NASASpaceFlight.com Forum
General Discussion => Q&A Section => Topic started by: Nomadd on 04/23/2018 01:43 am
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I'm trying to come up with an explanation for beo launch windows that can be done on a bar with bottles and shot glasses representing various celestial bodies.
Is there a name for the earth orbit that matches the solar equatorial plane?
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Yes indeed and did you know that you could set sail with Capitan Morgan and never reach dry land ??
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Is there a name for the earth orbit that matches the solar equatorial plane?
Ecliptic?
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I'm trying to come up with an explanation for beo launch windows that can be done on a bar with bottles and shot glasses representing various celestial bodies.
Is there a name for the earth orbit that matches the solar equatorial plane?
There's no particular name for the Earth orbit that lies in the ecliptic plane, AFIAK. It's not particularly useful, because it's entirely possible to launch into the ecliptic from a Earth orbit whose plane does not lie in the ecliptic. Because the solar orbital velocity at 1 AU is ~30 km/s, it's actually very difficult to launch out of the ecliptic without a gravity assist.
BEO launch windows are timed to either add to Earth's heliocentric velocity (to reach outer planets) or subtract from Earth's heliocentric velocity (to reach inner planets).
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I might need a few more glasses to picture it. It seems like being off the earth's equator plus the difference between equatorial and ecliptic could put your launch trajectory so far off the ecliptic, you'd lose a lot. I'm thinking of going from leo instead of direct because of refueling.
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I might need a few more glasses to picture it. It seems like being off the earth's equator plus the difference between equatorial and ecliptic could put your launch trajectory so far off the ecliptic, you'd lose a lot. I'm thinking of going from leo instead of direct because of refueling.
Try drawing some vectors in a heliocentric reference frame: one leaving LEO with a v_inf of 1 km/s going north, and one leaving LEO with a v_inf of 1 km/s going south (e.g. Earth escape in directions orthogonal to the ecliptic).
When you add in the vector component of the Earths 30 km/s heliocentric orbital velocity, you will see that those escape orbits are only at 2 degrees to the ecliptic!
It really doesn't matter what direction you are launching in... you aren't going "so far off the ecliptic", ever (without a major gravity assist).
You can launch in a direction that fails to add or subtract from Earth's orbital velocity in any significant way, which does actually "lose a lot" if you are trying to reach a higher or lower planet. But as long as you get the direction right, your LEO plane's angle to the ecliptic is mostly irrelevant.
Edit: I would note that "getting the direction right" is hard (the Earth heliocentric orbital velocity vector has to lie in your LEO orbital plane at the same time that you are in position to burn for escape), and that getting the declination correct makes this a lot easier. So it is certainly relevant in that way.
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Ok. I think I can get that across if I add some peanuts to the mix. Pluto is about the only sort of a planet far enough off the ecliptic to matter, and since you'll probably be borrowing some inertia from Jupiter anyhow, you can adjust then.