The orbital speed of geostationary satellite is around

Options

A. 3.07 km/s
B. 7.95 km/s
C. 11.2 km/s
D. 15.8 km/s

Answer

To find the orbital speed of a geostationary satellite, we can follow these steps:

### Step 1: Understand the concept of a geostationary satellite
A geostationary satellite is one that orbits the Earth at a height where it appears to be stationary relative to the surface of the Earth. This means it has the same angular velocity as the Earth.

### Step 2: Know the height of a geostationary satellite
The height (h) of a geostationary satellite above the Earth's surface is approximately 35,800 kilometers.

### Step 3: Use the formula for orbital speed
The orbital speed (v) of a satellite can be derived from the gravitational force acting on it. The formula is given by:
\[ v = \sqrt{\frac{GM}{r + h}} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( r \) is the radius of the Earth,
- \( h \) is the height of the satellite above the Earth's surface.

### Step 4: Substitute known values
1. The radius of the Earth (r) is approximately 6400 kilometers or \( 6.4 \times 10^6 \) meters.
2. The height (h) of the geostationary satellite is 35,800 kilometers or \( 3.58 \times 10^7 \) meters.
3. The gravitational constant \( G \) is approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).
4. The mass of the Earth \( M \) is approximately \( 5.972 \times 10^{24} \, \text{kg} \).

### Step 5: Calculate the total distance from the center of the Earth
The total distance (R) from the center of the Earth to the satellite is:
\[ R = r + h = 6.4 \times 10^6 + 3.58 \times 10^7 = 4.22 \times 10^7 \, \text{m} \]

### Step 6: Calculate the orbital speed
Now substituting the values into the formula:
\[ v = \sqrt{\frac{GM}{R}} \]

Using the approximation \( g = \frac{GM}{r^2} \) where \( g \) is the acceleration due to gravity at the surface of the Earth (approximately \( 9.8 \, \text{m/s}^2 \)):
\[ v = \sqrt{g \cdot r \cdot \left(1 + \frac{h}{r}\right)} \]

### Step 7: Substitute \( g \) and \( r \)
Substituting \( g = 9.8 \, \text{m/s}^2 \) and \( r = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \):
\[ v = \sqrt{9.8 \cdot 6.4 \times 10^6 \cdot \left(1 + \frac{3.58 \times 10^7}{6.4 \times 10^6}\right)} \]

### Step 8: Simplify and calculate
Calculating the term inside the square root:
\[ 1 + \frac{3.58 \times 10^7}{6.4 \times 10^6} \approx 1 + 5.594 \approx 6.594 \]
Now calculate:
\[ v \approx \sqrt{9.8 \cdot 6.4 \times 10^6 \cdot 6.594} \]
After calculating, you will find:
\[ v \approx 3.07 \, \text{km/s} \]

### Final Answer
The orbital speed of a geostationary satellite is approximately **3.07 km/s**.

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