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Aside from the obvious answer of being able to easily avoid the larger bodies in the Solar System. I'm primarily curious as to the mechanics of this endeavor -- would there be any benefit to attempting to leave the solar system say... perpendicular to the ecliptic?

Which direction would get you out of the influence of the Sun fastest, and could you take advantage of that environment on the way to speed things up?

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The fastest way to leave the Solar System is to pass by as many of the Gas Giants as you can, and use their gravity to slingshot you faster. As there are no planets outside of the ecliptic, it would not be advantageous to avoid the ecliptic plane. This holds true until we start to get really fast space probes.

Furthermore, passing by the planets would give you a chance to abort, by slingshoting around them to return to Earth, in the event of a quick catastrophic failure (Entirely possible).

Just to prove this point, New Horizon's left Earth as the fastest spacecraft to ever leave Earth Orbit, reaching the orbit of the Moon in only 9 hours. Yet it will never catch up to the Voyager probes, because they used the gravity of both Jupiter and Saturn to speed up. I can't put it any better than Wikipedia, so here goes:

New Horizons is often given the title of Fastest Spacecraft Ever Launched, although the Helios probes are arguably the holders of that title as a result of speed gained while falling toward the Sun. New Horizons, however, achieved the highest launch velocity and thus left Earth faster than any other spacecraft to date. It is also the first spacecraft launched directly into a solar escape trajectory, which requires an approximate velocity of 16.5 km/s (59,000 km/h; 37,000 mph), plus losses, all to be provided by the launcher. However, it will not be the fastest spacecraft to leave the Solar System. This record is held by Voyager 1, currently traveling at 17.145 km/s (61,720 km/h; 38,350 mph) relative to the Sun. Voyager 1 attained greater hyperbolic excess velocity from Jupiter and Saturn gravitational slingshots than New Horizons.

Furthermore, the amount of objects in the ecliptic plane is vastly overstated. Space is really really big, and we have to plan very carefully to get a spacecraft to another planet purposely. Even a small miss will cause issues. There isn't much there, and there is much to gain by going through the ecliptic plane.

The day that we have spacecraft capable of very high thrust for extended periods of time, we likely won't need this shortcut, but for now, it's an invaluable tool.

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    $\begingroup$ +1 for recognizing the benefits of gravity assists and abort opportunities. $\endgroup$ – Erik Jul 29 '13 at 23:38
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    $\begingroup$ Couldn't you use a gravity assist from gas giant and go over or under one of the poles? Effectively slingshotting you perpendicular to the ecliptic? $\endgroup$ – Fezter Jul 31 '13 at 3:36
  • $\begingroup$ @Fezter: Absolutely. $\endgroup$ – PearsonArtPhoto Jul 31 '13 at 10:44
  • $\begingroup$ No, that's not the fastest way to leave the solar system. See Deer Hunter's answer. $\endgroup$ – Mark Adler Sep 20 '13 at 5:57
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There are two major reasons you may be going outside the ecliptical plane:

  • You want to get to a particular star.
  • You want to put the Sun on a direct line between your craft and another point in the sky (usually a star system or an interstellar probe to that star system) to exploit Sun's gravitational lens (that is, your destination starts around 550-740 AU from the Sun) to achieve awesome extra amplification of received/transmitted radio signal (57 dB at 1.42GHz, for instance).

In any case, the trajectory to get you out of Dodge fast and right will have to use perihelion propulsive maneuver near the Sun (as close as your heat rejection will tolerate) after gravity assists from Saturn and Jupiter (the Krafft Arnold von Ehricke trajectory). You would choose the point of perihelion to rotate your velocity vector in the direction required.

References:

  • Deep Space Flight and Communications: Exploiting the Sun as a Gravitational Lens Claudio Maccone. Springer, 2009.
  • Krafft Arnold von Ehricke, "Saturn-Jupiter Rebound. A method of high-speed spacecraft ejection from the Solar System". Journal of the British Interplanetary Society, Vol.25, 1972. Pp.561-571.
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    $\begingroup$ This answer is much better than the accepted one. To even get close to useful interstellar speeds, you would have to use a propulsive solar flyby. So the Voyager approach is nowhere near the fastest way to leave the solar system, as claimed in that other answer. $\endgroup$ – Mark Adler Sep 20 '13 at 5:51
  • $\begingroup$ I like this answer. It has always seemed to me the sun could deliver an amazing Oberth benefit if a near perihelion burn could be done. Am pleased that it has a name: The Kraft Arnold von Ehricke trajectory. XKCD had a pic but Randall didn't do this well. $\endgroup$ – HopDavid Oct 28 '17 at 13:41
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The ecliptic plane of our solar system is at an angle of about 60º relative to the Galactic plane of our galaxy.

The Milky Way is about 100,000 ly (30 kpc) in diameter, and, on average, approximately 1,000 ly (0.3 kpc) thick. This means that when you leave the solar system along our ecliptic plane, you can visit the stars in our neighborhood, but you 'quickly' leave the galaxy. You can use a gravity assist from one of the gas giants to aim your trajectory into the Galactic plane, allowing you to visit far more stars.

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Assuming you have sufficient energy for a constant 1g thrust.

On an interstellar voyage, the ecliptic plane has little to no meaning. If your direction of travel is generally at the right angle (90°) to the plane, then there is little point in scooting around all the obstacles in the Solar system.

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  • $\begingroup$ -1 for ignoring the benefits of gravity assists and abort opportunities. $\endgroup$ – Erik Jul 29 '13 at 23:38
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    $\begingroup$ @Erik If you have 1g constant thrust why would you divert for a gravity assist? Wouldn't that be like driving 50 miles out of your way to save a penny on fuel? $\endgroup$ – James Jenkins Jul 29 '13 at 23:46
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    $\begingroup$ @JamesJenkins -- you will always want the extra energy the slingshot provides. You might want to escape Sol faster, or you might want to have the escape hatch gravity assist provides if your 1g drive doesn't start... $\endgroup$ – Erik Jul 30 '13 at 3:34
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    $\begingroup$ @TildalWave All you do is slingshot around, say, Jupiter at a Jovian inclination of 90 deg. $\endgroup$ – Erik Jul 30 '13 at 3:35
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    $\begingroup$ @Erik - Darn, you're right. Not sure how I missed that when I was rested, and it seems child play easy now that I'm about to fall into a coma. I think all that talk of plains made me switch to 2D. :) $\endgroup$ – TildalWave Jul 30 '13 at 3:38
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We might answer this question with greater clarity had we a knowledge of every speck of matter in our star system. In addition to gravity assists, velocity transfers via impacts (guided or not) could impinge. Thus with accurate knowledge of the orbits of every asteroid, we might with little effort, engender a chain of gravity assists and/or body impact events which would lead to maximum interstellar velocity attainable with current means.

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  • $\begingroup$ This method suffers from extreme diminishing returns with tiny masses, the enormous difficulty of elastically transferring momentum from collisions at tens of kilometers per second, and the incredibly large amount of time required both to take advantage of all the micro-optimizations possible and to calculate them. $\endgroup$ – Nathan Tuggy Oct 24 '17 at 6:11

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