Velocity of a geosynchronous satellite Assuming Earth radius = 6400 km and geosynchronous satellite altitude ≈ 36,000 km, estimate the orbital velocity in km/h.

Introduction / Context:
Geosynchronous orbits have a period equal to Earth’s rotation (~24 h). At the geostationary radius, a satellite’s tangential velocity follows directly from distance traveled in one orbital period.

Given Data / Assumptions:

• Earth radius R_E ≈ 6400 km.
• GEO altitude h ≈ 36,000 km.
• Orbital radius R = R_E + h ≈ 42,400 km.
• Period T ≈ 24 h (using 24 for simplicity; sidereal day is ~23.934 h).

Concept / Approach:
The distance traveled per orbit is the circumference: 2 * pi * R. The average orbital speed v is distance / time. A quick calculation provides the expected ~11,000 km/h velocity.

Step-by-Step Solution:
Compute orbital radius: R = 6400 + 36,000 = 42,400 km.Compute circumference: C = 2 * pi * 42,400 ≈ 2 * 3.1416 * 42,400 ≈ 266,000 km (approx.).Compute speed: v = C / T ≈ 266,000 km / 24 h ≈ 11,083 km/h ≈ 11,100 km/h.

Verification / Alternative check:
Using the sidereal day T = 23.934 h yields ~11,120 km/h, confirming the order of magnitude and the option rounding.

Why Other Options Are Wrong:

• 28,000 or 36,000 km/h: too high for GEO; such speeds correspond to lower orbits.
• 15,000 km/h: still high compared to the computed value.
• 7,900 km/h: closer to LEO orbital speeds in m/s (≈7.9 km/s), not km/h.

Common Pitfalls:

• Mixing units: km/s vs km/h.
• Forgetting to add Earth’s radius to altitude for orbital radius.

Final Answer:
11,100 km/h