EML2

[Hop_David]

Due to EML2's proximity to the moon, I'm putting this in the moon section. I hope to expand on this as I have time.

An orbit's characteristic energy is

.5v² - GM/r

Or sometimes C3 is used. C3 is twice the characteristic energy.

For this post, v is velocity wrt earth, GM is earth's gravitational parameter and r is distance from earth's center.



For earth surface I'm assuming from the equator which is moving about .5 km/s.

If nudged from the moon's Hill Sphere, an object from EML1 will fall into an ~100,000 km x 300,000 km ellipse. An EML2 object nudged from the moon's Hill Sphere will raise to an 440,000 x 1,800,000 km ellipse:



Characteristic energy is inversely proportional to semi-major axis. It can be seen from above pic that the ellipse from EML2 has a semi major axis about five times that EML1. So EML2 has about 1/5 the orbital energy of EML1. Both these have C3 much closer to zero than LEO or GEO. (LEO in this pic would be the size of a fly speck).

Being a Lagrange region, something parked at EML2 can stay there with very little station keeping.

Using a perilune burn, EML2 is about 9 days and 3.5 km/s from LEO:



EML2 is about 2.5 km/s from the lunar surface and about .4 km/s from an object in lunar DRO.

An object nudged from EML2 has an apogee of ~1.8 million km which is above earth's Hill Sphere (about 1.5 million km radius). Depending on timing of EML2 release and the nudge, the object might sail clear out of earth's influence into a heliocentric orbit. Or the the sun can bend the ellipse to a perigee deep in earth's gravity well:



Pellets in pic above are all nudged from EML2 with tiny burns varying by meters/sec. The blue pellets sail into heliocentric orbit.

Check out the #3 orange pellet -- it has a perigee deep in earth's gravity well. Using the sun to lower perigee, we can go from EML2 to a deep earth perigee with less than a .1 km/s nudge. An 6678 x 180000 ellipse has perigee velocity of 10.9 km/s. At this speed a .4 km/s suffices to send the pellet into an 11.3 km/s hyperbolic orbit for Trans Mars Insertion.

 

[gbaikie]

This seems to be about sending cargo [non crew] to Mars. Which is important- or most of the cost of manned
mars program.
But I like EML 1 or 2 in regards to getting crew fast to Mars. Which involves using oberth effect of swinging  near earth.
Can we have pretty pictures of that, too?

Edit: Oh, my mistake, I missed it, it seems this would take a long time for the crew- though I suppose one can dock with it on way in. Edit, edit, maybe crew could only go once through Van Allen belt on way to Mars- it's complicated though, not sure if possible or worth the risk. But with three passes thru belt could be launch crew to a highly elliptical and dock before passing thru Van Allen for second time, and third pass thru the belt on the outbound to leg to Mars.
In past I usually I think crew going to L-1 first. But thinking instead that crew launches from Earth and goes to say +120,000 km apogee and docks with in coming spacecraft before it reaches 30,000 km from Earth.

[sdsds]

Thanks very much for posting this! I have a question about the early part of your presentation, where you discuss the characteristic energy of various trajectories. It looks like your treatment includes only the gravitational potential due to the mass of the Earth. Is that realistic for Earth-Moon Lagrange points? What about the mutual potential energy of the object and the Moon?

[Hop_David]

Thanks very much for posting this! I have a question about the early part of your presentation, where you discuss the characteristic energy of various trajectories. It looks like your treatment includes only the gravitational potential due to the mass of the Earth. Is that realistic for Earth-Moon Lagrange points? What about the mutual potential energy of the object and the Moon?

Once nudged loose, a payload fairly quickly gains distance from the moon and the moon's influence becomes negligible.

Nudged from EML1 payloads will speed ahead of the moon. So the moon will exert a backwards tug for a time. So the actual orbit has a slightly lower apogee but not a whole lot.

Nudged from EML2 payloads will lag behind the moon. So the moon will pull it forward for awhile resulting in a slightly higher orbit.

I've attached a screen shot from my orbit sim where pellets are nudge from from EML. The nudges range from 3 to 4 meters/sec.

Compare these orbits to what I pictured above and they're not too far off.

In my OP is a screen capture of payloads nudged from EML2. But these are hard to compare since the sun distorts the orbits a lot after they pass the Hill Sphere. Still, I think you can you can tell the apogees are about 1.8 million km.

[Hop_David]

I like EML 1 or 2 in regards to getting crew fast to Mars.

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.

Let's imagine Impaler's MTV -- a 200 tonne Mars Transfer Vehicle (MTV) propelled by Hall Thrusters with acceleration of .000001 km/s^2. 11.6 days of acceleration gets 1 km/s delta V. It takes this ship about 3 months to spiral out of earth's gravity well. We don't want humans aboard, especially during the ~2 month spiral through the Van Allen Belts. It makes sense to send humans to dock with the MTV after it's climbed most the way out of earth's gravity well.

Radius of the MTV's spiral will eventually reach 327,000 km, the distance to the moon's Hill Sphere. So why not let the MTV spiral to EML1 and let the moon lend a hand getting the MTV to EML2? Park the MTV at EML2 where station keeping is inexpensive.

Send humans to EML2 to rendezvous with the MTV. From LEO to EML2 is 9 days and about 3.5 km/s.

As pictured in the OP, EML2 is pretty close to C3=0. But at a speed of 1.15 km/s, it doesn't enjoy much Oberth benefit. The MTV still needs to do another 3 km/s to reach a 1.52 aphelion (Mars' distance from the sun). At 1 km/s per 11.6 days, it will need about 34 days to achieve this 3 km/s.

Now let's imagine there's propellent and life support consumables at EML2. These might come from lunar cold traps or asteroids parked in a DROs. Stocking the MTV with EML2 water (for drinking, sanitation and radiation shielding) and EML2 oxygen to breathe would substantially reduce the gross lift off weight from earth's surface.

The MTV docks with a reusable Earth Departure Stage (EDS). This stage has a dry mass of 31 tonnes and carries 76 tonnes of lox/methane. At 107 tonnes the EDS is a little more than half the MTV's mass. Total mass at EML2 is now 307 tonnes.

