Eccentric anomaly as function of the eccentric and the mean anomaly

In summary, the conversation discusses two methods for approximating Kepler's equation, which is used to calculate the position of a planet in its orbit. The first method involves using a series expansion of the equation, while the second method involves making approximations and using trigonometric identities. Both methods lead to the same result, which is a power series with coefficients that can be used to approximate the equation. However, the first method results in unnecessary terms that can be dropped to simplify the solution.
  • #1
prankster1590
3
0
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?
 
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  • #2
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

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

Now, equating the same powers of \(\displaystyle e\) we obtain

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

exactly what we were looking for. Please let me know if I've omitted something you don't understand.
 
  • #3
Theia said:
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

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

Now, equating the same powers of \(\displaystyle e\) we obtain

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

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

Thank you very much
 
  • #4
prankster1590 said:
But I'm still trying to figure out how to get from \(\displaystyle \begin{align}a_3 &= -\tfrac{1}{2}a_1^2 \sin M + a_2 \cos M = \cdots = -\tfrac{1}{8} \sin M + \tfrac{3}{8} \sin (3M),
\end{align}\). 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 \(\displaystyle a_1 = \sin M\) and \(\displaystyle a_2 = \sin M \cos M\), then write everything in terms of sin and last part, use the expression of \(\displaystyle \sin ^3 M\):

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

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

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

where functions \(\displaystyle f_i(M)\) 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 \(\displaystyle e\). 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 \(\displaystyle e\) is smaller than about \(\displaystyle 0.66\).

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

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

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

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

where \(\displaystyle J_k(ke)\) is a Bessel function of the first kind. See e.g. wikipedia.
 
  • #5
Theia said:
In practise, yes. First, substitute the terms \(\displaystyle a_1 = \sin M\) and \(\displaystyle a_2 = \sin M \cos M\), then write everything in terms of sin and last part, use the expression of \(\displaystyle \sin ^3 M\):

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

[/URL]

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

FAQ: Eccentric anomaly as function of the eccentric and the mean anomaly

What is eccentric anomaly?

Eccentric anomaly is a mathematical parameter used to describe the position of a celestial body in its elliptical orbit. It is defined as the angle between the semi-major axis and the projection of the body's position onto the auxiliary circle.

How is eccentric anomaly related to eccentricity and mean anomaly?

Eccentric anomaly is directly related to eccentricity and mean anomaly through Kepler's equation, which states that the eccentric anomaly is equal to the mean anomaly minus the product of the eccentricity and the sine of the mean anomaly.

Why is eccentric anomaly useful in orbital mechanics?

Eccentric anomaly is useful in orbital mechanics because it allows for the calculation of the position of a celestial body in its elliptical orbit at a specific point in time. This information is crucial for predicting and understanding the motion of objects in space.

How is eccentric anomaly calculated?

Eccentric anomaly can be calculated using various methods, such as Kepler's equation or numerical methods. One common method is the Newton-Raphson method, which involves iteratively solving an equation until a desired level of accuracy is achieved.

Can eccentric anomaly be negative?

Yes, eccentric anomaly can be negative. This occurs when the celestial body is in the second or third quadrant of its orbit, where the eccentric anomaly is measured in the opposite direction from the positive x-axis.

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