Difference between revisions of "Launch Azimuth"
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where <math>i</math> is the desired orbit inclination, <math>\phi</math> is the launch site latitude, and <math>\beta</math> is the launch azimuth. Solving for azimuth: | where <math>i</math> is the desired orbit inclination, <math>\phi</math> is the launch site latitude, and <math>\beta</math> is the launch azimuth. Solving for azimuth: | ||
| − | <math>\beta=\ | + | <math>\beta=\arcsin\left(\frac{\cos(i)}{\cos(\phi)}\right)</math> |
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. | 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. | ||
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{| | {| | ||
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| − | |<math>\beta\!</math>||<math>=\ | + | |<math>\beta\!</math>||<math>=\arcsin\left(\frac{\cos(51.6^{\circ})}{\cos(28.5^{\circ})}\right)</math> |
|- | |- | ||
| − | | ||<math>=\ | + | | ||<math>=\arcsin\left(\frac{0.621148}{0.878817}\right)</math> |
|- | |- | ||
| − | | ||<math>=\ | + | | ||<math>=\arcsin\left(0.706800\right)</math> |
|- | |- | ||
| ||<math>=44.98^{\circ}\mbox{ or } 135.02^{\circ}</math> | | ||<math>=44.98^{\circ}\mbox{ or } 135.02^{\circ}</math> | ||
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{| | {| | ||
|- | |- | ||
| − | |<math>\beta\!</math>||<math>=\ | + | |<math>\beta\!</math>||<math>=\arcsin\left(\frac{\cos(51.6^{\circ})}{\cos(45.9^{\circ})}\right)</math> |
|- | |- | ||
| − | | ||<math>=\ | + | | ||<math>=\arcsin\left(\frac{0.621148}{0.695913}\right)</math> |
|- | |- | ||
| − | | ||<math>=\ | + | | ||<math>=\arcsin\left(0.892566\right)</math> |
|- | |- | ||
| ||<math>=63.20^{\circ}\mbox{ or } 116.80^{\circ}</math> | | ||<math>=63.20^{\circ}\mbox{ or } 116.80^{\circ}</math> | ||
Revision as of 12:04, 31 October 2006
The launch azimuth is the angle between north direction and the projection of the initial orbit plane onto the launch location.
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.
where is the desired orbit inclination, is the launch site latitude, and is the launch azimuth. Solving for azimuth:
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° (retrograde orbit), and that is due west.
Example: The International Space Station orbits at 51.6° inclination. A space shuttle launching from Cape Canaveral (latitude 28.5°) needs to travel at what azimuth?
As a launch on the southeastern course would overfly the Bahamas, Cuba, and South America, the northeastern course is always used. This is ocean all the way to Ireland.
A launch from Baikonur Cosmodrome (latitude 45.9°) to the same orbit requires an azimuth of
Launch again is always on the northeastern course. Both courses are over thousands of km of land, but some land belowe the course to the northeast has been reserved as the first-stage impact area.