Launch Azimuth

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.

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.

Relation between latitude and inclination

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.

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 often the vernal equinox. Since a spinning planet will rotate under most orbits, that might not be important, unless you want to match another body's orbit.

If the the Longitude of the ascending node or orientation of the orbital plane is important, then a spinning celestial body is more convenient to launch from because it often gives you two times each rotation to launch into a particular orientation. 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.

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 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 in Earth's case due east. The equator is a special case, you can launch anytime. There is only one solution if the inclination plus latitude exactly equals 180° (retrograde orbit), and that is the opposite of spin direction, or due west. The launch windows for polar orbits are always one half day apart, most other combinations will be less.

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?

$\beta \!$	$=\arcsin \left({\frac {\cos(51.6^{\circ })}{\cos(28.5^{\circ })}}\right)$
	$=\arcsin \left({\frac {0.621148}{0.878817}}\right)$
	$=\arcsin \left(0.706800\right)$
	$=44.98^{\circ }{\mbox{ or }}135.02^{\circ }$

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

$\beta \!$	$=\arcsin \left({\frac {\cos(51.6^{\circ })}{\cos(45.9^{\circ })}}\right)$
	$=\arcsin \left({\frac {0.621148}{0.695913}}\right)$
	$=\arcsin \left(0.892566\right)$
	$=63.20^{\circ }{\mbox{ or }}116.80^{\circ }$

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.

Rotation of the Earth

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.

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! Calculate this from the orbit altitude, or use 7.730km/s for a typical 300km circular orbit.

I show these 2D vectors using the notation ${\vec {v}}=\langle v_{x},v_{y}\rangle \!$.

${\vec {v}}_{inertial}={\vec {v}}_{earthrot}+{\vec {v}}_{rot}$
${\vec {v}}_{inertial}-{\vec {v}}_{earthrot}={\vec {v}}_{rot}$
${\vec {v}}_{inertial}=v_{orbit}\langle \sin(\beta _{inertial}),\cos(\beta _{inertial})\rangle$
${\vec {v}}_{earthrot}=\langle \cos(\phi ),0\rangle v_{eqrot}$	where $v_{eqrot}\!$ is the rotation speed at the Earth equator, given by $v_{eqrot}={\frac {2\pi r_{eq}}{T_{rot}}}$. For the Earth, $T_{rot}=86164.09{\mbox{s}}\!$ and $r_{eq}=6378137{\mbox{m}}\!$ so $v_{eqrot}=465.101{\frac {\mbox{m}}{\mbox{s}}}$

${\vec {v}}_{rot}$	$={\vec {v}}_{inertial}-{\vec {v}}_{earthrot}$
	$=\langle v_{orbit}\sin(\beta _{inertial}),v_{orbit}\cos(\beta _{inertial})\rangle -\langle v_{eqrot}\cos(\phi ),0\rangle \!$
	$=\langle v_{orbit}\sin(\beta _{inertial})-v_{eqrot}\cos(\phi ),v_{orbit}\cos(\beta _{inertial})\rangle \!$

$v_{rotx}=v_{orbit}\sin(\beta _{inertial})-v_{eqrot}\cos(\phi )\!$

$v_{roty}=v_{orbit}\cos(\beta _{inertial})\!$

Now we can find the rotating launch azimuth $\beta _{rot}\!$ and incidentally the total velocity needed (as well as the velocity saved by launching with the rotation of the Earth)

$\beta _{rot}\!$	$=\tan ^{-1}\left({\frac {v_{rotx}}{v_{roty}}}\right)$
	$=\tan ^{-1}\left({\frac {v_{orbit}\sin(\beta _{inertial})-v_{eqrot}\cos(\phi )}{v_{orbit}\cos(\beta _{inertial})}}\right)$
$v_{rot}\!$	$={\sqrt {v_{rotx}^{2}+v_{roty}^{2}}}$
	$={\sqrt {\left(v_{orbit}\sin(\beta _{inertial})-v_{eqrot}\cos(\phi )\right)^{2}+\left(v_{orbit}\cos(\beta _{inertial})\right)^{2}}}$
$\Delta v=v_{inertial}-v_{rot}\!$

Example (continued)

So, treating the $\beta \!$ 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:

$\beta _{inertial}=44.98^{\circ }\!$

$v_{orbit}=7730{\frac {\mbox{m}}{\mbox{s}}}\!$

$v_{xrot}=\!$	$v_{orbit}\sin(\beta _{inertial})-v_{eqrot}\cos(\phi )\!$
	$7730\sin(44.98^{\circ })-465\cos(28.5^{\circ })\!$
	$5464-409\!$
	$5055{\frac {\mbox{m}}{\mbox{s}}}\!$
$v_{yrot}=\!$	$v_{orbit}\cos(\beta _{inertial})\!$
	$7730\cos(44.98^{\circ })\!$
	$5467{\frac {\mbox{m}}{\mbox{s}}}\!$
$\beta _{rot}\!$	$=\tan ^{-1}\left({\frac {v_{xrot}}{v_{yrot}}}\right)$
	$=\tan ^{-1}\left({\frac {5055}{5467}}\right)$
	$42.76^{\circ }$	Launch to this azimuth
$v_{rot}\!$	$={\sqrt {v_{rotx}^{2}+v_{roty}^{2}}}$
	$=7446{\frac {\mbox{m}}{\mbox{s}}}$	This is how much speed your launch vehicle needs to produce
$\Delta v\!$	$=v_{orbit}-v_{rot}\!$
	$=7730-7446\!$
	$=284{\frac {\mbox{m}}{\mbox{s}}}$	This is how much speed you save by exploiting the rotation of the Earth