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The '''launch azimuth''' is the angle between north direction and the projection of the initial orbital plane onto the launch location. It is the compass heading you head for when you launch.
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It is the compass heading you head for when you launch.
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The '''orbital inclination''' is the angle between the orbital plane and the celestial body's reference plane. If the body spins then this is usually the equatorial plane. The intersecting line is called the line of nodes. See [[Orbital_elements|orbital elements]].
 
  
 
==Relation between latitude and inclination==
 
==Relation between latitude and inclination==
  
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Not all orbital inclinations can be reached directly from  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 without additional maneuvers.
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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.
  
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Another consideration could be the Longitude of the Ascending Node, or the angle between the ascending node and the vernal reference point. For Earth this is often the vernal equinox. Since a spinning planet will rotate under most orbits, that might not be important, and you can launch anytime you like.
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Using spherical trigonometry, we can calculate the launch azimuth required to hit any allowed orbit inclination.  
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Lets say though you want to match another body's orbit around the parent body. Then the Longitude of the Ascending Node (LoAN) or orientation of the orbital plane is important, and a spinning celestial body is more convenient to launch from because it often gives you two different times each rotation to launch into a particular orientation. Though if it rotates so slowly that you rather not wait that long, like Venus for example, then you must again resort to using multiple maneuvers to reach the desired orbit. If the body does not spin, then all the directly reachable orbital planes must contain the line from the launch location through the center of the body.
 
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Using spherical trigonometry, we can [http://orbiter-forum.com/showpost.php?p=51004&postcount=5 calculate] the launch azimuth required to hit any allowed orbit inclination.  
 
  
 
<math>\cos(i)=\cos(\phi)\sin(\beta)\!</math>
 
<math>\cos(i)=\cos(\phi)\sin(\beta)\!</math>
Line 21: Line 15:
 
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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As mentioned above, frequently there are two solutions or launch times per rotation for a given LoAN. At each of these you can chose between prograde or retrograde orbits. These will be northbound or southbound along the calculated azimuth line, with launch windows some time apart. There is only one solution if the inclination is precisely equal to the latitude, and that is the spin direction (or opposite for a retrograde orbit). In the Earth's case due east or west. Zero inclination or an equatorial orbit is a special case, you can launch anytime, because there is no ascending node to orient. The launch windows for polar orbits are always one half rotation apart, most other combinations will be less.
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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
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=== Example ===
 
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The International Space Station orbits at 51.6&deg; inclination. A space shuttle launching from Cape Canaveral (latitude 28.5&deg;) needs to travel at what azimuth?
 
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{|
 
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|-
 
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|<math>\beta\!</math>||<math>=\arcsin\left(\frac{\cos(51.6^{\circ})}{\cos(28.5^{\circ})}\right)</math>
 
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|-
 
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|                  ||<math>=\arcsin\left(\frac{0.621148}{0.878817}\right)</math>
 
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|-
 
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|                  ||<math>=\arcsin\left(0.706800\right)</math>
 
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|-
 
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|                  ||<math>=44.98^{\circ}\mbox{ or } 135.02^{\circ}</math>
 
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|}
 
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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.
 
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A launch from Baikonur Cosmodrome (latitude 45.9&deg;) to the same orbit requires an azimuth of
 
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{|
 
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|-
 
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|<math>\beta\!</math>||<math>=\arcsin\left(\frac{\cos(51.6^{\circ})}{\cos(45.9^{\circ})}\right)</math>
 
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|-
 
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|                  ||<math>=\arcsin\left(\frac{0.621148}{0.695913}\right)</math>
 
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|-
 
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|                  ||<math>=\arcsin\left(0.892566\right)</math>
 
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|-
 
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|                  ||<math>=63.20^{\circ}\mbox{ or } 116.80^{\circ}</math>
 
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|}
 
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Launch again is always on the northeastern course. Both courses are over thousands of km of land, but some land below the course to the northeast has been reserved as the first-stage impact area.
 
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== Rotation of the Earth ==
 
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The above is the correct azimuth, in inertial space. However, when you are sitting on the surface of the earth with your compass, plotting takeoff, you are rotating with the Earth. This rotation must be compensated for.
 
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[[Image:Launch Azimuth 1.svg]]
 
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The above triangle shows the geometry necessary for calculating the rotating frame launch azimuth. Of the three vectors above, two are known, the inertial vector and the earth rotation vector. The third is just the difference between those two vectors. Note that you have to know the speed of your target orbit! [[Front Cover Equations|Calculate]] this from the orbit altitude, or use 7.730km/s for a typical 300km circular orbit.
 
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I show these 2D vectors using the notation <math>\vec{v}=\langle v_x,v_y \rangle \!</math>.
 
