I realize that there are different orbital maneuvers for different goals (e.g. optimizing time, efficiency, cost, etc.). However, which variables are currently considered as the most important to be optimized? This leads to the question, which orbital maneuver is currently regarded as the most viable for a human mission to mars?

I'm aware that there is no best way to get there, just the most accepted way. Each maneuver will have its pros and cons. I'm just asking what is the current consensus on which orbital maneuver to perform?

EDIT: In order to clarify entirely what I mean, I am only concerned with transfer orbits such as Hohmann, bi-elliptical, etc. I am doing a school project that requires me to calculate the math behind an orbital transfer to mars for a human mission. It is nothing TOO technical but I still want it to be correct nonetheless, and I was curious to know which orbital transfer I should choose for the calculations.

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    $\begingroup$ I think closing this as "too broad" was a bit premature. The OP was asking about introductory college-level transfer orbits such as Hohmann, bi-elliptical, etc. The answer is "none of the above". That is, unless the OP has learned about Lambert's problem and porkchop plots. For the purposes of this question, that would have been more than enough. $\endgroup$ Mar 30, 2016 at 1:59
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    $\begingroup$ Related: When sending a probe to Mars, how is the optimal travel path calculated? on this site, and Why is there a gap in porkchop plots? at the physics.SE sister site. $\endgroup$ Mar 30, 2016 at 2:05
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    $\begingroup$ Doesn't this depend on other aspects of the mission? If i was on the ship, i'd want to get to Mars as fast as possible, to minimize my radiation dose. If i was sending the ship, i would cringe at the thought of the price tag for that extra speed. Also, what about abort options? $\endgroup$
    – kim holder
    Mar 30, 2016 at 2:52
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    $\begingroup$ Likely the trajectory "there" will be quite similar to Hohmann transfer, with some extra delta-V expended skewing it towards a shorter trip. The trip back would be either Hohmann-like (with similar skew) or Free Return, depending on both how long we'll be willing to let people stay in space and how much budget we get to allocate towards the trip. $\endgroup$
    – SF.
    Mar 30, 2016 at 14:36
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    $\begingroup$ @kimholder: The basic abort option after the transfer burn is not to land on Mars, just linger in the orbit for the transfer window back and use the fuel allocated for launch towards the return trip. $\endgroup$
    – SF.
    Mar 30, 2016 at 14:40

4 Answers 4


In order to clarify entirely what I mean, I am only concerned with transfer orbits such as Hohmann, bi-elliptical, etc.

It's none of those. Both are for transferring from one circular orbit to another, and with both orbits on the same orbital plane. Neither Earth's nor Mars's orbit is circular. Mars's orbit is notably non-circular, with an eccentricity of 0.0934. Mars's orbit is inclined with respect to Earth's orbit by 1.85 degrees. This is more than enough to make a Hohmann transfer a useless fiction. (Aside: Even if the orbits were circular and on the same plane, a bi-elliptical transfer wouldn't make sense for transfers from Earth to Mars. The orbits are too close to one another.)

What is done instead is to solve Lambert's problem, over and over and over. Given a central mass, a pair of distinct points in space $\vec r_1$ and $\vec r_2$ relative to that central mass, and a pair of points in time $t_1$ and $t_2>t_1$, Lambert's problem involves solving for the conic section that starts at $vec r_1$ at time $t_1$ and intersects $\vec r_2$ at time $t_2$.

Except for diametrically opposed points and radially aligned points, there are two such solutions to Lambert's problem. One, "the short way" (or sometimes, a Type 1 transfer), involves a change in true anomaly that is less than 180 degrees. The other solution, "the long way" (or sometimes, a Type 2 transfer), involves a change in true anomaly that is more than 180 degrees. Note: Some of these solutions might involve going faster than the speed of light. That's not a problem in Newtonian mechanics. It is a problem with regard to delta-V. Regarding those diametrically opposed points and radially aligned points: These are problematic with regard to Lambert's problem. The solutions are singular. The general approach is to ignore those points.

Suppose the point $\vec r_1$ represents a point on some initial orbit at time $t_1$, and point $\vec r_2$ represents a point on some target orbit at time $t_2$. With this, one can calculate the delta V needed at time $t_1$ to transfer from the initial orbit to one of those Lambert trajectories and the delta V needed at time $t_2$ to transfer from that Lambert trajectory to the target orbit. This is the delta-V cost for that particular trajectory. One of the two solutions will have a higher cost (typically much higher) than the other. We'll discard the high cost solution.

