Friday, 15 March 2013

Angle of Elevation of a Geostationary Satellite

In a previous post, many months ago, I calculated the height of a geostationary satellite using the laws of physics which relate to gravity and circular motion.  This time, I'll use that information to deduce the angle of elevation of a given geostationary satellite, but I'll take the simplified model where the satellite is at the same angle of longitude as the observer (i.e.on the same meridian).  The maths enters three dimensions when the observer is in Europe but the satellite is geostationary over Central America, instead of North Africa.

Drawing a simple diagram will help to outline the situation, and show how the key parts of the model fit together.

A = centre of the Earth
B = position of observer on the Earth's surface
C = satellite

Angle alpha is the angle of longitude of the observer (how far north, or south, of the equator they are).  For this example, I will be using a longitude of 50 degrees north (northern France/southern England).
The angle at B is 90 degrees (the angle between the radius from A and the horizon) plus the angle of elevation, beta, which we are looking to solve.

Lenth a is the straight-line distance from the observer to the satellite
Length b is the distance from the centre of the earth to the satellite, re (radius of earth) plus rs (altitude of satellite, measured from earth's surface)
Length c is the radius of the earth

We only know two of the lengths (b and c) and the included angle, alpha, so we must start solving the triangle by using the cosine rule:

In order to find angle B, and hence beta, we will first need to find length a.Substituting the known lengths and angle into the cosine rule, we get:

a2 = (re+ rs)2 + re2 - (2 x (re+ rs) x re x cos 50)

a2= 42,164,0002+ 6,378,0002- (2 x 42164000 x 6378000 x 0.6428)
a2= 1.473 x 1015 m2
a = 38,376,585 m

Now that we know all three sides, we can use the sine rule to determine an angle by knowing one other angle and the two opposite sides.  I will calculate angle C and then subtract A + C from 180 degrees to find C.

a / sin A = c / sin C

a = 38,376,585 m (as calculated above)
A = angle of longitude of the observer

c = distance from Earth centre to geostationary satellite, which was calculated previously as 42,164,000 m.
C = angle on diagram; angle at satellite between centre of Earth and observer on the ground.

So, by rearranging, we have

sin A) / a = sin C

If A = 50 degrees, then by substitution C = 7.3143 degrees

Therefore, B = 180 - (50 + 7.3143) = 122.6 degrees, and beta = 122.6 - 90 = 32.6 degrees.

There is an alternative route to finding angle beta, and that's by dividing the triangle ABC into two right-angled triangles by dropping a perpendicular from B onto the line AC, see below.  Angle Z = 90 degrees.

Firstly, calculate the distance BZ, which is common to both triangles ABZ and BCZ.  This can be done by simple trigonometry since angle Z is 90 degrees, and angle A is known (or determined by the observer):

sin A = BZ/ re  and where A = 50 degrees, BZ = 4,885,831 m.

Next, calculate AZ in the same triangle ABZ:

cos A = AZ/ re  and where A = 50 degrees, AZ = 4099699  m.

As we now know AZ, we can calculate CZ, and hence identify two of the sides of triangle BCZ.

CZ = AC (distance from centre of Earth to geostationary satellite) minus AZ
CZ = 42,164,000 - 4,099,699 m = 38,064,300 m

Finally:  angle B in triangle BCZ

tan B = CZ / BZ = 38,064,300 / 4,885,831  = 7.7909
B = 82.685 degrees

Now, we know that angle A = 50 degrees, so angle ABZ = 40 degrees.
Angle ABC = 40 degrees + 82.685 degrees = 122.685 degrees.
We want to know the angle between the observer's horizon and the satellite; since the angle AB and the horizon is 90 degrees ('the angle between a radius and a tangent is 90 degrees') this is simply 122.685 degrees - 90 degrees = 32.685 degrees... which agrees with the result from the first method.

QED :-)


  1. Thank you so much for this Precious article this article solved my problem ....... But one thing i found is a little error like from where this value comes ""C = 7.3143 degrees"" after continue efforts i found the mistake that you put c = 42,164,000 m in calculation but in real c = 6378000 so it will gives a 7.3143 degree...... Visit my website

  2. 'Angle alpha is the angle of longitude of the observer (how far north, or south, of the equator they are)' This should be latitude right?