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Does the delta-v requirement change depending on the angle a craft is leaving the ecliptic?

If a craft were to be leaving only a few degrees above the ecliptic, would it take less delta-v than if a craft was moving perpendicular to the ecliptic? Assume the craft is being launched from Uranus' moon Titania so it is moving in an orbit perpendicular to the ecliptic already.

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    $\begingroup$ Which starting conditions do you assume? On the ground? In earth orbit? And are you talking about the direction we leave relative to the earth moon system or the entire solar system? $\endgroup$ – lijat Jul 10 '19 at 14:18
  • $\begingroup$ Good edit! voting to re-open! $\endgroup$ – uhoh Jul 11 '19 at 3:54
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    $\begingroup$ I think you've just muddied the waters with the statement that you're on Titania. If you're on Titania, then your velocity vector changes from parallel to the ecliptic to perpendicular to the ecliptic, and back, twice every orbit. And there are points in Uranus's orbit (when the poles are facing the Sun) when you could use Uranus's orbital velocity to reduce the delta-v required. $\endgroup$ – Michael Seifert Jul 11 '19 at 21:44
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It depends on if you are doing a direct to leave the solar system or doing a flyby, but probably.

If you are directly leaving the solar system, then the close you are to the Earth's inclination around the Sun, the more your velocity will count. If you go completely perpendicular to that, it will take quite a bit more fuel, as you have to do an inclination change AND an escape velocity change at the same time.

If you are doing a flyby, this still mostly applies, but particularly a flyby around Jupiter can pretty easily accomplish both an inclination change and escape velocity at the same time.

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