How Can Fourier Series Help Solve Kepler's Equation?

Click For Summary

Homework Help Overview

The discussion revolves around the application of Fourier series to Kepler's equation, specifically exploring how to express the equation in a series format involving Bessel functions. The original poster seeks guidance on deriving a formal solution and expanding it in powers of eccentricity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the form of Kepler's equation and the appropriate use of Fourier series. The original poster attempts to expand a function involving sine in terms of Fourier-Bessel series and questions the validity of their approach. Others suggest that the original poster clarify their understanding and provide more background information.

Discussion Status

The conversation reflects a mix of confusion and attempts to clarify the problem. Some participants express frustration over the lack of background information provided, while others encourage the original poster to show their work to facilitate better assistance. There is no clear consensus on the correct approach yet, as multiple interpretations of the problem are being explored.

Contextual Notes

Participants note that the original poster did not provide sufficient background information, which is necessary for effective assistance. There are also mentions of homework guidelines that require showing work, which the original poster has not fully adhered to.

suckstobeyou
Messages
19
Reaction score
0
GIVE ME A HINT! Fourier series / Kepler's equation

By expanding [tex]e \sin\psi[/tex] in a Fourier series in [tex]\omega t[/tex], show that Kepler's equation has the formal solution
[tex]\psi = \omega t + \sum_{n=1}^{\infty}{\frac{2}{n}J_{n}(ne)\sin{\omega t}}[/tex]
where [tex]J_{n}[/tex] is the Bessel function of order n. For small argument, the Bessel function can be approximated in a power series of the argument. Accordingly, from this result derive the first few terms in the expansion of [tex]\psi[/tex] in powers of [tex]e[/tex].
:confused:
 
Last edited:
Physics news on Phys.org
whats the matter?

is this question too confusing or something, if you don't understand a part of it please let me know.:rolleyes:
 
Kepler's equation

Here's the Kepler's equation for those of you who don't know it:

[tex]\omega t = \psi - e \sin \psi[/tex]

and Fourier series I think should be in the following form:

[tex]\sum_{i=1}^{\infty}e\sin \psi[/tex]

ANY HELP WOULD BE GREATLY APPRECIATED, YOU DON'T HAVE TO GIVE ME THE ANSWER JUST ANYTHING THAT'S ON YOUR MIND WOULD BE HELPFUL. EVEN THE SLIGHTEST CLUE COULD HELP ME GREATLY.

THANKS...
 
Last edited:
suckstobeyou said:
is this question too confusing or something, (snip)

Where to begin?

* You didn't provide any background info in the opening post. Most of us who browse the Homework Help section are goofing off at work, and we don't have our old textbooks on orbital mechanics laying around.

* After doing some digging, I've discovered that the information that you did provide is wrong.

This:

[tex]\psi = \omega t + \sum_{n=1}^{\infty}{\frac{2}{n}J_{n}(ne)\sin(\omega t)[/tex]

should have been this:

[tex]\psi = \omega t + \sum_{n=1}^{\infty}{\frac{2}{n}J_{n}(ne)\sin(n\omega t)[/tex]

* You didn't show any work, which is required by the Physics Forums Guidelines, which you agreed to follow. This process works by show-and-tell. You show you work, we tell you where you went wrong.

See the notice at the top of this Forum: https://www.physicsforums.com/showthread.php?t=94380 It's there to be read, as are the Guidelines you agreed to.

Now on to your question.

Here's the Kepler's equation for those of you who don't know it:

[tex]\omega t = \psi - e \sin \psi[/tex]

OK, that helps things along.

and Fourier series I think should be in the following form:

[tex]\sum_{i=1}^{\infty}e\sin \psi[/tex]

You are just randomly guessing. No Fourier series looks like that.

You could try looking into Fourier-Bessel series, as it looks like you are supposed to expand [itex]e\sin(\omega t)[/itex] in the Bessel function basis. But you can't expect any help here without showing us something.
 
ok fair enough. I'm new to Fourier-Bessel series but I'll try my best...
[tex]f(x) = e\sin(\omega t)[/tex]
[tex]a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx = \frac{1}{\pi}\int_{-\pi}^{\pi}e\sin(\omega t) \cos(nx)\,dx[/tex]
[tex]b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx = \frac{1}{\pi}\int_{-\pi}^{\pi}e\sin(\omega t) \sin(nx)\,dx[/tex]
[tex]f(x) = \frac{1}{2}a_0 + \sum_{n=1}^{\infty}[a_n \cos(nx) + b_n \sin(nx)][/tex]
so that's the Fourier series. Should I solve for [itex]a_n[/itex] and [itex]b_n[/itex] and then substitude them in the series? Or am I completely doing this the wrong way?
I also don't understand how you changed [itex]e\sin(\psi)[/itex] to [itex]e\sin(\omega t)[/itex]
Tom Mattson said:
You could try looking into Fourier-Bessel series, as it looks like you are supposed to expand [itex]e\sin(\omega t)[/itex] in the Bessel function basis.
 
Last edited:
here's the exapnasion

[tex]f(x) = \frac{1}{\pi}\int_{-\pi}^{\pi}e\sin(\omega t)dt + \sum_{n=1}^{\infty}\frac{1}{\pi}\int_{-\pi}^{\pi}[e\sin(\omega t)\cos(nt)dt]\cos(nt) + \sum_{n=1}^{\infty}\frac{1}{\pi}\int_{-\pi}^{\pi}[e\sin(\omega t)\sin(nt)dt]\sin(nt)[/tex]

please help I haven't heard back from anybody since my last post! I'm doing as much as I can...
 
your incapacity to help out or solve this straight forward problem either classifies you as a bunch of psychopaths or outright retards
 
is anybody even looking at this post? what am I doing wrong?
 
  • #10
This thread is two years old?
 

Similar threads

Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K