EML2

[Impaler]

Would that be a constant acceleration?
If so it requires storing electrical power when at night.
Unless one is in a Sun-synchronous orbit [polar orbit].
Or it's getting power beamed from Earth or something.

Assuming constant power and ISP, acceleration would gradually rise from the vehicle mass dropping, as with every vehicle obeying the rocket equation.  But the change will be rather modest and can safely be ingnored for now.

I think that mass for energy storage is likely to be prohibitive, the vehicle will simply shut down propulsion when in the dark, this could significantly increase the time needed to spiral out, but the spiral out is an unmanned flight, almost no amount of time is too much, the only thing were worried about is Van-Allen radiation induced degradation of the vehicle, primarily the solar system.

A sun-synchronous orbit is an orbit that passes over a specific spot on Earth at the same local solar angle each day generally for the purpose of photo reconnaissance, it dose not expose the craft to more sunlight as far as I know.  A polar orbit is what we need and specifically a polar orbit who's plane is perpendicular to the rays of the sun such that the vehicle is going over the Earths terminator and is continually sunlit.

If an initial equatorial orbit is what we must start from (much easier launch and probably the only option for the larges launcher classes), then the process is more complex but it would still be do able.  To my knowledge know one has worked out how much time the shadowing of Earth causes but it should be modest in the long run.

Beamed power would require a world wide network of beaming stations, as well as a secondary receiver system on the vehicle, I'm very doubtful this would be worth the modest

[gbaikie]

Would that be a constant acceleration?
If so it requires storing electrical power when at night.
Unless one is in a Sun-synchronous orbit [polar orbit].
Or it's getting power beamed from Earth or something.

Assuming constant power and ISP, acceleration would gradually rise from the vehicle mass dropping, as with every vehicle obeying the rocket equation.  But the change will be rather modest and can safely be ignored for now.

Yes with Ion engine we could ignore the loss of fuel mass. But nice to know the total amount used
to get out of LEO. But how large the solar panels are and having them at low cross section to orbital path
and finally what percentage sunlight per orbit.

I think that mass for energy storage is likely to be prohibitive, the vehicle will simply shut down propulsion when in the dark, this could significantly increase the time needed to spiral out, but the spiral out is an unmanned flight, almost no amount of time is too much, the only thing were worried about is Van-Allen radiation induced degradation of the vehicle, primarily the solar system.

Well one will get periods of darkness if not in sun-synchronous orbit, even at 10,000 km or more above Earth, so batteries which handle say up 15 mins of darkness and allow full thrust might be good idea.

A sun-synchronous orbit is an orbit that passes over a specific spot on Earth at the same local solar angle each day generally for the purpose of photo reconnaissance, it dose not expose the craft to more sunlight as far as I know.  A polar orbit is what we need and specifically a polar orbit who's plane is perpendicular to the rays of the sun such that the vehicle is going over the Earths terminator and is continually sunlit.

If same local solar angle is dawn, then on the other side it's dusk- at the earth surface. But one will be in constant full sun in orbit.

If an initial equatorial orbit is what we must start from (much easier launch and probably the only option for the larges launcher classes), then the process is more complex but it would still be do able.  To my knowledge know one has worked out how much time the shadowing of Earth causes but it should be modest in the long run.

ISS at 400 km and as I recall it is about 40% darkness per orbit of about 90 min orbit.

[sdsds]

Would that be a constant acceleration?
If so it requires storing electrical power when at night.

Assuming constant power and ISP, acceleration would gradually rise from the vehicle mass dropping, as with every vehicle obeying the rocket equation.  But the change will be rather modest and can safely be ignored for now.

I definitely think the "constant acceleration" assumption is dubious for both reasons suggested (changing mass and shadowing of photovoltaic panels). But it's easy to model!

A way around the changing mass issue would be to assume a variable thrust. A way around the shadowing issue would be to get out of LEO more quickly, e.g. with a chemical propulsion burn right at the beginning. I attach the plot of a revised simulation which starts with a 1385 m/s "kick" burn and then 56 days of ion propulsion. (Note the blue circle of Earth at the center is no longer completely obliterated by the green of the ion-drive trajectory.)

[sdsds]

Very interesting! By gamma you mean flight path angle? What's phi?

One of the two angles is as you suggest the current direction of the velocity vector; the other is the current direction of the radial vector.

