Project Hail Mary - Why does a return trip to another star require 10x the fuel compared to a one-way trip?

In Project Hail Mary by Andy Weir, the protagonist states that sending a ship to another star and bringing it back would take ten times as much fuel as a one-way trip. The relevant quote is:

"Sending a ship to another star probably took an absurd amount of fuel. Sending that ship to another star and bringing it back would take ten times as much fuel."

I understand the basic idea that carrying extra fuel adds weight, which in turn increases fuel requirements. However, intuitively, I would have guessed the required fuel for a return trip would be closer to double the one-way trip's fuel.

Any insights into the physics and math behind this claim would be appreciated.

Let's keep it classical. The rocket change of speed calculation is

$$\Delta v =v_e \ln \frac{m_0}{m_f}$$ where $v_e$ is the exhaust speed, $m_0$ the initial mass and $m_f$ the final mass. That means that to have that change in speed you need the initial mass $$m_0=m_f\exp\frac{\Delta v}{v_e}$$

If you account for acceleration and deceleration you get a factor of 2 in the exponential

$$m_{0,1}=m_f\exp\frac{2\Delta v}{v_e}$$

where the subscript 1 means that you did one trip. If you do a round trip (2 trips) you get

$$m_{0,2}=m_{0,1}\exp\frac{2\Delta v}{v_e}$$

You want the ratio: $$\frac{m_{0,2}}{m_{0,1}}=\exp\frac{2\Delta v}{v_e}=10$$

So yeah in the classical ideal case it seems to depend on the exhaust speed vs cruise speed ratio, not much on the payload.

Note that the ratio is what enters in the efficiency: $$\eta=\frac{2 (\Delta v/v_e)}{1+(\Delta v/v_e)^2}$$ so it is directly tied to the rocket efficiency. If the efficiency is $\eta=1$ ($\Delta v \approx v_e$) you get already a factor of $e^2 \approx 7.4$.

Disclaimer: I realized that the calculation above uses the total weight, but one could compare the propellent weight $\Delta m = m_0-m_f$, providing the following ratio:

$$\frac{\Delta m_2}{\Delta m_1}=1+\exp \frac{2\Delta v}{v_e}$$

in that case you the ratio is about 8.4 at $\eta=1$.

Let's assume the star is similar to the sun, and let's ignore any planets where we could make (or buy ;-) fuel. Assume it takes x tons of fuel to send the ship to that star. Then it obviously will take the same x tons to return the ship back to earth - as you imply in your question.

However, as we can't make fuel over there, the return fuel must be carried to the star by the starship. This increases the mass of the ship, meaning it needs extra fuel to get there in the first place. All that fuel has to be accelerated at the start of the trip, and decelerated at the end. That takes a lot more fuel. That extra fuel in turn has to be accelerated with the ship, until it is burnt. And that is where the 10x figure comes from.

If we could make fuel at the star (as has been suggested for return trips to Mars), then your idea would be a lot closer to reality.

The reason why the required fuel is not merely doubled has to do with the fact that an increase in delta-V results in exponentially more fuel needed.

Assuming that the return trip would take the exact same path (so double delta-V) and all else being the same, using the ideal rocket equation, the required total mass of the ship would increase by a factor of e², which equates to 7.4 times as much fuel. The e² value is derived by using the Tsiolkovsky rocket equation, solving for the Total mass, and then noting that since the delta-V term is an exponent of e, that doubling delta-V would result in e² more total mass.

However it should be noted that this 7.4x value is under ideal circumstances where all you need to do is add more fuel, in actuality the required fuel would increase even further since that extra fuel would also require more mass for fuel tanks and structural support which in turn would require even more fuel to carry that mass and so on (this is known as the tyranny of the rocket equation), so a value of 10x instead of the ideal value of 7.4x seems entirely reasonable to me as a first order approximation.

The only way to reduce these requirements below the ideal 7.4x would be to either reduce the delta-V by taking a shortcut back on the return trip somehow (gravity-assist slingshot maneuver?), or to use an entirely different means of achieving that delta-V which does not use traditional rockets.

Space travel is hard.

Fundamentally the Hail-Mary has much better fuel than we have in reality (the astrophage which approaches anti matter levels of energy density), and has a much harder task than today's space ships (getting up to near light speed and back) so it is easy for that to just end up at 10x (of course Project Hail Mary is a fictional story and does not come with a data sheet²).

But to give an intuitive view, look at the Perseverance Rover; that is 1,025 kilograms delivered to Mars. The launch vehicle for that was an Atlas V. The Atlas V is 21,054 kg, but it holds 284,089 kg of fuel. So let's be kind and ignore the spaceship and just say it takes 284,089 kg of fuel to get 1,025 kg of "stuff" to Mars, or a 284:1 fuel-to-stuff ratio. So let's say you want to get the whole Perseverance Rover back from Mars again (perhaps to put in a museum). You might say, "put in another 284,089 kg of fuel and job's done". But that extra 284,089 kg becomes part of the stuff for the first leg. And each kg of stuff requires 284kg of fuel to get to Mars. So you'd actually need an extra 80,656,000 kg of fuel for the round trip.

So in real life, it takes 284x more fuel to do a round trip to Mars than to do it one-way.

Now I've made a lot of approximations and a round trip to a star isn't exactly the same as a round trip to Mars¹ but I hope you can see how 10x is not at all an exaggeration for dramatic effect. In fact, the space agencies of today would bite your hand off for 10x.

¹ I'd probably say that "you have to get your fuel up to near lightspeed and back down to stationary before you can use it for the return trip" is arguably much worse than "you have to get your fuel up to Mars before you can use it for the return trip"

² Project Hail Mary does have a ship diagram, though, and you can see it is "mostly fuel", so it has the same fundamental problem today's spaceships have.