So, before I get on with confessing my sins and explain what I've been lying about, I'm realizing I need to say a bit more about instabilities and why that horribly weird spiral trajectory — which I'm going to guess would have been especially horrifying for those Mercury and Gemini astronauts who all got their start as experimental aircraft test pilots; just mention "spiral dive" to any pilots you know and see what they have to say about it (hint: it's not something one usually lives to tell about) — and about why said trajectory is something to be embraced rather than feared.
Imagine a ball bearing rolling back and forth in a parabolic valley.
This scenario, the Simple Harmonic Oscillator, shows up all over the place, pendulum clock, weight on spring, child on a swing. It's like 90% of physics is about making as many situations as possible look like simple harmonic oscillators. Not suprising since this one of the easiest things to solve; we have this hammer; if we can make everything look like a nail, so much the better.
Qualitatively, there's only one solution:
x ∝ cos(ωt)
The ball just goes back and forth and back and forth and back and forth. Forever and ever; the epitome of stability. Once you know the frequency ω, you've got the whole story.
(I use the wonky "is proportional to" (∝) symbol so as not to have to be writing out lots of pointless constants. If, in what follows, you also want to imagine (t-t0) wherever you see t, feel free. So really what we're saying is x=x0cos(ω(t-t0)), the most general solution. Or you can just ignore the math altogether.)
And now we sneak in and make a little, teentsy sign change. What harm could it do?
The next morning, the security guards wake up and find that the parabolic valley has been turned upside-down/inside-out/whatever and we now have a parabolic hilltop (... along with Spock getting a goatee, Federation → Empire, Cat → Dog, etc...).
Whereas before, we had one simple solution, there are now nine (9) and they're all different qualitatively.
Instead of running away screaming, I'm going to make a table listing them all, so that you can get a sense of what we're up against. You may find it easiest to read them in a clockwise or counter-clockwise order.
|x ∝ −e−ωt
The ball has always been rolling in from the left; someday it may reach the top but we won't live to see it.
|x ∝ +sinh(ωt)
The ball approaches from the left,
makes it over the top,
and rolls down to the right
|x ∝ +e+ωt
The ball has always been rolling away to the right; if it was ever at the top, long ago, nobody remembers that far back.
|x ∝ −cosh(ωt)
The ball approaches from the left,
fails to get to the top,
rolls back to the left
|x = 0
The ball was, is, and ever shall be perfectly balanced at the top of the hill. Amen.
|x ∝ +cosh(ωt)
The ball approaches from the right,
fails to get to the top,
rolls back to the right
|x ∝ −e+ωt
The ball has always been rolling away to the left; if it was ever at the top, long ago, nobody remembers that far back.
|x ∝ −sinh(ωt)
The ball approaches from the right,
makes it over the top,
and rolls down to the left
|x ∝ +e−ωt
The ball has always been rolling in from the right; someday it may reach the top but we won't live to see it.
Tell me where you are and how fast you're going, and I'll tell you which box you're in.
The colorings are energy levels. All of the gray boxes have the same energy. Meaning if, when you're at some particular place and you have just the right velocity, you'll be in one of the gray boxes. If you're going faster, then you're in one of the blue boxes; if you're going slower, you're in one of the red boxes.
The gray boxes are essentially boundaries, drawing the fine line between success (blue boxes) and failure to cross over (red boxes). And you can skirt as close to them as you dare, so if, say, you're in one of the blue boxes, by reducing your energy you can make your trajectory be arbitrarily close to the trajectory in the gray box on either side. The closer you get, the more time it takes, so if you're going over the hill, you can arrange to take exactly as much time as you want by picking the right velocity/energy level.
In the center is chaos. An infinitesimal change to an x=0 scenario has eight possible outcomes. Rounding errors will ruin your day if you're not sufficiently clever.
Now, as I mentioned earlier, L1 is actually a saddle point. That is, assuming we orient our axes the right way, it's a parabolic hilltop in the x direction but in both of the y and z directions it's a parabolic valley. Parabolic valleys are places where we can park arbitrary amounts of energy while we're passing through (well okay, there are limits). In other words, if we're going through L1 from the Earth to the Moon or vice versa, we can make the transit take however long we want by changing the size of the spiral.
