Rotating LPe in non-circular orbits

[cymrych]

So I've been playing around with the latest AMSO release, trying to optimize my lunar orbits while maintaining close association to the historical timeline. Using a combo of LTMFD and IMFD, optimization over AMSO's standard 110km circular orbit with respect to historicity wasn't too difficult, with the exception of one detail.

In order to use only retrograde thrust for the 2nd LOI burn (the "circularization" burn which reduces the original LLO of 300+km x 111km to roughly 122 x 100km) requires, of course, a burn not at the periapsis in order to arrive at the desired PeA and ApA. However, this rotates the LPe.

Eventually, through some trial and error, I established that the magnitude of this LPe rotation is constant for a given set of burn variables. For example, in my experiments: a -41.6m/s burn retrograd with no inward or planar velocity set at about 128 seconds either before or after periapsis would result in the desired ApA and PeA with an LPe shift of +/-84*. This discovery let me adjust the final LPe as needed, and the mission went spiffily with the ApA and PeA exactly where I wanted.

My question is: how would I go about determining how far the LPe would shift for a different set of burn elements?

Conceptually, I understand of course why the LPe shifts, but for the life of me I can't figure out how the rotation magnitude is computed. As this is the single element I cannot account for directly with an mfd, I have to make a burn and see what results ... and it gets a little tiring making burns, jotting down resulting numbers, exiting out and restarting the sim, modify the initial inputs, burn and see the results, rinse and repeat until satisfied. Much simpler to just figure it out ahead of time!

 

[boogabooga]

Would this help?

http://www.orbithangar.com/search_quick.php?text=precession+mfd&submit.x=0&submit.y=0

 

[cymrych]

Hmm, sounds interesting. I'll give it a whirl.

Still interested in figuring out the math as well, you know, pencil and paper style, so if anyone can point me to some resources please let me know!

 

[Fizyk]

I thought about this problem for a while and here is what I came up with:

The orbit can be described with the equation:

[math]r(\phi) = \frac{p}{1+e\cos\phi}[/math]
[math]\phi[/math] here is the true anomaly.

What determines your orbit is actually three numbers:
1. Distance from the center [math]r[/math]2. Your velocity [math]v[/math]3. Direction of your velocity, which can be given by [math]r' = \frac{dr}{d\phi}[/math]
From [math]r[/math] and [math]v[/math] you can calculate total energy, which immediately gives you the semi-major axis [math]a[/math] by:

[math]-\frac{GM}{r} + \frac{v^2}{2} = -\frac{GM}{2a}[/math]
So [math]a = \frac{GMr}{2GM-rv^2}[/math]
You can connect that with [math]p[/math] and [math]e[/math]:

[math]a = \frac{1}{2} \left( \frac{p}{1+e} + \frac{p}{1-e} \right) = \frac{p}{1-e^2}[/math]
To get the last equation, we need to calculate explicit formula for [math]r'[/math]:

[math]r' = \frac{dr}{d\phi} = \frac{pe\sin\phi}{(1+e\cos\phi)^2} = r^2 \frac{e\sin\phi}{p}[/math]
Ok, so now you are in orbit described by [math](r,r',v)[/math] and make a [math]\Delta v[/math] prograde burn. What changes? In the simplest approximation you did the burn in one instant, so [math]r[/math] doesn't change, and neither does [math]r'[/math], since the burn was prograde. [math]v[/math] obviously changes, and so do [math]p[/math], [math]e[/math] and [math]\phi[/math] (because your periapsis shifted, and the new true anomaly will be with respect to the new periapsis). Let's denote their final values as [math]p_f[/math], [math]e_f[/math] and [math]\phi_f[/math].

