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I need some help rearranging some orbital mechanics formulas.

All images have been borrowed from http://www.braeunig.us/space/orbmech.htm which has a through treatment of orbital equations, but provides no further insight on my question.

Given the following:

Position equations

Apsis equations

I have $r$, $\phi$ and $R_{a}$ or $R_{p}$ (only one will be relevant in each instance). I need to derive $a$, $e$ and $v$.

I've had no luck at all trying to rearrange these formulas to produce those results. I can't even assert one way or the other if it is possible, however knowing the system they represent it seems like there should be a single solution in each instance.

Background:

The goal here is to calculate a two tangent transfer from an arbitrary elliptical orbit to a circular orbit. The outputs $a$, $e$ and $v$ give the core parameters of the transfer orbit. The supplied $r$ and $\phi$ are the altitude and flightpath angle at the point of the first tangential burn. These are preserved during the burn while $a$ and $e$ are changed. $R_{a}$ / $R_{p}$ is the radius of the destination orbit. Since the destination orbit is circular meeting it at either the apoapsis or periapsis of the transfer orbit will will be another tangential burn. The choice of which to use is dependent on whether we are ascending or descending.

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I believe that it's well known that you have to use approximation methods here. You might try asking it over on physics SE. –  Carl Brannen Apr 5 '11 at 0:59
    
That would be annoying, I already have an iterative solver for a more general version of which this is a subset. I was hoping that knocking out half the free variables would let me lose the iterative solver and all the performance issues that go with it. –  tolomea Apr 5 '11 at 1:12

1 Answer 1

I don't think you are going to find an algebraic solution. Those sines and cosines don't play nice. The numeric solution is well behaved.

However, if you want the most efficient transfer, it should be done at the apsides. Firing anywhere else wastes propellant rotating the orbit, and if you are going to a circular one you don't care. In particular, if the desired orbit is above your current periapsis, you should first adjust the apoapsis. In this case $\nu=0$, the trig functions go away, and all is well. When I asked one of our orbit people if there was an intuitive reason this order was right, he replied that you want to leave as much fuel as low in the gravity well as possible.

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Nice intuitive explanation of practical orbital mechanics! I'll vote it up when I get my votes back in 75 minutes. –  Carl Brannen Apr 5 '11 at 22:47
    
I presume by "adjust the apoapsis" you mean by doing a tangent burn at periapsis? That is kinda the core point, periapsis is both the lowest and fastest point. In addition to the observation about leaving spent fuel low in the gravity well reaction engines are more efficient at higher velocities due to the Oberth effect. –  tolomea Apr 6 '11 at 11:36
    
@tolomea: you are right. You do a tangent burn at periapsis that is sized to put the apoapsis where you want it. Here the required $\delta v$ is easy to calculate from the energy equation. Then when you get to apoapsis you fire again tangentially to raise the periapsis and circularize. Again the $\delta v$ can be calculated from the required energy. –  Ross Millikan Apr 6 '11 at 13:24

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