EML2 to TMI is about .9 km/s and exhaust velocity of lox/methane is 3.6 km/s. e^(.9/3.6) - 1 is .28. So getting this 307 tonne mass to TMI will take 68 tonnes of prop.

After TMI, the EDS separates from the MTV. It still carries about 9 tonnes of lox/methane. A .5 km/s braking burn slows it to an elliptical orbit with an apogee of about 1 lunar distance. After about 5 orbits the apogee nears the moon. At perilune .18 km/s injects it into a elliptical lunar orbit with apolune near EML2. At apolune .14 km/s parks it at EML2.

ELM2 to TMI is 9 days. TMI to the edge of the Hill Sphere is about 6 days. 34 - 15 = 19. 19 days isn't much of a savings but this is the outbound trip.

If I remember right, Impaler's MTV is reusable and returns to earth. Does it spiral back down to LEO? This would add another 7 km/s to it's delta V budget and 3 months to trip time. With the infrastructure I'm talking about it would only need to spiral to EML2 and then chemical rockets could take the astronauts from EML2 to LEO. To refuel the MTV, xenon would have to be delivered to EML2. But a xenon delivery tanker would be a small fraction of the mass of the MTV.

Impaler's MTV would be good for moving between heliocentric orbits but would suck at descending/ascending planetary gravity wells in a timely manner. It'd be well suited for traveling between EML2 and Deimos.

But I consider Mars colonization an unlikely fantasy if building space infra-structure gives no return on investment. Far more plausible in the near term is an asteroid retrieval vehicle as suggested in the Keck Report. At least Planetary Resources has a shot at enjoying ROI.

The Keck vehicles would take nearly two years to spiral from LEO to C3=0. For these more plausible vehicles the time savings could be huge.

[Hop_David]

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).

Attached is a screen capture from a spread sheet where the Earth Departure Stage would do a 3.1 km/s perigee burn. Since falling from EML2 takes .4 km/s, the EDS would need a total delta V budget of about 3.5 km/s. This wouldn't be a reusable EDS.

Coming to Mars, a 4.67 km/s peri-aerion burn would capture the MTV into a 3697x23459 km ellipse. This is atmosphere grazing with apo-aerion at Deimos height. Orbital period is 13 hours, so about 2 periaerion drag passes per day. More drag passes per day as apo-aerion is lowered. I believe aerobraking could circularize the orbit within a week or so.

This transfer would take 3.4 months or about 103 days.

I've also attached the spreadsheet. A user can play with the transfer orbit by changing aphelion and perihelion.

[rklaehn]

Due to EML2's proximity to the moon, I'm putting this in the moon section. I hope to expand on this as I have time.



Pellets in pic above are all nudged from EML2 with tiny burns varying by meters/sec. The blue pellets sail into heliocentric orbit.

Check out the #3 orange pellet -- it has a perigee deep in earth's gravity well. Using the sun to lower perigee, we can go from EML2 to a deep earth perigee with less than a .1 km/s nudge. An 6678 x 180000 ellipse has perigee velocity of 10.9 km/s. At this speed a .4 km/s suffices to send the pellet into an 11.3 km/s hyperbolic orbit for Trans Mars Insertion.

Question: For a transfer from EML2 to TMI without chemical propulsion, would there still be some benefit from doing an earth flyby (orange trajectory), or would you just go straight to interplanetary space (blue trajectory)?

For the earth flyby, you would get at least some oberth effect for the part of the acceleration done in the earth gravity well. Also, you get to do all your acceleration at 1AU. But I am not sure if this is worth the additional complication and delay.

The orange trajectory is about 100 days, right?

[Hop_David]

The orange trajectory is about 100 days, right?

I try to guesstimate these orbits using two body mechanics. First half of orange looks like a 450,000  x 1,800,000 ellipse which has a period of 140 days. Second half is a 6,678 x 1,800,000 ellipse which has a period of 100 days.

So I was going to tell you (100+140)/2 which is 120 days.

But then I thought why not focus the shot gun blast and do a screen capture as the pellets round perigee? The upper left of the square gives number of time increments. For this sim I had set the increments at 600 seconds.

The number in the upper left is 12111. 12111x600 seconds is 7266600 seconds or 84 days!

It looks like the sun has thrown these pellets back towards earth in a decidedly hyperbolic orbit.

To check I did another screen capture zoomed out. The way the pellets do a bee line out of earth's neighborhood suggest hyperbolas wrt earth.

Both screen captures attached.

[sdsds]

That's a really neat class of trajectories! Does your pellet simulation tell you the end state of the pellets (i.e. position and velocity), so you could calculate their specific orbital energies to see if they are greater than zero?

[Impaler]

Let's imagine Impaler's MTV -- a 200 tonne Mars Transfer Vehicle (MTV) propelled by Hall Thrusters with acceleration of .000001 km/s^2. 11.6 days of acceleration gets 1 km/s delta V. It takes this ship about 3 months to spiral out of earth's gravity well. We don't want humans aboard, especially during the ~2 month spiral through the Van Allen Belts. It makes sense to send humans to dock with the MTV after it's climbed most the way out of earth's gravity well.

Remember it was back and forth design, I provided the target acceleration rate, you specified ISP and targeted DeltaV to estimate 130 dry mass and 70 propellent, and a total thrust level of 200N, then I specified the X3 Hall thruster and the RAPDAR solar array which you found reasonable.  I believe the exact figures would work out too 100 tons payload, 22 tons for the dry propulsion systems, 70 ton of propellents, 8 tons for tank.  I do not recall if any propellent was allocated for returning from Mars.  Lastly I suspect a higher ISP then what was used may be desirable, though it will require more power to achieve the same acceleration the overall savings my be worth it.  Overall the vehicle concept is not unreasonable but could use more refinement and a more detailed analysis of it's propellent usage through the course of it's cycle between Earth and Mars.

Radius of the MTV's spiral will eventually reach 327,000 km, the distance to the moon's Hill Sphere. So why not let the MTV spiral to EML1 and let the moon lend a hand getting the MTV to EML2? Park the MTV at EML2 where station keeping is inexpensive.

Send humans to EML2 to rendezvous with the MTV. From LEO to EML2 is 9 days and about 3.5 km/s.

