This may be a stupid question but I didn't find much information online about it. The concept is simple, could we launch a vehicle into space from Earth, stop right on Mars trajectory and wait for the planet in order to land on the surface? Would this be the shortest way from Earth to Mars? Would it lower the fuel cost of the travel compared to the Hohmann transfer method?


7 Answers 7


Maybe some visual intuition for what actually happens in the Hohmann transfer helps?

enter image description here

It's already very close to what you are describing. In the top arc, the spacecraft (yellow), is going a bit slower than Mars (red), so it's indeed "waiting" for the planet to catch up to it.

It only touches the orbit of Mars in a point, but that's all we need if we time the launch right (in practice, the zone where the encounter happens is quite a lot wider than a point).

But we can't just stop at the top of that arc. Stopping cost just as much as accelerating in space. Coasting along is free.

That's what's nice with the Hohmann transfer, it's mostly free coasting, that when timed right happens to line up the orbits nicely very close to each other.

  • $\begingroup$ Also, in practice, there's a few correction burns along the way, so you get even more leeway. But if you miss that burn, you're screwed :) $\endgroup$
    – Luaan
    Jan 8, 2020 at 9:17
  • $\begingroup$ Doesn't Mars' gravity help with stopping? $\endgroup$
    – d-b
    Jan 8, 2020 at 20:43
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    $\begingroup$ @d-b: the simplified view is: when you enter the SOI of Mars, now you have a new problem: you are falling towards Mars. There are two possibilities: you hit Mars at orbital velocity and have a rapid unplanned disassembly, or you miss Mars. If you miss Mars, well, energy is conserved, so you will attain the same height on the way back up as you had when you started falling, and you will end up back in interplanetary space. $\endgroup$ Jan 8, 2020 at 23:16
  • $\begingroup$ @d-b: How to solve this conundrum? You arrive at Mars SOI on a trajectory such that you miss hitting Mars, and you burn your engines retrograde at the lowest point. Now you have less energy and you will not attain the same height on the way back up; you're in Mars orbit. Alternative plan: skim the atmosphere enough to slow you down without burning fuel; but now you have another problem; you're in an orbit where you are losing energy at each go round, and therefore will eventually hit the planet again. $\endgroup$ Jan 8, 2020 at 23:19
  • $\begingroup$ And the whole point of the Hohmann transfer orbit is precisely that it is the lowest-energy method of transit. Any other method - thus anything faster than it included - will require more energy (and so either an exponentially bigger rocket or a rocket with higher $v_\mathrm{exh}$, both of which are, of course, big engineering and technological problems). $\endgroup$ Jan 12, 2020 at 14:27

It depends whether or not you want to orbit or land softly upon Mars, or just hit it.

For the former, you have to match orbits with it, that probably means burning more fuel. For the latter, you can skip the Mars orbit injection and just crash. This is quite fuel efficient, especially as the reduced delta-V requirements mean that you need less fuel for the trans-Mars injection burn, too.

Regarding the rest of your question, though:

stop right on Mars trajectory

Stop how? You'll be in a heliocentric orbit once you've left Earth's sphere of influence; you can't just park... you're flying round the sun at quite a few kilometres per second. Getting out of that orbit requires a serious amount of fuel, and once you've burned that fuel you'll just fall back towards the Sun and an eventual firey demise. Even if you did deorbit yourself and timed it just right so that Mars caught up with you before you fell back, it will be barrelling towards you at about 24km/s and it'll hit you really hard. So you've still crashed, only you've crashed a lot harder, and you've needed a mindbogglingly powerful rocket with masses of fuel in order to do so.

(edit: you could, in fact, use a solar sail statite to hold your position once "stopped" in space, which solves the problem of falling back towards the sun, though not the whole "Mars hitting you at ninety thousand kilometres per hour" thing)

Would it lower the fuel cost of the travel compared to the Hohmann transfer method ?

There aren't many cheaper ways (in terms of fuel) to get a spacecraft out to Mars' distance from the sun, and the ways which are cheaper are also often a lot slower.

