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

[PeterDonis]

the mass ratio calculations don't actually seem to support the top speeds required for the courses described in the book.

Using the formulas I gave in post #30 just now, if we know the maximum rapidity $\alpha$, the proper time $T$ required to reach it at proper acceleration $a$ is given by

$$
T = \frac{\alpha c}{a}
$$

Plugging in $\alpha = \ln 21$ and $a = 1.5 \text{g}$, I get $T = 1.97$ years. That looks consistent with the time you gave in the OP.

 

[rocky]

Plugging in α=ln⁡21 and a=1.5g, I get T=1.97 years. That looks consistent with the time you gave in the OP.

Does this account for the need to slow down? Don’t we need to use the square root of the mass ratio?

 

[PeterDonis]

Does this account for the need to slow down?

No, you're right, it doesn't. But I suspect that that might be the underlying error in the calculations behind the scenario in the book, if the numbers you gave in the OP are correct.

Btw, where did you get the numbers in the OP (the 100,000 kg payload and 2,000,000 kg fuel mass, and the claimed trip times and accelerations)? Are they actually given explicitly in the book?

 

[PeterDonis]

Btw, an equation for the mass ratio (including the need to turn around and slow down) in terms of just the acceleration and the distance is:

$$
R = \exp \left[ 2 \cosh^{-1} \left( \frac{ad}{c^2} + 1 \right) \right] - 1
$$

This makes use of the equations $R = \exp \left( aT / c \right) - 1$ and $a T / c = \cosh^{-1} \left( ad / c^2 + 1 \right)$ from the relativistic rocket equation page. Note that the latter equation is for a $d$ of one-half of the total distance; in my equation above, I multiply this by $2$ and plug in $d / 2$ for the distance so that we can just use $d$ as the total distance to be covered (11.9 light years in the case of the Earth to Tau Ceti trip).

For $a = 1.5$ g and $d = 11.9$ ly, I get an $R$ from the above of $413.5$.

 

[rocky]

Btw, where did you get the numbers in the OP (the 100,000 kg payload and 2,000,000 kg fuel mass, and the claimed trip times and accelerations)? Are they actually given explicitly in the book?

Yes, those numbers came from the book. The full capacity fuel mass is 2,000,000kg. The dry mass of the ship is described as "around 100,000kg", but it would need to be about 20 times lighter for the mass ratio to work. The acceleration for the ship on the outbound journey might actually be 15 m/s^2, but it's generalized as 1.5g throughout the book.
And the times for the trips are all specifically mentioned in the book too:

• The trip from Earth to Tau Ceti took three years and nine months. And it was done by accelerating constantly at 1.5 g’s the entire time
• If I do the long trip home with just 1.33 million kilograms of fuel (which is all my remaining tanks can hold), the most efficient course is a constant acceleration of 0.9 g’s. I’d be going slower, which means less time dilation, which means I experience more time. All told, I’ll experience five and a half years on that trip.

The times for those courses mostly check out with my calculations, but even a perfectly efficient mass ratio for a photon rocket doesn't support the top speeds they would attain. My initial question was about why 0.9 g would be more efficient than 1.5g, when optimizing for time. The only answer that makes sense is that the engines are less efficient at 1.5g compared to 0.9g, but that doesn't really fit with how they work in the story. I ran some quick calculations for the 1.5g course with a lower maximum rapidity, and it looks like it would need to be below 70% of the max rapidity from the 0.9g course in order for the 0.9g course to get there faster.

For a=1.5 g and d=11.9 ly, I get an R from the above of 413.5.

That's about what I was getting as well.

 

[PeterDonis]

those numbers came from the book

Ok, thanks!

 

[PeterDonis]

even a perfectly efficient mass ratio for a photon rocket doesn't support the top speeds they would attain.

Thinking of it in terms of top speed puts in an extra step that's not necessary. See my post #34. The mass ratio (for a trip where you have to slow down and stop at the end) can be expressed solely in terms of the proper acceleration and the distance to be traveled. And that formula can be inverted to express the proper acceleration that can be sustained given a mass ratio and a distance to be traveled.

What that formula doesn't cover is the case where you coast for part of the time in the middle of the trip. That would require a more complicated formula, with an extra parameter to express how much of the trip is spent coasting, but it should be workable.