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{|
 
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|-
 
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|<math>\vec{v}_{inertial}=\vec{v}_{earthrot}+\vec{v}_{rot}</math>
 
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|-
 
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|<math>\vec{v}_{inertial}-\vec{v}_{earthrot}=\vec{v}_{rot}</math>
 
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|-
 
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|<math>\vec{v}_{inertial}=v_{orbit}\langle\sin(\beta_{inertial}),\cos(\beta_{inertial})\rangle</math>
 
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|-
 
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|<math>\vec{v}_{earthrot}=\langle\cos(\phi),0\rangle v_{eqrot}</math>||where <math>v_{eqrot}\!</math> is the rotation speed at the Earth equator, given by <math>v_{eqrot}=\frac{2\pi r_{eq}}{T_{rot}}</math>. For the Earth, <math>T_{rot}=86164.09\mbox{s}\!</math> and <math>r_{eq}=6378137\mbox{m}\!</math> so <math>v_{eqrot}=465.101\frac{\mbox{m}}{\mbox{s}}</math>
 
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|}
 
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{|
 
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|-
 
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|<math>\vec{v}_{rot}</math>||<math>=\vec{v}_{inertial}-\vec{v}_{earthrot}</math>
 
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|-
 
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|                          ||<math>=\langle v_{orbit}\sin(\beta_{inertial}),v_{orbit}\cos(\beta_{inertial}) \rangle- \langle v_{eqrot}\cos(\phi),0\rangle \!</math>
 
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|-
 
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|                          ||<math>=\langle v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi),v_{orbit}\cos(\beta_{inertial})\rangle \!</math>
 
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|-
 
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|}
 
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{|
 
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|-
 
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|<math>v_{rotx}=v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi)\!</math>
 
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|-
 
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|<math>v_{roty}=v_{orbit}\cos(\beta_{inertial})\!</math>
 
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|}
 
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Now we can find the rotating launch azimuth <math>\beta_{rot}\!</math> and incidentally the total velocity needed (as well as the velocity saved by launching with the rotation of the Earth)
 
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{|
 
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|-
 
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|<math>\beta_{rot}\!</math>||<math>=\tan^{-1}\left(\frac{v_{rotx}}{v_{roty}}\right)</math>
 
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|-
 
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|                          ||<math>=\tan^{-1}\left(\frac{v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi)}{v_{orbit}\cos(\beta_{inertial})}\right)</math>
 
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|-
 
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|<math>v_{rot}\!</math>||<math>=\sqrt{v_{rotx}^2+v_{roty}^2}</math>
 
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|-
 
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|                          ||<math>=\sqrt{\left(v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi)\right)^2+\left(v_{orbit}\cos(\beta_{inertial})\right)^2}</math>
 
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|-
 
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|<math>\Delta v=v_{inertial}-v_{rot}\!</math>
 
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|}
 
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=== Example (continued) ===
 
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So, treating the <math>\beta\!</math> from the first section as an inertial azimuth, we can calculate the rotating launch azimuth, the launch azimuth you actually aim for in your compass:
 
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Cape Canaveral to 300km orbit at ISS inclination:
 
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{|
 
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|-
 
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|<math>\beta_{inertial}=44.98^{\circ}\!</math>
 
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|-
 
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|<math>v_{orbit}=7730\frac{\mbox{m}}{\mbox{s}}\!</math>
 
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|}
 
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{|
 
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|-
 
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|<math>v_{xrot}=\!</math>||<math>v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi)\!</math>
 
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|-
 
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|                        ||<math>7730\sin(44.98^{\circ})-465\cos(28.5^{\circ})\!</math>
 
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|-
 
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|                        ||<math>5464-409\!</math>
 
−
|-
 
−
|                        ||<math>5055\frac{\mbox{m}}{\mbox{s}}\!</math>
 
−
|-
 
−
|<math>v_{yrot}=\!</math>||<math>v_{orbit}\cos(\beta_{inertial})\!</math>
 
−
|-
 
−
|                        ||<math>7730\cos(44.98^{\circ})\!</math>
 
−
|-
 
−
|                        ||<math>5467\frac{\mbox{m}}{\mbox{s}}\!</math>
 
−
|-
 
−
|<math>\beta_{rot}\!</math>||<math>=\tan^{-1}\left(\frac{v_{xrot}}{v_{yrot}}\right)</math>
 
−
|-
 
−
|                          ||<math>=\tan^{-1}\left(\frac{5055}{5467}\right)</math>
 
−
|-
 
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|                          ||<math>42.76^{\circ}</math>||Launch to this azimuth
 
−
|-
 
−
|<math>v_{rot}\!</math>||<math>=\sqrt{v_{rotx}^2+v_{roty}^2}</math>
 
−
|-
 
−
|                          ||<math>=7446\frac{\mbox{m}}{\mbox{s}}</math>||This is how much speed your launch vehicle needs to produce
 
−
|-
 
−
|<math>\Delta v\!</math>||<math>=v_{orbit}-v_{rot}\!</math>
 
−
|-
 
−
|                      ||<math>=7730-7446\!</math>
 
−
|-
 
−
|                      ||<math>=284\frac{\mbox{m}}{\mbox{s}}</math>||This is how much speed you save by exploiting the rotation of the Earth
 
−
|}
 
−
 
 
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[[Category: Articles]]
 
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[[Category:Tutorials]]
 

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