1

I am wondering about the travel times in The Expanse novels, and not for the first time.

I am currently at the beginning of book 4, Cibola Burn (2014), and the Rocinante has passed the ring, traveling to Illus/New Terra. Alex announces that they can arrive after a flying time of over 70 days on a high burn.

But here is the problem: The math does not seem to make any sense. Suppose they fly at a constant acceleration A for the first half of the trip, turn the ship around and fly at a constant deceleration -A for the second half of the trip (this is the standard travel mode in The Expanse, as is repeatedly said). Then after time t, they would have traveled the distance x(t) = 2A(t/2)^2. Suppose that A=g, the acceleration due to gravity on earth (which would certainly not be a "high burn" schedule), then they would travel 1199 AU in 70 days. (Just plug *"2 * (35 days)^2 * acceleration due to gravity in astronomical units" into Wolfram Alpha). On high burn, they would get much faster. It is not said how far out the ring is in the solar system of Illus, but 1199 AU is very far out.

The 18 month travel time for the research vessel to reach Illus from Luna are also not realistic. In this time, they would have traveled many thousand AU.

In novel 3, Abaddon's Gate (2013), it is mentioned that the Rocinante needs around 3 months travel time to reach the ring from Ceres. At 1g, the Rocinante would travel 1982 AU in this time, and still 660.6 AU at 1/3*g. But the ring is between the orbit of Neptune and Uranus, so certainly less than 30 AU out. So these numbers do not match up, even if the ring was at the opposite side of the sun from Ceres at the beginning of the journey. 660 AU is far outside the sun's heliosphere.

Also in novel 3, it is said that it would take "only" 12 days to reach Mars from Earth if the orbits are aligned, when really, it would take not even 2 days at 1g.

I am very confused by this, because I was under the impression that the scientific facts of the books are very well researched. I cannot understand why this rather simple math would be so far off. Am I missing something?

Improve this question
9
  • 2
    d = 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).
    – Ethan
    Commented Apr 14 at 18:44
  • 1
    Also 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.
    – Ethan
    Commented Apr 14 at 20:41
  • @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.
    – Gouvernathor
    Commented Apr 14 at 21:07
  • 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.
    – BradV
    Commented Apr 14 at 21:50
  • 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 2Ring
    Commented Apr 15 at 7:26

1 Answer 1

Reset to default
2

In the expanse the ships are limited not by their ability to burn but by their occupant's ability to withstand said burns. In Cibola Burn in particular we learn that Naomi cannot handle the gravity of a planet. Although I cannot recall immediately, this implies that the burn is much lower than 1g. In the final books there are references to a "comfortable" 1/3 g being the preferred burn.

Next, the gates do not orbit their stars (this is an important point in the last few books). Rather they are stationary. So not only do the ships have to essentially cancel the orbital speed entirely when approaching a gate, they have to get up to a reasonable orbital speed when leaving one. Consider that Earth orbits the Sun at 30km/s so any ship leaving Sol gate heading for Earth has a crazy amount of burning they need to do.

We know that the gates are often extremely far from their inner planets, and ilus most definitely is an inner planet of its star system.

Finally, in the last books all of the between gate travel has a major impact on their reaction mass. They spend a lot of time not burning to conserve it.

So all of this implies that the travel time from Earth to ilus is indeed quite long. For the Barbapicola it isn't unreasonable to assume that it has to burn even slower, especially when full of ore.

Improve this answer
2
  • 1
    30 kps are not crazy for ships with an Epstein drive. So in all I would call the numbers slightly fishy ..
    – o.m.
    Commented Apr 15 at 5:25
  • There's no room to maneuver in the pocket universe at high velocities meaning you'd be approaching the ring traveling fairly slow following a deceleration phase to be able to hit an arbitrary exit gate. Plus there's the potential for other traffic. Slowing down in the ring zone at high G with a fusion drive means a lot of radiation and potential for destroying other spacecraft with your drive exhaust. Then what happens if you're forced on the float with significant unshed velocity?
    – user16145658
    Commented Apr 16 at 9:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.