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Parametric Equations of Keplerian Orbit

  1. Apr 20, 2013 #1
    1. The problem statement, all variables and given/known data
    The figure illustrates a Keplerian orbit, with Cartesian coordinates (x,y) and
    plane polar coordinates (r,φ).
    F = -(G*M*m)/r^2
    The parametric equations for the orbit:
    r(ψ) = a ( 1 − e cos ψ)
    tan(φ/2) = [(1+e)/(1-e)]^(1/2)* tan(ψ/2)
    t(ψ) = (T/2π) ( ψ − e sin ψ)
    where ψ is the independent, parametric variable; a, e, and T are constants.

    Prove, directly from the parametric equations, that the angular momentum L is
    constant. Express L in terms of the constants of the orbit (and any other relevant parameters).

    2. Relevant equations
    L = r x p
    τnet = r x F


    3. The attempt at a solution
    I know that in order for L to be constant, its derivative (net torque) must be 0, but this isn't really getting me anywhere. I know torque = rxF and L = rmv, but quite frankly I can't seem to figure out how to do this using the parametric equations.
    What I have done using the parametric equations is this:
    L = m*r(ψ)*v(t) = m*(a*(1-e*cos(ψ)))*v(t)
    From here I'm having a difficult time finding an expression for v(t). I tried using the relations
    ω=((2*π)/T) and v = ω*r for the following work:
    t(ψ) = (1/ω)*(ψ-esin(ψ))
    ==> ω = (ψ-esin(ψ))/t(ψ)
    ==> ω = (ψ-esin(ψ))/((T/(2π))*(ψ-esin(ψ))
    ==> ω = (2π/T) (this obviously backtracked)
    ==> v = r*(2π/T)

    ∴L = m*(a*(1-e*cos(ψ)))*(a*(1-e*cos(ψ)))*(2π/T)
    or L = m*a2*(1-e*cos(ψ))2*(2π/T)

    I'm pretty sure this isn't right because it isn't going to be constant for different r values. Any help would be appreciated.
     

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    Last edited by a moderator: Apr 21, 2013
  2. jcsd
  3. Apr 20, 2013 #2

    mfb

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    The time-derivative of L = r x p is not r x F. Don't forget that r depends on time as well.
    Only if r and v are perpendicular to each other. In general, they are not.
     
  4. Apr 21, 2013 #3
    Ok, well my intuition tells me to make L=r(psi)mvsin(θ) where θ is the angle between the positive x-axis and the line from the focal point to the point in the orbit

    In my notes, we proved momentum was constant by making momentum L = m(x*vy - y*vx) and converting to polar coordinates. Does this equation apply to any elliptical orbit?

    As for the time derivative of L = rmv, I am just unsure of how to do this as I can't seem to figure out the correct expression for v.
     
    Last edited: Apr 21, 2013
  5. Apr 21, 2013 #4

    mfb

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    If you calculate the cross-product, this is exactly what you get. It looks tricky to get v_y, v_x based on the functions in the problem statement, however. L=r(psi)mvsin(θ) looks easier to evaluate.

    The formula is wrong for an elliptic orbit anyway, there is no need to get v.
     
  6. Apr 21, 2013 #5

    vela

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    Yes, it is. When you apply the product rule, the other term drops out because dr/dt is parallel to p.
     
  7. Apr 21, 2013 #6

    mfb

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    Oh, right.
    Good, now it makes sense again (as ##\dot{L}=\tau=r \times F##).
     
  8. Apr 21, 2013 #7
    Ok, I'm confused here. I'm continuing forward with L = mvr(ψ)sin(θ), but I don't understand why I don't need to get v, since v isn't constant? I know that when taking the time derivative of L one of the terms is r x p = 0 because they parallel
     
  9. Apr 21, 2013 #8

    vela

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    These relations only give you one component of the velocity, due to the fact that the angle ##\varphi## is changing. The other component arises because ##r## varies with time.
     
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