A new spaceport is proposed for Camden County, Georgia (USA). The latitude is 30.56N and the only available launch azimuth is 123 degrees. What is the resulting LEO without mid-course maneuvers? Can show me how you calculate this?


3 Answers 3


It's just a matter of some spherical trig.

$$ \cos(inclination) = \cos(lat)*\sin(azimuth) $$

So in your example, the inclination would be equal to: $$ \arccos(\cos(30.56)*\sin(123)) = 43.77\deg $$

To simplify the calculation, I assumed that your given azimuth was an inertial one. The calculations that take into account the rotation of the Earth and what the compass onboard would measure as the azimuth are in the link above.

  • $\begingroup$ should be the accepted answer.. $\endgroup$ Jun 17, 2021 at 13:19

To amplify on the correct answer of @M.A.H.:

To apply this general spherical triangle (https://en.wikipedia.org/wiki/Solution_of_triangles) to the particular problem of orbital inclination:

enter image description here

Point $C$ is the geographical North Pole; Point $B$ is the launch point, so the arc $C-B$ is the co-latitude of the launch point; $A$ is the point where the orbit track crosses the equator, so the arc $C-A$ is $90^o$ and the orbital inclination is $\alpha$. Once the launch azimuth is related to the angle $\beta$ (depending on the exact geometry) the triangle can be solved for $\alpha$, and also $\gamma$, the longitude of the ascending node of the orbit.

  • $\begingroup$ If my understanding of your description is correct, the orbital inclination in your diagram is 90 degrees, making this a polar orbit; even graphically, an orbit that passes directly over the poles will be a polar orbit. It would be nice to see how a non-polar orbit is mathematically related to the position B and the azimuth beta. $\endgroup$
    – pstatix
    Aug 12, 2022 at 15:27
  • $\begingroup$ The arc A-B (yellow) is the orbital track... $\endgroup$
    – DJohnM
    Aug 13, 2022 at 4:01

Orbital inclination will be equal to the latitude. The reason is simple - you just need that orbit plane will pass through the center of the Earth and your launch point

See explanation here: Of inclinations and azimuths

  • 3
    $\begingroup$ I understand from the referenced article that a due east azimuth would provide a 30.56 degree orbital inclination. But how does launching to a different azimuth, like 123 degrees for my example, change the OI? $\endgroup$
    – stevew
    Aug 8, 2016 at 11:07
  • $\begingroup$ You need to make your launch in the orbital plane - i.e. you can't launch with another azimuth $\endgroup$ Aug 8, 2016 at 11:16
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    $\begingroup$ This is incorrect. The only limitation on launch inclination is that you cannot launch directly into an inclination less than your latitude. You can certainly launch into higher by changing your azimuth. $\endgroup$
    – M.A.H.
    Aug 8, 2016 at 12:32
  • $\begingroup$ M.A.H., exactly as they do at Kennedy. From satobs.org: "Cape Canaveral is located at approximately 28.5 degrees North latitude. Launch azimuth is constrained at the Cape between 35 and 120 degrees due to land mass overflight restrictions. All launches from Cape Canaveral are therefore restricted in inclination between 28.5 and 59 degrees." So what I don't understand is how the orbital inclination is calculated knowing the azimuth. $\endgroup$
    – stevew
    Aug 8, 2016 at 20:58
  • $\begingroup$ @stevew my apologies, "This is incorrect" was in response to this answer. In my answer and the answer of DjohnM, I think we covered that calculation pretty well. The cosine of the inclination is equal to the cosine of the latitude multiplied by the sin of the azimuth. $\endgroup$
    – M.A.H.
    Aug 8, 2016 at 21:34

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