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Maximum gravity boost

  1. Nov 1, 2006 #1

    tony873004

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    What is the fastest speed relative to the Sun that Jupiter can eject an object onto a hyperbolic trajectory?

    I imagine you'd want the object to be travelling as fast as possible when it encountered Jupiter. This would give it a velocity at just under Solar escape velocity as it approached Jupiter.

    More important to the actual answer, I'd love to know a formula that took planet mass and planet-sun distance as inputs and produced a max ejection velocity.
     
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  3. Nov 1, 2006 #2

    DaveC426913

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    Unless you're satisfied with an answer expressed simply as a force, you would also have to factor in the mass of the vehicle in order to be able to calculate its final velocity.


    Also, what does planet-sun distance have to do with it? Or are you looking for a very general formula?
     
  4. Nov 1, 2006 #3

    tony873004

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    Thanks, Dave. Yes, a general formula is what I want. I could probably figure this out through simulations on a planet-by-planet basis, but I'm sure it is analytically computable with a single formula based on angular momentum. The formula would also need to contain a planet's radius as it represents minimum passage distance.

    Escape velocity from the Sun is directly related to distance from the Sun. So the planet must accelerate the object to the escape velocity for the planet's Sun distance.

    If Sun and Planet mass >> object mass, the mass of the object would have little to do with the maximum solar velocity Jupiter, or any planet, could accelerate it to.

    Interestingly, what ever this speed is, is also the maximum velocity an exo-solar object passing through our solar system could have and still be captured by a close passage to Jupiter, or other planet.
     
  5. Nov 2, 2006 #4

    DaveC426913

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    No. It has nothing to do with the planet mass : object mass ratio and everything to do with F=ma. If m[object] is large, then a[object] will be small.

    Mm. I see where you're going, the old time-reversability thing, but it's not the same.

    Remember, the object escaping started off with a rocket boost. So if you reverse the orbital mechanics for an incoming object, yes it would be captured by the Sun, but the numbers could be significantly different in terms of how fast it could be going and still remain in the system.
     
  6. Nov 2, 2006 #5

    tony873004

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    Only if force is constant in that equation will you have in inverse relationship between mobject and aobject But force is not constant here, just like force is not constant in the example of dropping two rocks from the roof of a building. The more massive rock lands with more force, but not more velocity since acceleration is constant.

    Similarly, acceleration is constant in the ejection problem, and force is not. So two spacecraft of differing masses on the same trajectory will get the same ejection boost from Jupiter assuming mcraft<<mJupiter.

    The capture trajectory would be the same as the escape trajectory up until the point of the rocket boost. But that's not what I'm after. I just mentioned it as a trivial fact. There's a discussion in my class about interstellar space travel and accelerating to 10% the speed of light. Some are suggesting a Jupiter boost. Intuitively, I know Jupiter doesn't have the potential to add this kind of acceleration to a spacecraft. But a formula in hand would be more convincing than just my intuition.
     
  7. Nov 2, 2006 #6
    In principle, isn't it just twice jupiter's orbital velocity (plus the object's initial velocity)?

    In jupiter's frame it's just like any other (hyperbolic) orbit, so the object will depart jupiter at whatever speed it approached jupiter. In the solar system frame jupiter isn't really perturbed and the best the object can do is if it both approaches from and departs in the direction jupiter is moving.

    As for physical constraints it seems that the radius (and mass) of the planet would limit the deflection you can get for a given initial velocity and that complete solar system dynamics would determine possible initial trajectories, launch windows, etc.. but there's plenty of room to move when you have [itex]\sqrt 2[/itex] times the escape velocity to start with.
     
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