MHB Eccentric anomaly as function of the eccentric and the mean anomaly

[prankster1590]

I'm reading, or more like deciphering a book about rocket science (Rocket propulsion and spaceflight dynamics from Cornelisse, Schoyer and Wakker). Most of my math skills are self taught from books and I usually figure things out but this one keeps haunting me.

It's some kind of series expansion of Kepler's equation to obtain an equation that aproximates E as function of M and e.

So Kepler's equation is written in this form: x = e sin (M + x) with x = E - M

x is expressed as power series in e

x = a₁e + a₂e² + a₃e³ + ...

Thus with the help of the double angle trig identity this becomes:

x = a₁e + a₂e² + a₃e³ + ...+ = e sin M cos x + e sin x cos M

Than the book says do a series expansion of sin x and cos x and substitute with the power series. It is not very clear but I assumed that they meant this:

e sin m (1 - x²/2! + x⁴/4! -...+) + e cos M (x - x³/3! + x⁵/5! -...+)

And than substituting the power series

e sin M [1 - (a₁e + a₂e² + ...+)²/2! + (a₁e + a₂e² + ...+)⁴/4!] + e cos M [(a₁e + a₂e² + ...+) - (a₁e + a₂e² + ...+)³/3! + (a₁e + a₂e² + ...+)⁵/5!]

This should give an equation which should allow me to calculate a₁, a₂ and a₃. By equation the coefficients of equal powers of e. But after weeks of trying and trying I still not have the solutions for a₂ and a₃.

The solutions should be: (but I don't know how)

a1 = sin M
a2 = 1/2 sin 2M
a3 = 3/8 sin 3M - 1/8 sin M

Can somebody shed some light on this for me. I read something about it having to do with the Lagrange inversion theorum but I'm not familiar with that.
Another derivation I found in Spherical astronomy by Smart is the following:

Keplers's equation is: E = M + e sin E

The assumption is that e is very small. The first approximation is therefore:

E₁ = M

The second approximation:

E₂ = M +e sin E₁ = M + e sin M

E₃ = M + e sin E₂ = M + e sin (M + e sin M)

With the angle sum trig identity this becomes

E₃ = M + e (cos M sin (e sin M) + cos (e sin M) sin M)

Since e is small this can be written as:

E₃ = M + e² cos M sin M + e sin M

And with the double angle identity this becomes:

E₃ = M + e sin M + 0.5 sin (2M)

So that was quite easy. But the next estimation. I don't get it.

E₄ = M + e sin E₃ = M + e sin (M + e sin (M + e sin M).

So I have tried all kinds tricks and identities. Like sin (3x)= whatever you make of it. But I never got the answer that I need. Which is:

E = M + e sin M - 1/8 e³ sin M + 1/2 e² sin (2M) + 3/8 e³ sin (3M)

or

E = M + (e - e³/8) sin M + 1/2 e² sin (2M) + 3/8 e³ sin (3M).

So how do I get to that answer?

 

[Theia]

I'm sorry you've waited so long.

As far as I can see, your solution (first method with series) is ok. But it looks like there are many terms that are not necessary. Let's drop them and see what is left. We have

$$a_1e + a_2e^2 + a_3e^3 + \cdots = e\sin M \left( 1 - \frac{a_1^2e^2 + \cdots}{2} + \cdots \right) + e\cos M \left( a_1e + a_2e^2 - \cdots \right) $$.

Now, equating the same powers of $$e$$ we obtain

$$\begin{align}
a_1 &= \sin M \\
a_2 &= a_1 \cos M = \sin M \cos M = \tfrac{1}{2} \sin (2M) \\
a_3 &= -\tfrac{1}{2}a_1^2 \sin M + a_2 \cos M = \cdots = -\tfrac{1}{8} \sin M + \tfrac{3}{8} \sin (3M),
\end{align}$$

exactly what we were looking for. Please let me know if I've omitted something you don't understand.

 

[prankster1590]

I'm sorry you've waited so long.

As far as I can see, your solution (first method with series) is ok. But it looks like there are many terms that are not necessary. Let's drop them and see what is left. We have

$$a_1e + a_2e^2 + a_3e^3 + \cdots = e\sin M \left( 1 - \frac{a_1^2e^2 + \cdots}{2} + \cdots \right) + e\cos M \left( a_1e + a_2e^2 - \cdots \right) $$.

