I had two questions regarding choosing a path to Mars. Firstly:

I wanted to have a look at the mathematics behind the reason as to why we choose Hohmann's transfer over a journey that is a straight line from Earth to Mars. Is moving at high speeds that require greater deceleration when entering Mars orbit the only problem that prevents us from doing this? I suppose the answer to this question is no. Then what else restricts us from performing a much simpler mission?


What is the math behind calculating the time taken to get to Mars with a specific orbit? So, what is the time difference between a certain Hohmann's transfer to Mars and moving in a straight line to the Red Planet (assuming we can solve the problems associated with a journey which takes a straight path)?

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    $\begingroup$ Does this answer your question? Why not travel to Mars in 2 months? $\endgroup$
    – asdfex
    Commented Dec 17, 2021 at 16:53
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    $\begingroup$ @Ryan the math for Hofmann transfers is easy to find on wikipedia. As for a "straight line" check out the trajectory for New Horizons, the fastest satellite ever launched. It's path from earth to mars orbital distance is very much still significantly curved. To even get it close to looking like a line would require orders of magnitude more deltaV. There is no rocket today or even in conception that would even come close. $\endgroup$
    – eps
    Commented Dec 18, 2021 at 0:58
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    $\begingroup$ Elon has said that the spaceX plan is to make the trip shorter by going initially faster than what would be required for a hoffman transfer (and thus requiring more fuel initially and to slow down for the orbital insertion than a HT would require [basically a tradeoff between fuel and mission length]). But it's still going to be a significantly curved path. This approach also increases the risks involved with a manned mission -- if the engines fail to light for the insertion burn the crew will almost certainly be lost to space. A HT gives you a free return if things go wrong (eg Apollo 13) $\endgroup$
    – eps
    Commented Dec 18, 2021 at 1:01
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    $\begingroup$ Disagree with the close vote; based on comments it's not a duplicate and answers there don't fully answer this. However the new and well-received answer does a very good job of answering this question, so voting to leave open! Many years ago I attended an outreach talk in Iowa by some folks from NASA. A young person asked as similar question: "Why don't you just fly straight across" and the speaker simply could not and did not answer. It's a great question and this is a great answer, and if this one is duplicated again we can now point back here. No need to block further answers. $\endgroup$
    – uhoh
    Commented Dec 18, 2021 at 1:30
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    $\begingroup$ @eps A HT only guarantees a free return if you're talking about orbits around a given body (ie: Earth-Moon gives you free return to Earth). In the case of an Earth-Mars, however, you get a free return that intersects Earth's orbit around the Sun, but you won't arrive at the right time to meet Earth when you get there (ie: the phasing is critical for transfers between bodies around a parent object). $\endgroup$
    – J...
    Commented Dec 18, 2021 at 17:44

4 Answers 4


Only answering your first question here, and in qualitative terms:

You can't travel to Mars in a straight line for the same reasons you can't throw a ball in a straight line: gravity. If you wanted to throw a ball & have it travel in a straight line, you'd need something fighting gravity the whole way: wings, rocket engines, or similar. To travel to Mars in a straight line, you'd also need to fight gravity the whole way. Since there's almost nothing in space to push against (no air), you can't use wings, so you'd have to use a rocket engine instead. The problem is, you'd need a tremendous amount of rocket fuel to do that, because every bit of rocket fuel you bring makes your ship heavier, which means you need more fuel & stronger engines, which makes your ship heavier...

So it turns out it's best (meaning cheapest, easiest, etc) to "throw" ourselves to Mars in the same way a ball is thrown: with one big push at the start, and then fly in a curved arc the rest of a way. This type of throwing, done most efficiently, is called a Hohmann transfer. Why do we need to be efficient? Because technology isn't the limiting factor on spaceflight, cost is.

So in short, yes, it could be done, but it would be exorbitantly inefficient & thus very expensive. Resultingly, it isn't.