As pictured in the OP, EML2 is pretty close to C3=0. But at a speed of 1.15 km/s, it doesn't enjoy much Oberth benefit. The MTV still needs to do another 3 km/s to reach a 1.52 aphelion (Mars' distance from the sun). At 1 km/s per 11.6 days, it will need about 34 days to achieve this 3 km/s.

Now let's imagine there's propellent and life support consumables at EML2. These might come from lunar cold traps or asteroids parked in a DROs. Stocking the MTV with EML2 water (for drinking, sanitation and radiation shielding) and EML2 oxygen to breathe would substantially reduce the gross lift off weight from earth's surface.

The MTV docks with a reusable Earth Departure Stage (EDS). This stage has a dry mass of 31 tonnes and carries 76 tonnes of lox/methane. At 107 tonnes the EDS is a little more than half the MTV's mass. Total mass at EML2 is now 307 tonnes.

EML2 to TMI is about .9 km/s and exhaust velocity of lox/methane is 3.6 km/s. e^(.9/3.6) - 1 is .28. So getting this 307 tonne mass to TMI will take 68 tonnes of prop.

After TMI, the EDS separates from the MTV. It still carries about 9 tonnes of lox/methane. A .5 km/s braking burn slows it to an elliptical orbit with an apogee of about 1 lunar distance. After about 5 orbits the apogee nears the moon. At perilune .18 km/s injects it into a elliptical lunar orbit with apolune near EML2. At apolune .14 km/s parks it at EML2.

ELM2 to TMI is 9 days. TMI to the edge of the Hill Sphere is about 6 days. 34 - 15 = 19. 19 days isn't much of a savings but this is the outbound trip.

I had assumed that hydro-lox would be the preferred propellent choice if some kind of lunar/asteroid ISPP is being used as water should be more abundant and easier to turn into propellents (simple electrolysis).  Carbon compounds while very likely to be present are likely to be in a whole slew of different forms requiring more complex synthesis. 

Admittedly hydrocarbons are far easier to store so perhaps this is your motivation, but if the source is the lunar cold-traps then presumably hydrogen can be stored their passively with zero boil-off and the propellent delivery could be done via a 'launch on demand' directly to the waiting vehicle at EML2.

Finally the dry mass of the stage seems rather high, you said this stage is intended only for use in space so it should be free of most parasitic mass other then insulation to keep boil off to a reasonable level during the ~2 weeks in which it actively has propellent, I would think that 10% dry mass would be a perfectly reasonable figure, unless their are other demands being put on it that I don't know of.

I suspect that a simple increase in the propellent tank for the MTV will be able to fully replace the EDS.  At 3000s the MTV would need 10% propellent fraction to do 3 km/s.  Thus our 200 ton vehicle could just be 220 tons at EML2.  Now more propellent at LEO would be needed for it to push that larger mass up to EML2, as the DeltaV from LEO to EML2 is 7 km/s for low thrust vehicles that means a an additional 20% on top of the additional propellent to get the additional propellent to EML2, which is 5 tons.  So in totality the MTV needs 25 tons more IMLEO in propellent (Xenon) to replace the EDS.  And while I think the EDS presented above is too conservative and could be considerably lighter it will not be so light as to cost competitive with a simple enlargement of the MTV tanks, to speak nothing of the infrastructure costs of establishing in space propellent production.

If I remember right, Impaler's MTV is reusable and returns to earth. Does it spiral back down to LEO? This would add another 7 km/s to it's delta V budget and 3 months to trip time. With the infrastructure I'm talking about it would only need to spiral to EML2 and then chemical rockets could take the astronauts from EML2 to LEO. To refuel the MTV, xenon would have to be delivered to EML2. But a xenon delivery tanker would be a small fraction of the mass of the MTV.

Impaler's MTV would be good for moving between heliocentric orbits but would suck at descending/ascending planetary gravity wells in a timely manner. It'd be well suited for traveling between EML2 and Deimos.

But I consider Mars colonization an unlikely fantasy if building space infra-structure gives no return on investment. Far more plausible in the near term is an asteroid retrieval vehicle as suggested in the Keck Report. At least Planetary Resources has a shot at enjoying ROI.

The Keck vehicles would take nearly two years to spiral from LEO to C3=0. For these more plausible vehicles the time savings could be huge.

Upon return to Earth gravity well I would bring the vehicle back to EML2 and separate the 100 ton habitat leaving it at EML2, then take the drive-section (sooo Star Trek) and nearly empty tank down to LEO, re-propellent (yes I made that word up) it in LEO with all the propellent needed for the next journey and fly back to EML2 to rejoin the saucer section, I mean habitat.  This is functionally equivalent to the tanker you described but avoids the cost of another vehicle.  Crew would return to Earth via a fast taxi craft similar to what brought them to the MTV initially.

With regard to descending/ascending into planetary gravity wells, yes it is quite slow compared to chemical propulsion, but we would only carry the heavy habitat up the Earth gravity well once and unmanned as you described earlier.  Second the decent into Mars gravity well takes only ~1/3rd as long so at most a month is added to the journey each way for this maneuver.  Lastly we can choose to lower the ISP of the thrusters and increase thrust at the cost of propellent to get this maneuver done faster, it would be a trade off between extra propellent mass vs the same mass in shielding, which exposes the crew to less radiation.

[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).

Attached is a screen capture from a spread sheet where the Earth Departure Stage would do a 3.1 km/s perigee burn. Since falling from EML2 takes .4 km/s, the EDS would need a total delta V budget of about 3.5 km/s. This wouldn't be a reusable EDS.

Coming to Mars, a 4.67 km/s peri-aerion burn would capture the MTV into a 3697x23459 km ellipse. This is atmosphere grazing with apo-aerion at Deimos height. Orbital period is 13 hours, so about 2 periaerion drag passes per day. More drag passes per day as apo-aerion is lowered. I believe aerobraking could circularize the orbit within a week or so.

This transfer would take 3.4 months or about 103 days.

I've also attached the spreadsheet. A user can play with the transfer orbit by changing aphelion and perihelion.