And since the various frequencies/periods stay roughly the same if we don't go too far out, that means we can spiral around as many times as we want while going through. But also, since we can mess with the y and z directions independently, that gives us even more choices re what direction we're going once we're out of the neighborhood of L1.
What we need is to build a map. Essentially, you can think of there being a (4-dimensional but never mind that) sphere of possible ways to park energy in the y and z directions. Imagine that sphere as being the center box in the table above (i.e., what the orbits would be if we weren't moving at all in the x direction).
Then you have the upper-right and lower left "Rolling Away" boxes which are now (5-dimensional) tubes leading from the sphere to elsewhere, and then the pair of "Rolling in" boxes, which are tubes from elsewhere back to the sphere. These bound the set of possible useful transit trajectories (the blue stuff) that take us from the elsewheres on the left (in the big hole where Earth is) to the elsewheres on the right (the moon and everything outside), which are what we want out of this.
At which point our agenda is simple (hahahahaha): Solve for where the sphere is. Figure out where the tubes go. Once we know where the tubes go, we know what our choices are.
The Actual Lay of the Land
Something is indeed rotten in Denmark and the core of it is that we're not actually doing the happy two-body problem that Newton solved, where angular momentum is conserved, everything has to move on conic-section-shaped trajectories, and ellipses are forever. Counting on our fingers, we see that Earth is one, Moon is two, and spacecraft makes three (3) bodies there, at which point there are no closed-form solutions, Newton gave up, Lagrange figured out a few things and then gave up.
There are lots of weird nooks and crannies, and we're in the process of stumbling onto one of them. Indeed the very fact that we can even have saddle points where chaotic things are happening is all part of why there can't ever be a closed form solution.
But now we need the Big Picture.
L1 is at minus 170km but "the top" is not at 0km. Why? Because, this whole time, I've actually been using a rotating reference frame where the earth and moon are fixed. Which means, among other things, there's centrifugal force to contend with, which gets stronger the farther out you go.
Meaning that 6000 km deep hole where the Earth is is not in the middle of a plane, but rather at the center of this parabola-shaped hill (well okay, parabola-of-revolution-shaped hill), which turns out to be a volcano with a 6000-km deep crater at the top of it. The circular rim of the crater is where the moon's orbit is; everything slopes downwards in all directions outside of it.
Things are further messed up because the moon is sufficiently big to put the earth off-center. That is, since earth and moon actually revolve around their common center of mass, the earth is displaced somewhat (4600km) in the direction opposite to the moon. Which then tilts that aforementioned circular crater rim; rather than being a constant −160½km altitude, the point opposite the moon on the rim (the L3 Lagrange point) is a bit lower (−161km) because it's closer to the earth.
Now if L3 is the low point on the rim, you might be thinking the place opposite it, where the moon is, should be the high point, but
- the moon is there, and
- the moon has its own gravity, which we have to add back (450km deep hole, remember?)
Where is the high point? Follow the rim 120° from L3 in the direction of the moon's orbit and you get to L5 and O'Neill's space colony. If you'd gone 120° the other way you'd have gotten to L4 instead. L5 and L4 are both at −160km and are the real (twin) hilltops. They are as high as you can go and there is no place that's 0km after all.
Things are actually further messed up because of the Coriolis force, which I haven't told you about, which happens to be crucial for understanding why L4 and L5 are stable even though they are hilltops which should otherwise be completely disastrous from a stability point of view. Fortunately, for L1 and L2, the Coriolis force only messes with the frequencies and tilts the various axes a bit; it doesn't change the overall qualitative picture, so I can skip that part.
(Nor was I never clear on why O'Neill preferred L5 to L4. Everything you can do with L5, trajectoriwise, you can do with L4; it's all symmetric, see. I'm also now wondering if the hilltop genuinely is the best place to be; it's actually the hardest place to get to in the Earth-moon system. There are so many tasks you need to do to maintain a space colony and keep everybody alive; station-keeping was never even remotely the biggest problem. Stability also means it's harder to leave, which will suck if you ever want to move the colony somewhere else. Though I suppose being at hilltop may reduce the probability that random rocks will arrive from infinity and ruin your day. That, to me, would be a much better selling point than stability — if it's actually true; haven't done the math on that one yet...)
