# Goldstein 3.3

1. Jul 15, 2009

### buttersrocks

1. The problem statement, all variables and given/known data
(Goldstein 3.3)

If the difference $$\psi - \omega t$$ in represented by $$\rho$$, Kepler's equation can be written:

$$\rho = e Sin(\omega t + \rho)$$

Successive approximations to $$\rho$$ can be obtained by expanding $$Sin(\rho)$$ in a Taylor series in $$\rho$$, and then replacing $$\rho$$ by its expression given by Kepler's equation. Show that the first approximation by $$\rho$$ is given by:

$$tan \rho_1 = \frac{e Sin(\omega t)}{1-e Cos(\omega t)}$$

and that the next approximation is found from:

$$sin^3(\rho_2 - \rho_1) = -\frac{1}{6}e^3 sin(\omega t + \rho_1)(1+e cos(\omega t))$$

2. Relevant equations

All shown above...

3. The attempt at a solution

Okay, the first part quickly pops out of the Maclaurin series for Sin(rho). The second part, however, I'm having some trouble with. I can think of many different ways this might be approximated and don't know which approach the book is looking for. If someone could set me on the right track, I'd very much appreciate it. (the e^3/6 is making this look like they want me to use the next term of the taylor series or something.)

Methods I can try:
Expand the taylor series around $$\rho=\rho_1$$ for the first however many terms. (While this may not be the method the book is looking for, it's probably going to give something more accurate than the maclaurin series with the same number of terms...)

Take the next term of the Maclaurin series. (I'm getting stuck when doing this.)

Something that strikes me odd, but perhaps he wants some sort of iterative approach using Newton-Rhapson at this point?

Thanks.

2. Jul 15, 2009

### TMFKAN64

That's not the expression for the second expansion I see in my version of Goldstein.

3. Jul 15, 2009

### buttersrocks

you probably have the first printing. If you check the errata:

http://astro.physics.sc.edu/goldstein/1-2-3To6.html

you'll see that the expression typed above is indeed correct and it is what is printed in my sixth printing goldstein.

4. Jul 15, 2009

### TMFKAN64

Yeah, I just checked a more recent printing, and indeed, it's been fixed to what you have there. Apologies...

That's pretty clearly the second term in the sine expansion. I haven't worked out how it gets that form yet though.

5. Jul 16, 2009

### buttersrocks

No worries. If you do manage to get in that form, I'd appreciate you setting me in the right direction. I can get almost there, but I can never get rid of terms involving $$\rho_2$$ that are not inside of the LHS sine.

Mainly, is it the second term of the Maclaurin series where I substitute in the first term for some constants, or do I re-expand the Taylor series about $$\rho_1$$ and take it to the cubic term. It's too much algebraic manipulation for me to sit down and beat it into submission without knowing that I'm doing what he actually expects.