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I don't need an exact number, also it can be to travel to the ISS or the moon, it's not important. I just want to get an idea of how much fuel is used in a «recursive» sort of way.

Edit: I am asking how much of the initial total amount of fuel at takeoff is used to lift fuel. I know fuel is consumed on the go, but that does not render the question unanswerable.

Edit2: By saying it was not important that the travel was to the ISS or the moon, I meant that any example could be taken, and not that it was irrelevant to the calculation.

Winston
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    This question can't be answered as posted. Fuel is continuously burned to generate the thrust that it takes to lift the remaining fuel. This means that there is no fixed "total" on which to base the percentage calculation on. – David White May 06 '19 at 18:16
  • Well, I thought it was obvious I was talking about the total fuel initially carried by the rocket during takeoff but I will edit. – Winston May 06 '19 at 18:19
  • The answer right now would literally still be "it depends on what they are planning to do". The % of fuel used to lift fuel will change depending on the destination. – JMac May 06 '19 at 18:44
  • All the fuel is used to "lift" the rocket, except that used for braking at the end... Or did you mean what percentage of fuel is used to get out of the atmosphere? –  May 06 '19 at 18:55
  • @SolarMike I think they mean something more convoluted than either. To bring up a payload, you will also need to provide fuel capable of giving the payload the required momentum. To bring up the fuel, you will also need more fuel. Therefore, for any rocket lift there will be some % of the energy spent (thus fuel burned) whose sole purpose is to lift the fuel you need to keep moving the payload. I don't know enough about rockets to say how you would calculate that, but I'm guessing it's recursive (as mentioned in the question). Can't really pin down a % without payload and distance though. – JMac May 06 '19 at 19:06
  • @JMac could just be geostationary orbit... but until the OP can give precision to the question... –  May 06 '19 at 19:08
  • @SolarMike They say "also it can be to travel to the ISS or the moon, it's not important." So presumably they just don't realize that without that information the answer is essentially "anywhere from practically 0% to practically 100%". – JMac May 06 '19 at 19:10
  • @JMac "until the OP can give precision to the question"... –  May 06 '19 at 19:11
  • @Exocytosis, the percentage will vary with each increment in height, up to the point where the fuel is totally used up. – David White May 06 '19 at 19:39
  • By saying it was not important that the travel was to the ISS or the moon, I meant that any example could be taken, and not that it was irrelevant to the calculation. – Winston May 07 '19 at 21:04
  • And David White the percentage I ask is relative to the initial fuel quantity. This does not vary with time, although it does vary with the destination. Another way to put it is: if fuel was provided by a side ship, how much fuel would it be needed to set e.g. a one metric ton satellite into geostationary orbit, or bring one metric ton to the ISS or whatever example that you like. I am asking for a number, not a formula (although the latter may justify the former). – Winston May 07 '19 at 21:12
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    Possible duplicate of Why are rockets so big? – Kyle Kanos May 15 '19 at 23:52

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I think the answer to your question lies in the rocket equation, which says

$$\Delta v = v_e\log M_i/M_f,$$

where $\Delta v$ is the total magnitude of changes in velocity during either a single manoeuvre or a whole voyage, $M_i$ is the initial mass, ie. sum total of payload and fuel, $M_f$ is the final mass, which would be your payload, and $v_e$ is the velocity of the exhaust, assumed constant.

So let $p$ be the proportion which is fuel, so $M_f = (1-p)M_i$. Then we have

$$\Delta v = v_e\log \left(\frac{1}{1-p} \right),$$

So you see why it's hard to answer you question as a percent. As $\Delta v$ increases, that is for farther or more complicated voyages, the proportion of fuel $p$ must increase as well.

If you'd like a differential relation between velocity and mass we can take $\Delta v \to dv$, $M_i \to M_f + dM$ in the rocket equation to obtain

$$dv = v_e dM/M.$$

(This is usually the formula which is derived first, by considering conservation of momentum.)

Ryan Thorngren
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