So to get out from L1, instead of having to climb 170km as you might have originally thought, it's looking like, depending which direction we go, we only have to climb 10km at the most.
But it gets better.
The presence of the moon actually cuts a huge notch in the crater rim. If we continue our hike along the rim from L3 past the peak at L5 we'll find ourselves headed decisively downwards. Then the rim wall splits, going around either side of the big hole where the moon is. Directly across the moon from where they split, the walls rejoin on the far side and the rim continues around up to L4. L1 is the saddle point on the inner wall; L2 (you knew there had to be an L2) is the saddle point on the outer wall. And that is the last of the flat spots; Lagrange proved that there could only be five and this is where he left things 200 years ago.
L2, at −169km, is a measly one (1) kilometer higher than L1. As long as you have at least 140 m/s (313 mph) of velocity when you get to L1, you'll have enough energy to get to L2. And everything I've said about L1 (i.e., that it's a saddle point, that there are tubes, etc...) is true of L2 as well.
So if you're stationary at L1, you just need to put on 140 m/s of ΔV. But it's actually easier than that. The moon is right there. L1 has two outgoing tubes, one headed back towards Earth, the other outward. L2 likewise has two incoming tubes. See where the outward bound L1 tube intersects L2's from-inwards tube. Find the pair of intersecting orbits that comes closest to the moon. That is where you want to do your burn and chances are it'll be a lot smaller than 140 m/s (because the deeper you are, the faster you're moving and the faster you're moving, the less ΔV you need to achieve a particular energy change, i.e., to gain that last kilometer).
Once you are at L2, you are definitively outside the crater.
At which point we switch to the Earth-Sun rotating frame, where there is an entirely different set of Lagrange points. As it happens, the Earth-Sun L1 and L2 points are each about 1.5 million kilometers from Earth, your being at the Earth-Moon L2 point means you're now moving in a 444,000 km radius circle around the earth — exactly where depending on the time of the month — at something like 1200 m/s which is 300 m/s faster than what you'd ordinarily need to stay in circular orbit around Earth at that distance.
Which means, once you get sufficiently beyond L2 and away from the moon's influence, you're being flung away. Depending on how you timed things — and you can hang out in the halo orbits as long as you need to in order to time things just right — you can arrange to get flung away in any direction you want; and you'll be left with enough energy to both get away from the moon and get up another 63km worth of wall (this "wall" now being the wall around the solar crater whose rim is where the Earth's orbit is). Which is good because the Earth-Sun L1 and L2 points are both only about 50km higher from where you are now.
Or you can view everything from a completely non-rotating frame and see that the Moon just gave you a big gravitational assist. And when you get to Earth-Sun L2, the Earth is going to give you one, too, if you've played your cards right.
Except that, once you know where the tubes are, it's no longer a matter of chance. That is, you know where the outgoing tube from Earth-moon L2 is and where the incoming tube for Earth-sun L2 is and thus where they intersect. You then have a bunch of trajectories you can use.
And from Earth-sun L1 or L2, we can similarly go all sorts of other places, Sun-Mars L1, Sun-Mars L2, Sun-Jupiter L1, Jupiter-Ganymede L2, and on, and on. All pairs of co-orbiting bodies in the solar system, sun-planet, planet-moon, etc. each have their own L1 and L2 points guarding the entrances to their respective craters. Since everything is time-reversible in classical mechanics, you have trajectories going both ways, i.e., for every weird spiral trajectory that sends you away from L1 or L2 off to wherever, there's a corresponding one that brings you back in. You can string these trajectories together playing mix and match with them, giving you a way to visit any planet or moon that you like — admittedly, these low-fuel trajectories tend to be really slow, but if you're an unmanned satellite, you don't care.
This is the essence of the Interplanetary Transportation Network.
(which might seem to be a conclusion, but I actually have more to say about rockets in Part 6)
I, of course, forgot to mention the really cool part of Farquhar's thesis, which was the proposal that Collins be put in a halo orbit at L2 behind the Moon — which, as noted above, is a measly 1km worth of additional energy/effort beyond my have-him-orbit-L1 plan.
It then so happens you can make the radius big enough so as to remain visible from Earth at all times.
And then we do Far Side landings with no gaps in communication.