  • 19
    $\begingroup$ Crashing a.k.a. lithobraking :) $\endgroup$
    – gerrit
    Jan 6, 2020 at 12:16
  • 1
    $\begingroup$ @gerrit Considering we're talking about a high velocity on a low-density atmosphere, it's going to be a lot of crashing I suppose. $\endgroup$
    – Mast
    Jan 6, 2020 at 13:16
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    $\begingroup$ @Hashim you'd "float" in the absence of any gravity fields. In our solar system, if you'd become stationary relative to the Sun (which is hard if starting from any planet's orbit you have a lot of speed to cancel) then you'd start falling straight towards the Sun and fall into it in due time. $\endgroup$
    – Peteris
    Jan 7, 2020 at 0:32
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    $\begingroup$ @Hashim Nothing in space is stationary. Everything is moving really quickly relative to at least one other object. If there's nothing near you in space, you appear to be floating but it's an illusion because space is so big. So, when you're floating, you're still moving really fast. $\endgroup$
    – CJ Dennis
    Jan 7, 2020 at 1:27
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    $\begingroup$ "reduced delta-v requirements" - pretty sure when you crash you lose all your delta-v, if you think about it... $\endgroup$
    – corsiKa
    Jan 7, 2020 at 4:55

could we launch a vehicle in space for Earth, stop right on Mars trajectory...

Yes, you could have a trajectory that came to a stop (briefly) in the path of Mars like how a ball thrown upwards stops (instantaneously) before falling down, except in this case you'd have to be moving directly away from the Sun on a straight-line trajectory with eccentricity = 1.

...and wait for the planet...

No, you can't wait there, the Sun's gravity is omnipresent. You will always be in some orbit, you just have to choose one and enjoy the ride!

...in order to land on the surface?

Also No. Well, if you have saved a lot of fuel and have some nice parachutes, maybe.

  1. Mars is coming at you at 24,000 m/sec
  2. Mars' gravity will add an additional 5,000 520 m/sec by the time you impact.

The 2nd item can be calculated as follows. Initial specific energy $v^2/2$ (2.88E+08 m^2/s^2) added to the gravitational specific potential energy difference between infinity and Mars' surface $GM/r$ (1.26E+07 m^2/s^2) put into $v = \sqrt{2E}$ adds about 520 m/s to the impact velocity assuming normal incidence and no atmosphere.

  • 5
    $\begingroup$ The relative velocity and escape velocity don't add linearly to get arrival speed at a planet. You can use the change in specific gravitational potential energy to work out the new specific kinetic energy and speed. For the object in question, that's about 1.26E7 J/kg, which, for an object moving at 24000 m/s relative, is a change in velocity of about 520 m/s, if i've pumped the numbers into Wolfram Alpha correctly. $\endgroup$
    – notovny
    Jan 6, 2020 at 10:58
  • $\begingroup$ @notovny yes that's of course correct. Please feel free to edit this and fix it. I'll try to get to it tomorrow, but you are welcome to edit. Thanks! $\endgroup$
    – uhoh
    Jan 6, 2020 at 11:11
  • $\begingroup$ @notovny I've made an edit, how's that? $\endgroup$
    – uhoh
    Jan 6, 2020 at 23:01

There is no "stopping" in space - no matter how far you are from other celestial bodies, the force of gravity will always be tugging on you, pulling you in some direction. Within the Solar System, that tug is typically pulling you toward the Sun, unless you happen to be quite close to another planet or moon. If you were to try to get to the same path as Mars orbit and then just "stop", you would get pulled out of position by the Sun's gravity. To remain in place, you'd have to constantly burn a rocket to counteract the Sun's gravity.

This would be similar to just hovering a rocket a few feet above the earth's surface - it's extremely fuel inefficient, as you need to burn fuel the entire time you're waiting. It's much more efficient to wait in a stable orbit, which does not require any fuel to maintain. This is essentially the Hohmann transfer mentioned in other answers, which allows a planet to "catch up" to a slower moving spacecraft that's coasting through space, (or for the spacecraft to catch up to the planet) and is timed such that the orbital paths intersect when both objects are in the same place.


You cannot "stop on Mars' trajectory" - that is not how physics works. If you want to go out as far as Mars' orbit, then expend energy to stop orbiting the sun, you will leave that orbit.

This will take a lot of energy, and you will not end up in the correct orbit.

Lowest dV is worked out through running multiple simulations - Hohmann Transfer seems to be lowest energy.

