Space Exploration Stack Exchange is a question and answer site for spacecraft operators, scientists, engineers, and enthusiasts. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Just to clarify... I've made some research and generally know what the case is, didn't mean to make the question sound stupid... :)

I know there is a thing called Hohmann transfer orbit, named after Walter Hohmann, who came up with the idea of it, and each and every spacecraft we've sent to Mars so far, has used this very method to reach Mars. On one side, the Hohmann transfer orbit is quite fuel efficient, compared to other methods, and for a unmanned probe, it wouldn't matter if the voyage to Mars is 7 months.

However, we are all aware of the hazardous environment during the trip, if you're a human. So a round-trip of 5 months is better than year and a half.

I know that in order to travel to Mars in a straight line we need to achieve a direct Earth-solar escape velocity... The probe New Horizons already did that.

If we launch in a straight line, and we've calculated the position of Mars in the time frame of arrival, we'll rendezvois with Mars.

However, we wouldn't be able to get into orbit around Mars in order to land, because our speed relative to Mars would be too high. So we need to do a deceleration fuel burn, to reduce speed.

The two combined (direct Earth-solar escape velocity and Mars orbit deceleration fuel burn) would require very large amounts of fuel, especially for a manned mission, which is heavy to lift. But it is achievable either with a Mars manned spacecraft assembled in LEO (much like the ISS), or with some other means.

So why not travel to Mars in a straight line, or am I missing something?

UPDATE (as to further clarify the question):

If the only obstacle in achieving this would be large amounts of fuel required, how much would it be, and what techniques we may apply in particular so that we minimize the amount of fuel required? What about the assembly of a Mars manned vehicle in LEO, how would that change things?

share|improve this question
I think you already have listed the reasons why we can't do that. Consider that for a Hohmann transfer to Mars, you need already 2000 times the mass of the payload as fuel, which is already quite expensive. For a "straight" orbit, the factor would be much higher. See… – oefe Dec 29 '13 at 15:06
oefe, fuel is not as expensive as you imply it to be, since it is only a small fraction of the cost of the vehicle. I heard Elon Musk stating that the fuel for Falcon 9 launch costs $200'000, which would make for 3 percent of the cost of the vehicle. I updated the question so it be more specific... – Martin Asenov Dec 29 '13 at 17:36
It's not the cost of the fuel alone. More fuel also means larger launch vehicles, which means more powerful engines, etc., so the cost of the entire project explodes. Similar with you proposal of in-orbit assembly, that would also drive the cost up. – oefe Dec 29 '13 at 17:45
It isn't the cost of the fuel, it's the cost of the fuel-tank that makes space-travel so hard. – john3103 Dec 31 '13 at 15:49
@oefe — A "straight" orbit ? – Nicolas Barbulesco Jun 7 '14 at 11:02
up vote 19 down vote accepted

The answer to your title question "Why not travel to Mars in 2 months?" has already been answered. Money. Lots and lots of money.

We should first note that the answer to a different question: "Why haven't people gone to Mars?" is much the same. Money. However the amount of money to go there in two months is many times greater than the money to go there in eight months. Or even six months. To go there in two months, you are riding up the very steep part of the exponential rocket equation.

Assembling the vehicle in LEO does not substantially change the total mass required in LEO. All that does is allow you to use more smaller launch vehicles instead of fewer large launch vehicles. (I say "all", but in fact it is a big cost improvement to not have to develop new giant launch vehicles.)

Keep in mind that the "only" in your "If the only obstacle in achieving this would be large amounts of fuel required" hides the important fact that you have to get the fuel to where it needs to be used. That's where all the money is spent.

To answer the different question in the body on how much more fuel and how to minimize it, see this paper for some investigations of fast transfers and minimizing their costs. In that paper, you can see for just the $\Delta V$ to leave Earth orbit and leave Mars orbit (so not including landing and taking off from Mars, or Earth for that matter), we go from $5.6\,\mathrm{{km}/s}$ to $28.6\,\mathrm{{km}/s}$! And that only goes down to about a 2.5 month trip time, not your two months.

Note that the velocity is in the exponent of the rocket equation. You are not even multiplying here, but raising the exponential term to the fifth power in order to get closer to your two months. What that exponential term was in the first place depends on the specific impulse of the rockets you're using. For typical chemical propulsion, the term was about 3.5 for a Hohmann transfer. It goes up to about 580 for a 2.5 month transfer! About 170 times the mass.

That means about 170 times the cost. There would be learning curve savings in the mass production of that many, that large launch vehicles, but we're still talking around one hundred times the cost. Even if I assume nuclear thermal rockets with a specific impulse of $1000\,\mathrm{s}$, then to get 2.5 month transfers would cost 10 times as much. Even if I imagine trying to mine Phobos for return fuel, we're still talking many times the cost. And this all glosses over the mass of the propulsion systems -- I assumed zero mass for them, where taking them into account, with staging, would reduce the effective specific impulse. I have also glossed over the cost of arriving at both bodies at much higher velocities, requiring much more massive entry systems or orbit insertion systems.

