Continuous low-thrust to Mars

keermalec · May 7, 2007

Hi, calculating a Hohmann transfer orbit with high-thrust acceleration at departure and arrival is pretty straightforward (http://en.wikipedia.org/wiki/Hohmann_transfer). However, when one tries to calculate a minimum transfer orbit using continuous low-thrust, things get much trickier. For example, getting from Earth to Mars using a Hohmann transfer orbit requires delta-vs of 2.93 km/s at Earth orbit and 2.64 km/s at Mars orbit. the trip takes 258 days and is a semi-ellipse with the Earth at perigee and Mars at Apogee. Now if I want to get to Mars using a continuous but low thrust as, for example, when using an ion drive, I suppose the total delta-v will still be 5.57 km/s but the duration and total angle accomplished around the sun depends entirely on the amount of thrust. Is it right to suppose that by continuously accelerating for 258 days to attain a speed of 5.57 km/s I will get to Mars, or is it more complex than that? 5.57 km/s in 258 days is about 0.00025 m/s/s acceleration, or 0.000025 Gs. Will I do a 180Â° turn around the sun at this thrust? If I have a higher thrust, will I do less than 180Â°? In the end the question is: if Earth is at angle A and Mars is at angle B, what is the thrust and total delta-v needed to get from Earth to Mars, sing continuous thrust? <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

origin · May 7, 2007

You cannot use the sun for a slingshot effect. The slingshot works because the space craft uses the orbital velocity of the planet for the acceleration. This is why venus is used because of its orbital velocity (plus its close). The Sun has essentially no velocity relative to Mars. <div class="Discussion_UserSignature"> </div>

keermalec · May 7, 2007

I have expressed myself wrongly: I do not intend to use the sun for a slingshot effect. By angle I mean the angle between earth's position and mars'. The trajectory is a spiral from Earth to mars: I do not pass close to the sun. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

kelvinzero · May 7, 2007

Not a clue! <img src="/images/icons/laugh.gif" /> I havent seen much maths here. Perhaps you could try a physics help forum such as: http://www.physicsforums.com/index.php

spacester · May 8, 2007

Hi Keermalec, good question! I had this in my archives, just waiting for you. I'm glad the link is still alive: kudos to Capt R.J. Stockermans at the Royal Military College in Kingston, Ontario. http://www.rmc.ca/other/usn/ionsupp/ionsupp.htm Continuous thrust transfers greatly complicate our orbital analysis of Part A of this module. In the classic two body problem, we have chosen the earth or the sun as our main source of gravitational attraction. Now our vehicle is under the influence of this force as well as a constant thrust. While this may appear to be straight forward, in general, there are no analytical, closed form solutions. Energy and momentum are no longer conserved in a given trajectory as we are adding thrust and thus, if we are expanding the orbit, we are adding energy to the orbit. For these reasons, for a first order solution, approximations are made and/or charts have been devised to aid in solving this problem. Three different approximate methods will be discussed here. /safety_wrapper> <div class="Discussion_UserSignature"> </div>

keermalec · May 11, 2007

Way cool Spacester! This is EXACTLY the answer I was looking for. Thank you very much! <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · May 12, 2007

I'm very happy to hear that! You're welcome. If you're going to be doing some actual calculations, I would be VERY interested in any details you'd like to share, and of course let me know if I can provide any assistance. At some point in the relatively near future, I should have a section on my site you can utilize to do things that can't be done on a forum, things like full answers to questions like this one. But in any case, I'm happy to point you in the right direction. I learned a LOT from that site, poke around and you'll see what a great resource it is. Oh, and I would note that the 258-day Hohmann transfer is a fiction. That is the time needed to transfer from a circular Earth orbit to a circular Mars orbit, but there is NO SUCH THING. It is pure fiction. You prolly knew that, but further details can be found on the Hohmann to Mars thread. <div class="Discussion_UserSignature"> </div>

keermalec · May 13, 2007

Spacester, I read thru your "Hohmann to Mars" post and was quite impressed by your precision. I get significantly different results when applying my understanding of a patched-conic approximation. My delta-vs are much higher than yours and so I am probably and hopefully wrong. Can you tell me what if anything is off in the following calculation: Transfer from LEO to LMO assuming all orbits are circular: LEO: 6'521 km Earth SOI: 922'388 km LMO: 3'490 km Mars SOI: 575'789 LEO to SOI-transfer: 3.18 km/s SOI-transfer to SOI: 0.58 km/s Escape velocity from SOI: 0.93 km/s Earth to Mars-transfer (Hohmann): 2.94 km/s Mars-transfer to Mars: 2.65 km/s Capture at Mars SOI: 0.38 km/s SOI to LMO-transfer: 0.24 km/s LMO-transfer to LMO: 1.4 km/s Total: 12.32 km/s or 10.68 km/s if we stop at Mars SOI. This is much higher than the 6 km/s or so you get. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · May 13, 2007

