Mars transfer window frequency

Question

According to this website I drew up a timeline of Mars transfer windows and the time you'd need to stay on Mars before the next return window:

Blue to red arrows are Earth to mars transitions, red to blue the opposite. Yellow boxes are amount of time on Mars

This seems to indicate that a mars trip would last around 2.5 years in total with 0.7 years transfer both ways and a little over a year on the surface. This seems somewhat short however. In popular science news they often talk about how "Mars [stays] are either short term (ten days or so on the planet) or long term (two years on the planet)" however this doesen't line up with my graph. Where is the discrepancy?

Answer 1

The short term Mars missions do a non-standard path to get back. I believe many of them actually use a Venus flyby to get back to Earth, which you can see a bit on this Wikipedia article. In any case, one can return more quickly than waiting for an optimal transfer window, but at a much higher cost, even if there isn't a Venus flyby in the process. The mission generally is called the "Opposition" mission.

From the Case for Mars, which briefly discusses the plan, here are some dates. Both missions take about 6 months to arrive. The Conjunction mission has 550 days on Mars, the Opposition mission has 30 days. The Conjunction mission takes about the same time to return to Earth, 6 months, while the Opposition takes about 14 months return time. It also requires about 2 km/s more delta v to return.

The "Conjunction" plan has been presented in many mediums as the preferred plan to get to Mars, due to its shorter mission duration. I believe the article you included is just flat wrong. The Case for Mars cites a 550 day on Mars mission being the standard. It seems you are showing there is about a 450 day window.

The journey probably won't occur at the exact optimum for a couple of reasons. One can do a faster mission to Mars by offsetting the mission time somewhat for only a small extra fuel cost. There are a number of "Free Return Trajectories" where one can return to Earth at little to no cost. The optimal of these will do the transfer orbit at a 2 year orbital period, which will cause it to return to Earth in 2 years if something fails. The time to get to Mars for such an orbit is 180 days. I believe these will get to Mars about a month earlier than the full Hohmann transfer orbit. Of course, the time on Mars varies depending on the exact orbital mission.

The other primary reason for not leaving at the optimal trajectory is you get there sooner than an optimal trajectory would do so. One could therefore land an unmanned spacecraft at the landing site if something should require that it be landed there, or could land it at the site of the next landing if that turns out to not be required.

Bottom line, the prevalence of the "Opposition" class mission in Mars planning probably lead to the error in the article you linked to. An opposition mission could be done, and would be almost a year shorter, but would spend most of that time in transit through the solar system, and less of it doing actual science.

Answer 2

It turns out you have a lot more flexibility in terms of transit times and surface stays than you'd think. I really like this plot by Casey Handmer.

Above the X-axis, it shows the transfer windows from Earth to Mars. Below, it shows the windows from Mars to Earth. By choosing a transfer option, you can draw a diagonal. The point where it hits the x-axis is the date on which you arrive. From there you can choose a new transfer option for the return. The red line is Casey's choices for very short stays.

You could reduce the transit times by picking points that are more in the red zones, though of course that takes more delta V. You could get a longer surface stay, such as half a year, by leaving from Earth earlier and/or from Mars later, at the cost of a longer Earth-to-Mars transit time and more delta V.

With the upcoming Starship, higher delta V options are becoming an option. It's more expensive, but something like sending extra tankers along to Mars might be worth it if it makes the trip better for humans. I think the articles you're referring to are based on minimizing the delta V, with a trajectory roughly like what Casey sketched in his plot.

Note: This graph is for free return trajectories, but only as a precaution; the delta V and time calculation is still including a landing and take-off.

Answer 3

You both are really right. A nominal Hohmann transfer takes 257 days (0.7 years). It is the least possible energy. However, Zubrin in the book cited recommends a 180 day transfer, which uses a little more propellent but offers a free return just in case something goes wrong. However, the free return means an abort and no Mars landing. If you do the shorter trip at a slightly higher energy cost the stay is 550 days to nominal return. The difference is you chopped 0.4 years off of the travel time and spend it at Mars instead. Therefore 550 days (1.5 years) instead of about 1.1 years on Mars, which is about what you were quoting.

(Free return means you just go right past Mars and your path coincides with where Earth will be a couple of years later. That is basically what Apollo 13 did to get back from their moon trajectory).