3
$\begingroup$

In the book, project Hail Mary, Eridians were not aware of relativity. It is claimed that, due to this, they calculated that they would need much more fuel than actually needed for interstellar travel. The reason being time dilation. I believe, this is missing an important aspect of relativistic mass. The ship should consume more and more fuel for progressive acceleration due to relativistic mass. So what is the net effect of relativistic mass, time dilation and length contraction on fuel consumption calculation? Would we end up calculating more fuel is required or less fuel is required if we didn't know relativity? Is the answer dependent on actual trajectory, acceleration etc?

Edit (additional information): What we know about their trip plan is that they had acceleration, constant velocity and deceleration phase. Also, the time to reach to destination(in ship's frame of reference) is not constant. They ended up reaching their destination earlier than planned AND with lesser fuel than planned.

$\endgroup$
1
  • 6
    $\begingroup$ Remember that not everyone has read every book in existence. Can you add more details on the plot? For example, are they traveling with constant thrust? Or with an acceleration burn, a coasting phase and a deceleration burn? Because with the latter scenario, the amount of fuel you need (in addition to reaching escape velocity) depends solely on how long you are willing to wait until you arrive. $\endgroup$
    – Philipp
    Nov 25, 2021 at 11:01

1 Answer 1

5
$\begingroup$

Here is a simplified calculation for an ideal rocket. The classical rocket equation to obtain a change $\Delta v$ in speed is $$\Delta v = v_\text{e} \ln \frac{m_0}{m_f} $$

where $v_e$ is the exhaust speed, $m_0$ is the initial mass and $m_f$ is the final mass $m_f<m_0$.

If you account for special relativity, you get

$$ \Delta v= c\tanh\left(\frac{v_\text{e}}{c} \ln \frac{m_0}{m_f}\right )$$

where $c$ is the speed of light. For small $v_e\ll c$, both equations are equivalent.

In general, $x>\tanh(x)$ for $x>0$. This means that for a smaller $m_0/m_f$ (less propellant) you get a larger change in speed with the classical equation.

A more detailed calculation would need to include type of rocket, trajectories, acceleration and much more.

Source: Tsiolkovsky rocket equation

A case for the Eridians

Maybe Eridians where not thinking in those terms. They just considered the following itinerary: accelerate the ship to speed $v_0$, turn down the engine, reach planet at constant speed.

If that is the case, maybe they thought they needed a large speed to get to destination fast. But the time needed could be slightly less due to length contraction.

They wanted to reach a distance $L$ at constant speed $v_0$ in a time $t$. Classically $t=L/v_0$, but in special relativity is $t=L\sqrt{1-v_0^2/c^2}/v_0$. That means that for a given speed, the observed time for the occupants of the ship is less than what is expected classically.

Conclusion: maybe Eridians thought they needed a larger speed than what is actually needed.

Disclaimer: I do not know nothing about the book. Yet without a precise itinerary/speeds/acceleration the conundrum is still unsolved. If the trip is as described above maybe you can use the rocket equation and the Eridians paradox holds (they used more than what they needed, see section below), but if the trip is different you need to compare how much fuel you need in comparison to the gain in time.

I am also neglecting the time to accelerate and I probably should account for deceleration in order to land but that is symmetric to the acceleration part. So I guess it is actually twice the fuel but the arguments below should still hold.

Some calculations

Suppose that the Eridians wanted to achieve distance $L=1\;\mathrm{ly}$ in $t=3\;\mathrm{years}$ (as perceived from inside the ship).

Classically they would need $v_0=c/3$, relativistically they need $v_0=c/\sqrt{10}=c/3.16...$.

Putting it into the rocket equation, setting $x=v_e/c \ln m_0/m_f$,we get that classically $x=1/3$, while relativistically $x=\tanh^{-1}(1/\sqrt{10})=0.3272$. Which means that they could have used slightly less fuel if they had accounted for relativity.

Let us write the equation for a given $L/tc$:

Classically $$x=\frac{L}{tc}$$ while relativistically

$$x= \tanh^{-1}\left[\frac{1}{\sqrt{1+1/(L/tc)^{2}}}\right]\approx \frac{L}{tc}\left[1-\frac{1}{6}\left(\frac{L}{tc}\right)^2\right]$$

The $x$ calculated relativistically is alway smaller than the expected classically. Which means that by actually taking into account time dilation, you always need less fuel! (according to the trip above).

The book was right I guess, (ideal) Eridians were indeed wrong.

$\endgroup$
5
  • $\begingroup$ This is a nice answer because $\Delta v$ is one of the most important parameters in space travel because it also takes into account possible manoeuvers one could make (at least from what I learned from kerbal space program). But this answer seems in contradiction with the question: the book predicted the Eridians would have surplus fuel while you predict they would fall short if I interpret it correctly. $\endgroup$ Nov 25, 2021 at 11:36
  • $\begingroup$ I do not know about the Eridians, maybe I misunderstood the question? If Eridians used classical mechanics they should have predicted less fuel. But if they used the right equations but did not account for time dilation then I guess they could have used less fuel as the path would have taken less time? $\endgroup$
    – Mauricio
    Nov 25, 2021 at 11:44
  • $\begingroup$ I don't know the book either but based on _"they calculated that they would need much more fuel than actually needed for interstellar travel." I would suspect that they predicted to need more fuel than was actually necessary based on special relativity. Maybe when you include time dilation this drops out but it's hard to tell without more context of their travel $\endgroup$ Nov 25, 2021 at 11:56
  • 1
    $\begingroup$ @AccidentalTaylorExpansion I expanded on the second point with estimations now. $\endgroup$
    – Mauricio
    Nov 25, 2021 at 12:34
  • 1
    $\begingroup$ @AccidentalTaylorExpansion actually the answer is not trivial, but you need always less fuel if you account for relativity. $\endgroup$
    – Mauricio
    Nov 25, 2021 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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