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Energy, Non-coplanar solar system orbits for space ships

  1. Sep 30, 2012 #1
    Hi all,

    As many of you have seen, I am thinking about spaceship trajectories. This next question falls under the same subject, but it is an entirely different question from the previous questions I have asked.

    Spaceships of the near future will fly in the plane of the solar system. Suppose if all the planets orbit the sun, they do it in the xy plane. The sun spins about its axis such that the sun's spin angular momentum vector points along the positive z-axis. Let's assume planet A is on one side of the sun, and planet B is on the other. I want to send a spaceship from planet A to planet B. I could do this two ways:

    1) I could send the ship so that it remains co-planar with the solar system. Trajectory 1 remains in the xy plane also. If Trajectory 1 were to trace out a circle, the normal vector of this area would point along +/- z_axis. This is how ships of today would go.

    2) Trajectory 2, the ship flies in an orbit where it still goes around the sun, but leaves the solar plane alltogether. Its orbit/trajectory is in the xz plane. At some point in this orbit, when the ship looks down many AU, it will see the sun's north pole. It will return to the solar plane when planet B is at the right place. If this trajectory were to be a circle, the normal vector to this circle would be +/- y_axis.

    Ignoring, for the moment, the obvious advantages of Earth's tangential pick-up velocity, gravity assists, and Oberth Effects as obvious advantages to choosing the ship's co-planer trajectory #1, does trajectory 2 cost anymore energy than trajectory 1? Is there any difference in any energy for a spaceship that remains co-planar with the solar system, and for a spaceship that exits the solar plane and returns to it later?

    Most basic, according to Newton's Law of Potential Energy for Gravity, only the radius matters, (-Gm1m2/r) but not the θ or ∅. HOwever, I know that if the sun spins in the xy plane, all orbits of planets and spaceships most naturally will settle in the xy plane after some time. This is also the same reason that accretion disks around stars and black holes are found in the xy plane of these objects, but not ever in the z-direction. That means, the minimum energy for such a system is to have everything travel in the xy plane. Thus, for anything to travel to in the xz plane, more energy would be required than to travel in the xy plane, but I can't see how, from Newton's Law of Potential Energy for gravity.

    I guess the basic question is this:

    Is there any difference in any energy for a spaceship that remains co-planar with the solar system, and for a spaceship that exits the solar plane and returns to it later?

    Gravity says "NO", but I think angular momentum of the main body providing the gravity says "Yes", but I don't know how.
     
  2. jcsd
  3. Sep 30, 2012 #2

    Dale

    Staff: Mentor

    The two orbits would have similar ΔE, but vastly different Δv. Even future spaceships cannot neglect momentum.
     
  4. Sep 30, 2012 #3
    You can go from an orbit around earth, to an orbit around Venus with less than 7 km/s delta-v

    http://www.projectrho.com/public_html/rocket/appmissiontable.php#id--Erik_Max_Francis%27_Mission_Tables--Delta_V_Required_for_Travel_Using_Hohmann_Orbits--Solar_System

    If you wanted to go in an orbit across northpole of the sun, you'd have to cancel the 30 km/s orbital speed of the earth and build up about 20 km/s orbital speed in a perpendicular direction, so you need about 36 km/s near the earth, and even more the match velocities with Venus, whose orbital speed is even higher.

    The best way to get in such an orbit would be a gravity assist from Jupiter, as done by the Ulysses spacecraft wich went to Jupiter to get in a polar orbit around the sun.
     
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