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-by-Step Solution:

1. **Understand the Definition of a Geostationary Satellite**:
- A geostationary satellite is one that remains stationary relative to a point on the Earth's surface. This means it orbits the Earth at the same rotational speed as the Earth itself.

2. **Identify the Relevant Forces**:
- The gravitational force acting on the satellite provides the necessary centripetal force to keep it in orbit. The gravitational force \( F_g \) can be expressed as:
\[ F_g = \frac{G M m}{A^2} \]
where:
- \( G \) is the universal gravitational constant (\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the Earth (\( 6 \times 10^{24} \, \text{kg} \)),
- \( m \) is the mass of the satellite,
- \( A \) is the orbital radius (distance from the center of the Earth to the satellite).

3. **Centripetal Force Requirement**:
- The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:
\[ F_c = \frac{m v^2}{A} \]
where \( v \) is the orbital speed of the satellite.

4. **Set the Forces Equal**:
- For the satellite to remain in a stable orbit, the gravitational force must equal the centripetal force:
\[ \frac{G M m}{A^2} = \frac{m v^2}{A} \]

5. **Cancel Out the Mass of the Satellite**:
- Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{G M}{A^2} = \frac{v^2}{A} \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ v^2 = \frac{G M}{A} \]
- Taking the square root:
\[ v = \sqrt{\frac{G M}{A}} \]

7. **Substituting Known Values**:
- The orbital radius \( A \) for a geostationary satellite is approximately \( 4.2 \times 10^7 \, \text{m} \).
- Substituting the values:
\[ v = \sqrt{\frac{(6.67 \times 10^{-11}) (6 \times 10^{24})}{4.2 \times 10^7}} \]

8. **Calculating the Value**:
- First, calculate the numerator:
\[ G M = 6.67 \times 10^{-11} \times 6 \times 10^{24} \approx 4.002 \times 10^{14} \]
- Now divide by \( A \):
\[ \frac{4.002 \times 10^{14}}{4.2 \times 10^7} \approx 9.52 \times 10^6 \]
- Now take the square root:
\[ v \approx \sqrt{9.52 \times 10^6} \approx 3086.84 \, \text{m/s} \]

9. **Convert to km/s**:
- To convert meters per second to kilometers per second, divide by 1000:
\[ v \approx 3.08684 \, \text{km/s} \approx 3.09 \, \text{km/s} \]

10. **Final Answer**:
- The orbital speed of a geostationary satellite is approximately **3.09 km/s**.