Departure orbit radius   1   A.U.   Speed   6.283185307   A.U./yr         
Destination orbit radius   1.524   A.U.   Speed   5.089643749   A.U./yr      4.740604351   AU/yr to km/s
Transfer orbit perihelion   0.7   A.U.                  
Transfer orbit aphelion   1.53   A.U.               2r1   2
Flight path angle at departure   0.36777381   Radians   21.07188715   degrees   Perigee burn (300 km altitude falling from high apogee)         
Vinfinity at departure   2.375992013   A.U./yr   11.26363808   km/s   4.888082825   km/s      
                        
Flight path angle at destination   0.067838707   Radians   3.88687157   degrees   Periaerion burn to a 3697x23459 km ellipse         
Vinfinity at destination   1.084314564   A.U./yr   5.140306341   km/s   2.568080059         
                        
Time of Flight   0.370249764   years   4.442997163   months   135.2337261   days      

This appears to have flight time of 4.44 month, but I would guess that's it's at top of the perihelion orbit, and so one would do a patched conic, like you do with most Hohmann transfer to mars and thereby shorten the travel time, plus that would be adding some orbital speed so one is going closer to Mars velocity [mostly regarding it's orbital vector].
Anyways it's less delta-v than I expected.   
And I think minimum travel time should be 3 months for crew.
Edit: I mean from the crew launching from Earth it should take 3 month or less to reach Mars [Though could  be less travel time but would take considerable more rocket fuel and it seem with enough shielding 3 months should fast enough so reduce microgravity effect and crew radiation exposure].                      

Addition:
An alternative just do 1.524 [or less] and would be using more delta-v at Mars distance. So use a lot less at Earth and keep it to used later [using cryogenic rocket fuel a bit of problem] Or you look out right side window and mars is say 5 million km to your right and without rocket power Mars would stay ahead of you, so you using using rocket power to catch up. So being in the inside track, the rocket power not go ahead, but outwards to towards Mars [you will get closer and Mars gravity is helping a little]. Or this is simply a bigger patched conic adjustment. And carrying more fuel should reduce radiation effect more.
And as long as one does lose the rocket fuel somehow, it tends to give more abort options.
So one ends up using about same total delta-v, and one is just using it later. And should be about same time involved [about 3 months].
[[Edit perhaps one could use ion propulsion instead carrying a lot chemical fuel for last part of leg- one more than say a week of time to do the burn or don't need high thrust. That gives bonus of giving crew access to a lot of electrical energy. Though there seems like there is something wrong, or because now I am beginning to think maybe one could do whole thing with Ion engines. This is too close to a hohmann transfer delta-v and it seems it should cost more. There is cost to changing vector. And/or one has a "loss" or exchange of potential energy in regards to Sun's gravity- and no doubt in my mind the Hohmann is the most efficient transfer [in terms of delta-v, obviously, not referring to the amount of time it takes to get somewhere].
Or the only "answer" would seem to be one taking a lot orbital energy from Earth's orbit.
Always thought is was sort of like gravity assist, but more of a powered gravity assist.

Question re: the 2.321668688 km/sec  how much Oberth effect is added to the trajectory?

Transfer orbit perihelion   0.87   A.U.                  
Transfer orbit aphelion   1.524   A.U.               2r1   2
Flight path angle at departure   0.222899017   Radians   12.77117293   degrees   Perigee burn (300 km altitude falling from high apogee)         
Vinfinity at departure   1.534703827   A.U./yr   7.275423642   km/s   2.321668688   km/s      
                        
Flight path angle at destination   0   Radians   0   degrees   Periaerion burn to a 3697x23459 km ellipse         
Vinfinity at destination   0.750540983   A.U./yr   3.55801785   km/s   1.511760053         

[Hop_David]

For the earth flyby, you would get at least some oberth effect for the part of the acceleration done in the earth gravity well.

Tried to do screen captures as red pellet #1 crosses lunar orbit on the way in and again as #1 crosses lunar orbit on the way out. My reflexes are slow so I reset time increment to 300 seconds. Capture attached.

4.86 days in cislunar space (the space between LEO and Lunar orbit). Rats, if crossing points are ~384,400 km, this looks like an ellipse with a mere 3 Lunar Distance (L.D.) apogee, still within the Hill Sphere. That doesn't make sense, maybe there's a flaw in my spreadsheets.

If crossing in the vicinity of lunar apogee looks like an ellipse with a 8 L.D. apogee or nearly double Hill Sphere radius. Hope that's the case. 

So going with an ellipse a = 4 L.D., e=.9916. This would spend about 40 days within earth's Hill Sphere. Average velocity would be about .9 km/s.

Going with Impaler's 1 mm/s^2 acceleration, 20 days from edge of Hill Sphere to perigee would give about 1.7 km/s.

So let's look at the kinetic energy for 1/2 mv²… Screen capture attached.

Looks like the Oberth benefit gave us 1.53 mega-joules per kilogram.

[Hop_David]

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).

Attached is a screen capture from a spread sheet where the Earth Departure Stage would do a 3.1 km/s perigee burn. Since falling from EML2 takes .4 km/s, the EDS would need a total delta V budget of about 3.5 km/s. This wouldn't be a reusable EDS.

Coming to Mars, a 4.67 km/s peri-aerion burn would capture the MTV into a 3697x23459 km ellipse. This is atmosphere grazing with apo-aerion at Deimos height. Orbital period is 13 hours, so about 2 periaerion drag passes per day. More drag passes per day as apo-aerion is lowered. I believe aerobraking could circularize the orbit within a week or so.

This transfer would take 3.4 months or about 103 days.

I've also attached the spreadsheet. A user can play with the transfer orbit by changing aphelion and perihelion.

