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Functions that define an orbit

  1. Dec 8, 2008 #1
    In a simple flat 2 dimensional plane of a two body orbit, both circular & elliptic.
    How do you write the functions wrt “t” time for:

    1) Force vector equation (that applies acceleration)
    2) Velocity vector equation
    3) Position vector equation

    That allows applying simple calculus integration to obtain #3 from #2 and #2 from #1
    Likewise simple calculus should show #1 as derivative of #2, and #2 as derivative of #3.

    Are polar coordinates preferred?
    or can X&Y coordinates work as well.

    If two masses are both large, (from equal to one 10% of other) is it necessary to use a barycentre with a “reduced mass”?
  2. jcsd
  3. Dec 8, 2008 #2

    D H

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    Things are much, much easier in polar coordinates. The forces are central, meaning they only depend on distance. The derivation is also much easier when expressed as in terms of a reduced mass with respect to the barycenter.

    Most sophomore/junior level classical mechanics texts cover this problem in detail.
  4. Dec 8, 2008 #3
    None that I've found - got a title, author.
  5. Dec 8, 2008 #4

    D H

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    Goldstein, Classical Mechanics; Marion, Classical Dynamics of Particles and Systems
  6. Dec 8, 2008 #5
    I find those books are only about 15 miles away from me, so I can get to them.

    Do you recall if they actually show plot-able functions with respect to time that calculus can convert to the other functions: I.E. Convert a function giving an acceleration plot to a velocity function. And then convert the velocity function to a position function that reveals a plot that turns out to be elliptical (circular in the special case).

    I’m not looking for the “http://en.wikipedia.org/wiki/Orbit_equation"” as on Wiki with explanations of how elliptical variables indicate an elliptic orbit that fit with the inverse square rule. Those equations already apply “e” for eccentricity with the elliptical conclusion already in place.

    I’m looking for the detailed force/acceleration function that allows calculus to derive (integrate to) the velocity and position functions that reveal a plot that matches the elliptical shape.
    These are what justified establishing the orbit equations.

    I didn’t think this would hard to find, but Books and Web Sites seem to only address the defined equations that result not the calculus that produced the result.
    The Amazon table of context for these books, show they might be the same way I cannot tell for sure.

    Do you or anyone know of a web link or summary that give the detailed calculus without assuming the ellipse to start with, but shows how the ellipse comes as a result.
    It seems so fundamental there should be one some where.
    Last edited by a moderator: May 3, 2017
  7. Dec 8, 2008 #6
    The following analysis comes from Fowles and Cassiday, Analytical Mechanics, 7th ed. LaTeX isn't liking my use of mathbb, so I'll use arrow notation for vectors. Watch the dots; they're hard to see.

    Let's say you have an equation of motion given by [tex]m\ddot{\vec{r}} = f(r) \hat{r}[/tex]; in polar coordinates this becomes [tex]m(\ddot r - r \dot \theta^2) = f(r)[/tex] and [tex]m(2 \dot r \dot \theta + r \ddot \theta) = 0[/tex]. Then angular momentum divided by mass [tex]l = r^2 \dot \theta[/tex] is conserved. Writing [tex]r = 1/u[/tex], we get
    [tex]\dot r = -\frac{\dot u}{u^2} = -\frac{\dot \theta}{u^2} \frac{du}{d\theta} = -l \frac{du}{d\theta}[/tex]
    since [tex]\dot\theta = lu^2[/tex]. Then
    [tex]\ddot r = -l \frac{d}{dt} \frac{du}{d\theta} = -l \frac{d\theta}{dt} \frac{d}{d\theta} \frac{du}{d\theta} = -l^2u^2 \frac{d^2u}{d\theta^2}.[/tex]
    Putting that in [tex]m(\ddot r - r \dot \theta^2) = f(r)[/tex] and manipulating a bit gives
    [tex]\frac{d^2u}{d\theta^2} + u = -\frac{f(1/u)}{ml^2u^2}.[/tex]

    In the case of gravity where [tex]f(r) = -k/r^2, k = GMm[/tex], this equation ends up being
    [tex]\frac{d^2u}{d\theta^2} + u = \frac{k}{ml^2}[/tex]
    which has general solution [tex]u = A \cos(\theta - \theta_0) + k/ml^2[/tex], which becomes
    [tex]r = \frac{\alpha}{1 + \epsilon \cos (\theta - \theta_0)}[/tex]
    where [tex]\alpha = ml^2/k[/tex] and [tex]\epsilon = Aml^2/k[/tex]. If [tex]\epsilon < 1[/tex], this is the equation for an ellipse with latus rectum [tex]\alpha[/tex] and eccentricity [tex]\epsilon[/tex].

    Unfortunately this doesn't give you r and [tex]\theta[/tex] in terms of t explicitly.
  8. Dec 9, 2008 #7
    Thanks that helps. I think this last point is the one I will continue to have trouble getting them in terms of t explicitly.
    I think I’m finding the standard derivations are based on making a good Physics Guess (angular momentum conserved) that allow easier math to be applied in order to confirm the ‘guess’ as a good one.

    A good guess, can short-cuts the more complex math demonstration of explicitly detailing a position plot based on the inversed square assumption that should reveal an exact match to an ellipse without using the “physics guess”.
    Evidently I’m looking for a more rigorous and complete mathematical calculus derivation than most texts are going to find necessary to demonstrate the basic point in physics as true.
    I’ll keep looking for the more detailed approach as I think there is value in it, or find someone to help me on the calculus - I’m no pro at it.
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