Describing elliptic orbit as a parametric function

[Tegewaldt]

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 ω_apo and ω_peri ?

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

 

[Janus]

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

r = \frac{a(1-e^2)}{1+ e \cos \theta}

 

[Tegewaldt]

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.

 

[Janus]

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

M= nt

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

Then solve
M-e \sin M -M
and
1-e \cos M

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 theta using

\cos \theta = \frac{\cos E -e}{1-e \cos E}

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.

 

[Tegewaldt]

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!

 

[D H]

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.

 

[Tegewaldt]

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.