Hmmm . . . well you don't 'stop at SOI'. You go straight from LEO to a Mars-transfer trajectory. The purpose of defining SOI is to be able to treat each of the three trajectories independently. The Hohmann trajectory you are calculating is from the center of Earth to the center of Mars, ignoring the gravity of each. So it's a mathematical abstraction to start with, and you use the additional mathematical abstraction of the SOI to get real results. On the journey to Mars you fine-tune your trajectory such that you cross the imaginary SOI sphere at just the right point so that Mars gravity bends your path (compared to if Mars wasn't there) and you fall into that gravity well, passing above the surface (on the far side of the planet) at the altitude of the periapse of your desired elliptical Mars orbit. This is exactly the same thing as a "gravity-assist" maneuver except we're going to burn our engines at the periapse point to slow down and get captured into orbit (also grav-assist typically happens on the near side of the planet IIRC). The fine-tuning of the trajectory on the way to Mars targets a "miss-distance" above the planet, ignoring the planet's gravity. Knowing your heliocentric velocity vector, targeting the "miss-distance" insures you make the periapse pass at the right altitude, which is called the "impact parameter". The SOI is ONLY a mathematical abstraction, used to make these calculations, nothing more than that. It is at the Martian SOI where you mathematically convert your heliocentric velocity vector into a Mars-centric velocity vector. This in turn lets you plot the hyperbolic orbit as you pass by Mars (if you don't burn engines to slow down). The impact parameter method just mentioned is a convenient method of doing that calculation. As far as leaving Earth, the SOI is where you convert from your Earth-centric velocity vector to a Heliocentric velocity vector. It's the same velocity, but relative t <div class="Discussion_UserSignature"> </div>

keermalec · May 13, 2007

Spacester: you are right again. Since posting my little note I was doing ressearch on just this and realising I had got it all wrong. I found a nice pdf which clearly explains the patched-conic approach and I am writing a short php script to calculate this right now. http://homepage.mac.com/hanspeterschaub/work/Papers/UnderGradStudents/ConicReport.pdf I now understand there are only two burns: one at LEO to attain excess escape velocity, and one at LMO, to slow me down to orbital velocity. My mistake was in thinking I had to transfer to a high orbit first, before injecting into a Hohmann transfer, and then doing those two steps on arrival at Mars too. Clearly a waste of fuel. Let me get back as soon as I have some results and maybe questions. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · May 13, 2007

Holy Cow! What a great reference! Why, back in the day, I had to walk uphill both ways and didn't have fancy-dancy references like that. <img src="/images/icons/laugh.gif" /> By the time you get done, I'm probably going to be the one asking YOU questions. If you want to get into Faster-than-Hohmann I can point you in the right direction. That site I linked to earlier talks about the one-tangent solutions, and I have some references on Lambert's Solution somewhere in my archives. <div class="Discussion_UserSignature"> </div>

spacester · May 13, 2007

I can't help myself, it's a pet peeve . . . that reference, as excellent as it is, treats Mars as in a circular orbit. My methods account for the real solar system with Mars in its actual elliptical orbit. I would dearly like to combine his excellent 4-body solution with a method that gives real-system results. As excellent as his work is, it still will not get you to Mars. <div class="Discussion_UserSignature"> </div>

MeteorWayne · May 13, 2007

Indeed, the orbit of Mars is the most eccentric (e=0.093, the largest of the "8" planets), except for Mercury which is a Newtonian and Einsteinian special case. <img src="/images/icons/smile.gif" /> The difference between Perihelion and Aphelion is nearly 0.3 AU. The difference in position and velocity cannot be ignored in even making rough calculations for Mars. The result is the propensity of intelligent humans to launch missions to Mars somewhere around every two years <img src="/images/icons/wink.gif" /> <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

keermalec · May 13, 2007

You are right, Reppert does consider Mars' orbit to be circular. Do you happen to know where I can get precise astronomical data on Earth and Mars Apoapsis and Periapsis? I mean dates as well as distances. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · May 13, 2007