Could you explain how you did that?

I'm only just learning how to use Mathematica and its framework for numerical solutions to systems of differential equations. That means I am still "doing it wrong" from the perspective of sophisticated Mathematica users. This link gives a view of what I'm currently doing, and what I should be doing instead:
 http://mathematica.stackexchange.com/questions/83101/how-to-extract-state-vector-after-ndsolve-for-subsequent-re-use

[gbaikie]

I've made a spreadsheet of Non-Hohmann transfers. Time can be shortened by shrinking the transfer ellipse's perihelion (which needs to be less than 1) or increasing transfer ellipse's aphelion (which needs to be more than 1.524).

Btw I like this program because can it provide graphics, though not sure how to show screen shots.
But still not sure it's accurate.
So put in Venus distance to get some idea about it.
So did .74 to 1.524.
Of course time travel would from Earth, but one think of a hohmann transfer from Venus and flying by Earth so it's outside gravitational sphere of influence.
So from Earth distance to Mars distance at Aphelion it takes "156.8512591   days" Or 5.2 months.
And roughly it takes 7.2 months for hohmann from Venus to Mars. Or from Venus going to Mars it seems takes about 2 months to reach Earth distance.
So that appears about right. And as graph shows it's traveling much shorter distance. Plus it slows as it gets closer to Mars distance.
Takes 5.2 months reach Mars distance and will take another 5.2 month to return back to Earth distance. And roughly earth take 6 month to reach same point. So if didn't stop at Mars and did nothing to change trajectory, Earth would pass the same point in it's orbital path about 4 month before spacecraft reached it

But point is delta-v needed. Now just eyeballing it looks like about 20 degree angle- oh, it's says it's 19.16 degrees.
As read it the return to Earth low orbit distance gives 10.33353758 km/sec, and one adds:
4.234240882 km.
So I am just going to add vectors, here:
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html
So 29.78 km/sec at 180 and 15.5 km/sec at 90, and gives 152.5 degree at 33.57 km/sec
and 180 - 152.5 is 27.5 degrees. So not doing 90 degrees [just picked as "round number"]
Say put in 120 and that is 160.31 degrees [close to 19.16 degrees] and 39.858 km/sec

So left Venus at it's Venus orbital speed plus delta-v to get to Mars:
Venus orbital velocity mean: 35.02 km
I don't think one needs to add more 4 km/sec delta-v to reach Mars from Venus with hohmann..
Though not sure what the number is off hand. And in addition it's going a bit slower by time it reaches Earth distance. Hmm.
Oh, it's going slow down leaving earth. I guess that why we got the other number:
"Periaerion burn to a 3697x23459 km ellipse   2.197776706 "
??Maybe sort of??. Or if was left at about 2 km/sec slower it makes more sense.   
So guess it's at least close.

I don't want to take 5.2 months to reach Mars. But I wanted to see how compares.   

But while we on topic, with patched conic added that trajectory to Mars could a lot less than
5.2 months- say roughly less 4 1/2 months. Which kind of interesting in terms of alternative
launch windows to Mars. Or in terms delta-v it seems comparable to 6 to 7 month trip to Mars that
one could launch 2 1/2 months earlier.

And one picked before was 0.7   A.U to 1.53 AU
slightly different with delta-v 4.888082825 km/sec
rather than this one of 4.234240882  km/sec.
And this not counting Oberth effect

And what I was thinking about before was something equal to between Venus and Mercury.
Let's try .6 to 1.53:
6.532076673 km/sec and 4.014760506 month

And try .55 to 1.53:
7.41168672 km/sec and 3.832299771 months
And try .5 to 1.53
8.339101035 km/sec delta-v and 3.665922263 month

So including Oberth effect the rocket delta-v should about 7 km/sec or less.

Now lets try the extreme of 2 months [what's practical possible is 2 to 3 months]
So first try .5 to 1.6:
So that is 31.48215411 degrees and
8.874201136 km/sec added and 2.847714055 months to aphelion
and with patched conic it should be about 2 month

And so seems to me the big difference probable amount needed for patched conic-
it's going be probable double as much has for 6 to 7 month hohmann tranfer.
it's seems more of vector and velocity difference. But for direct landing it seems
well within the ability of heat shield. Or Mars capture [not aerobrake- and we never done
a capture] would have quite advantage for doing 2 month trip.
So risky and cost more delta-v. So keep with 3 month trip time. It's good enough.

