Relativistic Space Travel: Optimizing Proper Time [Project Hail Mary]

[rocky]

TL;DR

Minor book spoilers. Optimizing for time in a fuel-constrained interstellar trip

Warning: Minor spoilers for the book Project Hail Mary ahead.

A ship has travelled to Tau Ceti (11.9 light years from Earth). The ship uses light as propulsion, and effectively converts the fuel mass into light energy. It carries 2,000,000 kg of fuel. During the journey it underwent a constant acceleration of 1.5g, speeding up until the midpoint, then slowing down in the second half. The ship experienced just under 4 years of proper time during this journey.

During a mishap, the ship jettisons 3 of its 9 fuel bays, reducing the total fuel capacity to 2/3 of the original amount. The ship refuels, and prepares to return to Earth. The pilot calculates that the most efficient (author's wording) course is a constant acceleration of 0.9 g, which will experience 5.5 years of proper time. However, they only have about 4 years worth of food. I am assuming that "efficient" is relating to time, since that is the limiting factor for the ship's occupant.

My question is: Why would the 0.9 g course be the most efficient? If they simply accelerate at 1.5 g up to the same maximum rapidity of this course, then coast in the middle before slowing down similarly, the proper time is significantly shorter. The attached image shows the math for the following three scenarios:

1. Constant acceleration at 1.5g, which takes 3.9 years.
2. Constant acceleration at 0.9g, which takes 5.5 years.
3. Acceleration at 1.5g up to the same top speed from scenario 2, coasting at that speed, then slowing again at 1.5g. This only takes 4.05 years.

Andy Weir's work is generally somewhat accurate, but I could not figure out why this course might make sense. I could only think of a few possible reasons for the planned 0.9g course, but none of them really hold up:

1. The engines are less efficient at higher acceleration. This is never mentioned, and doesn’t really make sense with how they work in the book.
2. The maximum thrust is impacted by the missing fuel bays. This would be a plausible explanation, but I don’t believe it is ever mentioned in the book.
3. The pilot simply didn’t think of this, and only calculated a constant acceleration course.

From a physics perspective, is there something that I'm missing here? If anyone is familiar with the book, was there some other reason, or does my proposed course make sense?

 

[rocky]

My image got compressed beyond readability, so here is the PDF version

 

[PeterDonis]

@rocky please post your equations directly in the thread using the PF LaTeX feature. (There is a LaTeX Guide link at the bottom left of every post window.) Posting equations and text in attachments is not allowed.

 

[rocky]

@rocky please post your equations directly in the thread using the PF LaTeX feature. (There is a LaTeX Guide link at the bottom left of every post window.) Posting equations and text in attachments is not allowed.

Apologies, here is the content of the attachment:

Relevant equations:
Proper time to travel a given distance at constant acceleration:
$\tau = \frac{c}{a}\cosh^{-1}\left(1 + \frac{a D}{c^2}\right)$
Rapidity:
$\phi = \frac{a \tau}{c}$
Velocity at a given rapidity:
$v = c \tanh{\phi}$
Distance traveled while accelerating up to a given rapidity:
$D = \frac{c^2}{a}(\cosh{\phi} - 1)$
Lorentz factor:
$\gamma = \cosh{\phi}$
Proper time experienced during a given Earth-time and Lorentz factor:
$\tau = \frac{t}{\gamma}$

Scenarios:

1. Constant acceleration of 1.5 g's.
2. Constant acceleration of 0.9 g's.
3. Acceleration of 1.5 g's up to the max rapidity of scenario 2, coasting, and negative acceleration of 1.5 g's.

