Consider a rocket launch from Kennedy, latitude 28.5 deg, to a target orbital inclination of 50.6 degrees for rendezvous with the ISS.

The launch azimuth for this orbital inclination is roughly 45 deg north of due east.

At T0, inertial velocity is roughly 400 m/s due east imparted by earth's rotation.

At orbital insertion, inertial velocity is roughly 7,800 m/s pointing northeast.

In between, the velocity vector must gradually transition between its initial eastward direction to its final northeastward direction.

This means the cross-product of the rocket's position and velocity vectors, r x v, which defines the eventual orbital plane, is constantly evolving all the way up to orbit.

So I want to say that at T0, r x v will indicate an inclination angle of 28.5 degrees, and that only at orbit insertion will r x v indicate an inclination of 51.6 deg...

Is this correct? Does it take the whole launch for the inclination angle to reach its target orbital value? (Pretend for a moment that the orbital inclination angle means something even before you reach orbit---does it take the whole launch for that value to stabilize?)

  • $\begingroup$ Technically speaking, this sounds right. I wouldn't call orbital inclination a useful number as long as you are in powered flight and not in an orbit. The transition between 28° and 51° is smooth, but depends on the flight profile. $\endgroup$ – asdfex Jan 28 at 8:44
  • $\begingroup$ @asdfex: Thanks for confirming! So it seems if you do a gravity turn without explicit yaw control, then you should expect to be slightly off from your target orbital inclination once you reach orbital insertion---unless during PEG orbital insertion you control not only for pitch but also for yaw. Does this seem right? $\endgroup$ – user36480 Jan 28 at 9:03
  • $\begingroup$ Not really. All you have to make sure is that the gained velocity during launch plus the initial velocity matches the orbit at the end of the burn. That is, you launch a bit further northwards than the plain formula cos(latitude)*inclination tells. $\endgroup$ – asdfex Jan 28 at 9:10
  • $\begingroup$ @asdfex: Oh so the cos(lat) equation doesn't account for earth's rotation? This might explain why I'm consistently a few degrees off. But then how do we calculate the launch azimuth in a way that factors in our initial eastward velocity? $\endgroup$ – user36480 Jan 28 at 9:38
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    $\begingroup$ Who dares joke about KSP? Yes, I'd simulate a launch there if I wasn't building my own simulation elsewhere ;-) $\endgroup$ – user36480 Jan 28 at 17:43

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