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Low energy MoonEarth transfer
by MontBlanc2012 11132017, 06:57 AM
Here (below) is a simple challenge designed to test your thinking about ways of making transfers from low lunar orbit to the Earth.
The scenario set out in the attached .scn file is as follows: A Delta Glider starts in a 20 x 20 km lunar orbit with a total 700 m/s of main fuel and 10 m/s of RCS. The challenge is to land anywhere on Earth. For the purpose of this exercise, achieving an Earth approach orbit with an Earthcentric periapsis < 70 km altitude constitutes a 'landing'. Now, at first blush, this sounds simple enough  after all, it is one of the first 'missions' that budding Orbinauts will set themselves. However, Apollostyle MoonEarth transfers typically require around 1,000 m/s of fuel. But in this challenge, the DeltaGlider starts with just over twothirds that amount. In fact, the fuel needed to lunar escape from a 20 km parking orbit is 692 m/s  so, you have barely enough fuel to 'escape' from the Moon let alone set up a low perigee encounter. Finding a MoonEarth transfer solution that meets the rather restricted fuel constraints of this scenarion requires a little 'out of the box' thinking. As will all problems of this kind, the solution is quite simple once you frame the problem in the rightway; and devilishly hard if you don't. But once you know what you are doing, it's quite easy to set up in Orbiter using standard tools. As a hint, you may find that TransX is not of much use in finding a solution. For those that easily manage to find the solution, well done!  and you may wish to pay appropriate hommage to Messrs. Lagrange and Belbruno. I'm happy to provide a solution upon request. As for the challenge scenario itself, here it is: Code:
BEGIN_DESC Delta Glider in a 20 x 20 km lunar orbit with a total 700 m/s of main fuel and RCS. The goal is to land anywhere on Earth. For the purpose of this exercise, achieving an Earth approach orbit with an Earthcentric periapsis < 70 km altitude constitutes a 'landing'. END_DESC BEGIN_ENVIRONMENT System Sol Date MJD 52006.7542552436 Help CurrentState_img END_ENVIRONMENT BEGIN_FOCUS Ship GLNT END_FOCUS BEGIN_CAMERA TARGET GLNT MODE Cockpit FOV 40.00 END_CAMERA BEGIN_SHIPS GLNT:DeltaGlider STATUS Orbiting Moon RPOS 1677740.022 137005.718 506935.076 RVEL 477.7254 65.8104 1598.8449 AROT 15.788 10.126 42.772 AFCMODE 7 PRPLEVEL 0:0.015058 1:0.004584 NAVFREQ 0 0 0 0 XPDR 0 HOVERHOLD 0 1 0.0000e+000 0.0000e+000 AAP 0:0 0:0 0:0 END END_SHIPS 
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11142017, 01:27 PM  #2 
Orbinaut

And just as a demonstration, here is a link to a YouTube video demonstrating the basic scheme (contains spoilers!):
https://www.youtube.com/watch?v=m4FD...ature=youtu.be Sadly, I fluffed the step back to real time upon Earth approach and ended up scooting by Earth with a ~ 70 km encounter. But, meh!, not a big deal. Last edited by MontBlanc2012; 11232017 at 07:23 AM. 
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11172017, 05:30 PM  #3 
thing

Hi,
I couldn't resist this one :) I managed to land close to Overberg base in SA after a fairly rough reentry (8g at one point) with 220g fuel remaining in RCS. I used the IMFD "Delta Velocity" + IMFD "Map  Plan View" to plan the escape burn. Set up a Prograde burn of 670m/s and tweak the burn time until you get reasonable solution, then tweak the dV. Here's my escape burn solution: Coming up on Moonflyby: MCC at first Apogee: Scraping the tanks for a last Perigee adjust burn before reentry: On the ground: Thanks for the challenge, I enjoyed it Tip o' the hat to Mssrs.Lagrange and Belbruno! Cheers, Brian 
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11212017, 01:51 PM  #4 
Orbinaut

Quote:
Although we can set up low energy transfers of this kind using IMFD, we can't ensure that these transfers are 'optimal'. However, I suspect that with a bit of tweaking with dates and prograde quantities, we an get to within 5  10 m/s of the optimal amount without much effort. You might want to try the inverse low energy transfer from low Earth orbit to lunar ballistic capture. The trajectory is essentially the timereversed, mirror image of the Moon to Earth transfer. I haven't tried this myself yet, but I'm sure it is doable in Orbiter. 
08272018, 08:36 PM  #5 
Orbinaut

