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 Nickname: Jack Crenshaw     Articles(30)    Visits(67555)    Comments(3)    Votes(5) Jack Crenshaw's column features algorithms and plug-and-play routines, along with explanations of how they work. Blog Archive: 2013 -  May.  2011 -  Jun.,  May.,  Mar. View All

Posted: 01:10:15 PM, 31/05/2011

# Fly me to the Moon (Part 2)

The three-body problem

One look at figure 1, though, tells you that this ain't your grandfathers ellipse. It's a strange and complicated trajectory, bending left, then right, then around the moon, in a sinuous path perhaps more familiar to a figure skater than an astronaut.

The trajectory is not an ellipse because this isn't a two-body problem, it's a three-body problem. As soon as the early astronomers—which included such mathematical giants as Gauss, Euler, Lagrange, and Poincaré—learned about Newton's triumph with the two-body problem, they probably said the 17th century equivalent of, "Hot doggies! Let's do it again, with one more body!" The idea wasn't based on just idle curiosity. Real three-body encounters exist in nature. A near miss of a planet by a comet is one; the motion of the Moon, as perturbed by the Sun, is another.

So they set out to solve the three-body problem. They didn't have much luck; the problem has proven to be intractable. Despite their best efforts and a lot of concentrated brainpower, they never found that elusive closed-form solution, and such a solution has since been shown to be impossible.

This is not to say that they didn't get any useful results. To help them study the problem, they used a time-tested technique: If you can't solve a given problem, try solving a different one. In this case, they reasoned that they might have better results if they made a few simplifying assumptions. The result is called the Restricted Three-Body Problem (RTBP). based on the assumptions that:

1. The mass of one body is negligible. Thus the remaining two bodies form a two-body system, orbiting their common center of mass.

2. The orbit is a circle The motion of the massless body is restricted to the plane of the other bodies' orbits.

The early astronomers were seeking a closed-form, analytical solution like Newton had found for the two-body problem. They didn't get that, but they were able to get a lot of insight into the motion. In particular, Lagrange was able to identify and study the five Lagrange equilibrium points, L1 through L5. Points L1, L2, and L3 lie along the Earth-Moon axis, and are unstable. Points L4 and L5 lie in the orbit of the Moon, spaced 60 degrees ahead and behind it. These points are stable; any masses that find their way there tend to stay there, drifting along in lazy orbits around their Lagrange points.

Still, we now know that the analytical solution to the RTBP doesn't exist. If you can't solve a dynamics problem analytically, you have only one recourse: Solve it numerically, by integrating the equations of motion numerically.

Those early guys could and did do just that, but it's not an exercise for the faint-hearted. If you try it, make sure you have several new pencils with large erasers. For all practical purposes, the RTBP had to wait for digital computers with enough horsepower to crank out the numerical solution. By a happy coincidence, such computers became available in the late 50's, about the same time I did. Indeed, it was a computer simulation of the RTBP that I used in most of my studies.

Looking at figure 1 again, one other aspect stands out: The trajectory is symmetric. The reason doesn't leap out at you, at least it didn't to me. It took me a few years to figure out why, but once you see it, it's obvious. You're not likely to see this explanation in any textbook, so remember, you heard it here first.

Here's the deal: Although the Moon influences the motion of the spacecraft throughout the flight, it has very little effect while the spacecraft is near the Earth. Near the Earth, we can expect the trajectory to look a lot like a two-body solution; that is, segments of an ellipse. But it's a long, skinny ellipse with a perigee near the Earth, and an apogee way out past the Moon.

Now, the parameters of most two-body orbits depend on two constants of the motion; the energy and the angular momentum. However, the eccentricity of our elliptical segments is high—about 0.97 or 0.98. That's very near the 1.0 of a parabola. And it turns out that the parameters of the parabola depend on only one of the constants: The angular momentum. Show me a parabolic orbit whose perigee has some specific value, and I'll tell you what its angular momentum has to be.

