Thursday, December 29, 2022

The Jacobi Integral is no invariant outside a system

It occurred to me that the Russell / Ocampo / Longuski cyclers are, also, defined in terms of a unit (inner) circle. I'll start with the Aldrin and its precious cargo of fuzzy li'l rodents.

Aldrin like all cyclers starts at {0,1} in a frame-of-reference heading down - turnwise - velocity {2π,0} per Earth year. It actually starts at the highest orbit which Earth-Luna share as we can get it; but nobody is calculating this. The liftoff, we assume, has one of Cole's gravity-deflection fields so its beginning vector in sidereal terms is {-0.9198936, 7.3104506}.

For awhile the Aldrin is scraping close to Earth so I'm starting my timer when (xy - sat).Length() is 0.161724165 AU. This {0.5881579, 0.81678015} allows 1.00650859 AU from the Sun so we're actually almost 1/6 our way to the equilateral - to leading L4. 2Ω rounds to 3.000 as we'd expect. Velocity {-5.30233, 5.0659566} but we'll take off the velocity-vector of the frame, which is {-5.1319814, 3.6955051} here. J = 2Ω - 1.38099813*1.38099813 = 1.093. That's the system invariant so I get to keep this. Right? RIGHT?!

Let's [eff around and] find out. With the higher potential that 1.36659813 AU gets me, still 0.3666 AU from home: 2Ω=3.331. Aldrin's speed in this part of the frame comes down to 3.86913 AU/year. J is now -11.64.

Yes, we've kept the Sun at the zero-point, not opposite Earth. But.. cyclers run a long time. Longuski et al. were planning on the synod, not the year; by preference two synods, like S1L1 or 2-3-1-5. Jupiter orbits at 11.86 Earth years so is noticeably budging our Sun over the spans in question.

A reminder that this is a cycler so will be coming back to Earth. At a high Vinf, granted; so it's not a free return-squared. Still. Should I not have had a low negative Jacobi, at the start?

I think once we've left a system, Jacobi just gives up and tells us we're on our own.

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