2-body problem - how to get the angle theta?

[alyx_vance]

Homework Statement

I'm trying to solve the 2-body problem analytically by following this book:
http://books.google.com/books?id=im...ential equations&pg=PA60#v=onepage&q&f=false"
(note: the book preview is not complete, but you can find pages 69-73 http://books.google.com/books?id=oNvFAzQXBhsC&lpg=PA58&pg=PA64#v=onepage&q&f=false", it's a different version of the book but this section is identical as far as i noticed)

On page 68, the last equation (2.35) gives me the solution to r, the magnitude of the displacement vector between the bodies. But how do I get the angle $$\theta$$? It's a function of time too, but all the equations with it contain also r, which I don't have yet computed.
I understand how to get the constants (probably) and how to convert between the various coordinate systems, but I just don't see a way to get the angle $$\theta$$. As far as I see it, r(t) depends on $$\theta$$(t) and $$\theta$$(t) depends on r(t), so I'm kind of stuck...

Homework Equations

Equation 2.35 in the first book link and 2.36 in the second.

The Attempt at a Solution

I wondered if maybe $$\theta'$$ is constant but it doesn't seem so. I think I am missing some basic concept :(

 

[ideasrule]

Acos(θ)+Bsin(θ) can be written in terms of a single cosine (or if you prefer, a single sine):

Acos(θ)+Bsin(θ)
=1/sqrt(A²+B²) * (cos(φ)cos(θ)+sin(φ)sin(θ))
=1/sqrt(A²+B²)*cos(θ-φ)

Since we get to define the starting point of θ, we might as well define it so that the phase φ is 0. In fact, this is how the orbit equation is usually written; I don't know why your book doesn't write the equation this way.

 

[alyx_vance]

I'm sorry but I still don't understand it :(
I get it that the Acos(θ)+Bsin(θ) can be rewritten using only a single cosine...but that doesn't tell me how to compute θ(t). Define it so that the phase is 0? How? Isn't it changing with time?

 

[ideasrule]

Oh, I'm sorry; I misunderstood your question. There is no explicit solution for θ(t). Theta (also called the true anomaly) can only be calculated using numerical methods like Newton's. For the solar system, this works well because the orbits are nearly circular and corrections are small compared to the angle calculated assuming a circular orbit (called mean anomaly).

 

[alyx_vance]

Oh. Wasn't there supposed to be an analytical exact solution to the 2-body problem? I am writing a n-body simulator and I wanted to compare the numerical integration methods to an exact solution in the case of two bodies...
Could you please give me some reference? I didn't see it mentioned anywhere that 2-body needs numerical methods. And is there some open-source software I could look into to see how it's done? It would save me some time, thanks.

 

[ideasrule]

There's an analytical solution to the two-body problem in the sense that:

1) You can write the distance r in terms of theta, the true anomaly
2) You can write down the formula for a converging series that gives the true anomaly

However, this doesn't mean that there's a closed-form solution (one with no series, just well-known functions) to the true anomaly. As it turns out, there isn't.

The solution for the true anomaly is essentially the solution to the inverse Kepler equation. See: http://en.wikipedia.org/wiki/Kepler's_Equation. Once you get the eccentric anomaly, true anomaly can be calculated using the much simpler equation here: http://en.wikipedia.org/wiki/Eccentric_anomaly#From_the_true_anomaly

In practice, however, that series isn't usually used. See here for practical numerical methods of calculating the true anomaly: http://www.stargazing.net/kepler/kepler.html

 

[alyx_vance]

Thank you very much, the last link looks helpfull.