Departure orbit radius   1   A.U.   Speed   6.283185307   A.U./yr         
Destination orbit radius   1.524   A.U.   Speed   5.089643749   A.U./yr      4.740604351   AU/yr to km/s
Transfer orbit perihelion   0.7   A.U.                  
Transfer orbit aphelion   1.53   A.U.               2r1   2
Flight path angle at departure   0.36777381   Radians   21.07188715   degrees   Perigee burn (300 km altitude falling from high apogee)         
Vinfinity at departure   2.375992013   A.U./yr   11.26363808   km/s   4.888082825   km/s      
                        
Flight path angle at destination   0.067838707   Radians   3.88687157   degrees   Periaerion burn to a 3697x23459 km ellipse         
Vinfinity at destination   1.084314564   A.U./yr   5.140306341   km/s   2.568080059         
                        
Time of Flight   0.370249764   years   4.442997163   months   135.2337261   days      

This appears to have flight time of 4.44 month, but I would guess that's it's at top of the perihelion orbit, and so one would do a patched conic, like you do with most Hohmann transfer to mars and thereby shorten the travel time, plus that would be adding some orbital speed so one is going closer to Mars velocity [mostly regarding it's orbital vector].
Anyways it's less delta-v than I expected.   
And I think minimum travel time should be 3 months for crew.
Edit: I mean from the crew launching from Earth it should take 3 month or less to reach Mars [Though could  be less travel time but would take considerable more rocket fuel and it seem with enough shielding 3 months should fast enough so reduce microgravity effect and crew radiation exposure].

Indeed. I attached an illustration a transfer orbit with a .7 A.U. perihelion and a 1.53 aphelion. Vinfinities are indicated in red.

The aphelion is just a tad over Mars' radius so the flight path angle is only about 4 degrees.

The perihelion is way below earth's radius so the transfer orbit's flight path angle is 21ş at intersection with earth orbit. The direction difference in these two vectors really jacks up Vinf. Ordinarily an 11.26 km/s vinf would be intolerable. But when dropping from a high apogee, perigee burn is only ~4.9 km/s. Kinda high but doable.

Turquoise is the area transfer orbit sweeps out from departure to destination, about 31.5% of the ellipse's area. The ellipse's semi-major axis is 1.115 A.U. It's period is 1.115^3/2 years which is 1.177 years. 31.5% of 1.177 years is 4.44 months.

If you want something close to Hohmann, set perihelion at .999 and aphelion at 1.525.

[Hop_David]

Admittedly hydrocarbons are far easier to store so perhaps this is your motivation,

Yes. I looked for short routes back to EML2 that wouldn't take extra delta V. The shortest one I could find was 50 days. 8 or 9 tonnes of propellent would need to be stored until the final perilune and EML2 parking burns.

but if the source is the lunar cold-traps then presumably hydrogen can be stored their passively with zero boil-off and the propellent delivery could be done via a 'launch on demand' directly to the waiting vehicle at EML2.

If memory serves cold traps are 40 K and we need 20 K to stop hydrogen boil off. I do think the cold traps would drastically reduce the difficulty of refrigerating hydrogen though.

I like to imagine lox/hydrogen ships for delivering stuff from lunar surface to EML2 and back. But for longer trips I'm thinking methane. Unless the ULA propellent depot boys (Zegler, Kutter, & Barr) manage to reduce boil off. I give them better than even odds if they are funded. But perhaps I'm succumbing to the space cadet optimism I'm prone to.

Finally the dry mass of the stage seems rather high, you said this stage is intended only for use in space so it should be free of most parasitic mass other then insulation to keep boil off to a reasonable level during the ~2 weeks in which it actively has propellent, I would think that 10% dry mass would be a perfectly reasonable figure, unless their are other demands being put on it that I don't know of.

Thanks, you're gracious to say that. I was trying to err on the conservative side. Also high dry mass ratios make the vehicle sturdier and thus more amenable to re-use.

I suspect that a simple increase in the propellent tank for the MTV will be able to fully replace the EDS.  At 3000s the MTV would need 10% propellent fraction to do 3 km/s.  Thus our 200 ton vehicle could just be 220 tons at EML2.  Now more propellent at LEO would be needed for it to push that larger mass up to EML2, as the DeltaV from LEO to EML2 is 7 km/s for low thrust vehicles that means a an additional 20% on top of the additional propellent to get the additional propellent to EML2, which is 5 tons.  So in totality the MTV needs 25 tons more IMLEO in propellent (Xenon) to replace the EDS.

As mentioned, infra structure capable of making propellent could also provide water and oxygen. Drinking water, water for cleaning, radiation shielding, and oxygen to breathe could make up nearly half an MTV's mass.

So 125 tonnes more in IMLEO.

And while I think the EDS presented above is too conservative and could be considerably lighter it will not be so light as to cost competitive with a simple enlargement of the MTV tanks, to speak nothing of the infrastructure costs of establishing in space propellent production.

My EDS has ~30% mass fraction which makes it very durable. It never has to endure atmospheric re-entry. Round trip delta V budgets are .18 km/s.

A 30% dry mass fraction is more robust than the Falcon booster, lower delta V budget and it doesn't have to endure a 2 km/s re-entry. Rendezvous with prop depots and space craft could be slow and easy, no zooming towards a barge at terminal velocity and dealing with 9.8 meter/s^2 acceleration. This EDS is much more amenable for re-use than a booster.

An upper stage with payload will have about a 10% dry mass fraction, about as sturdy as a soda can. Now you're going to slam this foil cylinder into the atmosphere at 8 km/s. 16 times as much kinetic energy as a Falcon booster re-entry. Much, MUCH more difficult to re-use than a booster.

I don't expect reusable upper stages. I do believe Musk will be able to re-use boosters but there will be refurbishment and transportation costs.

If earth is the sole source of propellent, I expect Musk to cut launch costs by half, but not more than that.

[Hop_David]

That's a really neat class of trajectories! Does your pellet simulation tell you the end state of the pellets (i.e. position and velocity), so you could calculate their specific orbital energies to see if they are greater than zero?

Sadly no. The two numbers in the upper right are number of time increments and orbital energy. Maybe I could do something with the second number but haven't figured out how.

My attempts at orbital simulations have been miserable failures. More knowledgeable folks have advised me I was using 1st order Runge Kutta methods when I should use 2nd or 3rd order.

The sims are JAVA that I stole from Bob Jenkins. Well, I didn't actually steal it as he gave me permission to use it. Jenkins' Java sims are one of my favorite toys.

Here is a page. My most recent operating system is paranoid of using JAVA from strangers. I had to jump through a few hoops setting system preferences so I could use my own pages, but it's doable.

[Impaler]

Unless the ULA propellent depot boys (Zegler, Kutter, & Barr) manage to reduce boil off. I give them better than even odds if they are funded. But perhaps I'm succumbing to the space cadet optimism I'm prone to.

