Posted: 06:34:55 PM, 02/06/2011

## Fly me to the Moon (Part 3)? ? |
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Click here to view the previous part. ? 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. ? In 1960, I was using a simulation of the RTBP to study the circumlunar trajectory. I found it pretty easy to get trajectories of the proper shape, and that allowed me to do a full parametric study over a wide range of parameters. I'm afraid I took this pleasant result very much for granted; it was my first lunar simulation, after all, and I didn't pick the coordinate system. ? I was more than a little surprised when we moved on to a three-dimensional, N-body simulation and found the task to be much harder. At the time, I figured that the difference was the three-dimensional nature of the simulation, and the more realistic orbits for the Moon and planets. Only now, at the end, do I realize that it's not the higher fidelity of the model, but the rotating coordinate system, that makes the difference. ? ??????????????????? It's the difference between the golf putt and the figure-skating maneuver. The rotating coordinate system takes timing out of the equation, and that's what makes the problem easier. ? One last point: Using the rotating coordinate system is easiest in the RTBP, where the rotation rate is constant, as is the radial distance between Earth and Moon. In such a case, it's easy to see that the Moon must stay pinned at that one spot on the x-axis. ? But switching to a more realistic model doesn't have to negate the value of the coordinate system. Even with the most accurate model for the motion of the Earth-Moon system (including solar perturbations), you can always define a coordinate system such that the x-axis is along the Earth-Moon axis, the Moon's velocity lies in the x-y plane, and the z-axis is normal to its angular momentum vector. If you also normalize the scale to the instantaneous Earth-Moon distance, there's the Moon again, pinned to a fixed point. ?
In ?
?
As you can see, transforming the coordinate system warps the trajectory quite a lot. Now the segments of the trajectory that are close to the Earth don't look much at all like parts of an ellipse. But they really are. That is to say, if you transform an elliptical orbit into the same coordinate system, it'll look very much like ?
The interesting part of ? Now, I said earlier that the Moon has virtually no effect on the motion, when the spacecraft is close to the Earth. The converse is also true; Near the Moon, the trajectory looks very much like a conic section. Not an ellipse, though. The relative velocity (which, you'll recall, is mostly due to the Moon's orbital motion) is too high for that. Instead, it's a hyperbola. If you want to go into orbit around the Moon, you need a decelerating burn, called Lunar Orbit Injection (TLI). ?
If you remember your geometry, you know that a hyperbola is asymptotic to two straight lines called, amazingly enough, the asymptotes. In ?
The hyperbolic character of the circumlunar trajectory near the Moon is apparent in ? This encounter between the Moon and the spacecraft is a special case of a swingby, or gravity assist, maneuver. JPL uses such maneuvers all the time, to boost a probe's velocity to reach the outer planets. But because our trajectory is symmetric, there is no boost; the velocity vector is bent from one asymptote to the other, but the magnitude isn't changed. ? In a way, the Moon only serves to reflect the velocity back toward the Earth. As we approach the Moon, the velocity is downward in the figure, and out away from the Earth. After the encounter with the Moon, the downward velocity hasn't changed, but the horizontal velocity has been reflected back toward the Earth. If we think about the trajectory purely in terms of the velocity vector, we've only rotated it through an angle of about 90 degrees. ?
I'm sure it will come as no shock to you that this angle is tightly connected to the miss distance. If we miss the Moon by a lot, the angle must be much shallower. If we miss it entirely, there is of course no deflection at all. To skim close to the surface, the trajectory must be bent as you see it in ? There's a very important implication of this relationship. If we truly want a low-altitude approach, which we did for Apollo, we have no choice but to let the trajectory bend through a fairly large angle. To do this, we must arrive at the Moon with a relatively high radial velocity; say about 800 m/s. This is important because it requires a higher velocity, and therefore more fuel, after the TLI burn. ?
I always find it interesting when a given constraint on a design drives other constraints which drive other constraints which drive... ? In the case of Apollo, we wanted the astronauts to have the best chance of coming back to Earth alive. To insure that, we wanted a free return, circumlunar trajectory. We wanted to do our TLI close to the Earth, and just skim the atmosphere on the way back. This forced the trajectory to be the symmetric figure-8 we've come to know and love. ? We wanted to pass close to the Moon. To do that, the Moon-relative hyperbola needed a fairly large deflection angle, which meant that the radial velocity had to be high, which meant that the velocity after TLI had to be larger than we might have liked. ?
When I first began generating trajectories for the Google Lunar X-Prize, I naturally started out thinking along the lines of my earlier designs. But I soon realized that the trajectories should be different, because the requirements are different (sound familiar?). The Apollo design was driven by the desire for a free return, in case the astronauts had to abort the LOI maneuver. In the X-Prize mission, that maneuver is essential. If the LOI fails, it's Game Over; the mission has failed. What happens to the spacecraft after that, we don't care. ?
Finally, because our trajectory doesn't need a return leg, the whole discussion about the shape of the Moon-relative hyperbola is moot. We no longer need a radial velocity when we get to the Moon. Instead of approaching the Moon from about a 45 degree angle, as in the trajectory of ?
? Because this trajectory has lower energy, it has a longer trip time: More like five days than three? ?
In the ? |

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