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Parabolic Two-Body Motion

  1. Jun 20, 2013 #1
    I have been working with elliptical motion for a while, so am aware that when one body revolves around another in an elliptical orbit it sweeps out equal areas in equal times (measured from a focus).

    However, I am not sure if the same applies to parabolic motion. If, say, a comet approaches and then departs from the Sun in a parabolic trajectory, does its motion sweep out equal areas in equal times as measured from the focus of its parabola?
     
  2. jcsd
  3. Jun 20, 2013 #2

    tiny-tim

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    Hi DuncanM! :smile:
    Yup! :biggrin: (and same for hyperbolic motion)

    Even works without the inverse square law …

    you can prove it using only that the force is central (ie, always towards a fixed point). :wink:
     
  4. Jun 21, 2013 #3
    Yes, it works for all values of the eccentricity (Circle, Parabola, Ellipse, Hyperbola).
     
  5. Jun 21, 2013 #4
    Thanks for the info.

    Are you good with conic sections? I am investigating the transition from elliptical to parabolic motion and could certainly use as much help as I can get. I am presently stumped.

    Here is what I am doing:
    I am investigating elliptical motion at a particular time and would like to determine its position over the range 0 < e <= 1 (e is the eccentricity of the ellipse). So I solve Kepler's Equation at this time for each value of e.

    For example, for circular motion, at θ = π/6, sin(θ) = 1/2.
    If the angular velocity of the motion is ω = 1, then the period T = 2π time units.
    So... θ = π/6 corresponds to a time T/12 (one-twelfth the period into the motion).

    What I am hoping to do now is to detect a pattern for elliptical motion for the time T/12--and I would like to include the point for e = 1.

    So I solve Kepler's Equation for several values of e:
    E - e sin(E) = π/6​
    The trouble starts when e = 1. This equation can actually be solved for sin(E), E, and cos(E)--all finite numbers.

    However, I can't find a way to transform this coordinate to a point on the conic section.

    Assuming the superposing circle is of radius a = 1, the x-coordinate is x = cos(E).
    For 0 < e < 1, the y-coordinate would usually be found as √(1 - e2) sin(E).
    However, in the present case, e = 1 so this method doesn't work.

    In fact, because e = 1, this motion is actually that of a parabola, not an ellipse. And this brings me to my problem. Is there any way to determine the point on the parabola that corresponds to the (cos(E), sin(E)) point found by solving Kepler's Equation for e = 1? In fact, I don't even know what kind of parabola it is. All I know is that, because e = 1, it is a parabola. Is there any property or restriction of a parabola that would help me define the parabola on which this point lies?

    Going back to fundamentals, the underlying conic that generates the parabola, does the cone axis coincide with the focus of the parabola?

    Even if I know equal area is swept out in equal time by both the ellipse and parabola, I still don't think that is enough info to define the particular parabola.
    x = cos(E), known.
    dA/dt = CONSTANT, known from elliptical motion.
    But neither the focus nor vertex are known because the parabola is unknown, so the origin of the coordinate system in the parabolic plane isn't even known.

    Is there anything I can use to figure out what parabola is defined by a particular point in time for e =1?
     
  6. Jul 1, 2013 #5
    Reinventing celestial mechanics, are we?
    Google these: true anomaly, mean anomaly, Barker's equation
    (apologies if you're way ahead of me)
     
  7. Jul 1, 2013 #6
    By the way, regarding the original question - "does a parabolic orbit sweep out equal areas in equal times?" - look up Newton's original explanation of how this works. It's based on the geometry of triangles, and makes it very obvious. A good summary of it appears in Moulton's "Introduction to Celestial Mechanics", an old book which can be found on Google books.
     
  8. Jul 13, 2013 #7
    It will.

    In fact, one can derive Kepler's Second Law from the conservation of angular momentum for a central force.
     
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