Now, equating the same powers of $$e$$ we obtain

$$\begin{align}
a_1 &= \sin M \\
a_2 &= a_1 \cos M = \sin M \cos M = \tfrac{1}{2} \sin (2M) \\
a_3 &= -\tfrac{1}{2}a_1^2 \sin M + a_2 \cos M = \cdots = -\tfrac{1}{8} \sin M + \tfrac{3}{8} \sin (3M),
\end{align}$$

exactly what we were looking for. Please let me know if I've omitted something you don't understand.

Hi. Thank you very much. I'm so happy :) Now I see what I did wrong. The part about equating the same powers of $$e$$ confused me a bit. But I'm still trying to figure out how to get from $$\begin{align}a_3 &= -\tfrac{1}{2}a_1^2 \sin M + a_2 \cos M = \cdots = -\tfrac{1}{8} \sin M + \tfrac{3}{8} \sin (3M),
\end{align}$$. I always have troubles with trigonometry. I just substitute a1 and a2 and see where the ships lands.

Thank you very much

 

[Theia]

But I'm still trying to figure out how to get from $$\begin{align}a_3 &= -\tfrac{1}{2}a_1^2 \sin M + a_2 \cos M = \cdots = -\tfrac{1}{8} \sin M + \tfrac{3}{8} \sin (3M),
\end{align}$$. I always have troubles with trigonometry. I just substitute a1 and a2 and see where the ships lands.

In practise, yes. First, substitute the terms $$a_1 = \sin M$$ and $$a_2 = \sin M \cos M$$, then write everything in terms of sin and last part, use the expression of $$\sin ^3 M$$:

$$\begin{align}a_3 &= -\tfrac{1}{2}a_1^2 \sin M + a_2 \cos M \\
&= -\tfrac{1}{2}\sin ^3 M + \sin M \cos ^2 M \\
&= -\tfrac{1}{2}\sin ^3 M + \sin M (1 - \sin ^2 M) \\
&= -\tfrac{3}{2}\sin ^3 M + \sin M \\
&= -\tfrac{3}{2} \cdot \tfrac{1}{4} \left(3\sin M - \sin (3M)\right) + \sin M \\
&= -\tfrac{1}{8} \sin M + \tfrac{3}{8} \sin (3M).
\end{align}$$

As for a method used here, there are some problems for general usage. The method you have described here assumes you can write the eccentric anomaly in form

$$E = M + \sum_{i = 1}^{\infty} f_i(M)e^i,$$

where functions $$f_i(M)$$ are some functions of mean anomaly and we found the first three of them by using the serie method above. Basically, what this means, we have assumed that there exists a Taylor serie for some range of $$e$$. And thus we should do very careful work to find out what is the radius of convergence of the serie. I'm not trying that calculation here, instead I take a literature value and say that it can be proved that this Taylor serie converges if the eccentricity $$e$$ is smaller than about $$0.66$$.

For more general use, one must write the serie in other way round, namely

$$E = M + \sum_{k=1}^{\infty} g_k(e) \sin(kM)$$.

In this form (Fourier serie) the serie does converge in all $$0 \le e < 1$$. Also, it can be shown that the functions $$g_k(e)$$ can be written as

$$g_k(e) = \frac{2J_k(ke)}{k}$$,

where $$J_k(ke)$$ is a Bessel function of the first kind. See e.g. wikipedia.

 

[prankster1590]

In practise, yes. First, substitute the terms $$a_1 = \sin M$$ and $$a_2 = \sin M \cos M$$, then write everything in terms of sin and last part, use the expression of $$\sin ^3 M$$:

$$\begin{align}a_3 &= -\tfrac{1}{2}a_1^2 \sin M + a_2 \cos M \\
&= -\tfrac{1}{2}\sin ^3 M + \sin M \cos ^2 M \\
&= -\tfrac{1}{2}\sin ^3 M + \sin M (1 - \sin ^2 M) \\
&= -\tfrac{3}{2}\sin ^3 M + \sin M \\
&= -\tfrac{3}{2} \cdot \tfrac{1}{4} \left(3\sin M - \sin (3M)\right) + \sin M \\
&= -\tfrac{1}{8} \sin M + \tfrac{3}{8} \sin (3M).
\end{align}$$

[/URL]

Wow. So simple. And I tried so hard. Lol. Many Many thanks to you.