Question

A geostationary satellite is 42000km above the center of earth. If the satellite latitude is 50o north with respect to the earth station and the spherical earth mean radius is 6371 km. Calculate the satellite elevation angle.

Answer 1

The diagram shows a vertical cross-section of Earth.

S is the Earth station

G is the geostationary satellite

C is the centre of earth.

We assume that the earth station lies on the equator. So, the stationary satellite lies at 50° latitude

According to the question, ∠GCS = 50°

Given,

CS = radius of earth = 6371 Km

CG = Distance of satellite from centre of earth = 42000 Km

θ is the required angle of elevation

From Triangle CNS,

cos 50° = CS/CN

=> CN = CS/(cos 50°) = 6371/(cos 50°) = 9911.52 Km

Also,

tan 50° = SN/CS

=> SN = CS tan 50° = 6371 tan 50° = 7592.66 Km

External angle = sum of opposite internal angles

=> ∠SNG = ∠NCS + ∠NSC = 50° + 90° = 140°

NG = CG - CN = 42000-9911.52 = 32088.48 Km

From law of cosines,

SG² = SN² + NG² - 2(SN)(NG)cos(∠SNG) = 1460592450

SG = 38217.70 Km

From law of sines,

=>

Putting in the values,

sin θ = 0.5397

=> θ = sin^-1(0.5397) = 32.66°

Hence, the required angle of elevation is 32.66°