Launch Azimuth

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Relation between latitude and inclination

Not all inclinations can be reached at a position on a celestial body. The problem is, that the launch location has to be a point inside the target orbit plane. So, if the latitude of a launch location is higher than the inclination, the orbit can't be reached directly.

Using spherical trigonometry, we can calculate the launch azimuth required to hit any allowed orbit inclination.

$\cos(i)=\cos(\phi )\sin(\beta )\!$

where $i$ is the desired orbit inclination, $\phi$ is the launch site latitude, and $\beta$ is the launch azimuth. Solving for azimuth:

$\beta =\arcsin \left({\frac {\cos(i)}{\cos(\phi )}}\right)$

This shows mathematically why the inclination must be greater than the launch latitude: Otherwise, the argument to the inverse sine function would be greater than 1, which is out of its domain. Therefore there is no solution in this case.

Also, note that frequently there are two solutions: one northbound and one southbound. There is only one solution if the inclination is precisely equal to the latitude, and that is due east. There is only one solution if the inclination plus latitude exactly equals 180