  • 1
    $\begingroup$ Hohmann Transfer is (usually) the lowest dV for direct transfer between orbits (eg. going from a circular orbit around the Earth to a circular orbite around Mars). You can still save quite a bit of extra dV if you don't have to match the target orbit - e.g. if you can use aerobraing to shed some extra velocity at the target planet. Of course, this is trickier than it sounds - you need to shed quite a bit of velocity to go from the transfer orbit to the target orbit, and you need to burn as low as possible to get as much dV as you can from the Oberth effect, so atmospheres are tricky. $\endgroup$
    – Luaan
    Jan 7, 2020 at 8:18

The usual answer would be that the Hohmann transfer orbit is already essentially there. But those are meant for lowest dV direct transfer - e.g. a straight shot from the Earth to Mars. They are extremely fuel efficient - the trip from Earth to Mars can take around 3.9 km/s dV - compare that to Mars' orbital velocity of about 24 km/s or the Earth's of 30 km/s. However, if you're in no particular hurry, we can do better.

The lowest dV transfers we have devised are on the so-called Interplanetary Transport Network. These only require you to reach a Lagrange point - and from there, you get everywhere you want to go for essentially no dV cost. We already had a few missions that used one such path for studying the Sun - the first was probably the Genesis mission. The path taken allowed the craft to reach its target point with measly 0.8 km/s dV, stay there for three years and then return back to Earth essentially for free. The fun thing is that for getting to Mars or Jupiter, you need essentially the same amount of dV - you're "coasting" all the way, with only very tiny tweaks to your path. The main drawback is that it's ridiculously slow. For a trip to Mars or Jupiter, you would still only need about 0.8 km/s dV (compared to the ~4 km/s dV to Mars or ~10 km/s dV to Jupiter for a Hohmann transfer). But it would also take you thousands of years.

On the other extreme, if you have a rocket powerful enough, you can ignore silly slow things like Hohmann transfers, and just go at constant acceleration all the way on a brachistochrone transfer. The payment is in ridiculous delta-V budgets. For example, a minimal transfer to Mars at 1g would only take about four days (compared to ~9 months for a Hohmann transfer), but would require about 3 000 km/s dV! Even a "meagre" 0.1g would mean just a couple days for the whole transfer, with the cost dropping to somewhere around 1 000 km/s. The slowest you can go on a brachistochrone is around 0.01g, which would make the Earth-Mars trip take about a month and need something like 400 km/s dV. Needless to say, we don't have any rocket engine that comes close to being able of doing a constant 0.01g trip to Mars.


Acceleration and Deceleration costs $\Delta v$ (Delta is scientific notation for "change of something" and V stands for Velocity) of which a spacecraft has a set budget (how long the engines can burn).

Here's an analogy: imagine that the spacecraft is attached by a thin string to your hand (representing Earth). You start spinning it around faster and faster, denoting going into a higher and higher orbit. If the spacecraft is going fast enough, the string will break from the tension(representing reaching orbital escape velocity). Any planet has a minimum escape velocity and going any faster, costs more $\Delta v$ (which is precious), so current spacecraft with a super constrained $\Delta v$ budget only ever reach the minimal escape velocity.

Once the spacecraft has broken free of the gravity well of its launch planet, its coasting along in an elliptical path around the Sun. Except that it needs to match its velocity (by expending precious $\Delta v$) close enough to that of its target planet to enter orbit around it (in the analogy, hook onto a string attached to that planet, without breaking it). For that, the spacecraft has to be approaching the planet as close to directly from behind as possible, since then the velocity difference is minimized.

Hohmann figured out a set of equations for how to calculate the dead astern approach for a given escape velocity, which is always constant for a planet, no matter the size or capabilities of the spacecraft.

  • $\begingroup$ Thanks for the edit @peterh, how do you input Δv in markdown? Just through Unicode? $\endgroup$
    – Eugene
    Jan 9, 2020 at 18:24
  • $\begingroup$ PeterH used Mathjax, but you can copypaste Unicode, too, like here: ≸ ≹ ☃. $\endgroup$ Jan 9, 2020 at 23:16
  • $\begingroup$ @Eugene Mathjax works on this way: type in $\Delta v$ and you will get $\Delta v$. $\endgroup$
    – peterh
    Jan 10, 2020 at 9:48
  • $\begingroup$ Thanks @peterh-ReinstateMonica, I wasn't aware of Mathjax. $\endgroup$
    – Eugene
    Jan 10, 2020 at 20:21

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