth. The average radius of the Earth is approximately:
\[ r \approx 6.4 \times 10^6 \, \text{m} \]

### Step 3: Substitute Values into the Orbital Velocity Formula
Now we can substitute the values of \( G \), \( M \), and \( r \) into the formula:
\[ v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.674 \times 10^{-11} \, \text{m}^3/\text{kg s}^2)(5.972 \times 10^{24} \, \text{kg})}{6.4 \times 10^6 \, \text{m}}} \]

### Step 4: Calculate the Value Inside the Square Root
Calculating the numerator:
\[ GM = (6.674 \times 10^{-11})(5.972 \times 10^{24}) \approx 3.986 \times 10^{14} \, \text{m}^3/\text{s}^2 \]
Now, substituting this back into the equation:
\[ v = \sqrt{\frac{3.986 \times 10^{14}}{6.4 \times 10^6}} \]

### Step 5: Perform the Division
Calculating the division:
\[ \frac{3.986 \times 10^{14}}{6.4 \times 10^6} \approx 6.22 \times 10^{7} \, \text{m}^2/\text{s}^2 \]

### Step 6: Take the Square Root
Now, taking the square root:
\[ v \approx \sqrt{6.22 \times 10^{7}} \approx 7900 \, \text{m/s} \]

### Final Answer
Thus, the orbital velocity of an artificial satellite in an orbit close to the Earth is approximately:
\[ v \approx 7900 \, \text{m/s} \]

---