I though the ULA integrated propellent management tech was a done deal, patented even.  It doesn't eliminate boil-off but it drops it a lot while also using the boil off as a power source and hugely reducing parasitic mass that is generally necessary to keep a stage 'alive'.  Add a bunch of sun-shades and your done, at least enough to do some kind of in-space tug which only holds propellent for a short time.  Now sending Hydrogen to Mars or creating a depot is another level of difficulty.

As mentioned, infra structure capable of making propellent could also provide water and oxygen. Drinking water, water for cleaning, radiation shielding, and oxygen to breathe could make up nearly half an MTV's mass.

So 125 tonnes more in IMLEO.

Life support closure is good and getting better all the time, see this paper on the current state of the art http://sites.nationalacademies.org/cs/groups/depssite/documents/webpage/deps_063596.pdf, it says the only 100 kg of make-up water is needed per person-year a very modest need (actually this may be out of date I recall that a very recent additional pyrolysis modules combines the CH4 and CO2 that are currently vented to make yet more water).  Conversely something like 700 kg of food, clothing and other manufactured goods are used so our consumables budget is already dominated by things that in-situ resource can not provide.  In fact it appears that learning to do laundry in space would reduce our consumable mass by more then getting water recycling to 100%, a very non-intuitive result.

Water is a good GRC shielding material per unit of mass, but because were effectively making it a structural part of the vehicle to remain there forever we only pay the cost to move this water from Earth once unlike propellents.  If the total habitat is going to mass 100 mT when fully loaded then I can't see more then a few tons of water for shielding because we should only try to shield from Solar Storm events, not GCR, even a full meter thickness of water will not do much to stop GCR's while having a tremendous mass.  Thus we would likely give our astronauts a lower total dose by just dropping the mass and going to the destination faster.  Overall the Radiation issue is one with a lot of zealotry around it driven by a combination of lack of data and Zubrin's constant attacks on NASA for not accepting higher risk.  I think the one thing people agree on is that were not going to use super massive thickness of shielding in space even if the mass was just sitting ready for us to use, we can't afford to push it around even with electric propulsion.

I don't expect reusable upper stages. I do believe Musk will be able to re-use boosters but there will be refurbishment and transportation costs.

If earth is the sole source of propellent, I expect Musk to cut launch costs by half, but not more than that.

I have the feeling your skepticism of upper stage reuse it driven by a need to justify the extraction of in-situ propellents.  In any case Musk already cut launch costs in half with F9, will cut them again with F9 Heavy, and will likely do it a third time with the 1st stage reuse that is almost assuredly going to happen.  Production of a super heavy class launcher even if it were to only have the 1st stage reused would likely drop costs again simply because so many launch costs do not scale with vehicle size making a larger launcher inherently more cost effective then a smaller one (so long as we compare commercial to commercial).

[sdsds]

Let's imagine Impaler's MTV -- [...] with acceleration of .000001 km/s^2. 11.6 days of acceleration gets 1 km/s delta V. It takes this ship about 3 months to spiral out of earth's gravity well. [...]

Radius of the MTV's spiral will eventually reach 327,000 km, the distance to the moon's Hill Sphere.

I'd like to contribute to this thread a (still incomplete) analysis of the spiral out from LEO to the cislunar vicinity. (Long term I would like to be using something open source, but the free Mathematica for Raspberry Pi was sufficiently tempting.) Taking a hint from a discussion related to this on stackexchange I use a state vector of (r, v, gamma, phi} to make the acceleration in the direction of the velocity vector easy. The results below start from 400x400 km LEO and use the 0.001 m/s^2 constant acceleration mentioned above.

After 77 days the spacecraft has completed more than 345 revolutions around the Earth, reached an altitude of 323,657 km and a velocity of 1233.58 m/s. (The semi-major axis of the resulting orbit is 446,048 km, which certainly gets it within the Moon's Hill Sphere.)

[EDIT: 75 days is apparently enough. Added below is a revised trajectory showing 75 days of propulsion spiral and the resulting orbit, and also the (average) orbit of the Moon.]

[gbaikie]

Let's imagine Impaler's MTV -- [...] with acceleration of .000001 km/s^2. 11.6 days of acceleration gets 1 km/s delta V. It takes this ship about 3 months to spiral out of earth's gravity well. [...]

Radius of the MTV's spiral will eventually reach 327,000 km, the distance to the moon's Hill Sphere.

I'd like to contribute to this thread a (still incomplete) analysis of the spiral out from LEO to the cislunar vicinity. (Long term I would like to be using something open source, but the free Mathematica for Raspberry Pi was sufficiently tempting.) Taking a hint from a discussion related to this on stackexchange I use a state vector of (r, v, gamma, phi} to make the acceleration in the direction of the velocity vector easy. The results below start from 400x400 km LEO and use the 0.001 m/s^2 constant acceleration mentioned above.

After 77 days the spacecraft has completed more than 345 revolutions around the Earth, reached an altitude of 323,657 km and a velocity of 1233.58 m/s. (The semi-major axis of the resulting orbit is 446,048 km, which certainly gets it within the Moon's Hill Sphere.)

So it's delta-v of 6.6529 km/sec ?
77 * 24 * 3600 = 6.6529 million seconds.

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.

Without storing energy, getting beamed energy, having nuclear power, or Sun-synchronous,
Then when at lower orbit one only gets solar energy for about 60% of 24 hr day.
If thrust only during sunlight it might get some weird orbit. Instead you could burn for say
10 min of the 90 min orbit, which should roughly give an elliptical orbit- raising the apogee.
So .6 of 90 mins is 54 mins. And could charge batteries for say 44 mins and use solar power
and battery power for the 10 mins of thrust.
So in beginning it's one only burn for 1/9th the time, but it should raise the orbital height of
apogee in a more efficient manner. Or roughly take instead 9 times longer is would be about 5 times
longer [with same thrust level].
But after raise the apogee high enough, one will have more of the 24 hour day in sunlight. So
it could become more significant once got to say 1000 km height. So one could then switch back to
constant thrust at time during more daylight- say when apogee is say somewhere around 1000 to 2000 km.

