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A previous question I asked here got clarification on the Earth-passing phases of the Lunar Cycler orbits that use the "backflip" maneuver. It came up in the book Artemis, and it used the name Uphoff-Crouch cycler (Uphoff & Crouch were the authors of the linked paper), so I will use that here. From the previous question, the key information was:

Earth-return trajectory that has a period of 1/2 month (or 1/3 month is some cases) in order to ensure return to the Moon after 2 (or 3) revolutions of the Cycler in its Earth-return orbit".

(bold emphasis mine)

What does the BackFlip lunar cycler do in its pass by Earth?

I started trying to run through the math, and the computations are put in this worksheet:

https://github.com/AlanCoding/Lunar-Cycler/blob/master/Lunar%20Cycler%20math.ipynb

I came to an impasse on the 3-orbit variety. I'll try to express the problem in light math here.

Say the moon's orbit is roughly circular, Ra=Rp. The elliptical transfer orbit is high elliptical, Rp=0 (we can relax both of these with numbers, later). For the 3-pass case, the transfer orbit's period must divide the moon's orbit's period by 3... but at first glance, it looks like it can't do that.

Starting with the equation for orbital period...

  • a - semi-major axis
  • T - orbital period

math

Problem: the factor above is below unity for f=3. Simply, 2/3^(2/3)=0.96

This is saying that the apogee of a 3-pass transfer orbit would not reach the moon. Was it just carelessness that the Uphoff-Crouch paper assumed a 3-pass orbit would work? I did play with numbers breaking the assumption of circularity of the moon's orbit, and making perigee of the transfer orbit equal to Earth's radius, and that changed 0.96 to 0.94, so that's not helping.

Surely, a 3-pass orbit would be superior to a 2-pass orbit (which I have still to prove to myself works), because it would be lower-energy.

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Set your time and length units to LD and lunar period and you can lose the 2 pi and mu.

$T=a^{3/2}$

So if you wanted an ellipse whose period was 2/3 that of the moon you'd have

$2/3=a^{3/2}$
$2/3^{2/3}=a$
$a=.763$

.763 of an LD is .763 * 384400 = 293352 km.

If perigee were 6678 km then apogee would be (2*293352)-6678. Which is 580026 km.

If you wanted a lower apogee you could raise the perigee.

If you wanted an elliptical orbit whose period is 1/3 that of the moon's you'd have:

$1/3=a^{3/2}$
$1/3^{2/3}=a$
$a=.481$

.481 of 384400 is 184800 km

Again apogee would be 2a - perigee. If perigee were 6678 km, apogee would be 362922.5 km.

In my opinion this apogee is too close to the moon. If the cycler goes deep inside the lunar Hill sphere, lunar perturbation will wreck the cycler orbit (I believe). But again apogee could be lowered by raising perigee.

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