THE authority is JPL's Horizons It's delightfully old-school. I use the telnet method, but it takes some getting used to. Hmm . . . I'm not getting it to connect right now . . . I'll try their new html interface, looks nice. Change to Ephemeris Type: Elements Change to Sun-centered Accept defaults Generate ephemeris Ah yes, the old familiar output . . . but not packed into a little text box that was confusing at times. NICE! I tell ya, you kids these days have it so easy. ROTFL BUT you still gotta know what's going on to answer your question. To find periapse, um, oh wow check out THIS reference. Wow, I was going to put together a sheet like this 'one of these days'. Periapse = a (1-e) Apoapse = a (1+e) where a = semi-major axis e = eccentricity which are given on the JPL output. what is interesting is that the eccentricity is actually an ephemeri (!!) To find the dates, you need to find when the true anomaly is zero (periapse) and pi (apoapse). I can't do everything for you or you won't learn <img src="/images/icons/wink.gif" /> <img src="/images/icons/laugh.gif" /> so I'll let you take it from there. edit: actually, True Anomaly is given in degrees, and if you read it carefully you find that we got lucky as to when we asked this question: Apoapse is coming right up. <div class="Discussion_UserSignature"> </div>

keermalec · May 13, 2007

OK I think I got it working now: the results seem much more plausible. I am still considering a circular orbit for mars simply in order to simplify my calculations but I will try to correct that in the near future: Transfer from LEO to LMO assuming all orbits are circular: LEO: 6'521 km Earth SOI: 922'388 km LMO: 3'490 km Mars SOI: 575'789 LEO to ESOI: Duration: 3.70814564129 days Delta-v: 4.03881502893 km/s ESOI to MSOI: Duration: 258.238215599 days MSOI to LMO: Duration: 4.50235684139 days Delta-v: 0.941269252633 km/s Total trip: Duration: 266.448718082 days Delta-v: 4.98008428156 km/s Now the interesting part is this: if we use continuous low thrust to accelerate from LEO to Earth SOI and to decelerate from Mars SOI to LMO, here are the specs: Acceleration from LEO to ESOI: (4.03881502893 km/s in 3.70814564129 days) 1.26061780842E-005 m/s/s Deceleration from MSOI to LMO: (0.941269252633 km/s in 4.50235684139 days) 2.41969271593E-006 m/s/s Now assuming a 100-ton ship equiped with an NSTAR-type ion-drive, necessary solar panels for power at Mars, 4.98 km/s maximum speed, ISP 3'127s, and acceleration 1.26061780842E-005 m/s/s gives: Fuel: 14.9838118457 ton s Fuel tank: 1.4084783135 tons Thruster: 0.549171948075 tons Solar panels: 0.803392716214 tons Structure: 3 tons Payload: 79.2551451765 tons With this solution we still get to Mars in 266 days, but 79% of ship mass is payload! Of course it would have to replenish its Xenon fuel supply at Mars, or use some other fuel such as Nitrogen, or carry 15 tons of fuel for the return trip as part of its payload. In any case, the efficiency of using an ion-drive seems to far outperform any chemical rocket scheme. Even if Mars were at 1.7 AU the payload portion would still be above 74%. Apart from the circular martian <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · May 14, 2007

Very interesting, nice work. Darn it, yes, I do see problems here. The numbers look good as far as they go, but you're not accounting for gravity losses. And you haven't demonstrated that you can get captured in Mars orbit, even a highly elliptical one. Anytime you accelerate in a gravity field, on a practical basis, you have gravity losses. So you can't use Hohmann numbers for ion drive. From earlier in the thread: Earth to Mars-transfer (Hohmann): 2.94 km/s Mars-transfer to Mars: 2.65 km/s So I assume you calculated the dV needed to go from LEO to ESOI in terms of orbital energy - which is the same as going via Hohmann if the impulses are instantaneous - and added that to 2.94 to get 4.0388: about 1.1 km/s, which looks like a good number to my eye. But Hohmann assumes one big short-duration burn at the start and one at the end. You're firing your neat little engine for 3.7 days, accumulating gravity losses the whole time. You're going to need to refer to those charts linked to earlier to account for that. I'm afraid the gravity losses are going to add up to a substantial number. Ditto at Mars, climbing down the gravity well there. You have a great idea here, don't get me wrong. But the dV to get from LEO to ESOI with the sqrt(C3) = 2.94 km/s needed for the interplanetary Hohmann transfer is going to be higher than the 1.1 km/s ideal. What is very clever with your approach is that you still get to coast all the way from Earth to Mars, avoiding the much higher gravity losses of an interplanetary continuous thrust flight. But then again, you're going to have gravity losses at Mars, firing the engines for 4.5 days. In fact, thinking about it, you're potentially going to go flying right past Mars with these puny little engines. 'Potentially', because not only can you aerobrake, but you're coming in with very low excess hyperbolic velocity relative to Mars. The point is that you don't have <div class="Discussion_UserSignature"> </div>

