More on Orbits, Part 3

So far we haven't accounted for the eccentricity of the orbit. We can get this number from the appropriate line in the almanac file. A value of zero means the orbit is circular. As the value moves towards 1, the orbit is more and more enlongated. GPS satellites have very circular orbits, with values less than 0.02. Comets typically have values much closer to 1.

From mean anomaly, we find the eccentric anomaly E, from mean anomaly M that we found in the previous post, and eccentricity ε.

M = E - ε sin E

This equation needs to be solved iteratively. Newton's method can be used to get a very accurate value after a few loops, by finding the root of the equation:

f(E) = E - ε sin E - M = 0

The derivative with respect to E is f '(E) , or: 1 - ε cos E

We can start with a first guess E₀ = M. Take E₁ = E₀ - f(E₀)/f '(E₀)

So E₁ = E₀ - (E₀ - ε sin E₀ - M) / (1 - ε cos E₀)

Then find E₂ with E₁ in the above function, and so on. You can check E_n in the original equation and keep going until E_n causes M to converge to the correct value.

E gives us the angle from the center of the ellipse to the satellite. Note that the center is the point between the two foci of the ellipse. The earth will be at one focus, not at the center.

At this point we can get the radius r which represents the distance of the satellite from the center of the earth.

r = a * (1 - ε * cos E)

Finally, to get the angle of the satellite position relative to earth, we convert to true anomaly ν.

cos ν = (cos E - ε) / (1 - ε * cos E)

Now we have the position in polar coordinates (r, ν) of the satellite in the orbital plane. There is still a little more work to get a current latitude and longitude position.