An artifical satellite cicles round the earth at a distance of 3400 km. Calculate the orbital velocity. Given the radius of the earth is 6400km. G` = 9.8 ms^(-2)`.

Explanation

To calculate the orbital velocity of an artificial satellite circling the Earth at a distance of 3400 km, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Distance of the satellite from the Earth's surface (h) = 3400 km
- Radius of the Earth (R) = 6400 km
- Acceleration due to gravity (g) = 9.8 m/s²

2. **Convert the Distances to Meters**:
- Convert the radius of the Earth and the height of the satellite from kilometers to meters:
\[ R = 6400 \text{ km} = 6400 \times 10^3 \text{ m} = 6.4 \times 10^6 \text{ m} \]
\[ h = 3400 \text{ km} = 3400 \times 10^3 \text{ m} = 3.4 \times 10^6 \text{ m} \]

3. **Calculate the Total Distance from the Center of the Earth**:
- The total distance (r) from the center of the Earth to the satellite is the sum of the Earth's radius and the height of the satellite:
\[ r = R + h = 6.4 \times 10^6 \text{ m} + 3.4 \times 10^6 \text{ m} = 9.8 \times 10^6 \text{ m} \]

4. **Use the Formula for Orbital Velocity**:
- The formula for the orbital velocity (V) at a height \( h \) is given by:
\[ V = \sqrt{\frac{g \cdot R^2}{r}} \]
- Substitute the known values into the formula:
\[ V = \sqrt{\frac{9.8 \text{ m/s}^2 \cdot (6.4 \times 10^6 \text{ m})^2}{9.8 \times 10^6 \text{ m}}} \]

5. **Calculate the Orbital Velocity**:
- First, calculate \( (6.4 \times 10^6)^2 \):
\[ (6.4 \times 10^6)^2 = 40.96 \times 10^{12} \text{ m}^2 \]
- Now substitute this back into the equation:
\[ V = \sqrt{\frac{9.8 \cdot 40.96 \times 10^{12}}{9.8 \times 10^6}} \]
- Simplifying gives:
\[ V = \sqrt{40.96 \times 10^6} \text{ m/s} \]
- Finally, calculate the square root:
\[ V \approx 6400 \text{ m/s} \]

### Final Answer:
The orbital velocity of the satellite is approximately **6400 m/s**.