Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Describing elliptic orbit as a parametric function

  1. Nov 23, 2015 #1
    Hi PF

    I've beent rying to model the lunar orbit around the sun (cardioide) as a parametric function, but have run into a problem.

    f(t) = r(t) :

    x = a cos(ωt)
    y = b sin(ωt)
    z = k t

    The angular frequency ω as well as the distance from to the center varies around the orbit.

    Is there some way to express ω(t), for instance as a harmonic approximation between ωapo and ωperi ?

    i've been fiddling with the Vis Viva equations and something about the Mean motion and eccentric anomaly, but am unsure if my approach is at all possible.

    Thanks a lot for your time and attention!

    -Tegewaldt
     
  2. jcsd
  3. Nov 23, 2015 #2

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you are trying to plot it as a function of time, then there is no direct solution for an elliptical orbit. It can only be done by iteration.
    On the other hand, it is possible to plot it it as a function of angle from periapsis.

    That is found by

    [tex] r = \frac{a(1-e^2)}{1+ e \cos \theta}[/tex]
     
  4. Nov 25, 2015 #3
    What would such an iterative approach look like?

    So far i've tried Kepler II to define the area swept over by a line from the sun to the earth during orbit. By using this argument to solve the radius as a function of time, a numerical approach returns values varying slightly from the expected ones, and periodic with π rather than 2π.
     
  5. Nov 25, 2015 #4

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    First you find the Mean anomaly as a function of time ( as measured from periapsis)
    this works out to

    [tex]M= nt[/tex]

    where n is equal to 2 pi divided by the period of the orbit.

    Then solve
    [tex] M-e \sin M -M[/tex]
    and
    [tex]1-e \cos M[/tex]

    Divide the answer from the first equation by the answer of the second and subtract this result from M.
    This gives you the first estimate of the Eccentric anomaly.

    Substitute this for M in the above process and repeat until the difference in the final answer differs from the previous answer by less than your required precision.

    example:
    Assume M= 1.3 (in radians) and e= 0.5
    The first equation gives -0.481779
    and the second gives 0.86625
    The division of the first by the second gives
    -0.556166
    1.3-(-0.556166) = 1.856166

    The first equation with this new value of M gives -.479779
    and the second gives 1.52253
    the division gives -0.315119
    1.856166-(-.315119)=2.17678

    repeat as above.

    The next several iterations give:
    2.36033
    2.49193
    2.59326
    2.67438
    2.74085
    2.79621
    2.84283
    2.88243
    2.91627
    2.94534
    2.97041
    2.9921
    3.010911
    3.02726
    3.04149
    3.05390
    3.06474
    3.07420
    3.08248
    3.08973

    Notice how the differences between successive answers decreases. Once it becomes small enough for your purposes, you have an acceptable value for E.
    Now solve for the true anomaly [itex]theta[/itex] using

    [tex] \cos \theta = \frac{\cos E -e}{1-e \cos E}[/tex]

    This, along with the formula I gave in my last post gives you everything you need to plot an elliptical orbit as a function of time.
     
  6. Nov 25, 2015 #5
    Would this method work for the aforementioned example with the moon orbiting the earth while orbiting the sun (the latter considered a fixed point)?
    -Thank you for the detailed explanation by the way, very helpful indeed!
     
  7. Nov 25, 2015 #6

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Before you go too far on this, @Tegewaldt, you should realize that you are not going to see what you expect to see, which is a cardioid (or more generally, an epicycloid). Instead, what you will see will be very close to an ellipse. Suppose you plot the orbit on a display comprising 2000x2000 square pixels. The deviation between the Moon's orbit and that of the Earth/Moon barycenter is at most two or three pixels.

    Even more importantly, an epicycloid is not a convex figure. There are pairs of points on an epicycloid such that a line segment between those points contains points that are on the exterior of the orbit. This is not true for the Moon's orbit about the Sun. If you pick any two points on the Moon's orbit about the Sun and draw a line segment between those points, every point on that line will be on or inside the orbit. There are no epicycles in the Moon's orbit about the Sun. The Moon's orbit is convex.
     
  8. Nov 25, 2015 #7
    The intended purpose of such a funktion is to predict lunar phenomena as accurately as possible, so the above shouldn't be a problem, but I realize that the "visual result" May not be very interesting.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Describing elliptic orbit as a parametric function
  1. Elliptical Orbits (Replies: 15)

  2. Elliptical orbits (Replies: 2)

  3. Elliptical Orbits (Replies: 2)

  4. Elliptical Orbit (Replies: 5)

Loading...