Orbital period of geosynchronous “system” satellites Geosynchronous communication satellites that appear fixed over one longitude complete one orbit around Earth in approximately how many hours?

Introduction / Context:
Geosynchronous satellites match Earth’s sidereal rotation rate so that, viewed from the ground, they revisit the same position over the equator each day (geostationary if zero inclination and eccentricity). This principle is central to fixed satellite services and broadcast distribution.

Given Data / Assumptions:

• Altitude ~35,786 km above Earth’s equator.
• Negligible eccentricity and inclination for geostationary case.
• Sidereal day period ~23 h 56 min (~24 h commonly stated).

Concept / Approach:

Kepler’s third law sets orbital period versus semi-major axis. At GEO altitude, the orbital period equals Earth’s rotational period, hence about 24 hours. This synchrony keeps the satellite’s sub-satellite point nearly constant in longitude.

Step-by-Step Solution:

Verification / Alternative check:

Any deviation from 24 h produces an apparent drift in the sky; station-keeping maintains the near-constant position.

Why Other Options Are Wrong:

12 h corresponds to MEO like GPS; 6 h/1 h are LEO/MEO-like times, not GEO.

Common Pitfalls:

Mixing solar day (24 h) with sidereal day (≈23 h 56 min); for general purposes, 24 h is acceptable.

Final Answer:

24 hours