If one could get constant thrust that would get you 300,000 km quicker. But if had only about 60% sunlight
at lower orbit then 10 min burn should rise almost as fast [or if had more thrust [the ion rocket could provide more thrust if provided with more electrical power solar + battery] then could be faster.  But even 10 min thrust period doesn't get one higher quicker orbit in time it would require less total delta-v [use less Xenon].

[Hop_David]

Let's imagine Impaler's MTV -- [...] with acceleration of .000001 km/s^2. 11.6 days of acceleration gets 1 km/s delta V. It takes this ship about 3 months to spiral out of earth's gravity well. [...]

Radius of the MTV's spiral will eventually reach 327,000 km, the distance to the moon's Hill Sphere.

I'd like to contribute to this thread a (still incomplete) analysis of the spiral out from LEO to the cislunar vicinity. (Long term I would like to be using something open source, but the free Mathematica for Raspberry Pi was sufficiently tempting.) Taking a hint from a discussion related to this on stackexchange I use a state vector of (r, v, gamma, phi} to make the acceleration in the direction of the velocity vector easy. The results below start from 400x400 km LEO and use the 0.001 m/s^2 constant acceleration mentioned above.

After 77 days the spacecraft has completed more than 345 revolutions around the Earth, reached an altitude of 323,657 km and a velocity of 1233.58 m/s. (The semi-major axis of the resulting orbit is 446,048 km, which certainly gets it within the Moon's Hill Sphere.)

[EDIT: 75 days is apparently enough. Added below is a revised trajectory showing 75 days of propulsion spiral and the resulting orbit, and also the (average) orbit of the Moon.]

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

Could you explain how you did that? I would like to run similar analysis but presently don't know how.

There's a free version of Mathematica?

[sdsds]

Responding to the last topic first:

There's a free version of Mathematica?

Yes. See Stephen Wolfram's blog post from November 21, 2013:

Putting the Wolfram Language (and Mathematica) on Every Raspberry Pi

This was only a little bit interesting back then, when the available Pi (model 1) had about as much computational power as a cell phone. But now that a quad-core Raspberry Pi 2 is available for $35, and the "raspbian" operating system is easy to install and includes Mathematica...? For those comfortable using  the X Window System (and even "ssh -X") it is pretty enticing.

I've only been using it for a week or so, and have been meaning to start a, "Flying Mathematica to Cis-Lunar Destinations" thread, but you beat me to it with this thread!

[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 180ş 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

[Hop_David]

Here's a couple screen capture of starting out in olive orbits and eventually leaving earth's Hill Sphere. Same orbits, different zooms.

[sdsds]

Here's a couple screen capture of starting out in olive orbits and eventually leaving earth's Hill Sphere. Same orbits, different zooms.

Is the place where the paths of the pellets diverge so greatly near the SEL-1 point? Or SE-L2?

[Hop_David]

Here's a couple screen capture of starting out in olive orbits and eventually leaving earth's Hill Sphere. Same orbits, different zooms.

Is the place where the paths of the pellets diverge so greatly near the SEL-1 point? Or SE-L2?

I don't remember where I set the barycentric longitude of the sun for this sim.

It's a little hard to tell which part of the Hill Sphere the pellet apogees are at since they take about 45 or 50 days to get there over which time the Hill Sphere rotates 45 or 50 degrees. Also these pellets spent about 10 days in the earth and lunar realms before exiting to the exterior realm.

It seems like aiming at different parts of the Hill Sphere will subject the pellets to different winds.

Aiming at SEL1 or SEL2 seems to boost apogee, often resulting in escape.

Aiming at 45ş angles seems to subject pellets to a sideways wind that rotates line of apsides. If the sideways push is prograde, apogee is boosted. If the push is retrograde, apogee is lowered.

Aiming at 90ş and 270ş (leading and trailing points of Hill Sphere) seems like I'm shooting against the wind.

Borrowing a Wikipedia illustration for tidal force, I believe these are the winds:

[Hop_David]

I've put a lot of this stuff in a blog post: EML2

[Burninate]

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.

I assumed this too, but batteries sized for the LEO day-night cycle are actually not huge relative to the other parts of the system, especially modern lithium ions.  At 1C, a conservative 1kg LiPo charges at around 150 watts, and holds 150 watt hours.  5C-10C charging is state of the art, but only a ~1.33C charge rate (a 45 minute charge) will switch the bottleneck over to energy density.  To run a ~6.9kw NEXT thruster (58kg) alongside its battery charging system, you need maybe 15kw peak power draw, which at 150w/kg is 100kg of solar panels.  A ~46kg battery(with unknown mass charger) can power a ~6.9kw NEXT thruster weighing 58kg for one half of an orbit at full depth of discharge.  System mass 202kg.

The alternative is a 58kg thruster and 6.9kw of solar panels at ~46kg, system mass 102kg, providing half of the impulse for the early part of the spiral, but stung with unclear penalties afterwards, as apoapsis is in shadow a larger portion of the time.

Depth of discharge and cycle life (allegedly ~1000 cycles for Tesla now) becomes the limiting factor in a battery system based on this, I think.  Generally you should assume half depth of discharge (by virtue of twice as much battery mass as the minimum) doubles cycle life.  There should be metrics available to compare other battery chemistries on the basis of total lifetime kwh per kilogram.  A battery system doesn't need to last forever, either, the benefit fades as the percentage of time spent in shadow drops.  At Earth surface, you get 50%.  At 0.1 Earth Radius (0.1Rⴲ = 637km) altitude, you're down to 36.3%.  At 0.2 Rⴲ, 31.4%.  At 0.5 Rⴲ, 23.2%.  At 1 Rⴲ, 16.7%.  At 5.62 Rⴲ (geosynchronous orbit), you're down to 4.8%.

And that's presuming things line up.  The farther out the inclined orbit, the greater percentage of the year it will not suffer Earth eclipses at all.

[sdsds]

The images below recreate the first section of the Farquhar three-impulse transfer from LEO to EML2. The first is an overview, shown in the rotating frame of reference used here:
http://clowder.net/hop/TMI/FarquharRoute.jpg

The second zooms in on the Earth-departure burn. No real surprises there. The third shows the same trajectory as the first, but in a non-rotating (inertial) reference frame. The spacecraft is going quite slowly (128.44 m/s) near the apogee of its highly elliptical orbit; the Moon in its circular orbit is rushing towards it at 1011.56 m/s.

