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I found this image showing the allowable launch azimuths for Vandenberg AFB and I was wondering why a 180° launch azimuth doesn't match a 90° inclination orbit. This led me to question the equation cos(i) = sin (A) . cos (L). Is there something missing on it? Does the rotation of the earth have any impact on the inclination of an orbit reached by a 180° launch azimuth?

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If the Earth did not rotate, a 180° launch azimuth would result in a 90° inclination orbit.

To compensate for the rotation, the azimuth should be bigger than 180°.

If the rocket were launched from the north or south pole, the azimuth would be exactly south or north, respectively.

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    $\begingroup$ … but then, when you're launching from the north pole, all directions are due south. :) $\endgroup$
    – TooTea
    Mar 9, 2022 at 11:41
  • $\begingroup$ You can't reach lower inclination orbits that your starting point without some (pretty energy intensive) plane change maneuvres. So yes, from the poles you'll always reach a polar orbit. So there, "something going south" might have another meaning. $\endgroup$
    – TrySCE2AUX
    Mar 11, 2022 at 13:39
  • $\begingroup$ Makes sense! So to calculate something like this I have to add two different vectors, right? One corresponding to the velocity of the surface of the Earth at a given latitude and another with the velocity needed to get to the desired orbit? $\endgroup$
    – Matt
    Mar 11, 2022 at 16:31
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Earth rotation.

I have encountered that plenty of times in KSP: when you switch to orbital frame of reference, your velocity marker is moved east because you have been moving with the surface.
You need to launch slightly west to get 90deg polar orbit

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