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    $\begingroup$ There is also the option to go so crazy fast to Mars that your hyperbolic trajectory is broadly indistinguishable from a straight line, but that runs into your fuel argument again. Spacecraft that have several hundred km/s of delta v to dump on an interplanetary jaunt aren't available for the foreseeable future. $\endgroup$
    – notovny
    Commented Dec 17, 2021 at 19:42
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    $\begingroup$ @notovny Indeed, one could argue that radio signals sent to and from Mars take that "crazy fast" approach. (Then again, there's also the option of redefining "straight line" so that it matches your actual trajectory, which probably applies even better! :-p) $\endgroup$
    – David Z
    Commented Dec 18, 2021 at 1:38
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    $\begingroup$ @notovny you could wait for the appropriate time and then start with a ~30 km/s burn to cancel Earth's orbital velocity, then burn into a radial trajectory to reach Mars, then a ~24 km/s burn to match up with it. Or you could if orbits really were 2D and you didn't have to concern yourself with the gravity of the planets, in reality it'd still be slightly curved...and only almost-straight in the frame of the sun. $\endgroup$ Commented Dec 18, 2021 at 2:58
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    $\begingroup$ From general relativity, mass curves spactime, and a mass in freefall follows a a stright path (called a geodesic) though this curved space. So with the Hohmann transfer we have the Earth, Mars, and the spacecraft all following straight paths. $\endgroup$
    – Jasen
    Commented Dec 18, 2021 at 7:55
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    $\begingroup$ We clearly need the Epstein drive. $\endgroup$ Commented Dec 18, 2021 at 10:47

Disclaimer: this answer makes some coarse simplifications in order to keep it focussed on the crucial points. If you feel like pointing out something about orbit eccentricities or whatever in the comments, by all means go ahead, but I won't reply.

You can travel to Mars in a straight line, it's just not efficient.


This is perhaps what most people think how moving between different-height orbits should work: you fly away from the sun in a straight line. After all, our problem is that we're too low, too close to the sun, and we want to get further away, so surely the way to accomplish that is to fire our thrusters to give us a heading straight away from the sun? Then we'll move in a line until our kinetic energy is exchanged for potential energy. Easy.

Well, this would actually be the best way if we had started from a platform that was somehow moored at 1 AU from the sun, and tried to get to another platform moored at 1.5 AU. But clearly, no such platforms exist – anything just sitting there would quickly fall into the sun itself. (Short of being fixed to a Dyson sphere or something).
Instead, we start from Earth which is buzzing around the sun at 30 kms, and before we could start heading out towards Mars radially we would need to brake from this velocity – but braking takes just as much fuel as accelerating to 30 kms from a standstill. So that's crazy expensive – 30 kms is a huge amount of $\Delta v$, and that's before we would even start our height-changing maneuver. And then once at Mars' orbit, we would need to accelerate again to match its velocity, before we could land there.

For the “lifting” itself, there are different ways you can do it. The most efficient way (like we cared about efficiency...) is to do it all in one quick burn at the start to give you enough kinetic energy so the radial velocity will zero again just when you reach Mars. Any other option will either waste time (and time=fuel=money if you're about to fall into the sun...), or require wasting even more $\Delta v$ on braking down the radial velocity at Mars.

Total cost: 30 kms + √((42 kms)2 - (34 kms)2) + 24 kms79 kms

That's absurdly expensive.

Fortunately, braking down from Earth's orbital velocity is not only unnecessary, it's actually completely counterproductive because that kinetic energy can be used to get to a higher orbit.


The next best idea you might have is to just accelerate outwards without braking from orbital velocity first. But that means the sun's gravity will not only slow down your outbound speed like it did in the radial case, but also bend your trajectory. For small orbital changes this is actually not noticeable. This is how for instance a spacecraft maneuvers right before docking to the ISS: you want to move the last meters closer, you simply fire the thusters in the opposite direction as one would intuitively expect, and that sends you towards the station. Never mind that the orbit bends a bit in the few seconds before you actually get there.