The thing  that was throwing me is the lower delta-v options look a lot better than thought
they would.
And wasn't considering how added something like patch conic and  could make
it an alternate to hohmann transfers [for cargo or 4 to 5 month crew transfer].
Plus with ion boosted with chemical it could work for crew and/or faster cargo.

[sdsds]

Before leaving the topic of spiraling out from LEO, I want to present the empirical observation that the time to reach the lunar vicinity is inversely proportional to the (assumed constant) acceleration provided by the drive. That is, if you cut the acceleration in half it takes twice as long to get there. Another way to say that is the product of the acceleration and the time required is roughly constant. Here's my data, and three plots of how the ascents proceed over time. Note they result in different coasting orbit eccentricities.

Aion -> 0.004, Tdrive -> 18.234, Product -> 0.072936
Aion -> 0.001, Tdrive -> 74.95, Product -> 0.07495
Aion -> 0.0005, Tdrive -> 151.3, Product -> 0.07565

[Hop_David]

Bi-Elliptic Transfers

If the semi major axi of departure and destination orbits differ by a factor of 11.94 or more, bi-elliptic transfer costs loss than Hohmann.

Radius of EML2 is 450,000 km, LEO is about 6738. 450,000/6738 = ~67. So EML2 certainly qualifies as a beneficiary of bi-elliptic.

From LEO a 3.1 km/s burn takes a payload to just a hair under escape. This can describe a multitude of orbits!



As you can see, a burn to a 1.8 million km apogee isn't much more than a burn to get apogee to EML1. I chose a 1.8 million km apogee because perigee of a 450,000 x 1,800,000 ellipse moves about 1.19 km/s, same altitude and speed as EML2. So at perigee the payload can slide right into EML2.

What's needed is an apogee burn to raise perigee to 450,000 km. A 6738x450,000 km orbit moves ve-e-e-ry slow at apogee, about .04 km/s. A 450,000x1,800,000 orbit doesn't move much faster, .3 km/s at apogee. So only a .26 km/s apogee burn is needed to get to EML2.



Hmmm. 3.1555+.26 is about 3.4 km/s. Better than a Hohmann but about the same as Farquhar's 9 day route.

But recall the apogee is beyond earth's Hill Sphere. If we time launch just right, the sun can provide the apogee delta V for us.

Here is a route I found with my shotgun orbital sim:

.

For this particular sim I put in user ability to set position of the sun as the sun's position makes a huge difference when you have apogees near or past the Hill Sphere. New moon is when barycentric longitude is zero. Setting sun's barycentric longitude at 180º gives a full moon.

I launched from LEO with a ~3.11 burn. The payload passes by the moon on the way out. I used the lunar gravity assist to boost apogee and rotate line of apsides. At apogee the sun does the work of lifting perigee. You can see the pellets slide right into an EML2 location running in circular path along side the moon.

Here's the same path zoomed out:



This 74 day route from LEO to EML2 takes 3.11 km/s

[sdsds]

I think you have found a special case of the ballistic lunar transfers described by Parker.
http://ccar.colorado.edu/nag/papers/AAS%2006-132.pdf

Your case is special because you use the lunar swing-by on the outbound leg. I think the opportunities for that occur every month, when the Sun and Moon are in the right relative positions. Is that how you see it?

Parker, using his magical mathematics, can do his version any day of the week.. He departs from LEO to the vicinity of SEL1, and then by computing backwards through time he finds an inbound trajectory from SEL1 that "just happens" to lead to an orbit around EML2. Patching the two trajectories together takes an arbitrarily small delta-v, since the inbound trajectory is so chaotically sensitive to tiny changes in its initial conditions.

At least that's my non-mathematician's view of what he's doing.

[sdsds]

Attached are three plots generated in Mathematica, showing a simulated slow departure from EML2. The first uses an inertial frame centered at the barycenter of the Earth-Moon system. The second uses a rotating frame in which the Moon appears essentially fixed. (Marks along the trajectory are approximately 1 day apart.) The third plots the distance from the barycenter as a function of time.

