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I am looking into some Lambert's solver-type research and came across this wonderful formula on Wikipedia at the bottom of the Lambert's Solver page, here. It has a parameterization gamma, a normal vector N, and then eccentricity and semilatus rectum equations. But they're all a function of position alone, assuming you know two positions:

enter image description here

Besides the change in length notation halfway through the eccentricity vector it seems like a killer set of equations since semilatus rectum and eccentricity are both functions of two radii only. It seems almost too good to be true, so I tried looking at the sources for the article and wasn't able to find this derivation or even these equations. I'm left scratching my head wondering how this came about.

Does anyone have any kind of derivation for it? Or know what's going on? Let me know if you have any information as I'm super interested in this parameterization that doesn't even require the gravitational parameter. Thanks!

Edit: After a good night's rest I'm realizing I wasn't clear with what I'm looking for. I'm trying to figure out how they managed to get an eccentricity vector with no velocity and no gravitational parameter. I am also looking for what reference frame this was derived in/if it can be used in an ecliptic and not a perifocal frame. I imagine it can be but that's kind of why I was looking for a source so I can follow and make sure. I looked through space stack exchange too and couldn't find anything. Please let me know if anyone knows anything!

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  • $\begingroup$ it is possible that there's some insight available already here in Space SE. This returns 41 hits, perhaps some are helpful. space.stackexchange.com/search?q=%22eccentricity+vector%22 $\endgroup$
    – uhoh
    Commented Jul 31, 2023 at 9:27
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    $\begingroup$ That does seem super weird. Given two position vectors, and the placement of the central body, I think that isn't enough information to staple down the magnitude or direction of the eccentricity vector. $\endgroup$
    – notovny
    Commented Jul 31, 2023 at 17:08
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    $\begingroup$ Ah, I see. It only gives you the eccentricity vector without gravitational parameter or velocity in two special cases (extremal and parabolic). $\gamma$ would require more parameters outside of those special cases. $\endgroup$
    – notovny
    Commented Jul 31, 2023 at 21:59
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    $\begingroup$ And thinking about it a little further, there isn't a unique parabola that's defined by a focus and two position vectors, so the $\gamma_p$ that produces a parabola seems to be just picking one that satisfies the requirements and is easy to calculate. Still not sure what's the uses is of finding the extremal case of $\gamma$ $\endgroup$
    – notovny
    Commented Aug 2, 2023 at 11:28
  • $\begingroup$ Thanks @ notovny. That makes sense and I figured my way out of needing the equation for the problem I've been trying to solve. Passing curiosity, though, do you know what the extremal case means? is it the minimum energy ellipse? or something else? I tried googling it and wasn't able to find anything. If you have any insight let me know, otherwise this will just be a passing oddity in my search. Thanks! $\endgroup$
    – Michael Gunnarson
    Commented Aug 2, 2023 at 19:30

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As I understand it, the initial formulation of Lambert's problem is entirely geometrical. In essence, provided the orbit is Keplerian and you know r1 and r2, the unoccupied orbital focus F2 is confined to one of two hyperbolic curves between r1 and r2, whereas the occupied one (i.e. F1, the central body) sits on the other curve and thus defines the shape of the hyperbola. Nothing in this construction requires you to know the dynamic properties of the gravitating body. enter image description here

The equations you list above basically allow you to pick one possible location for F2 from along the hyperbolic line of all possible F2s using the parameter gamma.

The eccentricity vector meanwhile, points from the central body to the periapsis, and has magnitude equal to the eccentricity. If the location of F2 has been picked, then the direction of the eccentricity vector is basically parallel to F2 - F1, the magnitude of which also gives twice the focal length. The semi major axis likewise can be derived from half of the sum of the distances from the two foci to either of the r points. With these in hand, calculating the eccentricity is basically the focal length / the semi major axis, which then completely nails down the eccentricity vector. All these manipulations only exploit the geometrical properties of ellipses, and don't require any dynamic or even kinematic calculations.

It's important to keep in mind that this doesn't actually solve Lambert's problem, since that is about finding the unique orbit that satisfies a particular transit time. That still requires iteration and also dynamic properties of the central body, and the point of this parametrisation is that you vary this gamma parameter as the independent variable in your iteration process to optimise for the transit time you ultimately want.

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