Question

The height of a geostationary satellite is _________

• A. 1000km
• B. 36000km
• C. 650km
• D. 2000km

Answer 1

The Correct Option is B

We need to determine the height of a geostationary satellite above Earth's surface, given the correct answer is approximately 36,000 km.

1. Geostationary Orbit Characteristics:
A geostationary satellite orbits Earth in the equatorial plane with an orbital period equal to Earth's rotation period, approximately 24 hours (86,400 seconds). This ensures the satellite remains fixed relative to a point on Earth's surface.

2. Orbital Dynamics:
For a circular orbit, the centripetal force required for the satellite's motion is provided by Earth's gravitational force:

\( \frac{m v^2}{r} = \frac{G M m}{r^2} \)
where \( m \) is the satellite's mass, \( v \) is the orbital velocity, \( r \) is the distance from Earth's center, \( G = 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \) is the gravitational constant, and \( M = 5.972 \times 10^{24} \, \text{kg} \) is Earth's mass. Simplifying:

\( v = \sqrt{\frac{G M}{r}} \)
The orbital velocity is also related to the period \( T \):

\( v = \frac{2 \pi r}{T} \)
Equating:

\( \sqrt{\frac{G M}{r}} = \frac{2 \pi r}{T} \)

3. Solving for Orbital Radius:
Square both sides:

\( \frac{G M}{r} = \frac{4 \pi^2 r^2}{T^2} \)
Rearrange:

\( r^3 = \frac{G M T^2}{4 \pi^2} \)
Using \( G M \approx 3.986 \times 10^{14} \, \text{m}^3 \text{s}^{-2} \), \( T = 86,400 \, \text{s} \), \( T^2 \approx 7.465 \times 10^9 \, \text{s}^2 \), and \( 4 \pi^2 \approx 39.478 \):

\( r^3 = \frac{3.986 \times 10^{14} \times 7.465 \times 10^9}{39.478} \approx 7.531 \times 10^{22} \, \text{m}^3 \)
\( r = (7.531 \times 10^{22})^{1/3} \approx 42,210 \, \text{km} \)

4. Calculating Height:
The height \( h \) is the altitude above Earth's surface, so subtract Earth's mean radius (\( R_e \approx 6,371 \, \text{km} \)):

\( h = r - R_e = 42,210 - 6,371 \approx 35,839 \, \text{km} \)
This value is often rounded to approximately 36,000 km in standard references due to slight variations in constants (e.g., \( R_e \approx 6,378 \, \text{km} \) or refined \( G M \)) or simplified calculations.

Final Answer:
The height of a geostationary satellite is approximately \( 36,000 \, \text{km} \).