The simulation assumes the Moon and Earth are in circular orbits around their mutual barycenter with periods of 27.321661 days (i.e. a sidereal month). Earth and Moon masses are approximate; the radius of the Hill sphere for the simulated Moon (6.45729*10^7 meters), and thus the location of EML2, was empirically determined based on the standard approximation.

The simulation is as yet two-dimensional, and does not yet take into account the influence of the Sun.

[sdsds]

http://clowder.net/hop/TMI/FarquharRoute.jpg

Quick review of the Farquhar route:
.15 km/s drops payload at EML2 to an 111 km perilune.
.18 km/s perilune burn drops to a near earth perigee.
.5 km/s perigee burn does TMI. (11.3 hyperbola velocity - 10.8 perigee velocity)

EML2 to TMI is about 9 days and .9 km/s.

I'm not sure the trajectory shown in that diagram is reversible. Can you make it happen in your pellet simulator?

[redliox]

How about arriving from a Hoffman transfer (presumably Mars) and trying to capture into EML1/2?  Is there any difference there if you do it by ion drive versus leaving?

[Hop_David]

Attached are three plots generated in Mathematica, showing a simulated slow departure from EML2. The first uses an inertial frame centered at the barycenter of the Earth-Moon system. The second uses a rotating frame in which the Moon appears essentially fixed. (Marks along the trajectory are approximately 1 day apart.) The third plots the distance from the barycenter as a function of time.

The simulation assumes the Moon and Earth are in circular orbits around their mutual barycenter with periods of 27.321661 days (i.e. a sidereal month). Earth and Moon masses are approximate; the radius of the Hill sphere for the simulated Moon (6.45729*10^7 meters), and thus the location of EML2, was empirically determined based on the standard approximation.

The simulation is as yet two-dimensional, and does not yet take into account the influence of the Sun.

Wow, looks like you Mathematica models can do the same stuff as my models based on Jenkins' Java.

In your sim you've evidently nudged towards the moon's center and/or did a braking burn. This will have the effect of sending a payload into the moon's realm. Nudging away from the moon's center and/or stepping on the gas has the effect of sending pellets into the external realm.

I just did some similar sims. In my first run I noticed pellets 10 and 11 straddled EML1. So I thought it'd be fun to aim for EML1 by successively tightening my shotgun blasts.

[Hop_David]

I'm not sure the trajectory shown in that diagram is reversible. Can you make it happen in your pellet simulator?

I've already found paths from EML2 to low earth altitude. See orange pellet number 3 in my OP.

As for orbits being reversible? For a long time I had made that questionable assumption. Then I learned that DROs are more stable than prograde lunar orbits so I'm no longer sure.

This made me quite anxious as I've been assuming we could use the Farquhar route from as well as to EML2. If someone's using a bad model, it is necessary to fess up or credibility is lost. But I really, really hate eating crow.

So I played with my orb sim. A burn whose components are .1 km/s down and .1 km/s braking will have the effect of sending an EML2 payload to a perilune quite close to the moon's surface. This vector has a norm of about .14 km/s. At perilune the pellet is moving retrograde wrt earth at ~2.3 km/s. An .18 perilune burn gives an .83 Vinf wrt the moon, what's needed to drop to a perigee near earth. I was pleasantly surprised to find my .14 and .18 km/s burns are similar to Farquhar's. I haven't been able to try the .18 km/s perilune burn though. After Jenkins' pellets are set in motion, I don't know a way to do additional burns enroute.

On the other hand I seem to recall Belbruno would send a bunch of pellets away from a specified destination. He'd watch which paths intersected earth and then would play the same paths backwards to find routes from earth to the destination.

So as the questions are the paths time reversible? I don't know. That retrograde lunar orbits behave differently than prograde has made my head spin. But perhaps that doesn't invalidate time reversibility as playing it backwards would make the moon's orbit retrograde as well as making the DROs prograde.

[Impaler]

How about arriving from a Hoffman transfer (presumably Mars) and trying to capture into EML1/2?  Is there any difference there if you do it by ion drive versus leaving?

A continuous electric propulsion systems wouldn't be on a Hohmann trajectory.  Depending on how long you desided to make the trip you took and how much deceleration you did if any you could have almost any incoming velocity relative to the destination planet.

I think their may be potential for a ballistic capture to EML2 upon return from Mars, it would save propellent at the very end of the journey where the backwards compounding is highest.  The longer time needed to effect capture would need to be eliminated from the crews radiation dosage by use of a taxi craft to retrieve the crew as soon as they near enough.

