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Halo Orbits

  1. Sep 16, 2005 #1

    tony873004

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    Is a Halo Orbit, such as the one employed by the SOHO spacecraft, a powered orbit (requires corrective burns)?. L1 is unstable, but everything I Google about Halo Orbits seem to suggest that the spacecraft happily orbits the L1 point in a Halo Orbit without corrective burns. Does it do this naturally, or is it a consequence of the correction burns that must be made to keep it from drifting from the L1 position? How about Lissajous Orbits? Are they powered?
     
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  3. Sep 16, 2005 #2

    pervect

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    Halo orbits require corrective burns.

    Consider the following diagram:

    Earth.........soho....................Sun

    Let the x-axis run across the page from left-right, and the y-axis run up and down. Let the z axis be orthogonal to both the x and y axes (out of the planea of the paper).

    Let the coordinates of L1 be x=0,y=0,z=0

    The position of the soho satellite will be stable in the y-axis, because there will be a net restoring force. Think of it as being like the restoring force on a very large pendulum. The same restoring force will exist on the z-axis as well. The position of lowest potential energy is at y=z=0. Increasing y or increasing z increasess the energy of the Soho satellite, because it gets further away from both the Earth and the Sun, and E=-GM/r, so high r means more energy. Thus the satellite naturally tends to fall towards the position of lowest potential energy, which occurs at y=z=0.

    The position of the soho satellite will be unstable along the x-axis however. Moving closer to either the earth or the sun (positive or netagive x) will decrease the potential energy of the satellite.

    Halo orbits work by taking advantage of the restoring force along the y and z axes. Powered burns are used to correct for the instability along the x-axis. There is an effective "spring constant" that tends to pull the satellite towards its equilbrium position at L1 (y=z=0) - a force which is IIRC proportional to distance. (You can explicitly add up the potential energies and take the derivatives with respect to y and z to find the magnitude of the force if you want). The halo orbit thus involves an orbit in the y-z plane (a circular orbit is the simplest conceptually), whereby this restoring force provides the centripetal force required to keep the satellite in the circular orbit.

    Your gravity simulator program can probably also be used to illustrate the "restoring force" in the y-z plane if you can prevent the instability in the x-axis from being a problem.
     
  4. Sep 16, 2005 #3

    tony873004

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    Thanks, pervect! That's the clearest explanation I've heard. I can do corrective burns in gravity simulator, so I might have a little fun playing with this over the weekend.

    Would I just ignore y & z, and only monitor the the spacecraft's position on the x-axis, vs.the position of L1 on the X axis, and when the spacecraft exceeds a certain threshhold, perform a burn to get it back in an acceptable range?
     
  5. Sep 16, 2005 #4

    pervect

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    That would be one way of doing it. You could make the corrective acceleration proportional to the distance offset. There will be some minimum value of the proportionality constant required for stability, above this increaing the constant will increase the "stiffness" of the system. However, the stiffer system will put more demands on your integrator, requiring a smaller time step, so you don't want it _too_ stiff.

    You might also think about including some damping, without it you'll have an undamped vibration along the x-axis.

    THere may be an easier approach. You can just recompile your simulator code to make x fixed. Don't bother to calculate it, just set it to the desired value. If you desire, you could add code to check that the total x-force was near zero (as should be the case).
     
  6. Sep 18, 2005 #5

    tony873004

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    I did it! I maintained L1 for an entire orbit burning only exactly towards or away from the Sun. It took me three tries. Things go badly quiclky without attention.

    Now my question is this: I started out with 0 z-axis, and as expected, it stayed put on the z axis. But it oscillated on the y-axis as I didn't have my oribital speed perfect to begin with. Had this speed been perfect to begin with, it seems like it would have stayed put there too. So why does SOHO utilize an orbit that oscillates on both the y and z axis, rather than zeroing out one or both of these components? Maybe the y can't be zeroed because of the eccentricity of Earth's orbit, but it seems that z should be easy to zero-out.

    Any idea how often SOHO performs its burns? Maybe I'm just not Googling the right terms, but I can't find this info.
     
  7. Sep 18, 2005 #6

    pervect

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    I *think* the reason SOHO is offset is so that our receivers on earth don't have to deal with radio noise from the sun when receiving signals from the satellite.

    If soho were right at L1, it would be directly on a line between the Earth and sun, and it would be necessary to point our telescopes and antennas directly towards the sun to communicate with the satellite.

    In the Halo orbit, we don't have to deal with this noise/blinding effect.

    At least that's my guess.
     
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