To solve the problem of calculating the orbital velocity and period of revolution of an artificial satellite orbiting the Earth at a distance of 3400 km, we can follow these steps:
### Step 1: Determine the total distance from the center of the Earth to the satellite.
The radius of the Earth (R) is given as 6400 km. The satellite is orbiting at a distance of 3400 km above the Earth's surface. Therefore, the total distance (r) from the center of the Earth to the satellite is:
\[
r = R + \text{height of the satellite} = 6400 \, \text{km} + 3400 \, \text{km} = 9800 \, \text{km}
\]
### Step 2: Convert the distance from kilometers to meters.
Since we need to use SI units, we convert the distance from kilometers to meters:
\[
r = 9800 \, \text{km} \times 1000 \, \text{m/km} = 9.8 \times 10^6 \, \text{m}
\]
### Step 3: Calculate the gravitational force acting on the satellite.
The gravitational force (F) acting on the satellite can be expressed using the formula:
\[
F = \frac{GMm}{r^2}
\]
Where:
- \( G \) is the gravitational constant, \( G \approx 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) is the mass of the Earth (approximately \( 5.97 \times 10^{24} \, \text{kg} \))
- \( m \) is the mass of the satellite (which will cancel out later).
### Step 4: Use the formula for orbital velocity.
The orbital velocity (v) of the satellite can be calculated using the formula:
\[
v = \sqrt{\frac{GM}{r}}
\]
### Step 5: Substitute the values into the orbital velocity formula.
We can substitute the values of \( G \) and \( M \) into the equation:
\[
v = \sqrt{\frac{(6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2)(5.97 \times 10^{24} \, \text{kg})}{9.8 \times 10^6 \, \text{m}}}
\]
Calculating this gives:
\[
v \approx \sqrt{\frac{3.986 \times 10^{14}}{9.8 \times 10^6}} \approx \sqrt{4.06 \times 10^7} \approx 6377 \, \text{m/s}
\]
### Step 6: Calculate the period of revolution.
The period of revolution (T) can be calculated using the formula:
\[
T = \frac{2\pi r}{v}
\]
Substituting the values we have:
\[
T = \frac{2\pi (9.8 \times 10^6 \, \text{m})}{6377 \, \text{m/s}}
\]
Calculating this gives:
\[
T \approx \frac{6.16 \times 10^7}{6377} \approx 9660 \, \text{s}
\]
### Final Results:
- **Orbital Velocity (v)**: Approximately \( 6377 \, \text{m/s} \)
- **Period of Revolution (T)**: Approximately \( 9660 \, \text{s} \) or about \( 2.68 \, \text{hours} \)
