Exactly as stated in the title. I guess it is not a trivial subject but I believe a rough algorithm or process description may be provided.

  • $\begingroup$ From Earth to the next planet (Venus or Mars), it's rather trivial and boring. $\endgroup$
    – ott--
    Commented Mar 19, 2016 at 20:19
  • $\begingroup$ I wouldn't call it "trivial and boring", Earth moves around the sun at 30 km/s and mars on average, 24 km/s. In a sense, the craft needs to slow down, but Earth's escape velocity will help with that. Generally speaking, the path to Mars is a spiral of sorts or "U shape", and there are optimal launch times every couple years or so. The calculations from earth towards mars are pretty complicated, but I'm not expert in this enough to explain it well. $\endgroup$
    – userLTK
    Commented Mar 19, 2016 at 21:59

2 Answers 2


To define "optimal" you need an objective function that you are maximizing or minimizing. What is your objective function?

For real Mars missions, the objective function can be quite involved, since many factors are considered. Let's assume an impulsive trajectory (i.e. very close to on target to Mars immediately after launch). Then there is the mass that can be delivered to the target, which is a function of the launch energy and departure declination. There is the arrival velocity, which determines the orbit insertion propellent for an orbiter, or the heat shield capability and aspects of the entry trajectory for a lander. There is the approach declination, which determines what orbits you can get into with one burn, or what landing site latitudes you can access. There is the visibility of the insertion burn or entry from Earth for telecommunications during critical events, so that you have data if something goes wrong. There is Mars relay orbiter coverage constraints for the entry and landing event, again to get more data in case something goes wrong. You will need to define a few weeks of available launch days (usually three weeks) to allow for weather, range, launch vehicle, or spacecraft delays. Over that launch period, you will need to satisfy all the other constraints. You may want to have the arrival day be the same for all launch days, in order to simplify planning. You may want to not have that arrival day be on Super Bowl Sunday, so as to get better press coverage and to not annoy the crew. (I seriously took that into account once.)

I could go on. Does this answer your question?


To address the comment on optimizing for a specific parameter, e.g., $C_3$, the process for an impulsive trajectory is to make a porkchop plot. Like this:

porkchop plot

For a given departure date from Earth and arrival date at Mars, there is one short, prograde orbit that connects them (see Lambert's problem). You then make a contour plot over a range of departures and arrivals of the parameters of interest. In the plot above, the blue contours are the injection energy. You can see two local minima, both around a $C_3$ of 16, for the 2005 opportunity. Though as noted, the $C_3$ doesn't tell the whole story. The departure declination, if greater than the launch site latitude, will reduce the injected mass at the same $C_3$.

  • $\begingroup$ You missed one point: How fast do you need to get it there? If need to get it there with minimal flight time, you presumably want to use the Hohmann transfer window, which only happens every couple of years. If you do not need to minimize flight time, or you need to use less fuel/delta-v for some reason, you could use a low energy transfer (aka weak stability transfer) instead. If you needed to get to Mars 'as fast as possible', which one you use is going to depend on how the planets are positioned. $\endgroup$
    – Kaine
    Commented Mar 22, 2016 at 1:08
  • $\begingroup$ As I said, I could go on. $\endgroup$
    – Mark Adler
    Commented Mar 22, 2016 at 2:48
  • $\begingroup$ Well yes, I should have defined the 'optimal'. I wanted to miminize fuel - energy needed. $\endgroup$
    – freefall
    Commented Mar 22, 2016 at 12:04
  • 2
    $\begingroup$ Those are two different things, but anyway whatever it is about the trajectory that you want to optimize, what you do is solve Lambert's problem for a matrix of departure and arrival dates and make a contour plot of the objective function. Then find the extremum in the contour plot. $\endgroup$
    – Mark Adler
    Commented Mar 22, 2016 at 16:21

Let me add one thing both Mark and Kaine omitted in their answers.

With "optimal" defined as "minimizing fuel" you use engines of extreme Specific Impulse, and these tend to have minimal thrust. They can't just perform a simple "transfer burn" - a couple of minutes of acceleration. They need to accelerate for hours or even months. So, instead of just entering the transfer orbit they need to spiral out of Earth's gravity well.

AFAIK there are no simple porkchop plots for these transfers - they are calculated individually, through numerical optimization of trajectories created through computer simulation.

  • $\begingroup$ AFAIK, no Mars missions to date have used anything other than chemical rockets for transfer. $\endgroup$ Commented Mar 24, 2016 at 13:53
  • $\begingroup$ @RussellBorogove: We're sending increasingly heavy payloads to Mars, and in case of these the acceleration times on ion engines become quite excessive. Ion engines are a fairly new invention, so simply by the time they entered the equation, the time-fuel-payload trade-off for Mars became too unfavorable for that use. Plus we need to remember that each mission bears a per-day cost (of maintaining the ground control) on top of per-launch cost, so for excessively long missions the upkeep costs may outrun the launch costs. (looking at you, Opportunity!) $\endgroup$
    – SF.
    Commented Mar 24, 2016 at 14:09
  • 1
    $\begingroup$ @RussellBorogove The ISRO's Mangalyaan (M.O.M.) probe used exactly that type of maneuver to get to Mars. I don't know what you call it other than a "Mangalyaan maneuver", which does have a nice ring to it. AFAIK you calculate your window same as any other, setting your 'departure' to same day as your last burn needed to break Earth's escape velocity, meaning your launch has to happen weeks/months in advance to allow time to perform all the orbit-raising maneuvers prior to the normal "departure window". $\endgroup$
    – IT Bear
    Commented Dec 5, 2016 at 21:14
  • $\begingroup$ It's looking like there will be an electric propulsion transfer to Mars coming up, by ESA's Earth Return Orbiter for Mars Sample Return. $\endgroup$
    – Mark Adler
    Commented Jun 2, 2021 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.