Now do this over and over and over again, but with different departure times $t_1$ and arrival times $t_2$. Generate a contour plot with departure dates and on one axis and arrival dates on the other, and eventually you'll come up with something like this, from
"Porkchop" is the First Menu Item on a Trip to Mars
: Porkchop plot for transfer from Earth to Mars in 2005

This is a porkchop plot. The blue contour lines show the energy (here in terms of C3) needed to transfer from Earth to Mars in 2005. (Note: Since we're interested in getting from Earth to Mars (or back), it makes more sense to use C3 rather than delta V.) Note that the transfer orbits are markedly non-Hohmann. A Hohmann transfer would take about 275 days. The optimal transfers instead take about 200 days (short way) and 400 days (long way).

I am doing a school project that requires me to calculate the math behind an orbital transfer to mars for a human mission.

You didn't mention what level of school this project is for. If the above is over your head, you can ignore all that and use the fiction of Hohmann transfers. If it's not over your head, this is exactly what is done to calculate transfer orbits to Mars at JPL.

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    $\begingroup$ No slight intended by my "over your head" remark. $\endgroup$ Apr 4, 2016 at 19:11
  • $\begingroup$ Is using only one transfer orbit always optimal (let's say in terms of delta-v)? $\endgroup$
    – JiK
    Oct 15, 2017 at 18:58

The best trajectory depends on many factors, most importantly things like: when you want to get there, how long you want to stay, and whether you want to optimize for time or for ∆v expenditure.

Optimizing for time reduces radiation exposure and consumables requirements for manned missions; optimizing for ∆v gives you more payload for a given size of launcher.

NASA has a sweet trajectory browser that lets you explore your options with given constraints.


The best from my understanding is a 2 year free return trajectory, with launch windows fairly close to the time of a Hohmann transfer. It involves a departure velocity of 5.08 km/s, 2 years to return to Earth, and 180 days to transit to Mars. Source is "The Case for Mars".

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    $\begingroup$ A two year free return trajectory is not anything close to Hohmann. A Hohmann transfer to and from Mars (which do not exist) would take about nine months to get there and nine months to return. $\endgroup$ Mar 30, 2016 at 2:03
  • $\begingroup$ Fair enough, I only meant that due to the alignment. Edited my answer accordingly. $\endgroup$
    – PearsonArtPhoto
    Mar 30, 2016 at 2:33
  • $\begingroup$ Why doesn't a Hohmann transfer to/from Mars exist? $\endgroup$ Mar 30, 2016 at 16:39
  • $\begingroup$ Hohmann does exist, just not a Hohmann with free return... $\endgroup$
    – PearsonArtPhoto
    Mar 30, 2016 at 17:03
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    $\begingroup$ A Hohmann transfer does not exist from Earth to Mars (or back again). A Hohmann transfer is between two circular orbits on the same orbital plane. Neither Earth's nor Mars's orbit is circular, and the orbital planes are inclined with respect to one another. $\endgroup$ Mar 30, 2016 at 19:51

enter image description here

There are good answers here already, so I am just adding context / information about porkchop plots and the difference between the "short way" and the "long way" when it comes to Lambert's problem.

The plot above is of the 2020 launch window from Earth to Mars. Note that this is a delta V plot, not a C3 and V infinity plot, which splits the 2 burns in a Lambert's transfer (the initial burn to leave Earth and the final burn to arrive at Mars).

Some context for the porkchop plot: the x-axis represents Earth departure times, starting from some epoch (in this case 2020 July 1), and the y-axis represents Mars arrival times starting from some epoch (in this case 2020 November 1). The pink contours on the plot show trajectories with equal delta V. The blue contours show travel times (in days) for trajectories along each line. We can then take a look at what the Mars 2020 spacecraft used.

Mars 2020 departed from Earth 2020 July 30, which on this plot would be at x = 30. It arrived at Jezero Crater on Mars on 2021 February 18, which is y = 110 in this plot. This point is within the delta V = 6 km/s contour, which is the lowest on this whole plot. That is why porkchop plots are a good tool for figuring out first order estimations for interplanetary missions (in this case we are ignoring flybys).

Some context on the short way and the long way. As was explained in another reply, the short way is when the change in true anomaly of the trajectory is less than 180 degrees. On a porkchop plot, these trajectories are the ones that are under the gap in the plot. Here is a 3D plot of what that would look like (this is roughly the trajectory that Mars 2020 flew): enter image description here

The long way is a trajectory where the change in true anomaly changes by greater than 180 degrees. These trajectories would be above the gap in the porkchop plot, and they look something like this: enter image description here

In practice, it wouldn't really make sense to use the long way if the delta V would be the same for the short way, since it adds more travel time. Unless maybe there was an asteroid or something of interest along the trajectory that would make it a good idea.


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