But Mars is kinda a bit more that a few meters away from Earth[citation needed], so in this case the bend would be noticeable. Now, you could theoretically prevent that by not firing in merely a short burn at the start, but instead steadily at always the same acceleration as the sun's gravity imposes on you, thus exactly cancelling the curving effect. As a result, you would move outbound in a straight line tangentially to Earth's orbit. The time-dependent distance to the sun would be $$ R_\text{tngt}(t) = \sqrt{r_{\text{Earth}}^2 + (t\cdot v_\text{Earth})^2} $$ Thus the travel time would be $$ T_\text{tngt} = \frac{\sqrt{r_{\text{Mars}}^2 - r_{\text{Earth}}^2}}{v_\text{Earth}} \approx 5.77\cdot 10^6\:\mathrm{s} $$ or 67 days. The fuel for that ride takes $$\begin{align} \Delta v_\text{tngt} =& \int\limits_0^{T_\text{tngt}}\!\!\!\!\mathrm{d}t \: \frac{G\cdot m_\text{Sun}}{(R_\text{tngt}(t))^2} \\=& \frac{G\cdot m_\text{Sun}}{r_{\text{Earth}}^2} \cdot \int\limits_0^{T_\text{tngt}}\!\!\!\!\mathrm{d}t \: \frac1{1 + (t\cdot \frac{v_\text{Earth}}{r_{\text{Earth}}})^2} \\=& \frac{G\cdot m_\text{Sun}}{v_{\text{Earth}}\cdot r_{\text{Earth}}} \cdot \int\limits_0^{T_\text{tngt}\cdot \frac{v_\text{Earth}}{r_{\text{Earth}}}}\!\!\!\!\mathrm{d}\tau \: \frac1{1 + \tau^2} \\=& \frac{G\cdot m_\text{Sun}}{v_{\text{Earth}}\cdot r_{\text{Earth}}} \cdot \arctan\left(T_\text{tngt}\cdot \frac{v_\text{Earth}}{r_{\text{Earth}}}\right) \\=& \frac{G\cdot m_\text{Sun}}{v_{\text{Earth}}\cdot r_{\text{Earth}}} \cdot \arctan\left(\sqrt{\frac{r_{\text{Mars}}^2}{r_{\text{Earth}}^2} - 1}\right) \end{align}$$ which comes out to 25.5 kms. That's not quite as insane as the radial approach, but still already much worse than a Hohmann transfer, and we're not finished yet: we still need to match Mars' orbital velocity, which involves getting rid of our radial velocity component and adding prograde velocity.

The latter bit indicates what we're doing wrong all the time: letting the sun bend our trajectory is actually a good thing, and what we want is prograde acceleration. The consequence is exactly what you do in a Hohmann transfer: don't bother with radial maneuvers at all, but only use two prograde burns and a freefall elliptical trajectory in between.

  • $\begingroup$ You calculated with only the mass of the Sun, it is much bigger than that of the Earth and of Mars. But close to Earth or Mars, their gravitational forces to the spaceship are bigger than the force of the Sun. So the straight line is possible only at some distance to Earth and Mars. $\endgroup$
    – Uwe
    Commented Dec 18, 2021 at 23:26
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    $\begingroup$ "we still need to match Mars's orbital velocity" - unless you're extremely accurate and have a very good heat shield! (And a landing site you don't care too much about.) $\endgroup$
    – Cadence
    Commented Dec 18, 2021 at 23:42
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    $\begingroup$ For completeness, adding the delta-v for a Hohmann transfer would make it even clearer that we are taking about very different numbers… $\endgroup$
    – jcaron
    Commented Dec 19, 2021 at 14:17

To paraphrase Doc Brown, you're not thinking fourth dimensionally. The term "straight line" might seem like a straight-forward concept (no pun intended), but we don't live in a Euclidean universe. A four dimensional trajectory creating a straight line when projected onto your two dimensional computer screen is not "straight" in terms of having zero spacetime curvature. Apart from during burns, all spaceflights travel along geodesics, which is what mathematically are considered "straight", somewhat analogously to how great circle routes look curved when projected onto two dimensional map, such as with Mercator projection, but are the closest thing to "straight" on a globe. And it's rather difficult to compare the time difference between following a path with a curved projection, versus one with a straight one, because a straight one would require massively more power, and of course you'll get there faster if you have more fuel to burn. For any fixed amount of fuel, the fastest path is going to be "curved" when projected onto Euclidean space, although it will get "straighter" as you put more and more fuel in.


Imagine a spinning CD and an ant going from the center to the edge of the CD in a straight line. From his point of view, it seems like a straight line, but if you plot his trajectory on a static piece of paper, it's a curve because the CD is spinning while the ant is going in a straight line.

The ant would need to expend extra effort to counter this rotation of the CD if you want the plot on the paper to look like a straight line.

  • $\begingroup$ It's easy to forget that, even after you've left earth's orbit, you're still orbiting the sun at nearly the same velocity the earth was. All the velocity you had on earth due to its orbit doesn't magically go away once you're in space. $\endgroup$ Commented Aug 14, 2022 at 1:01

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