posted by
damerell at 04:21pm on 03/12/2015
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I thought I'd found a bug in The Martian but
davidgillon pointed out I hadn't. This is preserved for curiosity's sake.
All figures come from here and appear to be KORECT although they are sometimes in stupid 18th-century units.
The final scenes of the book depend on the premise that the _Hermes_ is approaching Mars at a greater speed than that intended for rendezvous with the MAV, and so the MAV must have its weight drastically reduced in order to have enough delta-v to reach that speed. This doesn't quite work so there is clever improvisation etc etc.
However, the Hermes has an ion drive. This type of drive has a very high specific impulse (the amount of shove you get per kg of reaction mass) but a very _very_ low thrust to weight ratio. We are told, in fact, the Hermes can accelerate by 2 mm/s/s/.
The rendezvous is at 100km above Mars, where the orbital velocity is c. 3.5 km/s. The Hermes is arriving at 5.36km/s. Rather than the MAV getting up to that speed, can the Hermes slow down?
Normally, the idea that a space mission can find an extra 2km/s or so of delta-v is ridiculous. However, it's not so ridiculous with an ion drive where the quantity of reaction mass is small... and especially not because nearly all ion thruster designs use xenon as reaction mass and the Hermes has a fission reactor onboard which generates small quantities of stable isotopes of xenon.
(There is an assumption here that NASA built a capability into the Hermes to capture and use that xenon. However, the Hermes is meant to be by far the most capable and expensive spacecraft they ever built - and if the ship can capture the xenon it generates, it both has more flexibility for mission profile changes and NASA doesn't have to spend so much money sending up xenon to resupply it.)
The first objection I got was that if the Hermes slows down, it will delay the rescue mission long enough for Watney to die. However, I make the deceleration period 10.5 Sols and since our average speed during that period is circa 4.4km/s we need to travel for an extra 2.4 Sols at 5.36 km/s to make up the loss - of course, this isn't totally accurate because Mars doesn't stay put but it should be close enough.
So, anyone, tell me why I have not broken the Martian?
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All figures come from here and appear to be KORECT although they are sometimes in stupid 18th-century units.
The final scenes of the book depend on the premise that the _Hermes_ is approaching Mars at a greater speed than that intended for rendezvous with the MAV, and so the MAV must have its weight drastically reduced in order to have enough delta-v to reach that speed. This doesn't quite work so there is clever improvisation etc etc.
However, the Hermes has an ion drive. This type of drive has a very high specific impulse (the amount of shove you get per kg of reaction mass) but a very _very_ low thrust to weight ratio. We are told, in fact, the Hermes can accelerate by 2 mm/s/s/.
The rendezvous is at 100km above Mars, where the orbital velocity is c. 3.5 km/s. The Hermes is arriving at 5.36km/s. Rather than the MAV getting up to that speed, can the Hermes slow down?
Normally, the idea that a space mission can find an extra 2km/s or so of delta-v is ridiculous. However, it's not so ridiculous with an ion drive where the quantity of reaction mass is small... and especially not because nearly all ion thruster designs use xenon as reaction mass and the Hermes has a fission reactor onboard which generates small quantities of stable isotopes of xenon.
(There is an assumption here that NASA built a capability into the Hermes to capture and use that xenon. However, the Hermes is meant to be by far the most capable and expensive spacecraft they ever built - and if the ship can capture the xenon it generates, it both has more flexibility for mission profile changes and NASA doesn't have to spend so much money sending up xenon to resupply it.)
The first objection I got was that if the Hermes slows down, it will delay the rescue mission long enough for Watney to die. However, I make the deceleration period 10.5 Sols and since our average speed during that period is circa 4.4km/s we need to travel for an extra 2.4 Sols at 5.36 km/s to make up the loss - of course, this isn't totally accurate because Mars doesn't stay put but it should be close enough.
So, anyone, tell me why I have not broken the Martian?
More importantly...
(no subject)
If I'm understanding you correctly, you're saying additional deceleration is possible using the additional xenon generated by the reactor. My immediate reaction is that you're overestimating the amount generated. I think you need to quantify the xenon produced by the reactor versus the xenon consumed by the additional acceleration.
This also assumes the Hermes has the power to sustain the additional ion thruster use. That needs to be justified. if your reactor has too much power, then that's weight that can be designed out to give a cheaper mission.
And the one that sneaks up on you over the longer term:
What does slowing down do to the long term mission. Do they catch the MAV, but miss Earth....
(no subject)
From the linked paper, it does not seem unreasonable to suggest we might get an Isp of 4000s and hence an exhaust velocity of 40km/s or so. For the Hermes to accelerate at 2 mm/s/s the thrusters must exert a force of 220N. Hence we consume .55 grams of xenon per second, wanting about 6 tonnes of xenon for the maneuver. (Do I get about the same answer from conservation of momentum? Yes, I do. Hooray!)
And... no, you're right. The xenon production of the reactor is miniscule compared to that. Good catch.
I think they're OK in terms of the long-term mission since I had them decelerating into orbit; they can then accelerate again (using the xenon they, sigh, have not got) into something like the Earth-catching trajectory in the book.