Scenario 1:
Proper time to travel 5.95 light years (half of 11.9 ly, the distance to Tau Ceti) with constant acceleration of 1.5 g's.
$\tau_{half} = \frac{c}{1.5g}\cosh^{-1}\left(1 + \frac{(1.5g) (5.95 ly)}{c^2}\right) \approx 1.95 y$
Total trip duration:
\tau_{total} = 2 * \tau_{half} \approx 3.9 y

Scenario 2:
Proper time to travel 5.95 light years (half of 11.9 ly, the distance to Tau Ceti) with constant acceleration of 0.9 g's.
$\tau_{half} = \frac{c}{0.9g}\cosh^{-1}\left(1 + \frac{(0.9g) (5.95 ly)}{c^2}\right) \approx 2.76 y$
Total trip duration:
$\tau_{total} = 2 * \tau_{half} \approx 5.5 y$

Scenario 3: Using the same fuel limitation from scenario 2, but accelerating at 1.5 g's until reaching maximum rapidity.
Maximum rapidity from Scenario 2:
$\phi_{max} = \frac{(0.9g)(2.76y)}{c} = 2.5633$
Rearrange to get the proper time to reach max rapidity with 1.5 g acceleration.
$\tau_{burn} = \frac{c}{a}\phi = \frac{c}{1.5g}2.56 = 1.66 y$
Distance traveled during each burn:
$D_{burn} = \frac{c^2}{1.5g}(\cosh{2.56} - 1) \approx 3.6ly$
Distance to coast:
$D_{coast} = 11.9ly - 2 (3.6ly) = 4.7ly$
Maximum speed:
$v = c \tanh{2.56} = 0.988c$
Earth time to coast 4.7ly at 0.988c:
$t = \frac{4.7ly}{0.988c} = 4.76y$
Lorentz factor at a rapidity of 2.5633:
$\gamma = \cosh{2.5633} = 6.53$
Proper time experienced while coasting:
$\tau_{coast} = \frac{t}{\gamma} \approx \frac{4.76y}{6.53} \approx 0.73y$
Total proper time for Scenario 3:
$\tau_{total} = \tau_{burn} + \tau_{coast} + \tau_{burn} \approx 1.66y + 0.73y + 1.66y = 4.05y$

 

[Ibix]

Could be optimising for other things - fuel consumption is the obvious bound. Can the ship burn at 1.5g for long enough to follow your proposed plan?

 

[rocky]

Could be optimising for other things - fuel consumption is the obvious bound. Can the ship burn at 1.5g for long enough to follow your proposed plan?

My proposed plan assumes it can reach the same rapidity as the calculated plan. Basically, I’m wondering why the author proposed a plan with a lower acceleration. It kind of seems like a plot hole, since the ship is capable of higher acceleration, and the fuel constraint doesn’t seem to increase the journey duration that much.

 

[Ibix]

Ah yes, sorry - mixed up what you were proposing.

I haven't checked your numbers but your approach is correct. A higher acceleration means that at any point on the journey your speed is never less than, and often greater than, it would be at the same place on a lower acceleration profile. So you minimise elapsed time by maximising acceleration (true with Newton, even more so with Einstein).

Either there's some other constraint (no prolonged periods of zero g allowed? Engine efficiency is higher at lower thrust?) or it's a mistake.

 

[rocky]

Ah yes, sorry - mixed up what you were proposing.

I haven't checked your numbers but your approach is correct. A higher acceleration means that at any point on the journey your speed is never less than, and often greater than, it would be at the same place on a lower acceleration profile. So you minimise elapsed time by maximising acceleration (true with Newton, even more so with Einstein).

Either there's some other constraint (no prolonged periods of zero g allowed? Engine efficiency is higher at lower thrust?) or it's a mistake.

Thanks! I couldn't think of any constraint that made sense in the book. I don't believe engine efficiency was mentioned. The way the engine works in the book is essentially by converting mass to light energy and shining that behind the ship. I was just hoping to confirm that my math made sense, and see if anyone else had noticed this in the book.

 

[pervect]

My gut reaction is that it's unrealistic to believe one can maintain such a high acceleration for such a long time. But that's not your question. I did notice that you didn't have the mass ratio equation in your set of equations, https://www.desy.de/user/projects/Physics/Relativity/SR/rocket.html gives the simple formula M/m =exp(aT/c)-1, where T is proper time. A would be in the neighborhood of 1.5 light years/year^2, and T was about 4 years of proper time, so the author appears to be assuming a mass ratio of about e^6, so about 400 or so. Which seems pretty ambitious. And a bit unnatural to have a fixed acceleration rather than a thrust limited one, especially considering the very high mass ratio.

