Would it be possible for there to be an almost torchship (essentially a very, very weak one) which could fly almost brachistochrone trajectories? So instead of a full on brachistochrone, where you burn straight for the target planet then flip half way and decelerate, an almost-brachistochrone where you burn for a bit towards the planet then spend some time on the float before decelerating.
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$\begingroup$ Does a brachistochrone trajectory mean it gets you there quickest given the energy spent? $\endgroup$– WillkDec 25, 2018 at 1:07
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$\begingroup$ What do orbits have to do with brachistochrones? And since when are spaceships burning "straight for the target planet"? Orbital speeds are huge, and the distances in space are even larger. You can never ignore the presence of the big body in the center when planning a course for a spaceship. And you have nothing to slide on. And gravity is never uniform on a trip between celestial bodies. So, no brachistochrones in space... $\endgroup$– cmaster - reinstate monicaDec 25, 2018 at 1:10
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$\begingroup$ A brachistochrone trajectory requires a torchship; a ship with ridiculously, unreasonably good drive, for more information please visit the Atomic Rockets page for more info. . . projectrho.com/public_html/rocket/torchships.php $\endgroup$– Cosmic OrreryDec 25, 2018 at 1:29
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$\begingroup$ Mark Olson's answer below gives additional details, but the direct answer to you question is "yes." Essentially, you're still performing a Hohmann transfer, just along a shorter path. $\endgroup$– DanDec 25, 2018 at 5:39
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It appears to me that by "brachistochrone" you mean a least-time trajectory, i.e., the trajectory which given the delta-v the ship is capable of, is the shortest transit time path.
If you have a ship that accelerates at 1g to the half-way point and then flips over and decelerates at 1g, Mars is 2-5 days away (depending on the relative positions of Mars and Earth) and you can pretty well ignore the orbital velocities of the planets and just "head for Mars".
If the ship is capable of 1g for less time -- say just a day, total, then you accelerate towards Mars for a half-day, coast at around 400 kps, and then a half-day out from Mars, decelerate back down to zero velocity. This would be a much slower trip? Not that much slower -- the coast time would typically be in the 5-10 day range. (Planetary movement is still barely worth worrying about.)
If you 'only' have the capability of 2 hours of 1 g acceleration (1 hour at the start and another hour at the end), you hit a top velocity of around 35 kps, and the coast time is now maybe 50 days, and you need to take orbital motions carefully into account.
Bring the available acceleration time down another factor of ten and you pretty much have an ordinary chemical rocket!
There is a family of shortest time trajectories starting with 1g for the whole trip and ending at 1g for a few minutes. The former is a full-up torch ship; the latter is a Falcon 9 Heavy. At one end the trajectory is pretty much a straight line and at the other it's a Hohmann orbit and all the others are in between.
So an in between torchship would have an in between trajectory -- straighter than a Hohmann orbit, but curvier than a 1-g-all-the-way orbit.
answered Dec 25, 2018 at 4:34