Since [math]r[/math] didn't change, we get:
[math]\frac{p}{1+e\cos\phi} = \frac{p_f}{1+e_f\cos\phi_f}[/math]
[math]r'[/math] didn't change, either, so we have:
[math]\frac{e\sin\phi}{p} = \frac{e_f \sin\phi_f}{p_f}[/math]
Change in [math]v[/math] changed [math]a[/math]:
[math]a_f = \frac{GMr}{2GM - r(v+\Delta v)^2}[/math]
It will be better to express it in terms of [math]\frac{1}{a}[/math]:
[math]\frac{1}{a_f} = \frac{2GM - rv^2 - 2rv\Delta v - r(\Delta v)^2}{GMr} = \frac{1}{a} - \frac{2v\Delta v + (\Delta v)^2}{GM}[/math]
Since [math]a = \frac{p}{1-e^2}[/math], we get:
[math]\frac{1-e_f^2}{p_f} = \frac{1-e^2}{p} - \frac{2v\Delta v + (\Delta v)^2}{GM}[/math]
So we have 3 equations for 3 unknowns [math]p_f, e_f, \phi_f[/math]:
[math]\frac{p}{1+e\cos\phi} = \frac{p_f}{1+e_f\cos\phi_f}[/math][math]\frac{e\sin\phi}{p} = \frac{e_f \sin\phi_f}{p_f}[/math][math]\frac{1-e_f^2}{p_f} = \frac{1-e^2}{p} - \frac{2v\Delta v + (\Delta v)^2}{GM}[/math]
They should be solvable once you put some numbers in, but I didn't manage to solve them in a general case. Once you get the new values, [math]\phi_f-\phi[/math] will be the angle by which the periapsis will shift.

I hope that helps

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EDIT: I managed to simplify the problem further by using the conservation of angular momentum.

Let's note that [math]\frac{dr}{rd\phi} = \frac{r'}{r} = \tan \alpha[/math], where [math]\alpha[/math] is the angle between the velocity vector and the direction perpendicular to the radius, which means that:
[math]J = |\vec{r} \times \vec{v}| = rv\cos \alpha = \frac{rv}{\sqrt{1+\tan^2 \alpha}} = \frac{r^2v}{\sqrt{r^2+r'^2}}[/math](J here is angular momentum per unit of mass).

After burning, we have [math]J_f = J+\frac{r^2\Delta v}{\sqrt{r^2+r'^2}}[/math]
It is actually very easy to calculate [math]p_f[/math] from [math]J_f[/math]. Let's see what is the connection between [math]J[/math] and [math]p[/math] in a general case.

By conservation of angular momentum, we have:
[math]J = v_{min}r_{max} = v_{max}r_{min}[/math]
So:
[math]v_{min}\frac{p}{1-e} = v_{max}\frac{p}{1+e}[/math][math]\frac{v_{min}}{1-e} = \frac{v_{max}}{1+e}[/math]
From conservation of energy we get:
[math]\frac{v_{max}^2}{2} - \frac{GM}{r_{min}} = \frac{v_{min}^2}{2} - \frac{GM}{r_{max}}[/math]
[math]\frac{v_{max}^2}{2} - \frac{GM(1+e)}{p} = \frac{v_{min}^2}{2} - \frac{GM(1-e)}{p}[/math]
[math]\frac{v_{max}^2}{2} - \frac{GMe}{p} = \frac{v_{min}^2}{2} + \frac{GMe}{p}[/math]
After substituting [math]v_{min} = v_{max}\frac{1-e}{1+e}[/math]:

[math]\frac{v_{max}^2}{2}\left( 1- \frac{(1-e)^2}{(1+e)^2} \right) = \frac{2GMe}{p}[/math]
After simplifying:
[math]\frac{v_{max}^2}{(1+e)^2} = \frac{GM}{p}[/math]
[math]\frac{v_{max}}{1+e} = \sqrt{\frac{GM}{p}}[/math]
Finally, we get:
[math]J = \frac{v_{max}p}{1+e} = \sqrt{GMp}[/math]
[math]p = \frac{J^2}{GM}[/math]
So:
[math]p_f = \frac{J_f^2}{GM}[/math]
This should simplify the problem a lot