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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.

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    $\begingroup$ 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. $\endgroup$ – David White May 6 at 18:16
  • $\begingroup$ Well, I thought it was obvious I was talking about the total fuel initially carried by the rocket during takeoff but I will edit. $\endgroup$ – Exocytosis May 6 at 18:19
  • $\begingroup$ 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. $\endgroup$ – JMac May 6 at 18:44
  • $\begingroup$ 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? $\endgroup$ – user207455 May 6 at 18:55
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    $\begingroup$ Possible duplicate of Why are rockets so big? $\endgroup$ – Kyle Kanos May 15 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.)

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