There are no problems with the longer transfer time that can't be solved with much, much less mass. E.g. more supplies, more radiation shielding, centrifuge for gravity, etc. So bottom line, it would be many times the cost to go there significantly faster, with no apparent benefit other than reduced boredom.

Considering that we haven't sent anyone at even one times the cost, the answer to your question should now be crystal clear.

share|improve this answer
I believe you answered my question mentioning the specifics I needed. Thanks. The paper you referenced is a great read, I will definitely take a look on it. – Martin Asenov Dec 29 '13 at 19:13
Basically, I wanted to know if it's technically possible, because everything I could find on the Internet never mentioned if it could, it just wasn't getting that far. Instead what followed was "you have to use Hohmann transfer orbit", without saying why. But I wanted to know why – Martin Asenov Dec 29 '13 at 19:16
@Mark Adler The link in the question is now broken. Do you have an updated link? – Vedant Chandra Jan 30 at 7:22
Updated link... – Mark Adler Jan 30 at 7:33 Is closely related to the paper linked. It may be interesting as well. – Hohmannfan Jan 30 at 14:43

In another question - Proportions of a space-craft as compared to fuel - I explored the Tsiolkovsky rocket equation and the tyranny thereof. In the related NASA link, Don Pettit explains that, even using the Hohmann transfer orbit between Earth and Mars which is the most fuel-efficient, it takes about as much change in velocity to break LEO and head out toward Mars as it does to go from the launchpad to LEO. Where does that energy come from? Well, it comes from shoving superhot gases out your tailpipe.

So, given a spaceship roughly the mass of the Space Shuttle orbiter (just as an example; the Mars vehicle may be larger, maybe smaller), which is 110 tonnes plus up to 24 tonnes cargo, to get that into LEO we normally burn about 725 tonnes of LH2/LOX fuel through the main engines, plus about a thousand tonnes of perchlorate fuel in the SRBs. Gross liftoff weight of the Shuttle is 2 thousand tonnes, of which the orbiter vehicle and cargo is a maximum of 6.7% of gross liftoff mass. There's some mass inherent in the support structure of the tank and SRBs than is accounted for here, but not nearly as much as you'd think; the SLWT external tank, for instance, has an empty mass of only 3.7% of its loaded mass.

Once the Shuttle's in LEO, that's typically it; it carries a small amount of fuel for its deorbit burn and for maneuvering, but the main engines are cold for the entire remainder of the trip. However, we're now talking about getting that mass out of LEO and on a transfer orbit to Mars. That actually requires about the same delta-V as getting to LEO and therefore about the same amount of fuel (we have to burn a little more fuel to get out of the atmosphere because of drag; the Tsiolkovsky equation assumes an "ideal" - drag-less - rocket). So, take that 135 tonnes, strap another 1725 tonnes of fuel and maybe 140 tonnes of support structure on it, and blast off again from LEO to parts unknown.

... But wait, where'd that fuel come from? We have to get the fuel out of Earth's gravity well. That requires us to lift 1725 tonnes of payload into LEO on subsequent launches in order to "refuel" our orbiting craft. Assuming we can use a Shuttle-derived vehicle, like the Space Launch System, to do that with a similar per-launch total payload as launching the Shuttle (it will eventually be possible; SLS Block II is spec'ed to carry 130 tonnes of payload to LEO and we can probably improve on that), and also assuming the spacecraft we launch can keep its fuel tank instead of jettisoning it as the Shuttle does, it would take about 13 launches (those launches consisting of basically a big fuel tank on top of a bigger fuel tank) to get enough fuel into orbit to send our craft on its way to Mars. Each of those launches would burn 1725 tonnes of fuel to get 135 tonnes into orbit, for a total fuel cost of 22,425 tonnes.

... But wait, we want to be able to get back from Mars. Well, that requires a similar delta-V as getting there in the first place. So, we need 1725 tonnes of fuel to get our 135-tonne spacecraft out of Mars orbit and back down to Earth. How does that fuel get to Mars? It rides with the spacecraft. And that means that we need more than 1725 tonnes of fuel to break Earth orbit. In fact, it needs the same 22,425 tonnes that we calculated we'd need in order to get the 1725 tonnes of fuel from Earth to LEO, which will now be used to break Mars orbit. This 22,425 tonnes will be used to break Earth orbit and get the 1860 tonnes of vehicle and return fuel out to Mars.

And that fuel now needs to be lifted out of Earth's gravity into LEO so it can be strapped on the back of our much-larger rocket. To lift 22,425 tonnes of fuel, 135 tonnes at a time, up to LEO would require 166 more launches, in addition to the 13 needed for the return fuel, plus the one launch for the actual vehicle, for a total of 180 launches from Earth's surface to LEO in order to get this ship in orbit and fueled for its departure. In other words, we'd need more SLS launches carrying just fuel than total combined Saturn V and Shuttle launches (which are the only two vehicles we've ever launched with the lift capacity even close to getting the job done). Each of those launches needs 1725 tonnes of fuel, for 310,500 tonnes of fuel burned just getting the vehicle and fuel into LEO. Then the ship itself will burn 22,425 tonnes of fuel getting to Mars, then 1725 tonnes getting back, for a total fuel expenditure of about 335,000 tonnes. So, fuel costs aren't too bad; for LH2/LOX, at an 11%-89% mixture and today's prices (\$5.50/kg LH2, \$.20/kg LOX), we're looking at around \$250 million in raw fuel cost, plus off-gas losses (liquid hydrogen and oxygen don't just sit around in liquid form at room temp).

