Both Mars One and NASA have suggested sending uncrewed supply ships to Mars as part of a manned mission or colonization. These payloads can take all the time they like, not being constrained by human radiation exposure time limits. Instead, what you want to minimize is the total $\Delta v$ needed to reach Mars.

I can think of two possibilities that have already been used:
1) MOM-style multiple burns at perigee, culminating in trans-Martian injection.
2) Lunar slingshots, perhaps in multiples.

It's possible these two approaches can be combined. There might also be other possibilities, such as starting with a low-energy transfer to EML-2, which is apparently the lowest $\Delta v$ destination in local space.

How do you minimize $\Delta v$ to get to Mars?

EDIT: As Mark Adler points out, it may very well be more efficient to use electric propulsion, which would require higher $\Delta v$ but would result in lower launch mass. However, to avoid making this question too broad, let's assume you're using a high-thrust motor. Perhaps later I'll figure out how to properly ask the underlying question, "What's the cheapest way to send 1000 kg to Mars?"

  • $\begingroup$ MOM style phasing maneuvers do not decrease Δv required, but are done due to limitations of the engines to give all the Δv required in a finite burn. The total Δv added up in all subsequent burns will add up ti same amount. Or is it not? $\endgroup$ Dec 27, 2014 at 18:10
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    $\begingroup$ @KuldeepBarad the multiple burns all take place at perigee, the fastest part of the orbit, and thus take advantage of the Oberth effect $\endgroup$ Dec 27, 2014 at 18:16
  • $\begingroup$ This perspective of phasing maneuvers and decrement in total Δv due to subsequent burns at higher speed than single one was totally out of my perspective even after having known the Oberth effect. Thanks! $\endgroup$ Dec 27, 2014 at 18:25

4 Answers 4


How do you minimize Δv to get to Mars?

The answer is simple: Wait until 2018 or 2035. Those are the local minima in the Δv needed to get to Mars. The required minimum Δv varies by a large amount. There's a local minimum roughly every two years where transfers to Mars become feasible, but even this quantity varies considerably. There's a ~15 year variation on top of this ~2 year cycle.

Getting to Mars can cost a lot if you are extremely persistent about getting to Mars right now. The Δv goes down drastically if you are willing to wait a couple of years. The Δv drops by another factor of two on top of that already drastic decline if you are willing to wait a couple of decades.

  • $\begingroup$ I didn't realize that 2-year cycle Hohmann transfer minima varied by that much. I assume that any more exotic trajectory would also somehow be pegged to that cycle. $\endgroup$ Dec 12, 2014 at 22:42
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    $\begingroup$ @JerardPuckett - The problem with those exotic trajectories is that this requires the spacecraft itself applying the necessary Δv. With a "standard" trajectory, the departure Δv is provided by the launch vehicle and its upper stage, and the arrival Δv is provided mostly by Mars' atmosphere. The spacecraft itself needs very little Δv capability -- but only if the spacecraft doesn't have to follow some exotic trajectory. $\endgroup$ Dec 13, 2014 at 0:38
  • $\begingroup$ This energy minima is related to 'Opposition' of Mars with Earth, i suppose, which occurs about every 780 days. $\endgroup$ Dec 27, 2014 at 18:28
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    $\begingroup$ It might be useful to add a link to NASA's Trajectory Browser. For example, this query shows your mentioned dates quite well IMHO, or you could select to include all trajectories and then the graph becomes really funky in colors and oscillation :) $\endgroup$
    – TildalWave
    Dec 28, 2014 at 0:56

I don't think that's really the question you want to ask. The lowest energy trajectories are ones like you suggest, which achieve most of the energy to get to Mars with short impulsive burns very close to Earth. That requires chemical or nuclear thermal propulsion systems that can expend much of the propellant over a very short time.

What you want for Mars cargo are low-thrust trajectories that allow you to use much more efficient electric propulsion systems. Those take more energy and $\Delta V$ overall than those that can use the Oberth effect, but despite that require lower launch mass due to the very high $I_{sp}$.