[sdsds]

A burn whose components are .1 km/s down and .1 km/s braking will have the effect of sending an EML2 payload to a perilune quite close to the moon's surface.

Excellent! I tried it (using -104.3 m/s each so the delta-v is Farquhar's 147.5 m/s). My implementation is new and untested so take the results with a grain of salt. Attached are a 5-day plot and a 16 day plot, in the rotating frame. (Tick marks are at ~12 hours for the first; ~2 days for the second.)

On the stickier (but fascinating) question of reversible paths in the three-body problem, maybe they are reversible outside the Hill Sphere, but not inside?

[Hop_David]

I attach the plot of a revised simulation which starts with a 1385 m/s "kick" burn and then 56 days of ion propulsion. (Note the blue circle of Earth at the center is no longer completely obliterated by the green of the ion-drive trajectory.)

If heading for the moon it's not necessary to cut off to an ellipse with an L.D. apogee.

When nudging a pellet away from the lunar Hill Sphere at EML1, it tends to fall into what I call an Olive Orbit. A 100,000x300,000 orbit with the apogee maybe 20,000 km below EML1 altitude.

Conversely if we start with an olive orbit, doing a ballistic slide into EML1 is only a matter of timing. Once at EML1, there are very low delta V paths to EML2.

I wonder if there's a way to set the ion path so that at some point it has the same altitude, flight path angle and velocity as a point on the olive orbit?

I put in a screen capture of an olive orbit and the EML1 circular path superimposed over your kicked ion path.

[sdsds]

I wonder if there's a way to set the ion path so that at some point it has the same altitude, flight path angle and velocity as a point on the olive orbit?

Yes I think this might be possible. In particular the solver provided in Mathematica can accept either initial condition or end condition constraints. There might be subtleties though when it is asked to handle a mixture of both.

The simplest case that might give us an answer would be providing all end condition constraints (i.e. the position and velocity we wanted to reach). The solver than automatically "runs time backwards" for the amount of time specified.

The plots below show such a solution, but in a different situation. They show a 44 day trajectory which coasts into and then stops at EML2 without requiring any propulsion.

[EDIT: added a third plot, showing a 10 day trajectory that requires a 10 m/s burn upon arrival.]

[sdsds]

Perhaps the trouble with reversibility here is that we're looking at a plot drawn with a rotating reference frame. We can integrate backwards in time just fine: the spacecraft and the Moon will both "go backwards" reversibly. But we can't have the spacecraft go one direction in time and the Moon go the other....

[sdsds]

Attached below are three plots of yet another coasting trajectory. This one starts by approaching EML2 from the exterior realm. It then passes through EML2 moving at only 15 m/s, so stopping there would be easy. But instead of stopping it continues on past the Moon and the vicinity of EML1 and enters the interior realm. In addition to demonstrating there are at least two ways of coasting up to EML2 (one from the vicinity of the Moon, and one from the exterior), these plots demonstrate the Mathematica solver's ability to start with known conditions at any point along the path. 

[Hop_David]

Attached below are three plots of yet another coasting trajectory. This one starts by approaching EML2 from the exterior realm. It then passes through EML2 moving at only 15 m/s, so stopping there would be easy. But instead of stopping it continues on past the Moon and the vicinity of EML1 and enters the interior realm. In addition to demonstrating there are at least two ways of coasting up to EML2 (one from the vicinity of the Moon, and one from the exterior), these plots demonstrate the Mathematica solver's ability to start with known conditions at any point along the path.

Yes a very high orbit can drop into the moon's Hill Sphere and drop down into earth's realm.

After falling out the bottom of the moon's Hill Sphere, it usually drops into what I call the Olive Orbit, about a 100,000 x 300,000 ellipse. The Olive Orbit will continue to loop around the earth 4 or 5 times relatively unmolested by the moon. Then the 4th or 5th circuit the moon will sneak up behind him and yank him back wards at apogee resulting in an even smaller olive. This happens again after another 5 or 6 circuits.

Playing it backwards, I suspect an 80,000x280,000 ellipse would be sufficient to eventually fall through the moon's Hill Sphere at EML1, then out EML2. Once departing out EML2 it's possible to sail right out of SEL2 or SEL1