keermalec · May 14, 2007

NOTE: THE CALCULATION RESULTS SHOWN IN THIS POST ARE ERRONEOUS, AS NOTED IN A FOLLOWING POST. OK, according to this reference http://www.rmc.ca/other/usn/ionsupp/ionsupp.htm The delta-v required to change orbit using continuous low-thrust is dv = sqrt(GM/a0) - sqrt(GM/a) where a0 and a are the semi-major axis of departure and arrival orbits and dv is irrespective of acceleration Using this equation we get the following delta-vs Parent body Original orbit (km) Final orbit (km) GM (km3s-2) delta-v (km/s) V at final orbit (km/s) duration (days)* duration (s)* Acceleration (m/s) Earth 6'521.00 922'388.06 398'600.00 7.16 0.66 17.05 1'473'120.00 4.86E-06 Sun 1.51E+08 2.27E+08 1.33E+11 5.50 24.19 258.06 22'296'384.00 2.47E-07 Mars 3'490.00 575'789.31 42'828.00 3.23 0.27 75.20 6'497'280.00 4.97E-07 Total 15.90 km/s 350.31 days * duration calculated for Hohmann transfer Remember using continuous low thrust we match orbital velocities at at the destination orbits, so delta-vs presumably include all necessary gravitational losses Presumably also, the orbital velocities at ESOI and MSOI can be subtracted from the delta-v required to go from ESOI to MSOI, thus reducing the total delta-v to 14.97 and changing our table to Parent body Original orbit (km) Final orbit (km) GM (km3s-2) delta-v (km/s) V at final orbit (km/s) duration (days)* duration (s)* Acceleration (m/s) Earth 6'521.00 922'388.06 398'600.00 7.16 0.66 17.05 1'473'120.00 4.86E-06 Sun 1.51E+08 2.27E+08 1.33E+11 4.57 24.19 258.06 22'296'384.00 2.05E-07 Mars 3'490.00 575'789.31 42'828.00 3.23 0.27 75.20 <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · May 14, 2007

Dang, Keermalec, you're GOOD! I salute you! Those numbers look good. I don't have time this morning to look it over closely but those numbers look very good. It does seem counter-intuitive that this is faster than Hohmann, but I can see how that would be possible. 50% payload! Sweet! IINM, chemical propulsion can do no better than appx 34% I am really looking forward to giving this a very close look. In any case I SO agree with your final paragraph! I've long been a champion of 'high thrust astrodynamics' but at the same time I've never gotten into continuous thrust for exactly the reasons you suppose. But if you have found a way to deliver 50% payload fraction from LEO to LMO, and it looks like you have, I am a convert. Very very cool. <div class="Discussion_UserSignature"> </div>

keermalec · May 14, 2007

Ahem, looks like I goofed... I messed up the acceleration numbers by a factor of 1'000. Divided speed in km/s by time in s to obtain acceleration in m/s... Realised my error as I was delving deeper into the formulas for low thrust transfer, and things were not working out as I expected. Will post new corrected data soon. Sorry for posting the wrong good news. I must now admit continuous low thrust WILL take much longer to get to Mars than Hohmann. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

keermalec · May 14, 2007

And here it is, alas. To get to mars using current NSTAR technology will take 13 years (...) and delivered payload will only account for 18% of total mass. The problem is the mass of the thruster itself, which is very high compared to its thrust. Do note that if we do build large ion-drives their mass per Newton of thrust will probably be much lower than that of the 40 kg NSTAR referenced here, but how much is beyond my knowledge. Once the specs of the NEXT ion-drive (twice the thrust of the NSTAR) are out I can do this calculation again. However, note that even if the mass of the thruster was divided by 5, trip duration would only be reduced by a third... Revised accelerations Parent body Original orbit (km) Final orbit (km) GM (km3s-2) delta-v (km/s) V at final orbit (km/s) duration (days)* duration (s)* Acceleration (m/s) Earth 6'521.00 922'388.06 398'600.00 7.16 0.66 2'348.37 202'899'247.26 3.53E-05 Sun 1.51E+08 2.27E+08 1.33E+11 4.57 24.19 1'500.00 129'600'000.00 3.53E-05 Mars 3'490.00 575'789.31 42'828.00 3.23 0.27 1'059.37 91'529'884.05 3.53E-05 Total 14.97 km/s 4'907.74 days * duration calculated for Hohmann transfer Integrating this into the design of an ion-drive transporter would give: Delta-v: 14.97 km/s Acceleration: 3.53E-05 m/s/s ISP: 3127 s ** Thruster mass: 40.34 kg ** Thrust: 0.0926 N ** Power needed: 2288 W ** Solar panel perf: 90 W/kg at Earth ** Dist from sun: 1.52 AU Fuel: 38.60% Fuel tank: 3.63% Thruster: 15.08% Solar panels: 21.96% Structure: 3.00% Payload: 17.73% ** From Deep Space 1 specs <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

