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Two body problem with additional force

  1. Jun 21, 2011 #1
    Hey folks I'm trying to find some info on this scenario I'm working on.

    Consider a particle orbiting a star with eccentricity e, basically standard two body problem. I have an approximation for the particles distance from the sun at some time t. I would now like to introduce another force, F, that acts on the particle via the line joining the star and particle. I know the equation that determines this force and it is dependent on the distance between the star and particle.

    I would like to calculate the distance shown below

    [PLAIN]http://img842.imageshack.us/img842/6876/unledkdq.jpg [Broken]

    with the particle beginning at pericentre and completing one half of a full orbit. The black line is the standard 2-body orbit with no force and the red line is the (exaggerated) new path the particle will follow.

    I thought this would be quite easy to do but because of the eccentricity it has become far too complicated. The distance from star to particle is always changing and so the acceleration is always changing too. Do I have to start right back at the equations of motion and throw in this extra force or can I modify what I know given that the eccentricity and force is small? The force is proportional to 1/d^2 where d is distance from the star so any attempt to solve a differential equation and find d as a function of time proves impossible.

    If I have to start back at the equations of motion, how would I do that given that I don't know F as a function of time?

    I think I may have to make some assumptions to make this easier (which could perhaps be justified given that the eccentricity and force are small)...

    Assumption..?

    Particle takes the same time to orbit the star (at least for small timescales) so only need to think about how far it travels from 'original' orbit.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 21, 2011 #2
    Applying an arbitrary force in the radial direction will destroy the elliptical orbit; it will certainly not instantaneously alter the orbital parameters (as your diagram illustrates).

    The two orbits in your diagram is consistent with an instantaneous impulse delivered in the direction of motion at periastron.
     
  4. Jun 21, 2011 #3
    Hmm I kinda had a feeling that would be the case.

    Any idea what the actual effect would be? Would even a tiny force ruin the orbit?
     
  5. Jun 21, 2011 #4
    The elliptical orbit is a very special property of the inverse square law force (note: I believe a hook's law potential, and then one other integer power-law force---which I can't remember, will also produce ellipses), and in general, an arbitrary force will ruin it. If the magnitude of the force is small (compared to the gravitational force), it will have a small effect; if large, then a large effect...

    If your additional force were a weak one; i.e.
    [tex] \frac{F_\textrm{new}}{F_\textrm{grav}} \ll 1 [/tex]
    Then you could do a perturbative analysis to see what the effects are.

    Otherwise you would have to start from the differential equations of motion (as you suggest) and maybe find some sort of analytic solution, or more likely integrate numerically to find out the result---this isn't hard if you have any programming experience.
     
  6. Jun 22, 2011 #5
    Yeah perturbation analysis is what I'll have to do. I already did that in finding the distance from the sun in terms of the time (with eccentricity taken as the small parameter).

    This though is less obvious and I can't see how to do a perturbation analysis without going back to the equations of motion. I usually look for something to expand but I can't do that here and I'm not sure how to mix acceleration (or force) as a function of distance with distance as a function of time.

    So really I can't see how to even begin to set up a perturbation method for this one that doesn't involve trying to solve d(t)'' = 1/d(t)^2 which is far too complicated or just going back to the equations of motion.

    (I should say, the only perturbation methods I have even had to use are ones of the form [tex]x_0 + \epsilon x_1 + \epsilon^2 x_2 + ...[/tex]
    which usually arise from either some form of expansion or just plugging that into the equations of motion and equating the orders. What would the 'x' terms be in this case? It's just not obvious to me.)
     
  7. Jun 22, 2011 #6

    gneill

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    Staff: Mentor

    So this additional force is a central force (acting on the line between the particle and the star), and further it's an inverse square force, inversely proportional to the square of the distance between the particle and the star?

    Since that's the same as what gravity does, all it's really doing is adding a bit to the gravitational force, or "adjusting" G slightly...
     
  8. Jun 22, 2011 #7
    Wow never looked at it this way before that's far simpler than what I'm trying to do. So going back to the 2-body equations of motion is perhaps the way to go but it won't be as difficult as I imagined it would be. When you say adjust G slight do you mean actually adjust the value of the gravitational constant G?

    Edit: Oh I think I see what you mean...
     
  9. Jun 22, 2011 #8

    gneill

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    Staff: Mentor

    You can adjust G, multiply it by some appropriate factor, or perhaps replace it with a sum, like G + GF.

    If you simply replace it with a new appropriate value (call it G' perhaps), then all the usual equations pertaining to 2-body motion under the influence of gravity will still apply without change.
     
  10. Jun 22, 2011 #9
    Yeah that seems like it'll work.

    So, the solution to the equation of motion for the 2-body problem is given by the usual [tex]\frac{h^2}{\mu (1 + e \cos(\theta))}[/tex]
    where we replace [itex]\mu[/itex] with [itex]\mu'[/itex] where we have modified the G that appears in [itex]\mu[/itex] by something appropriate.
     
  11. Jun 22, 2011 #10
    That's the key question: is this an inverse-square law force? I was under the impression Deadstar meant some arbitrary force, F(r)
     
  12. Jun 22, 2011 #11
    Yeah it is, I said it in the OP but guess it wasn't exactly clear it was the perturbing force.

     
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