Onto your actual question. I believe you are correct to think that if the rocket has the capability of accelerating at 1.5g, it would be a faster trip to use the full available acceleration, then coast. This would require powering off the engines to coast, of course. It's easy to imagine some authorial fiat where one need to keep the engines running (perhaps for power generation or whatnot). But apparently the author didn't do that. I haven't read the book.

I haven't worked through the numbers in any detail, but your observation seems OK to me. You can compare your equations to the reference I gave - I didn't do a close comparison.

 

[Ibix]

And a bit unnatural to have a fixed acceleration rather than a thrust limited one, especially considering the very high mass ratio.

Depends on the limiting factor. If the (magic star-eating protozoa powered) engine can provide very high thrust then the limiting factor is the ability of the ship and crew to stand up to the acceleration. In that scenario fixed thrust of ~1g might actually make sense.

 

[DaveC426913]

Why guess when you can just ask him?

 

[DaveC426913]

I just pulled up the relevant passage and he does say he has enough food to survive the trip home at full fuel but nowhere near enough to the last five and a half years of a slower trip.

The implication does seem to be that the engine is more fuel efficient at .9g than it is at 1.5g.

 

[rocky]

After reviewing the book a bit, I found the mass of the ship, but now I'm even more confused.
The ship weighs 100,000kg, and carries 2,000,000kg of fuel. With a perfect fuel efficiency, that mass ratio can only get up to 0.91c if it needs to slow down as well.
$v=c\frac{R^2-1}{R^2+1}=c\frac{20}{22}\approx0.91c$
But that's lower than the velocity that would need to be reached on the initial journey in order to make it in the described time.

 

[DaveC426913]

After reviewing the book a bit, I found the mass of the ship, but now I'm even more confused.
The ship weighs 100,000kg, and carries 2,000,000kg of fuel. With a perfect fuel efficiency, that mass ratio can only get up to 0.91c if it needs to slow down as well.
$v=c\frac{R^2-1}{R^2+1}=c\frac{20}{22}\approx0.91c$
But that's lower than the velocity that would need to be reached on the initial journey in order to make it in the described time.

Are you accounting for the loss of a third of the ship's fuel on the outward bound leg?

 

[rocky]

Are you accounting for the loss of a third of the ship's fuel on the outward bound leg?

I’m not sure I understand the question. Doesn’t that equation give the max speed that can be attained with a given mass ratio, accounting for the need to slow down? The ship refuels after the journey, it doesn’t carry fuel to make a return trip.

 

[PeterDonis]

Are you accounting for the loss of a third of the ship's fuel on the outward bound leg?

As I understand it, the loss is not "on the outward bound leg", it's after arrival at Tau Ceti but before the ship refuels. And it's not a loss of a third of the ship's fuel, but of a third of the ship's fuel capacity. The ship's fuel tanks are basically empty when the mishap happens--but when it has to refuel to return to Earth, it only has 2/3 of the fuel capacity that it did before.

 

[PeterDonis]

Doesn’t that equation give the max speed that can be attained with a given mass ratio

But the mass ratio is different on the return trip than it was on the outbound leg, because of the loss of 1/3 of the ship's fuel capacity. The payload mass is still the same--100,000 kg--but with only 2/3 of the fuel. So the mass ratio is only 2/3 of what it was on the outbound trip.

 

[PeterDonis]

The ship weighs 100,000kg, and carries 2,000,000kg of fuel.

Is this on the outbound trip, or the return trip?

 

[DaveC426913]

I’m not sure I understand the question. Doesn’t that equation give the max speed that can be attained with a given mass ratio, accounting for the need to slow down? The ship refuels after the journey, it doesn’t carry fuel to make a return trip.

Just wanted to ensure you were accounting for all relevant factors in the trip.

 

[Ibix]

After reviewing the book a bit, I found the mass of the ship, but now I'm even more confused.
The ship weighs 100,000kg, and carries 2,000,000kg of fuel. With a perfect fuel efficiency, that mass ratio can only get up to 0.91c if it needs to slow down as well.
$v=c\frac{R^2-1}{R^2+1}=c\frac{20}{22}\approx0.91c$
But that's lower than the velocity that would need to be reached on the initial journey in order to make it in the described time.