keermalec · May 14, 2007

With a XIPS-25 type ion-drive the journey is brought down to 10 years, with 18% payload, and with a solid-core nuclear engine (in continuous thrust) it can be made as short as you wish (by increasing the size and thrust of the thruster), except payload will not go above 16% of inital mass. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · May 14, 2007

Hang on, all is not lost here, at least in my mind. I wasn't sure if we were on the same page yet, but now I know we weren't quite. You are proposing that we spiral out from LEO to ESOI, and then spiral out to MSOI, and then spiral in to LMO. IOW, if the engines died when we got to ESOI, we would be orbiting the Sun at only slightly further than Earth. We would have C3=0, so we need continuous thrust to get to Mars. I was supposing we spiral out from LEO to ESOI such that we have enough sqrt(C3) to get to Mars without firing engines after passing ESOI. I wasn't at all sure that was possible, and that's what I was going to check out. Thinking about it as I write, I'm pretty sure you can't do it that way. The thing I am now interested in is the possibility that we hit ESOI at 0 < sqrt(C3) < 2.94 and on the way to Mars we coast part time, fire engines part time, and arrive at Mars with no more velocity to kill than aerobraking can provide. It can be assumed that we are talking cargo transport, not crew. Crew needs to go much faster than Hohmann, IMO: that's one of the things I 'preach' about. So if Cargo transfer takes only somewhat longer than Hohmann, but at much higher payload fraction, that's a very attractive proposition. I'm pretty sure you would have to leave earlier, however. Unless you take so long that you leave during one launch period and arrive during the following one, appx 2-1/2 years later. Anybody want to give us $20,000 so we can work this stuff out full time? <img src="/images/icons/laugh.gif" /> <div class="Discussion_UserSignature"> </div>

keermalec · May 14, 2007

I get your drift spacester. If we were to design a ship that would 1. accelerate to ESOI in a spiral (7.16 km/s) plus an additional 2.29 km/s to send it on a Hohmann transfer to MSOI 2. Decelerate at LMO using atmospheric breaking (0.9 km/s) The total necessary delta-v to be induced by the ion-drive would be 9.45 km/s: Delta-v: 9.45 km/s Acceleration: 5.47E-05 m/s/s ISP: 2800 s Thruster mass: 10.4 kg Thrust: 0.063 N Power needed: 1300 W Solar panel perf: 90 W/kg at Earth Dist from sun: 1.52 AU Fuel tank mass: 0.09 x propellant mass Fuel: 29.12% Fuel tank: 2.74% Thruster: 8.86% Solar panels: 28.43% Structure: 3.00% Payload: 27.86% the long part of the trip would be the six years from LEO to ESOI, at 5.5 x 10-5 m/s/s. Then 260 days in Hohman transfer and a few days as the ship falls in mars' gravity well to slow down in its top atmosphere. This solution may be slow but it gets 28% of payload to Mars orbit. An N2O4/MMH ship on a standard 6 km/s Hohmann path would deliver 9% and a LOX/LH2 would deliver 22%. The big difference also being that your ion-drive ship can be used several times. Not so with high impulse thrusters which are usually useable only once. So if there is a way to produce ion-drive fuel on Mars (N2 atmosphere mining?) then the same ship can do the trip there and back, in a 14-year cycle... <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · May 15, 2007

Yeah, this doesn't seem to be working. Very nice effort, though. You've shown a lot of integrity here in questioning and correcting your own work. I still think there are untried approaches, that when taken in the context of an overall exploration / development / settlement / colonization strategy, can do great things. I've got one idea in particular for cargo to Mars, and maybe soon I'll give it the analysis and write-up it needs. N2 at Mars? There is some in the atmosphere (2.7%) and there is ammonia ice (NH3) at the North pole, IIRC. <div class="Discussion_UserSignature"> </div>

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