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prankster1590
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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?
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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