[sdsds]

The next plot shows what happens over the next 9.1 hours. The spacecraft does get out of the way of the on-rushing Moon, but its trajectory is altered by the Moon's gravity. The second plot shows what would happen after the lunar encounter were there no propulsive maneuver there. The direction of travel of the spacecraft is substantially altered! The final plot shows the same thing, but it is not as evident in the rotating reference frame.

[sdsds]

The plots below show a "quick path" home from EML-2. The total elapsed time is 10.8 days. The total delta-v is 583 m/s. The trajectory departs EML-2 with a 55 m/s burn at an angle of 148.5 degrees from the line between the Earth and Moon. It travels in the general direction of the Moon, looping around a bit until it reaches an apolune at 6.6 days. There a 158 m/s retrograde burn leads to a deep perilune at 8.3 days. A 370 m/s burn at perilune leads to a trajectory that intersects the Earth at a relative speed of 11,134 m/s.

Tiny navigation differences could lead instead to a "near miss" of the Earth, with a similar relative speed at an arbitrarily chosen perigee.

[gbaikie]

The plots below show a "quick path" home from EML-2. The total elapsed time is 10.8 days. The total delta-v is 583 m/s. The trajectory departs EML-2 with a 55 m/s burn at an angle of 148.5 degrees from the line between the Earth and Moon. It travels in the general direction of the Moon, looping around a bit until it reaches an apolune at 6.6 days. There a 158 m/s retrograde burn leads to a deep perilune at 8.3 days. A 370 m/s burn at perilune leads to a trajectory that intersects the Earth at a relative speed of 11,134 m/s.

I suppose that the 370 m/s burn could not be done with low thrust Ion rocket?
But other burns would have enough time to use a Ion rocket?

[sdsds]

Yes, I think that's correct on both. One interesting aspect is that the large perilune burn sets a lower bound on the thrust required for the entire round trip, once the Earth-departure burn is accomplished (presumably with a dedicated stage). For the other two burns, it sure seems like they could be replaced with some sort of continuous, low-thrust propulsion. Figuring out the "control law" that would make that thrust always be in the optimal direction seems like quite a challenge, though!

[sdsds]

Figuring out the "control law" that would make that thrust always be in the optimal direction seems like quite a challenge, though!

Bounded momentum model on the combined two body (Earth Moon) projective gradient.

That's easy for you to say!

I think by optimal you might mean, "has greatest effect" on energy, looking at something like the Jacobi integral. Decreasing the energy would effectively put the particle down into one of the wells of the effective potential. This would be going "always down hill" on the energy manifold. That gets someplace with optimal efficiency. I don't see how to make certain that someplace is the place we want to go.

For planar motion there is a four-dimensional phase space, (x, y, vx, vy). Consider starting at (x₁, y₁, vx₁, vy₁) with the desire to get to (x₂, y₂, vx₂, vy₂). I don't see an easy solution to that!

[sdsds]

On the topic of effective potential, I really like the plot attached below from http://www.cds.caltech.edu/~marsden/volume/missiondesign/KoLoMaRo_DMissionBook_2011-04-25.pdf showing the location of the Lagrange points. But it uses µ = 0.3 to make the image easy to see. For Earth/Moon,  µ = 0.01215. Attached second is Mathematica's version of the plot, using the realistic value for µ.

[Space Ghost 1962]

Figuring out the "control law" that would make that thrust always be in the optimal direction seems like quite a challenge, though!

Bounded momentum model on the combined two body (Earth Moon) projective gradient.

That's easy for you to say!

I think by optimal you might mean, "has greatest effect" on energy, looking at something like the Jacobi integral.

Yes.

Decreasing the energy would effectively put the particle down into one of the wells of the effective potential. This would be going "always down hill" on the energy manifold. That gets someplace with optimal efficiency. I don't see how to make certain that someplace is the place we want to go.

Due to symmetry of the created manifold (being a 2 or N-body problem),  going "always down hill" on the gradient means that before the next "fold" (or point of inflection where you might do a maneuver), you're either adding to or subtracting from the kinetic energy of such (e.g. you get the clue to the direction and magnitude, but you lose the forward/backward direction (!).

For planar motion there is a four-dimensional phase space, (x, y, vx, vy). Consider starting at (x₁, y₁, vx₁, vy₁) with the desire to get to (x₂, y₂, vx₂, vy₂). I don't see an easy solution to that!

When one collapses/"folds" a higher dimensional space to a fewer (in this case a plane), the "folding" (or projection) can be chosen to occur in only certain ways, many/most of which have infinite solutions (or "poles"). Yes, no obvious solutions.

However, inspection of the way it collapses yields (in this case) two which are orthonormal with a resultant in the plane. Again, differing in forward/backward direction. Not reducible in analytic but selectable.

Hope I'm not being too cryptic.

[sdsds]

Plodding forward with plots of the effective potential, here's a zoomed view of the vicinity around the Moon. Although it's pretty, it's hard to glean much from this other than seeing that the saddle points at EML-1 and EML-2 do exist. The second attached plot shows a contour map of the region. This makes the saddles and their locations more evident. Finally a cross section along the y=0 line, which shows exactly where the saddle points are located, and their relative heights.

[sdsds]

As for orbits being reversible?

I'm still wondering whether the "reverse" of the Farquhar route exists. Does the "mirror image theorem" suggest that the corresponding path from EML-2 to Earth would be on the opposite side of the Earth-Moon line?

EDIT: Apparently it's the "Theorem of Image Trajectories." See attached.

EDIT 2: Attached is a Mathematica "notebook" in .pdf form showing a route much like Farquhar's but in reverse. Total delta-v to a 230 km perigee 9 days after leaving EML-2 is less than 350 m/s.

[sdsds]

Necroposting to link this orbit animation from Shane Ross.
https://twitter.com/i/status/1889503469360263344

[sdsds]

Does anyone happen to know a name that could be used as a google search term for an orbital trajectory like the one shown in the first attached image? Or have a link to someplace where it's described?

The trajectory is balanced between corresponding halo orbits around L1 and L2 with low energy transfers available to and from each (example in second image.)

[sdsds]

And here's the low energy transfer to the corresponding L1 halo orbit.