I'm a programmer trying to model satellite orbits. What formula can I use to determine the latitude of a satellite after a quarter orbit, given the initial launch latitude and inclination? For example, given a start of [30,0] and an initial inclination of 60 degrees, what would be [x,90]?


Here's the code in action if it helps to clarify things.


The graphics library takes care of the map projection, which is why I only need to get the latitude. The crux of it are these lines here. It's a very VERY rough approx that doesn't model itself on sinusoidal motion but the mathematician in me knows that it's total bunk.

//The launch latitude var start = 30; //The launch direction var degree = 60; //The degree as a percentage from vertical var degPerc = Math.abs(degree) / 90; //Whether the launch is oriented North or South var direction = (degree >= 0) ? 1 : -1; //The latitude encountered after a quarter orbit var mid = (start + ((90-start) * 2 * degPerc)) * degPerc * direction;

  • $\begingroup$ As well as conservation of energy, have you used the fact that the angular momentum of the satellite about the centre of the planet will stay constant? This is because gravitational attraction is a central force. $\endgroup$ – Farcher Feb 2 '16 at 8:07
  • $\begingroup$ Yes. I've tried modeling it around a simple sinusoidal wave finding the height at π/2 and then multiplying by 90 again, but it keeps breaking when combining high initial inclination and latitude (like 80° at 80°) $\endgroup$ – mdupont Feb 2 '16 at 8:29

The trajectory will not be sinusoidal.
Depending on the initial speed it will be one of elliptical , parabolic and will "reach" infinity with zero speed or hyperbolic and will "reach" infinity with a finite speed. So the trajectory will very much depend on the initial speed of the satellite.

So how have you set up the computation?

  • $\begingroup$ I'm assuming a perfectly circular orbit, so the satellite would be traveling at constant speed. The purpose of the visualization is merely to demonstrate that rockets fired farther from the equator have less available orbits due to the increased minimum net angle of inclination. $\endgroup$ – mdupont Feb 2 '16 at 9:02

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