# Simplified calculation of the amount of fuel required for a trip — are my calculations faulty?

I've been always fascinated with how easily scifi characters travel around the Solar system and sometimes the galaxy. They just hop into a spacecar and go wherever they want.

So I've come up with a thought experiment that reproduces such a trip with modern technology. Only existing means of propulsion are allowed (i. e. no hyperdrive, lightfold, wormholes, etc), but existing technology can be imagined as idealized, perfected.

The goal is to make a roundtrip to the outskirts of the solar system. Let's imagine that we would like to reach a Voyager, chip a tiny piece off it as a souvenir and bring it back home. This is what a scifi hero could do in a single episode!

The "Swordfish II" customized MONO racer fueling up on Ganymede, Jupiter. It's imagined to be capable, all by itself, of rising off Ganymede and flying around the moons of Jupiter, as well as rising from Earth to the orbit around Earth. It only needs assistance for crossing the Solar system in like a day.

Here are the "givens":

1. We want to make the trip in a reasonable amount of time. Like weeks, not years.
2. Since that would cause life-threatening overload (G factor), we're gonna send an automatic probe.
3. The probe weighs exactly 1 kg, which includes hull, electronics, manipulator, engine, batteries, etc — everything except fuel.
4. We're assuming that the probe is travelling a straight line in open space with no gravity acting on it.
5. We do not account for the weight of fuel tanks hull. Let's assume that fuel tanks are made of fuel and are consumed efficiently.
6. The engine is capable of efficiently burning huge amounts of fuel in small amounts of time so that it reaches cruising speed quickly and spends most of the travel time flying with inertia.
7. We do not account for the weight of the souvenir we're chipping off Voyager. Or let's say we're making a close-up film photo of it and carrying the film there and back, weight of the film accounted in the weight of the probe. The important part is to stop at Voyager and then stop back at Earth.
8. At the start of the flight, the probe and its target are stationary in relation to each other.
9. I'm not specifying the parameters of the engine, the specific distance, time restrictions, etc. Those numbers are arbitrary and Voyager is just a legend (that might not even work out, I hope I'm not violating the speed of light...).
10. ⚠ Now the tricky part. Let's say that I have accounted of all those factors and calculated that in order to reach the initial speed necessary to meet time restrictions, the probe will need to burn 1000 kg of fuel. This is a "given".

But the problem is that the probe will have no fuel to stop. It will fly past the Voyager, being able neither to carefully pinch a fragment off Voyager, nor return back home.

In other words, the trip has four legs:

→ →
← ←

1. accelerating towards Voyager,
2. decelerating,
3. accelerating towards home,
4. decelerating.

And my calculation of 1000 kg of fuel only accounted for the first leg, which would result in the loss of the probe! In order to stop the probe and complete the second leg, it would need another 1000 kg of fuel, but there's no gas station in the middle.

Our only option is to take extra fuel with us, which will increase the weight of the payload.

So the question of my thought experiment is: given that 1000 kg of fuel are needed for a 1 kg probe to complete just the first leg of the trip in desired time, how much fuel does the probe need to take in order to complete four legs?

My understanding is that this problem can be solved with a simple proportion:

               Payoad    Fuel
Current leg    A         B
Next leg       A+B       ?


And the general answer is X = (A+B)*B/A.

Here's the specific solution and answer, rounded down to the order of magnitude:

               Payoad    Fuel
One leg        1 kg      10^3 kg
Two legs       10^3 kg   10^6 kg
Three legs     10^6 kg   10^9 kg
Four legs      10^9 kg   10^12 kg


So my answer to the thought experiment is that more than a trillion (10^12) kg of fuel will be necessary for the described roundtrip of a 1 kg probe. That's more than the weight of 166 pyramids of Giza!

That also compares to roughly 1% of crude oil reserves on Earth. I can imagine that getting this much weight to orbit would require burning ALL of Earth's oil.

The question for this StackOverflow post: is my way of thought correct? Can the payload-to-fuel ratio of 1-to-1000 (provided as a "given" for the first leg of the trip) be extrapolated like this with a simple proportion?

My friends say that this problem cannot be solved without the knowledge of the parameters of the engine, etc. They also say that my computation requires having a multiply more engines for each subsequent leg.

But I believe that the parameters of the engine would only be necessary to learn how much time it will take for the probe to complete four legs. But that's not part of the question. Time has been accounted in the "given" for the first leg and is no longer a concern.

They also say that since the ship gets lighter as it burns fuel, this problem can not be solved with a simple proportion, and some complex formulas are needed.

