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It doesn't have to be an exact answer, but something close to reality would be nice.

Specifically I want to know what the travel time to Mars would be using constant firing low-thrust engines.

What would the travel times be using something like 0.0004, 0.0008, 0.0016 or 0.0032 m/s²?

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  • $\begingroup$ you can try to do a little research yourself first. Search for "Low energy transfer" or "Mars ballistic capture" and terns discussed and articles linked in Interplanetary Transport Network and Low energy transfer. $\endgroup$
    – uhoh
    Commented Jun 7, 2016 at 2:06
  • $\begingroup$ 1. I did do a lot of research. 2. NONE of the links or terms you provided deal with TIME to Mars, they all talk about dV. 3. Objects under constant low-thrust to Mars do not follow ballistic trajectories. $\endgroup$
    – Martin Clemens Bloch
    Commented Jun 7, 2016 at 5:35
  • $\begingroup$ So for example, with your lowest acceleration, starting from LEO it will take you roughly six months of spiraling outward just to get high enough where the moon might be of some assistance, before you can even start your journey. You need to add some more information on your specifics. You may find that the interplanetary part is only a fraction of your total mission time from LEarthO to LMarsO. Oh, it's ballistic capture, not trajectory. $\endgroup$
    – uhoh
    Commented Jun 7, 2016 at 10:19
  • $\begingroup$ How do you arrive at that number? With that acceleration you should get 6km/s in 6 months... that seems like a lot. According to my wiki research that is close to the delta v for going to Mars from LEO. I'm not saying you're wrong, but how do you arrive at that number - "6 months"? $\endgroup$
    – Martin Clemens Bloch
    Commented Jun 7, 2016 at 18:34
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    $\begingroup$ @MartinClemensBloch The delta V for ion spirals is a lot more than delta V for Hohmann transfers from LEO. See Mark Adler's answer: space.stackexchange.com/questions/8420/… With no Oberth benefit it takes about 7 km/s to get from LEO to escape velocity. Then it will take another 6 km/s to get from 1 A.U. to a 1.52 heliocentric orbit. Then it would take another 4 km/s to climb down Mars' gravity well to low Mars orbit. $\endgroup$
    – HopDavid
    Commented Jun 8, 2016 at 4:04

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To calculate delta V of low thrust ion spirals, you subtract speed of departure orbit from speed of destination orbit. See Mark Adler's explanation.

Time spent is delta V/acceleration.

LEO is ~7.7 km/s.

At the edge of earth's Hill sphere, escape velocity is about .7 km/s

So 7 km/s to climb out of earth's gravity well from LEO.

Earth heliocentric orbit is about 30 km/s

Mars heliocentric orbit is 24 km/s.

So 6 km/s to get from earth to Mars heliocentric orbits

Escape velocity at the edge of Mars Hill Sphere is .3 km/s

Low Mars Orbit velocity is 3.4 km/s

So about 3 km/s to climb down Mars gravity well.

7 + 6 + 3 is 16 km/s. 16 km/s to get from LEO to LMO via ion engines. In meters, that's about 16,000 meters/sec.

(16,000 m/s) / (.0004 m/s^2) = 40 million seconds = 463 days.

For the other accelerations, the 1 A.U. to 1.52 A.U. heliocentric trip takes less time than Hohmann and delta V will be higher than 6 km/s. I can't give you the times for the other accelerations without investing more time and effort than I can afford at the moment.

As you can see, climbing in and out of planetary gravity wells takes more delta V (and therefore time) than doing the heliocentric transfer orbit. Which is why I advocate berthing a Hermes like craft at EML2 between trips. At the Mars end of the trip, Deimos might be good place to berth an ion propelled craft.

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  • $\begingroup$ Thank you for your answer, I added a bit to your delta Vs and then extrapolated to the other accelerations. Its not correct I guess, but now I should be in the right ball park. $\endgroup$
    – Martin Clemens Bloch
    Commented Jun 9, 2016 at 20:41
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Maybe you watched the movie "The Martian"? Well, most of that was pretty accurate (most egregious silliness was the duct tape and tarp hole repair, but I digress). The point mentioning that is that they were using a lot of computer time to come up with orbital solutions, and IIRC those were ballistic. (Ballistics is much much much easier than non-ballistic dynamics.) So, you are asking a really difficult question. What you want to do is to arrive at Mars' orbit exactly when Mars is there and with exactly zero velocity, right? Can you see how that depends on when you start? (Earth is orbiting the Sun, so when you start will define what time you need to get to that zero velocity just when Mars passes by. You can't just pick "any old time" of the year because depending on what year, Mars could be anywhere in its orbit. It would be a shame if you missed Mars by a day, and had to wait another whole Martian year — 686 days.)

OK now that I've quashed your dreams of an easy answer, have you considered Orbiter, a (free) software simulator? It's not a game, but you can pilot your own spaceship to Mars.

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    $\begingroup$ It's a good point - it's possible the question should include "...assuming this million dollar interplanetary mission isn't profoundly silly and starts at a bad time." Every time I see a link to Orbiter I get excited until I go there and am reminded that "Orbiter is a free and realistic space flight simulation program for the Windows PC.' I haven't even heard the term the Windows PC in a decade. $\endgroup$
    – uhoh
    Commented Jun 9, 2016 at 3:55
  • $\begingroup$ Tyson's trailer for The Martian describes a 124 day trajectory from low earth orbit to Mars orbit: youtube.com/… . Which is flat out impossible if Hermes' acceleration is 2mm/s^2. It would take more than 40 days to spiral from LEO to earth C3 = 0 $\endgroup$
    – HopDavid
    Commented Jun 9, 2016 at 19:40
  • $\begingroup$ Thank you for your answer. I totally appreciate the difficulty, that's why I had to ask in the first place :) $\endgroup$
    – Martin Clemens Bloch
    Commented Jun 9, 2016 at 20:39

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