  • $\begingroup$ Yes, the low thrust vs. high thrust dilemma was in the back of my mind while I was formulating the question, but I didn't want to make the question too broad. Ultimately, what you're trying to minimize is really the cost-per-kg of getting payload to Mars. $\endgroup$ Dec 12, 2014 at 17:29
  • $\begingroup$ Though this does not now answer the edited question, I will let it be. See David Hammen's answer. $\endgroup$
    – Mark Adler
    Dec 12, 2014 at 22:19

According to this recent Scientific American article, there is a cheaper way to get to Mars than the traditional Hohmann transfer. The mission design relies on a ballistic capture to do away with the Mars orbital insertion burn.

The premise of a ballistic capture: Instead of shooting for the location Mars will be in its orbit where the spacecraft will meet it, as is conventionally done with Hohmann transfers, a spacecraft is casually lobbed into a Mars-like orbit so that it flies ahead of the planet. Although launch and cruise costs remain the same, the big burn to slow down and hit the Martian bull's-eye—as in the Hohmann scenario—is done away with. For ballistic capture, the spacecraft cruises a bit slower than Mars itself as the planet runs its orbital lap around the sun. Mars eventually creeps up on the spacecraft, gravitationally snagging it into a planetary orbit.

enter image description here
Structure of the ballistic capture transfers to Mars.

As this adds at least several months to total travel time, probably not the best bet for manned missions, but for pre-positioning your 100 metric tons of water & construction equipment in Mars orbit, it should do nicely.

I have a couple of caveats about the Scientific American article. The author says that "ballistic capture" is synonymous with "low energy transfer," whereas AFAIK the former is a merely subset of the later. Also, he seems to imply that this approach does away altogether with the several-days-every-two-years launch window, but the paper actually only states that the launch window is more flexible.

The underlying paper is Edward Belbruno & Francesco Topputo, Earth–Mars Transfers with Ballistic Capture (2014)

Further reading:
Walter Hohmann, The Attainability of Heavenly Bodies (Washington: NASA Technical Translation F-44, 1960) Koon, Lo, Marsden & Ross, Dynamical Systems, the Three-Body Problem and Space Mission Design (2006)

  • $\begingroup$ Considering Earth-moon external transfer time around 3-6 months in low energy transfer, such mars trajectories would more than just months, i suppose. $\endgroup$ Dec 27, 2014 at 18:15
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    $\begingroup$ @KuldeepBarad there may very well be much longer, lower Δv solutions, but this one is only about orbital insertion. Trans-Martian injection & cruise phases are not addressed. $\endgroup$ Dec 27, 2014 at 18:31
  • $\begingroup$ You know, I had this open since xmas, but didn't get around to it. :) $\endgroup$
    – TildalWave
    Dec 27, 2014 at 19:02
  • $\begingroup$ @TildalWave great, an online link to a picture :) $\endgroup$ Dec 27, 2014 at 19:21
  • $\begingroup$ @JerardPuckett That's the spirit! Happy "Bill Lumbergh" hat hunting! ;) $\endgroup$
    – TildalWave
    Dec 27, 2014 at 19:24

In addition to possibilities outlined in your post, you can also take advantage of the fact that (gross simplification ahead) orbits around Lagrange points are possible, but very unstable, letting you "choose the instability you want" and leave the Lagrange point in almost any direction, without much fuel needed. The downside is that those trajectories can take very long. See wiki: Interplanetary Transport Netowrk and Low energy transfer for further details.

  • $\begingroup$ The EML points may very well be useful to achieve trans-Martian injection. What about SEL and SMarsL points? Can they be used to decrease the $\Delta v$ budget required by a simple Hohmann transfer? $\endgroup$ Dec 12, 2014 at 22:47
  • $\begingroup$ According to the wiki articles mentioned in the reply, they could be. But I do not know the actual numbers. $\endgroup$ Dec 12, 2014 at 23:22

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