Sorry for necroposting.
I've been having fun with GMAT for last days learning this tool and trying to solve these old Orbiter challenges. Looks like here we can use Earth gravity to minimize deltaV when we escaping the Moon and then we can use a gravity assist maneuver near the Moon to fly away from the Earth to give the Sun a chance to return us back home. For example, this is 664.5 m/s trajectory, calculated in GMAT (a single prograde burn at t+3900 sec).  P.S. Don't try to use these numbers in Orbiter with this scenario, I've mixed up frames of reference It turned out that there are even better solutions without grav.maneures. Eg. 650 m/s at t+530s. Last edited by Ajaja; 08292018 at 12:28 PM. Reason: edited 
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08282018, 10:35 AM  #6 
thing

Beautiful! Must get GMAT back on board :)

08282018, 12:20 PM  #7 
Orbinaut

Very nice.
As with BrianJ, it looks as if I'm going to have download GMAT. 
09262018, 06:02 PM  #8 
Orbinaut

Thanks MontBlanc2012 for this beautiful challenge. This one definitely deserves to be included in challenges of the next Orbiter update.
I had seen this challenge when it was first posted and in my initial attempts, I barely made it back to Earth using all the fuel at my disposal  even the RCS. Then I saw Ajaja's post about the GMAT 650 m/s solution, so I had to try again and see if I could reproduce it with IMFD. After a few minutes of playing around with the prograde dV and time variables, it seemed that 647.18 m/s was the "magic number". Even 0.01 m/s less than that and IMFD can't find any solution that gets back to Earth. But the 645.18  646 m/s range seems "special". Those solutions end up in a nice 3 Moon encounters trajectory that ends up with a straight path to Earth's core! I tried it first "just as is". Single burn from the Moon, no corrections, just to see where I'd end up. The result was very promising, I missed the Earth by "only" ~30M meters in the retrograde direction. So, in the next attempt, I performed the same burn but this time aimed for a 30M PeA in the prograde direction. Result: smack right in the center of Earth, using a single burn of 645.5 m/s. "If PeA = 30M Pro" gets me to the center, surely 36M meters will get me where I want, right?" Wrong! I tried all the burns that started from 35M and ended up to 105M meters above Earth's surface (adding 10M in each attempt) and they all ended up hitting the Earth near the center. In all the attempts, the difference in the dV was miniscule, but after that 3rd encounter with the Moon, you can see the PeA relative to Earth shrinking until it gets near the center. Below is a video of one of those attempts and the best dV result I had. (Being extremely careful with the RCS used for orientation and using only a single burn). Thanks again for this wonderful challenge! 
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09292018, 11:27 AM  #9 
Orbinaut

Quote:
1. An impulsive burn to transfer the vehicle from Low Lunar Orbit to a planar Lyapunov orbit around EM L1  (i.e., the Lagrange point between the Earth and the Moon). Approach to the Lyapunov orbit is along the stable manifold of the trajectory. 2. Because the Lyapunov orbit is unstable, as it approaches the Lyapunov orbit, it transfers quickly from the stable manifold to the unstable manifold of the same orbit. This unstable manifold enables a transfer to EM L2  (i.e., the Lagrange point on the far side of the Moon.) This transfer trajectory is ballistic and appears to require approximately three reasonably close passes of the Moon (as per Ajaja's suggestion in another thread.) 3. Approach to EM L2 is roughly along the stable manifold of the L2 point. Again, and because L2 is unstable, as it approaches L2, it quickly transfers to the unstable manifold leading away from the EarthMoon system. With correct timing, this unstable manifold connects with the stable manifold of the SunEarth L1 or L2 (not sure which). 4. Finally, as it approaches SE L1/L2, it transfers to the unstable manifold which again, with correct timing leads back to a close encounter with Earth. This strategy for solving the problem  i.e. to target EM L1, then a ballistic transfer to EM L2, leading to a transfer to SE L1/L2 and then back to Earth  appears to be the leastcost solution to the problem. The logic of this transfer mechanism warrants further explanation, but I'll need to assemble some graphics to elucidate the mechanism (and to show why there might be a lower bound to the dV required to implement this strategy.) 
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09292018, 11:54 AM  #10 
Orbinaut

