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

Question

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.

• How is the "ten times as much fuel" figure calculated?
• Is it based on realistic approximations, or is it an exaggeration for dramatic effect?
• Could this multiplier vary significantly based on specific assumptions (e.g., exhaust velocity, payload weight, efficiency)?

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

Answer 1

Let's keep it classical. The rocket equation for the change in speed is given by

$$\Delta v =v_e \ln \frac{m_0}{m_f}$$ 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 discardable weight $\Delta m = m_0-m_f$, given by the following ratio:

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

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

Answer 2

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.

Answer 3

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.

EDIT Just to make it clear, the 10x figure is an "order of magnitude" approximation. T make a real estimate, far more details about the trip are needed.

Answer 4

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.

Answer 5

10x is figuratively, and it's very optimistic.

Less precise than the other answers, but examples and rules of thumb: current rocket mass and payloads (to LEO, anyway) splits roughly as 90% fuel / 5% structural mass / 5% payload.

If you want to go furhter, payload is less than 2%.

If you want to do a round trip, essentially you have to launch a rocket up there that allows for that one return trip, so the payload of the outbound launch is that whole return rocket. That means that the rocket ship for the outbound launch has to be more than 50x the size / mass of the return ship.

Edit: I'm aware this is a simplistic approach, eg assuming symmetry, no gravity assist / swing-by maneuvers etc.

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Examples (numbers from Wikipedia):

Vehicle	launch mass [kg]	payload [kg]	percent	Launch Mass / payload
Saturn V	2'965'000	141'136 (to LEO)	4.8%	21.0
		52'759 (TLI)	1.8%	56.2
Ariane 5	780'000	10'500 (to GTO)	1.3%	74.3
Space Shuttle	2'030'000	27'500 (to LEO)	1.4%	73.8
		4'940 (to GTO)	0.2% hahaha	410.1
Titan IIIE /Centaur	632'000	15'400 (LEO)	2.4%	41.0
		1'500 (to Jupiter)	0.24%	410
		722 (Voyager 2)	0.12%	820

some comments on the examples:

• The Space Shuttle has such a low ratio to LEO because it's always carrying that cool snazzy re-entry capable space plane as well.
• Even at 0.12% payload, the Titan III rocket was only enough to produce roughly 2/3 of the solar system escape velocity. Actually exceeding that velocity was only possible using the swing-by maneuvers, main contributions from Jupiter and Saturn.

Answer 6

There's some ratio $r$ of total mass to payload for a one-way trip. That is, if the payload has mass $p$, there's some number $r$ such that the total mass is $rp$. If you want a two-way trip, then $rp$ is that mass you need to get back, but in the outward trip, that mass of $rp$ is payload. So in the outward trip, the total mass is $r^2p$.

For instance, suppose $r$ is $10$, and your payload is $1 000$ tons. Then to go one way, your ship would be $10 000$ tons. That is, you need $90 00$ tons of fuel to get the payload back. When you start the return trip, you have $9 000$ tons of fuel, and $1 000$ tons of payload, for a total mass of $10 000$. But when you start out, that $10 000$ of mass is essentially all payload; all of the fuel in it is for the return trip, and is dead weight on the trip out. So you need $90 000$ tons of fuel to get that payload out. Thus, the fuel for a round trip is $10$ times the fuel for a one-way trip.

Answer 7

How is the "ten times as much fuel" figure calculated? Is it based on realistic approximations, or is it an exaggeration for dramatic effect?

10x is not a number meant to be taken as calculated, not even as a number. It is not calculated in any way whatsoever, it could be an exaggeration but (of the top of my head) 10x is also somewhere in the realistic ballpark, see below.

Could this multiplier vary significantly based on specific assumptions (e.g., exhaust velocity, payload weight, efficiency)?

Precisely, without very concrete mission specs, it is impossible to calculate the increase even to the magnitude. A short selection of a myriad of questions that affect the factor:

• Does the fuel need to be decelerated over there? That increases the factor.
• Or can we doe something like an aerobreak? That lowers the factor.
• Do we just want to barely go into an orbit? Lower factor
• Do we actually want to do something in that star system? (like visit planets, try different orbital heights etc)? That MASSIVELY increases the factor because each time your return-fuel has to be moved as well.
• What kind of fuel are we talking about? Does it have significant mass on it's own (like IRL chemical rockets) or negligible (dark matter for example, where the containment vessel is the weight and that scales logarithmically)

And so on.

However: it will never be even close to 2x. 2x would be if we refuel there (then it would be exactly 2x, assuming no in-situ fuel plant is carried AND all life-support can be resupplied as well and refurbishments over there, aka we need a full-on base at the target, almost a brand new replacement vessel).But - assuming the sci-fy fuel in question has a somewhat relevant mass like chemical rockets (where fuel is a majority of the mass in almost all missions), the demand rises exponentially, because your return-trip fuel (and other supplies like food, spare parts etc.) basically become additional cargo on the first leg of the trip. Cargo that needs space (that needs structure with weight) and additional fuel. And because of the rocket equation, double the (useful) mass does not just mean double the fuel, but an exponential increase. (Again, assuming they don't have a magic fuel with nearly negligible weigth)

In short: Even if we could refuel at the destination, we'd already be slightly over 2x. But we can't so we have to get everything for the return trip there, decelerate it, maintain it and accelerate (most of) it for the return trip.