• Isn't this the process that Nasa and other use to workout the fuel payload required for all the trips they achieved? So, these formulae must exist? – user207455 Jul 5 '19 at 7:42
• What is this "film" of which you speak? And why does it weigh anything? In our world, images are recorded in solid state memory devices that weigh the same whether they are empty or full of hours of HD video. – Oscar Bravo Jul 10 '19 at 11:05
• The "film" is only necessary to justify the return flight, instead of simply beaming the image back at the speed of light. – lolmaus - Andrey Mikhaylov Jul 11 '19 at 12:44
• In other words, I need a reason to get the probe back home. The film does weigh something, but this weight is part of the probe weight of 1 kg and does not change when photo is taken. – lolmaus - Andrey Mikhaylov Jul 11 '19 at 12:55
• Now you know why they use gravitational boost from massive objects like Jupiter – Robert DiGiovanni Nov 26 '20 at 20:29

Voyager is about 122 AU out so to get there in, say 2 weeks, is an average speed of about 60 AU/week. Light covers an AU in 8.3 minutes so $$c$$ is about 1200 AU/week. Your probe won't go anywhere near light-speed so you can do the whole thing using Newtonian mechanics.

You accelerate linearly for 1 week, turn round then decelerate to a stop after another week. Then do the same coming home, arriving back four weeks after you left. Your average speed is 60 AU/week and peak speed at turnaround is 120 AU/week.

Start at the arrival home. 1Kg payload slowing down from 120 AU/week over the course of a week using 999kg of fuel. Now repeat the calculation for leg 3; accelerating from the distant point to the mid-point where you arrive with 1000 kg. Therefore you had to start at Voyager with $$10^6$$kg of fuel. Repeat again for leg 2; slowing down from mid-point on the way out - you had to start with $$10^9$$ kg. One more time for the initial leg - you had to start with $$10^{12}$$kg of fuel.

That's not so bad - if your fuel is as dense as water, you could fit in a spherical tank about 1.2km in diameter.

• Ok - so you have an exhaust velocity of about 4,300 km/sec (anti-matter engine? nice choice...) I've updated my answer. – Oscar Bravo Jul 11 '19 at 15:31
• I used this calculator: omnicalculator.com/physics/ideal-rocket-equation. But really it's just proportion - each of the four legs is kinematically identical and each time the ratio of initial to final mass is 1000:1. Basically, each kilo of payload takes 1000kg of fuel to move it one leg. – Oscar Bravo Jul 15 '19 at 8:07
• I'd like to see your ship just before launch. Your 1kg payload, the size of a milk carton, would sit on top of a beach-ball sized (1.24m) fuel tank for leg 4. This would be on top of a house-sized (12.4m) tank for leg 3, on top of a city-block sized (124m) leg 2 tank on top of the leg 1 tank, the size of a small moon (1.24km). – Oscar Bravo Jul 15 '19 at 13:59
• About the engine: to get the delta-V you want, you need an enormous exhaust velocity in the engine. Chemical fuel doesn't even come close. For example, with an exhaust velocity of 5000 m/s (quite good for a chemical rocket), a ton of fuel on a 1kg payload gets you a delta-V of about 35 km/s. – Oscar Bravo Jul 15 '19 at 14:03
• Well, yes. Time and distance. You want to get to Voyager (120 AU away) in - let's say - two weeks. That's an average speed of 60 AU/week. If you accelerate constantly for a week then decelerate constantly for another week, your peak speed is 120 AU/week. So that's your $\Delta V$. – Oscar Bravo Jul 16 '19 at 13:49

The specifications for your probe are unrealistic, such a probe could never be built. It might be possible to build a probe which would eventually overtake a Voyager, but even a one way journey would take centuries. You need to remember that the voyager was launched in 1977,so has more than 42 years start on your chaser probe, which would take a few years to build and couldn't be built to your specifications. Not only would the Voyagers have nearly 50 years start, they are also moving very fast. Scientists waiting for your probe to return would have to have been born centuries after you have died. As you no doubt realise, having caught up with a Voyager, the chaser probe would then have to decelerate and turn about, increasing the necessary fuel load.

• The Op does list two deceleration phases, as in 2 and 4.... – user207455 Jul 5 '19 at 10:27
• The whole point of my thought experiment is to point out how unrealistic such a roundtrip is. But not in the part of "such probe could never be built". One of the "givens" is that the probe has a perfect jet engine with an indestructible nozzle. I don't care about time either: it's not specified and is out of scope of the problem. The question is only about how fuel requirement is calculated. – lolmaus - Andrey Mikhaylov Jul 5 '19 at 10:38
• You can hit Voyager in 17 hours (!) if you go with a constant acceleration of 1g, not accounting for the gravitational pull of the Sun. You'd be at the speed of 600'000 m/s when you cover 122 AU. Of course, such a travel would require colossal amount of fuel (like a Moon worth of fuel), and maintaining constant acceleration is hardly feasible. But that's a thought experiment! In my case, the acceleration in the beginning of travel is gonna be tiny, which in turn will drastically increase the amount of time spent. But this can be mitigated by adding more engines and dropping them off as you go. – lolmaus - Andrey Mikhaylov Jul 11 '19 at 13:26