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Nine months is what google maps said to get to Mars. I would like to know if fuel was no expense how long could we get to Mars other than the method below? Is the 9 months and the method just what our current technology is capable of?

What if we launch in the opposite direction to meet Mars head on?

To clarify I am not asking: Why not travel to Mars in 2 months? nor "why not travel to Mars in a straight line". I would like to know if there are other ways to Mars than the picture shows? But I do like the idea of having a spaceport in the other question to build a launcher.

enter image description here https://image.gsfc.nasa.gov/poetry/venus/q2811.html

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closed as unclear what you're asking by Russell Borogove, Nathan Tuggy, Brian Tompsett - 汤莱恩, Organic Marble, Muze the good Troll. Mar 17 at 16:42

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Related, possible duplicate: space.stackexchange.com/questions/14580/… $\endgroup$ – Organic Marble Mar 16 at 17:20
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    $\begingroup$ Possible duplicate of Why not travel to Mars in 2 months? $\endgroup$ – Russell Borogove Mar 16 at 17:25
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    $\begingroup$ "if fuel was no expense" do you mean, if somebody subsidized propellants on the ground so there was no cost for any amount of fuel loaded on the rocket, or do you mean "all of the costs associated with carrying fuel on the voyage are ignored"? If the latter, it would be easy to get to Mars in a day or two, accelerating at 3+ g the whole way, and the path taken would be very nearly irrelevant. $\endgroup$ – Nathan Tuggy Mar 16 at 17:57
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    $\begingroup$ @SteveLinton I think your answer wasn't so bad to own-delete it. Wouldn't be deleted, I had voted it up. $\endgroup$ – peterh says reinstate Monica Mar 16 at 18:44
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    $\begingroup$ Actually with the usual transfer Mars catches up with you from behind. The leading side of Mars hits you in your trailing side. $\endgroup$ – Mark Adler Mar 17 at 3:58
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For some reason, I saw, or read this question as simply asking "How fast can you get to Mars if fuel is no expense". So this answers that question. I don't mind deleting it if that is considered appropriate.

Anyway, it depends what you mean by "if fuel is no expense". If you are assuming something more or less along the lines of current rockets, but are willing to use more fuel, and so carry less payload than is usually considered, there are trajectories that get you to Mars in three or four months within the capabilities of rockets currently being designed. The don't look very different from your picture, except that the transfer orbit actually reaches out past the orbit of Mars, although the rocket doesn't do that part of the orbit because it arrives at Mars first. Chemical rockets have nowhere near the capability to counter Earth's orbital motion (30 km/s roughly) and go in "the opposite direction" for instance. The usual way to represent the options for this kind of interplanetary trajectory is called a "pork chop plot" and shows options in terms of departure date, arrival date (or journey length) and delta-V required. There are numerous answers on this stack exchange which include examples of them.

If on the other hand, you imagine a "magical" rocket, powered by fusion or antimatter which can sustain a 1g acceleration continuously (much more would be uncomfortable for the crew) then the fastest trajectory looks very much like a straight line, and takes about 40 hours at closest approach (just use $s = \frac{1}{2} a t^2$ separately on each half of the journey. The motions of the planets and the Sun's gravity will be a small correction to this.

Finally if you just want to ship very small robust cargo items (such as individual protons or neutrinos) you could send them at close to the speed of light, for an absolute minimum delivery time of about 7 minutes.

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Here's a quick visual analogy.

Imagine you want to "hop a train"; jump onto a flatcar as it is rolling along the tracks.

The traditional approach is to run in the same direction the train is going, and then when you get beside the car make sure your speed matches, grab it, and jump on. It's dangerous, but very doable.

Your suggested approach is to run in the opposite direction the train is going, and then when you get beside the car, either grab it and jump on, or quickly reverse your speed and direction and board as in the traditional approach.

Neither of those non-traditional approaches would be pretty.

EDIT: After posting that, I realized that a much more graphic analogy would be running along the tracks to catch up to the train and pulling oneself onto the caboose, as compared with running toward the approaching train and jumping onto the front of the engine.

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To reach one body in space from another, you need to consider their speeds, not just their positions. To leave Earth going the other direction, you need to cancel out Earth's orbital velocity around the sun (about 30km/s), and then when you reach Mars you need to turn around and match Mars's velocity. This is much more expensive in fuel than more direct flight, and most importantly, no faster. The flight time is due to the need to reach a position much further from the sun, not because we have to catch up to Mars.

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    $\begingroup$ I think the 29.8 km/s orbital velocity of the Earth don't need to be canceled, instead the escape velocity of the Earth sould be paid (11.2km/s), plus the $\approx$ 2km/s gravity / drag loss, plus the gravitational potential between the Earth and Mars should be handled by initial speed (some km/s). Arrival on Mars wouldn't happen very softly, but deceleration could happen also by drag, requiring no fuel. $\endgroup$ – peterh says reinstate Monica Mar 16 at 20:13
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    $\begingroup$ @peterh That’s the usual Hohmann transfer. OP is explicitly asking about going in the opposite direction, i.e.via retrograde solar orbit. $\endgroup$ – Russell Borogove Mar 16 at 20:42

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