The outline makes perfect sense of how the flight "felt".
Especially the part leaving the EML2 and coasting to SEL1 (it was the L1 in this case). It felt like there was a giant funnel in that region of space and I was getting to it at just the right speed to have it grab me and throw me back in towards Earth. What was really surprising was the range of the different dV and time combinations that produced the exact same result. To someone untrained (like me), it would seem that there is only one extremely special combination of burn variables that get you to that region, but I was happily surprised to find out otherwise. Looking forward to reading more on this from you when you have the time. 
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10012018, 05:10 AM  #11 
Orbinaut

Quote:
Here, the Earth is located at the coordinate origin (0,0); and the Moon is located at the point (1,0). The two red dots are the collinear Lagrange points EM L1 and EM L2. The green oval is a planar Lyapunov orbit about L2. The blue line is the solution trajectory path starting from Low Lunar Orbit and ending 2.5 M seconds (or 28 days) later as it departs the EarthMoon system via L2 on its way to SE L1 and then back to Earth. Having plotted this graph, it seems clear that the planar Lyapunov orbits around EM L1 are not being used here. The key, really is the small planar Lyapunov orbit around EM L2 shown in green. The stable manifold of this orbit traces back in time (via a few laps around the Moon) to a low passage around the Moon. On the other hand, the unstable manifold of this Lyapunov orbit leads to SE L1 and then back to the Earth. In short, the green planar orbit provides the connection between lunar departure and Earth arrival. Now, in your solution, you don't spend much time 'on' the Lyapunov orbit: as you approach the orbit, you transition quickly to the unstable manifold and depart the EarthMoon system for SE L1. However, one could imagine using the Lyapunov orbit as a kind of 'parking orbit' where you approach the orbit ballistically. This would look something like: Once on the Lyapunov orbit, and with the help of some minimal stationkeeping, one could then choose a departure time along the unstable manifold leading to SE L1. Of course, with a single burn in Low Lunar Orbit, it isn't possible to carry out this sequence of operations, but using Lyapunov orbits (and other centre manifold solutions) helps provide structure to setting up and executing transfers of this kind. As for the lower bound cutoff, this is what I think is going on: In rotating coordinates, there is a conserved quantity (often called the Jacobi constant) for any given trajectory. This functions like conserved energy in inertial coordinates and can be used to define forbidden regions where the vessel cannot go. In the above diagram, the forbidden region for some trajectory the EarthMoon system is shown in grey. For low enough values of the Jacobi constant, one gets a bottleneck close to L2. Any trajectory with that Jacobi constant has to pass through the bottleneck if it is to make it 'outside' of the EarthMoon system. The smaller the Jacobi constant, the tighter the bottleneck. At some critical value of the Jacobi constant, the bottleneck closes entirely and access to the 'outside' is forbidden. My guess is that your 645.18 m/s solution is close to the point where the bottleneck closes leaving just a fairly small neck through which the trajectory must pass. Reduce the dV further and any trajectory that does make it 'outside' no longer has the correct timing to reach SE L1.  Post added 100118 at 05:10 AM  Previous post was 093018 at 10:58 AM  Out of curiosity, I took a closer look at my 700 m/s example solution. Plotting in rotating coordinates in the same way that I did for dgatsoulis' 645.2 m/s, one gets the following: And for comparison, here is dgatsoulis' solution again: My 700 m/s solution just takes advantage of the dynamics around the SE L1/L2 Lagrange point  and makes no use whatsoever of the EM L1/L2 Lagrange points. It's a fair bet that the best lowenergy escape roots from Low Lunar Orbit (LLO) back to Earth follow the dgatsoulis model  i.e., a connection from LLO to EM L2; then to SE L1/L2; and then back to Earth rather than my more simple solution of a direct transfer from LLO to SE L1/L2 and then back to Earth. (Although it is a different solution  and I don't have the details to plot it properly, plot it  Ajaja's 655 m/s solution seems to be much the same as dgatsoulis' in that it too makes use of the dynamics around EM L2.) 
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