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I already know about the six Keplerian Elements that are fundamental to a planet's orbit around the Sun, and for a few weeks now, I thought I'd got it down. Then, I wanted to have some fun, and test out an Orbital Mechanics Simulator (https://orbitalmechanics.info/) that I'd found on the web. Then, if you click the button that says "Add Launch", it gives you slides for three launch elements. I knew (at least I thought I knew) what Ω meant, I believe it is the argument of latitude of the satellite, but then I see two other angles, ψ and λ. I google up "Three Launch Elements", but I only get links about making a product succeed. Then, I be more specific, and I add satellite to the front, and the first article looked relevant, but I get a resource on a book that isn't what I want to know about, at least based on the title of it. Then, I go, start over, and google "Omega, Psi, Lambda", and, as it turns out, Google gave me a bunch of photos of a club called Ω,Ψ,Φ (Notice: the Club Name has Φ, not Λ). What do these launch elements mean, and how do you calculate them?

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    $\begingroup$ "I google up "Three Launch Elements", but I only get links about making a product succeed." lol, +1 for that alone $\endgroup$ Apr 27 at 17:36
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    $\begingroup$ Yeah. Sometimes Google fails you, and this is definitely one of those times. $\endgroup$
    – Questioner
    Apr 27 at 17:41
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The Help for Orbital Simulator gives the following for Launches:

  • $\phi$ is the latitude of the launch site.
  • $\lambda$ is the longitude of the launch site.
  • Ω is the ascending node of the orbit.

Those three items determine the inclination of the orbit.

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    $\begingroup$ How do they determine the inclination of the orbit? $\endgroup$
    – Questioner
    Apr 27 at 14:44
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    $\begingroup$ You have to make two big assumptions for those three parameters to determine the inclination of the resulting orbit: 1) the vehicle launches at an azimuth of 90°, i.e. due east, in the direction of planetary rotation, and 2) the vehicle's thrust vector has zero out-of-plane component. That last requires some attitude control as the vehicle moves toward its descending node (assuming launch from the northern hemisphere). Launches from KSC routinely demonstrate that going to a wide range of inclinations, both higher and lower than the launch site latitude, is possible. $\endgroup$ Apr 27 at 15:47
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    $\begingroup$ @TomSpilker I agree with point 2) (no out-of-plane thrusting during launch). But three points define a plane (Earth's center, latitude and longitude of launch site ~ point on orbit, and latitude and longitude of the ascending node), so those define the inclination. At least to the simple assumptions that the website makes. Right? In other words, changing the longitude of the ascending node changes the inclination. (I would think the inclination is normally known and the ascending node is calculated, not the other way around.) $\endgroup$
    – JohnHoltz
    Apr 27 at 16:26
  • $\begingroup$ @JohnHoltz Changing the longitude of the ascending node indeed changes the orientation of the orbit plane, but it doesn't change the angle between the planet's equatorial plane and the orbit plane, so the inclination doesn't change. If you're talking about inclination in a heliocentric reference frame it would indeed make a difference, but not in a planetocentric reference frame. $\endgroup$ Apr 27 at 17:51

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