Satellite communications coverage computation Assuming a minimum elevation angle of 5°, what percentage of Earth’s surface remains within the line-of-sight (LOS) coverage of a single geosynchronous satellite? Provide the closest value and consider standard Earth radius and geostationary altitude.

Introduction / Context:
In satellite communications, a key planning task is estimating what fraction of Earth’s surface can be “seen” by a geosynchronous satellite above a specified minimum elevation angle. This determines how many satellites are needed for regional or near-global coverage and informs link budgets, gateway siting, and service availability.

Given Data / Assumptions:

• Geostationary orbit height H ≈ 35,786 km above Earth’s surface.
• Mean Earth radius Re ≈ 6,378 km.
• Minimum ground elevation angle e = 5° (to account for terrain, clutter, and atmospheric effects).
• LOS coverage is modeled as a spherical cap on Earth.

Concept / Approach:

The coverage half-angle on Earth’s center, called the geocentric angle ψ, depends on orbit geometry and the elevation constraint. For a satellite at distance Re + H from Earth’s center, the common approximation for the cosine of ψ with an elevation limit e is:

The visible area is a spherical cap with area fraction f over Earth given by:

Step-by-Step Solution:

Verification / Alternative check:

For e = 0°, cos(ψ) ≈ 0.1512 giving f ≈ 42.44%. The elevation mask of 5° changes the value only marginally, confirming that ~42.4% is a robust estimate.

Why Other Options Are Wrong:

25% and 30.3% significantly under-estimate GEO coverage. 50% over-estimates; no single GEO spacecraft can view half of Earth considering curvature. 38.5% is closer but still low for e = 5°.

Common Pitfalls:

Confusing ground elevation angle with look angle at the satellite, or using flat-Earth geometry. Also, forgetting that the area fraction for a spherical cap is (1 − cos(ψ)) / 2 leads to large errors.

Final Answer:

42.4