Difference between revisions of "Launch Azimuth"

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The '''launch azimuth''' is the angle between north direction and the projection of the initial orbit plane onto the launch location. It is the compass heading you head for when you launch.
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free t-mobile ringtones] The '''launch azimuth''' is the angle between north direction and the projection of the initial orbit plane onto the launch location. It is the compass heading you head for when you launch.
  
 
==Relation between latitude and inclination==
 
==Relation between latitude and inclination==
Line 15: 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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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.
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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° inclination. A space shuttle launching from Cape Canaveral (latitude 28.5°) needs to travel at what azimuth?
 
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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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|-
 
−
|                  ||<math>=\arcsin\left(\frac{0.621148}{0.878817}\right)</math>
 
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|-
 
−
|                  ||<math>=\arcsin\left(0.706800\right)</math>
 
−
|-
 
−
|                  ||<math>=44.98^{\circ}\mbox{ or } 135.02^{\circ}</math>
 
−
|}
 
−
 
 
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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.
 
−
 
 
−
A launch from Baikonur Cosmodrome (latitude 45.9&deg;) to the same orbit requires an azimuth of
 
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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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|-
 
−
|                  ||<math>=\arcsin\left(\frac{0.621148}{0.695913}\right)</math>
 
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|-
 
−
|                  ||<math>=\arcsin\left(0.892566\right)</math>
 
−
|-
 
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|                  ||<math>=63.20^{\circ}\mbox{ or } 116.80^{\circ}</math>
 
−
|}
 
−
 
 
−
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.png]]
 
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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}=<v_x,v_y>\!</math>.
 
−
 
 
−
{|
 
−
|-
 
−
|<math>\vec{v}_{inertial}=\vec{v}_{earthrot}+\vec{v}_{rot}</math>
 
−
|-
 
−
|<math>\vec{v}_{inertial}-\vec{v}_{earthrot}=\vec{v}_{rot}</math>
 
−
|-
 
−
|<math>\vec{v}_{inertial}=v_{orbit}<\sin(\beta_{inertial}),\cos(\beta_{inertial})></math>
 
−
|-
 
−
|<math>\vec{v}_{earthrot}=<\cos(\phi),0>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}=6371010\mbox{m}\!</math> so <math>v_{eqrot}=464.581\frac{\mbox{m}}{\mbox{s}}</math>
 
−
|}
 
−
{|
 
−
|-
 
−
|<math>\vec{v}_{rot}</math>||<math>=\vec{v}_{inertial}-\vec{v}_{earthrot}</math>
 
−
|-
 
−
|                          ||<math>=<v_{orbit}\sin(\beta_{inertial}),v_{orbit}\cos(\beta_{inertial})>-<v_{eqrot}\cos(\phi),0>\!</math>
 
−
|-
 
−
|                          ||<math>=<v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi),v_{orbit}\cos(\beta_{inertial})>\!</math>
 
−
|-
 
−
|}
 
−
{|
 
−
|-
 
−
|<math>v_{rotx}=v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi)\!</math>
 
−
|-
 
−
|<math>v_{roty}=v_{orbit}\cos(\beta_{inertial})\!</math>
 
−
|}
 
−
 
 
−
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)
 
−
 
 
−
{|
 
−
|-
 
−
|<math>\beta_{rot}\!</math>||<math>=\tan^{-1}\left(\frac{v_{rotx}}{v_{roty}}\right)</math>
 
−
|-
 
−
|                          ||<math>=\tan^{-1}\left(\frac{v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi)}{v_{orbit}\cos(\beta_{inertial})}\right)</math>
 
−
|-
 
−
|<math>v_{rot}\!</math>||<math>=\sqrt{v_{rotx}^2+v_{roty}^2}</math>
 
−
|-
 
−
|                          ||<math>=\sqrt{\left(v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi)\right)^2+\left(v_{orbit}\cos(\beta_{inertial})\right)^2}</math>
 
−
|-
 
−
|<math>\Delta v=v_{inertial}-v_{rot}\!</math>
 
−
|}
 
−
 
 
−
=== Example (continued) ===
 
−
 
 
−
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:
 
−
 
 
−
Cape Canaveral to 300km orbit at ISS inclination:
 
−
{|
 
−
|-
 
−
|<math>\beta_{inertial}=44.98^{\circ}\!</math>
 
−
|-
 
−
|<math>v_{orbit}=7730\frac{\mbox{m}}{\mbox{s}}</math>
 
−
|}
 
−
{|
 
−
|-
 
−
|<math>v_{xrot}=\!</math>||<math>v_{orbit}\sin(\beta_{inertial})-v_{eqrot}\cos(\phi)\!</math>
 
−
|-
 
−
|                        ||<math>7730\sin(44.98^{\circ})-465\cos(28.5^{\circ})\!</math>
 
−
|-
 
−
|                        ||<math>5464-409\!</math>
 
−
|-
 
−
|                        ||<math>5055\frac{\mbox{m}}{\mbox{s}}\!</math>
 
−
|-
 
−
|<math>v_{xrot}=\!</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>
 
−
|-
 
−
|                          ||<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:Tutorials]]
 

Revision as of 11:01, 10 October 2007

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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.

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

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