If you are intending to stop dry and you are defining $R$ as the fuelled/dry mass ratio there shouldn't be squares on the $R$ in your peak speed formula. You do seem to have applied this correctly given your $10^5\mathrm{kg}$ dry mass and $2\times 10^6\mathrm{kg}$ fuel mass, so presumably the squares are typos.

 

[rocky]

If you are intending to stop dry and you are defining $R$ as the fuelled/dry mass ratio there shouldn't be squares on the $R$ in your peak speed formula. You do seem to have applied this correctly given your $10^5\mathrm{kg}$ dry mass and $2\times 10^6\mathrm{kg}$ fuel mass, so presumably the squares are typos.

Yes, sorry I didn’t clarify. R being the mass ratio required for each burn, so R^2 is the mass ratio for the whole trip.

 

[rocky]

Is this on the outbound trip, or the return trip?

Outbound trip. So to recap: ship mass 100,000kg, fuel 2,000,000 kg. Constant acceleration/deceleration at 15m/s^2 over a distance of 11.9 light years. It's supposed to take just under 4 years of proper time (which checks out for that acceleration and distance), but from what I can tell that would require the ship to be moving faster than that mass ratio would allow.

 

[PeterDonis]

R being the mass ratio required for each burn, so R^2 is the mass ratio for the whole trip.

No, it's not, because first, as I've pointed out, $R$ is not the same for both legs because the fuel capacity is reduced for the return trip, and second, because you're refueling for the return trip, so you don't have to carry the fuel for the return trip on the outbound leg.

 

[PeterDonis]

Outbound trip. So to recap: ship mass 100,000kg, fuel 2,000,000 kg.

Are you sure that mass ratio is for the outbound trip and not the return trip?

 

[rocky]

No, it's not, because first, as I've pointed out, $R$ is not the same for both legs because the fuel capacity is reduced for the return trip, and second, because you're refueling for the return trip, so you don't have to carry the fuel for the return trip on the outbound leg.

Sorry, I wasn't clear there. That's meant to be the mass ratio for the outbound trip, I just wanted to clarify that it includes acceleration and deceleration. Am I using the wrong formula?
The return trip would have a reduced mass ratio: 1,333,333kg fuel for a 100,000kg ship, although the ship weight might be reduced somewhat with some fuel bays removed. I don't think that was mentioned in the book.
However, I was just focusing on the outbound trip now, because I'm not sure that fuel ratio can reach the required velocity to make the trip as described (constant acceleration/deceleration of 1.5g to travel 11.9 light years, in under 4 years of proper time).

 

[Ibix]

Am I using the wrong formula?

If $R$ is the ratio of fuelly fueled mass to mass at turnover, which is also the ratio of mass at turnover to dry mass, then $R^2=21$ is correct given your numbers.

I haven't run the proper time figures given a continuous acceleration profile with a peak speed of 0.91c. Given his line of questioning I suspect @PeterDonis has, and has found that the mass figures you are quoting are consistent with the numbers for the return journey.

 

[PeterDonis]

Given his line of questioning I suspect @PeterDonis has

Actually, I haven't, because I wasn't clear on the problem specifications until now; my line of questioning has simply been to get those clear so we all are analyzing the same thing. I had thought the 2 millon kg fuel mass was for the return trip (because it makes the numbers nice and round--3 million kg for the outbound trip and 2 million for the return), but I wasn't sure, and as it turns out, rightly so.

 

[pervect]

My recollection is that there is a simple exponential relationship between mass ratio and maximum rapidity, regardless of the thrust profile.

Something like w = ln(M), with a perfectly efficient photon rocket, where w is the rapidity and M is the mass ratio of the rocket. However, I haven't been able to find a good source for this and my recollection might be wrong or incomplete.

Basically, in a one dimensional problem, rapdity adds linearlly.

My main source though, Baez's old relativistic rocket FAQ, https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html, (or one of it's variant sites), doesn't seem to have this relationship.