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Satellite Orbital Geometry Problem

Solve the spacecraft orbital geometry shown in the figure using the Law of Sines. Then find the percentage of earth coverage for a geosynchronous altitude and for an altitude of 300km which is often the orbital trajectory of the US space shuttle. Solve for elevation angles of 0 and 10 degrees.

 Solution using the Law of Sines: sin (90 - γ - φel) / (re) = sin (90 + φel) / (re + h) Where: sin (90 + α) = cos (α) sin (90 - α) = cos ( - α) = cos (α) So: cos (γ + φel) / (re) = cos (φel) / (re + h) Solving for gamma, our solution is: γ = cos-1[ (re)(cos (φel)) / (re + h) ] - φel

Next, we were asked to determine the percentage of signal coverage from two different orbits, a geo orbit of 35,872km and a space shuttle orbit of 300km.

% of Earth coverage = 0.5 (1 - cos (γ))

h = 35872km and 300km

el = 0 deg and 10 deg

For Elevation = 0 degrees: a Height of 300km = 2.25% coverage, and a Height at Geo = 42.46% coverage

For Elevation = 10 degrees: a Height of 300km = 0.74% coverage, and a Height at Geo = 34.10% coverage

Satellite key words: orbit, geometry, transmission coverage, spaceshuttle orbit, RF transmission, orbit, GEO, geosynchronous, earth station, station keeping, orbital parameters, signal path, atmospheric attenuation, path length, line of sight, LOS, ES, uplink, downlink, relay, handoff, handshaking, data pipe, angular dimensions.

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