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2d = 1/2 A t^2 So you're off by at least a factor of two to start with. Beyond that I suppose you have to guess at what "high burn" means to people who live in low gravity. (I have not read the books).– EthanCommented Apr 14 at 18:44
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1Also note that IRL "high burn" refers to fuel consumption rather than acceleration per se. So maybe in the Expanse a high burn flight plan refers to one that uses a lot of fuel compared to some hypothetical alternative flight plan. This would not necessarily imply that the mean acceleration during travel is > 1G.– EthanCommented Apr 14 at 20:41
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@Ethan having read the books, it does mean a high-g burn. And having read the following books in the series, I didn't make a deep analysis of it but in particular in book 9 there seems to be big inconsistencies in travel times through systems.– GouvernathorCommented Apr 14 at 21:07
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the equation you cite/use has the specific stipulation that it is only good for straight line motion. There are virtually NO 'straight line' travels between orbital bodies. The overly simplified flight paths commonly used to show things in movies and TV do not do justice to the actual curved path traveled. ALSO... I'd ask you be be very certain that you are using the proper units of measure for all factors. A single misapplied unit really bungles everything! Visit the space exploration section to seek understanding of transfers.– BradVCommented Apr 14 at 21:50
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1@BradV Those continuous burn trajectories are very different to our beloved Hohmann and bi-elliptic transfers. The acceleration dwarfs the Sun's field strength. Eg, at 1 AU, the Sun's gravity is ~ 0.00593 m/s^2 = 0.0006047 g. So if you're pulling 1/3 g you can virtually ignore the Sun's gravity.– PM 2RingCommented Apr 15 at 7:26
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