However, total launch costs are a big deal. The SLS, if it hits its cost goals, will be about half a billion dollars a launch. 180 launches to get the vehicle and fuel into LEO represents a cost of about \$90 billion just to get the materiel into space. Actually designing and building what we're launching could exceed a trillion dollars, given that it can not fail; if the crew has a "problem" halfway out to Mars, like the one Apollo 13 did, the chances of them getting back to Earth safely are nil.

share|improve this answer
thanks for the calculating the numbers. good read! – Martin Asenov Dec 31 '13 at 17:31
Has your answer on the linked question been deleted, or something? I don't see it. – Michael Kjörling Oct 30 '15 at 13:28

Like @oefe noted in the comments, you've already covered the reasons why not use direct launch to Mars pretty well in your question, so I'm going to assume you just haven't taken enough time for all these rather difficult concepts to clarify in your own mind, and I'll point you to a nicely written and relatively easy to understand description for it, that should help with that. From NASA Jet Propulsion Laboratory's Basics of Space Flight, Section I, Chapter 4. Interplanetary Trajectories:

When travelling among the planets, it's a good idea to minimize the propellant mass needed by your spacecraft and its launch vehicle. That way, such a flight is possible with current launch capabilities, and costs will not be prohibitive. The amount of propellant needed depends largely on what route you choose. Trajectories that by their nature need a minimum of propellant are therefore of great interest.

Hohmann Transfer Orbits

To launch a spacecraft from Earth to an outer planet such as Mars using the least propellant possible, first consider that the spacecraft is already in solar orbit as it sits on the launch pad. This existing solar orbit must be adjusted to cause it to take the spacecraft to Mars: The desired orbit's perihelion (closest approach to the sun) will be at the distance of Earth's orbit, and the aphelion (farthest distance from the sun) will be at the distance of Mars' orbit. This is called a Hohmann Transfer orbit. The portion of the solar orbit that takes the spacecraft from Earth to Mars is called its trajectory.


                                             enter image description here

                                                 Earth to Mars via Least Energy Orbit

Quote and image source: NASA Jet Propulsion Laboratory's Basics of Space Flight, Section I, Chapter 4. Interplanetary Trajectories

And so on and I suggest reading the whole lot. It's the easiest and still sufficiently complete description of interplanetary spaceflight economics that I can think of, and sometimes all it takes is reading or hearing it put in different words to really understand it. One other resource that I suggest is also reading Emily Lakdawalla's blog post on Why are MAVEN and Mars Orbiter Mission taking such different paths to Mars? But I'd start with the NASA JPL one first, since it deals with the concepts, then move on to an actual example as described in Emily's blog.

So, as you see, this all boils down to economics. For example, the Mars One project is expected to use 4 separate launches of Delta IV Heavy (four of the most powerful rockets currently available) to launch required parts into Low-Earth Orbit (LEO), use in-orbit assembly and then launch all of that into a Hohmann transfer orbit towards Mars. If they attempted to launch all required parts into a more direct trajectory, they would have to use many more launches (it's the lifting capacity that's dragging you down and you'd require more fuel to reach escape velocity for a more direct trajectory, requiring greater delta-v, which adds more mass which requires more thrust, ad nauseum), and do that at the same time for in-trajectory assembly, otherwise you end up with parts indefinitely chasing each other. To put it differently, we simply don't have lifting capacity available for a more direct trajectory to Mars launch. And even if we had, we'd still be able to launch a lot greater mass towards Mars by requiring less delta-v by using Hohmann transfer trajectory.

share|improve this answer
Thank you, but the information you reference does not answer that particular question I'm asking. I updated the question so it be more clear. Thanks for the links, they're quite good though! – Martin Asenov Dec 29 '13 at 17:38
Yes the operation is surely going to be costly. A Mars landing would easily become the greatest achievement of mankind in the 21st century, so I guess some big money could be spent on this one. Economics put aside (although important for sure), I'm particularly asking if it is technically possible to send humans to and land them on Mars in 2 months for the sake of as low as possible radioactive exposure? If I understand you correctly, no, because we would not be able to accelerate enough the fuel needed to accelerate... :) – Martin Asenov Dec 29 '13 at 18:20
@MartinAsenov No, it is possible, but you're wasting a lot of resources doing that. It's not just financial economy, it's the mass economy too. Still, if that wasn't a problem (let's pretend for a minute), then yes, more direct trajectory of course is possible. It doesn't make any sense though, because if you can afford to launch so many launchers (you still need to get to Mars same mass to sustain 4 astronauts) at the same time, you could as well build in LEO a better protected habitat for a lot more of them by using HTO. – TildalWave Dec 29 '13 at 18:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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