Problem in Derivation of Goldstein

I'm writing this in connection with najkim's answer above, just because I noticed something when some terms were simplified using the first identity

For the 2nd part, when najkim gets to the part

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

The left hand side of this equation can be further simplified to

$sin{rho_2} - e sin{(omega t + rho_1)}$

And from the first identity, $tan{rho_1} = frac{e sin{omega t}}{1-e cos{omega t}}$, we get

$e sin{omega t}cos{rho_1} = sin{rho_1} -e cos{omega t}sin{rho_1} $

which leads to

$ sin{rho_1} = esin{(omega t + rho_1)}$

So shouldn't the final expression be

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

In general there are many ways to numerically approximate a transcendental solution. In Goldstein, some recommended approximation methods seem strange, but the first time you solve a problem, it's best if you first stick to Goldstein's recommendations.

Successive approximations to $rho$ can be found by expanding $sin rho$ in its Taylor series and replacing $rho$ its expression (the Taylor series) given by Kepler's equation

This method of approximation does indeed seem odd, but you should probably go through with it. Write: $$ begin{align*} sin rho_1 &approx rho_1 \ &= esin(omega t + rho_1) \ &= esin(omega t) cosrho_1 + e cos(omega t) sinrho_1 \ Longrightarrow sinrho_1 (1 - e cos(omega t)) &= e sin(omega t) cosrho_1 \ Longrightarrow tanrho_1 &= frac{e sin (omega t)}{1 - e cos(omega t)} end{align*} $$

Once you solve it this way, we can try your approximation scheme as well, which is an excellent one. When you expand $sin rho_1$, via a polynomial expansion, you should also expand $cos rho_1$. In general, when you approximation a function applied at a small input value, do it for all occurrences of this small variable. Therefore, you should modify it to:

$$ {rho} = esin({omega}t)(1 - frac{rho^2}{2} + frac{rho^4}{24} - cdots) + ecos({omega}t)({rho} - frac{rho^3}{6} + frac{rho^5}{120} + cdots) $$

When you take the first order approximation here for $ rho_1 ll 0.1$, you should drop all terms of order $mathcal{O}(rho^2)$ and higher, and find: $$ begin{align*} rho_1 &approx e sin(omega t) + e cos(omega t) rho_1 \ Longrightarrow rho_1 &= frac{e sin(omega t)}{1 - e cos (omega t)} end{align*} $$ For very small values of $rho_1$, we have that $tan rho_1 approx rho_1$, so we get something that looks along the lines of the desired result.

For the second part of the question, there is a major typo in the book. Please consult Goldstein Errata. You should get: $$ begin{align*} ( sin(rho_2 - rho_1)) &approx ( sinrho_2 - sinrho_1) \ &= -frac{1}{6}e^3 sin^3 (omega t + rho_1) end{align*} $$

There are two ways of arriving at a decent second-order approximation. As before, you can expand using Goldstein's approximation method. Observe that because $sin rho $ is a concave down in the positive reals near zero, our previous approximation is an overestimate. When $rho < 0$, the function $sin rho$ is concave up near zero, and our approximation is now an underestimate. We can try and compensate for this by subtracting the cubic term:

$$ begin{align*} sin rho_2 &approx rho_1 - frac{rho_1^3}{6} \ &= e sin ( omega t + rho_1) - frac{e^3}{6} sin^3 ( omega t + rho_1) \ Longrightarrow sin rho_2 - e sin(omega t + rho_1) &= - frac{e^3}{6} sin^3 ( omega t + rho_1) \ Longrightarrow sin rho_2 - e sin(omega t) cosrho_1 - e cos(omega t) sin rho_1 &=- frac{e^3}{6} sin^3 ( omega t + rho_1) end{align*} $$

Then, from the earlier equation:

$$ tanrho_1 = frac{e sin (omega t)}{1 - e cos(omega t)} Longrightarrow e sin(omega t) cos rho_1 = sin rho_1 - e cos (omega t) sin rho_1 \ $$

Substitute this result back into the previous equation to find:

$$ ( sinrho_2 - sinrho_1) = -frac{1}{6}e^3 sin^3 (omega t + rho_1) $$

Finally, we can once again try your method as well, dropping all terms $mathcal{O}(rho^3)$ and higher:

$$ begin{align*} {rho} &= esin({omega}t)(1 - frac{rho^2}{2} + frac{rho^4}{24} - cdots) + ecos({omega}t)({rho} - frac{rho^3}{6} + frac{rho^5}{120} + cdots) \ &approx e sin(omega t)(1 - frac{rho^2}{2}) + e cos(omega t )rho end{align*} $$

We move all terms onto one side of the equation and normalize $rho^2$ to find: $$ 0 = rho^2 + (frac{2}{e} csc(omega t ) - 2 cot (omega t)) rho - 2 $$ which can be solved for $rho$ using the quadratic formula.

$$ rho = { cot (omega t) - frac{csc (omega t)}{e} pm sqrt{frac{ csc^2(omega t)}{e^2} - frac{2 cot(omega t) csc(omega t)}{e} + cot^2(omega t) + 2}} $$ Yikes!