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Assume delta-v isn't a problem and circular orbits. EDIT: Assume that you're already in orbit so you don't have to shift a massive load of fuel up, and the absolute ideal is something that has a calculable delta-V like a Hohmann transfer.

If there was some kind of industrial accident on Mars and we needed to leave right away, is there always a burn we can perform to get to where we want to be?

I can imagine that we could move into an orbit closer to the sun to 'catch up' Mars or further away than it to slow down, but would this be significantly quicker than just waiting for our ideal launch window?

Is there a way to optimise the arrival, similar to a Porkchop Plot?

Background - I'm thinking about a management game with orbital mechanics where there's a clear 'optimal' time to travel - this is the equivalent of the launch window - but there's an ability to travel at any time if you're willing to pay the extra cost for the rocket fuel.

So ideally the 'optimal' case is something like a Hohmann or Bi-elliptic transfer, and then the further you move away from your launch window, the more dV (and such the more cost) it will be.

Clearly it's a game so I can throw realism out the window, but I was wondering if there's a specific realistic trajectory that a rocket would take if an emergency mission needed to be mounted - plus, waiting two yeard for the optimal trajectory to restock your Mars base would be a bit of a drag.

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Hi Matt, you might want to reword your question, if delta-v isn't a problem then you can pretty much go anywhere anytime you want. You might want to give a little more background info upfront and restate it so that the assumption is that your interested in keeping normal physics, but you can assume idealized conditions. –  user11547 Feb 5 '13 at 11:01

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I think you are describing the bi-elliptic transfer, where you raise the apoapsis above the target and then do a retrograde burn there to lower the periapsis. Technically this can save a bit of $\Delta v$ compared to a Hohmann transfer because the $\Delta v$ is spent more efficiently (the Oberth effect), but it also takes longer. If dv really isn't a problem you will always make the best time with a brachistochrone trajectory: accelerate continuously towards the target until you reach the halfway point, then flip around and deccelerate till you reach the destination at a standstill. This is really a generalised brachistochrone since you need to take into account the $1/r^2$ variation of gravity in the solar system. You can also take advantage of aerobraking at the target (if there is an acceptable atmosphere) to push the turnaround point even closer to the target and cut some time off the trip. Tweak as necessary - the idea still works. Porkchop plots are used to minimize the energy/$\Delta v$ requirement since in the real world lifting un-needed rocket fuel out of the Earth's gravity well is an expensive waste.

EDIT: I read an article from NASA describing work using genetic algorithms to optimize these sorts of continuously accelerating transfers. Sadly I don't have the reference any more, but its probably possible to find online.

EDIT 2: You can find a lot of information relating to real and fictional technologies from a hard sci-fi writing perspective at Atomic Rockets. If you are prepared to throw realistic $\Delta v$ requirements out the window then the only limitation on mission time is the acceleration your squishy human cargo is able to endure. For instance, if you pack your people in acceleration couches for the duration you can accelerate at, lets say, 3g long term (I'm not volunteering for this). An accelerate halfway/flip over/deccelerate mission takes (in the first approximation neglecting gravity) takes a time

$$ t = 2\sqrt{\frac{d}{a}} $$

to go a distance $d$ at an acceleration of $a$. To get a rough worst case figure take Earth and Mars at their furthest opposition $ d \approx 2.7 \mathrm{AU} $ so

$$ t_{max} \approx 65 \mathrm{hr}\ !!!$$

or if you only subject your passengers to 1g

$$ t_{max} \approx 113 \mathrm{hr}\ !!!$$

Clearly this is nothing like a real world mission profile. The reason is that the $\Delta v$ requirement for the 1g mission is

$$ \Delta v = a t = 4\times10^6 \frac{\mathrm{m}}{\mathrm{s}} $$

For the 3g mission it goes up to $6\times10^6 \frac{\mathrm{m}}{\mathrm{s}}$. You need a rocket engine with an exhaust velocity of the same order of magnitude if you want to achieve anything like a reasonable cargo/passenger capacity. This is, ahem, not currently feasible. But if you're willing to posit it then you hardly need to worry about complications like the sun's gravity - you can basically go in a straight line! (If you want to check my numbers, please do. I'm not entirely sure I believe them!)

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Hey - I think this is great, it's almost there. I've added a bit of background above as well. If I understand correctly the brachistochrone trajectory could be calculated for an interplanetary burn and it would essentially be the shortest route, using the gravity of the sun to 'pull' the ship in after its burn? –  Matt Kemp Feb 5 '13 at 14:01
@MattKemp See my edit. Hope it helps - if not let me know! –  Michael Brown Feb 5 '13 at 14:37
A brachistochrone is a shortest time trajectory. It is more complicated here than the case in the wikipedia article since the acceleration is changing in both magnitude and direction, but the principle is the same. –  Michael Brown Feb 5 '13 at 14:39
I should also mention the Kzinti lesson: "a reaction drive's efficiency as a weapon is in direct proportion to its efficiency as a drive." Or equivalently Jon's law for sci-fi writers: "Any interesting space drive is a weapon of mass destruction." :) –  Michael Brown Feb 5 '13 at 14:49
Great, thanks - it also includes one specific tidbit: "A Hohmann orbit is the maximum transit time / minimum deltaV mission. A "Brachistochrone" is a minimum transit time / maximum deltaV mission" which neatly explains why they don't degrade nicely. What would be the 'midpoint' to accelerate to? Would you constantly accelerate in the direction of the target, or towards the position of the target that you expect to intercept it at? –  Matt Kemp Feb 5 '13 at 15:29

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