But that's exactly what the situation has to be for the circumlunar mission. On the outbound leg, we get the best efficiency when the TLI burn starts as close to the Earth as possible, just outside the sensible atmosphere. On the inbound leg, we need a trajectory that just grazes that same atmosphere. For all practical purposes, the two perigees are the same, so the angular momenta must also be the same.

This requirement places a special constraint on the encounter with the Moon. Whatever the Moon does to our trajectory, it cannot be allowed to change the angular momentum. As we'll see in a moment, this requires that the trajectory near the Moon must be symmetric as well.

All of this places an interesting constraint on the point of closest approach (I call it perilune, but strictly speaking, it should be periselene). You can put the perilune anywhere you like, as long as it's lunar latitude 0, longitude 180. I'll bet no one ever told you that before.

There's one more important point about figure 1: It's a LIE. Or, at the very least, highly misleading. The trajectory of figure 1 is supposed to represent an inertial coordinate system, "non-rotating relative to the background of fixed stars."

But look again at the figure. What's that little gray circle you see, out there to the right? You know, the one labeled "Moon"? It's just sitting there, isn't it, like the bullseye at a shooting gallery. But the real Moon isn't just sitting there; it's moving in an orbit of its own. It's moving pretty fast, in fact, at over 1,000 m/s (2,200 mi/hr). At this point in its trajectory, our spacecraft is close to apogee so it's moving much more slowly around 200 m/s.

When we think casually about a lunar encounter, most of us have a mental image of a sort of rendezvous, much as we might use to rendezvous with a geosynchronous satellite. But this is the wrong model. In reality, it's more like this: We put ourselves in the path of the Moon, wave our arms, go "Boogah Boogah," and dare it to run over us. Which it will absolutely do, unless we dodge it deftly. It's more like a matador's pass than a rendezvous.

Now perhaps you see the real issue with figure 1; an issue that makes the problem seem easier to solve than it really is. If the Moon were, in fact, fixed in space, the problem would be static. The only variable would be the initial position and velocity of the trajectory. You can think of the problem as be analogous to a golf putt on a particularly diabolical, warped and convoluted green. The target—the hole—is fixed, so to hit it, you only need to adjust your stroke.

But since the Moon is moving, you not only have to hit the ball just right, you have to do it at the right time.

Instead of that golf putt, the situation is more analogous to a figure-skating exhibition, where two performers glide smoothly out to the corners of the rink, far away from each other. Then at some prearranged time, they curve back to center ice, execute a waltz-like twirl, and part again.

The timing of the circumlunar trajectory is now the issue. If you get it right, you get that lovely little twirl that you see in figure 1. Get it wrong, and the Moon is simply not there when you arrive. To get it right, you have to lead the Moon, much as a hunter shooting at a clay pigeon. The typical flight time for an Apollo mission was about 3 1/2 days. The Moon moves about 13 degrees per day, so the rendezvous point should be about 45 degrees ahead of the Moon's current position. The geometry shown in Apollo 13 simply can't work.

Pin down that Moon

I must tell you in all candor, when I first started trying to generate a good lunar trajectory, I struggled for quite awhile before any of my attempts got within 20,000 miles of the target. It's a tricky thing. Worse yet, when you see that you've missed the desired trajectory, it's not immediately obvious what you should change to get closer.

Now we understand that the little gray circle of Figure 1 isn't really there; it only comes whizzing by just in time to dance with us. And for that reason, timing becomes an issue. So we ask ourselves: Is there something we can do to fix that?

Turns out, there is. What if we define a new coordinate system, a rotating one, that rotates at the same rate as the Moon moves around the Earth. If we do it right, we can arrange for the x-axis to always lie along the Earth-Moon axis. It's the same situation that we appeared to have in figure 1, but didn't.

This approach is, in fact, the same one adopted by Euler, Lagrange, and company. In their formulation of the RTBP, they wrote the equations of motion in the rotating system, which meant that they had to add centrifugal and Coriolis terms to the equations. We don't really have to do that. It's easier to compute things in an inertial system; we only need to use the rotating one during input and output transformations.

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