The point here is that the rocket weight should never include fuel. It can take off from any departure base of your choice. The fuel provided by the side ship would then be used only to lift the rocket structure, including engines and payload, for a total fixed weight (unlike a normal rocket that loses weight through fuel use).

This is the corollary question of the one I asked before, asking what percentage of fuel is used just to lift fuel in a rocket. But reading comments, I feel people might be more comfortable answering this less general question. The point is to understand how much fuel is spent only to carry fuel itself in rocket-based transportation. Again, although formulas are valuable, I am asking for numbers.

EDIT: I don't care about how much fuel the accessory ship needs to lift itself, nor any perturbation introduced to the main rocket by the side ship. Please imagine the fuel is supplied by a magic ship. I did write that it was autonomous, and I meant by that that you should not concern yourself with it. The only thing I want to know, is how much fuel is used only to lift the 10 tons and not to lift any fuel at all (i.e. leave that to the autonomous magical side ship).

  • $\begingroup$ You may have missed an important fact about real rockets: a lot of the mass of the rocket is their to contain and manage the fuel (that is the primary reason for stages—you get rid of some of that parasitic weight once its work is done). The ideal case is one where there is no superstructure at all and you only put the payload in orbit (payload including crew, life-support and what not). And while it is possible to compute quantities like the one you propose they are not very useful: we already engineer toward optimal performance with the available tech. $\endgroup$ – dmckee May 8 at 0:09
  • $\begingroup$ They are useful to me. I am not pretending it to be useful to space exploration. Now regarding fuel management, I do not forget about it, I just ask to remove the fuel from the total weight, without precising exactly what is lifted. Lastly, I am well aware that stages would not separate in the case I am suggesting, but I don't see how I could do otherwise while trying to get the data I am looking for. $\endgroup$ – Exocytosis May 8 at 0:14
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    $\begingroup$ Possible duplicate of Why are rockets so big? $\endgroup$ – Kyle Kanos May 15 at 23:52

The core question here is legitimate, though it has some edges of ill-definedness that make the specifics hard to answer.

  • Is aerodynamic drag a factor? The overwhelming majority of orbital rockets (i.e. all except a small class of air-launched rockets) are launched from reasonably close to sea level, and aerodynamic drag from the thick atmosphere at the launch site is a substantial consideration in the launch profile. The saying "most fuel is spent lifting other fuel" is mostly true $-$ but there's a nontrivial fraction (in the 5% regime, or so, if I understand correctly) that's spent shoving air out of the way of the rocket.

    This is reflected in several of the existing answers, including aspects such as "two ships make more drag than a single body", which are absolutely correct. However, if I understand the gist of your question correctly, this question lives in a world of spherical cows in vacuum, so let's ignore aerodynamic drag.

  • The answer depends on the type of fuel. Different fuels have different efficiencies, and they produce different amounts of thrust per kilogram of expelled propellant. The core concept here is that of specific impulse: the total impulse provided by a unit mass of propellant. (This is then generally divided by $g=9.8\:\rm m/s$, because much of this part of rocket science was developed by 1950s US scientists who didn't know how to use MKS units, and normalizing by $g$ helps compare results between MKS and Imperial unit systems.) Also, to make things worse, the specific impulse depends on the aerodynamic conditions the engine is in, as well as the design of the engine itself, and for any given engine it varies somewhat over the launch profile.

    Still, as an initial approximation, let's take $I_\mathrm{sp} \approx 300\:\rm s$ as a reasonable ballpark figure to use. (This means: we assume the fuel and oxidizer are such that burning one kilogram of the mixture will produce an impulse of $F\:\Delta t = 300\,g\,\mathrm{s} \approx 3,000 \:\rm N\:s$.)

The final piece of the puzzle is the orbit itself, which is (for the purposes that you're asking about) fully characterized by its $\Delta v$ (delta $v$), i.e. by the total change in velocity that any rocket needs to exert to get to that orbit. For the ISS, a suitable ballpark for the required $\Delta v$ budget is about $10\:\rm km/s$.

To put those two together, you multiply the $\Delta v$ by the mass you want to get to orbit to get the total impulse needed, and then you divide by the specific impulse to get the mass of propellant that you need to consume. For your case, that reads $$ m_\mathrm{prop} = \frac{m_\mathrm{payload} \:\Delta v}{g \:I_\mathrm{sp}} = \frac{10,000\:\mathrm{kg} \times 10\:\mathrm{km/s}}{9.8\times 300\:\mathrm{kg\:m\:s^{-1}/kg}} \approx 34,000\:\mathrm{kg}. $$ That is, about 34 tons, as a rough estimate.

Though of course, having said all of the above: This answer is completely useless and it is basically just fantasy. The so-called 'tyranny of the rocket equation' is real, and your way to avoid it (i.e. to pretend that it doesn't exist) is essentially useless in the real world. Rockets need to lift their own fuel to work, and that takes (exponentially) more fuel. Deal with it.

  • $\begingroup$ @Exocytosis You're welcome to flag any comments; you can find the moderator replies to your flags here. If you have specific technical objections to this answer (of the sort that would lead to a legitimate downvote), or you find aspects unclear, you're obviously welcome to voice them. $\endgroup$ – Emilio Pisanty May 12 at 18:37

More than would be required for a rocket with fuel. Your question cannot be answered with a definitive answer. There are too many variables to include. Quora says that 9.39 lbs of fuel is needed to lift 1 lb of equipment. Not sure if this is still true. But this should give you an idea on what is needed. But again this does not take into account variables such as atmospheric pressure, launch locations, atmospheric drag, etc.

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    $\begingroup$ Why do you think that it takes more fuel to lift a rocket without fuel than one with fuel? $\endgroup$ – G. Smith May 7 at 23:57
  • $\begingroup$ Maybe because of stage separation, but I am curious to hear his explanation too. $\endgroup$ – Exocytosis May 8 at 0:01
  • $\begingroup$ Rick, are you sure these numbers do not include lifting fuel itself? $\endgroup$ – Exocytosis May 8 at 0:09
  • $\begingroup$ You have to lift rocket a as well as the rocket supplying the fuel. The 9.39 is not including fuel. It is what is required to lift 1 pound of cargo to low earth orbit. The largest amateur rocket motor weighed 435 pounds and lifted 289 lbs of material and reached an height of 72 miles. $\endgroup$ – Rick May 8 at 1:53
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    $\begingroup$ I am not asking about fuel used to lift the accessory ship. I will edit the question as it is not clear enough, apparently. $\endgroup$ – Exocytosis May 8 at 20:10

If fuel was provided by an autonomous side ship, how much fuel would be required for a 10 metric tons rocket to reach the ISS?

Zero. (oops, sorry, read the sense of the question wrong) The same or more as if it was in a single rocket.

The vast majority of the energy needed to lift a rocket is the energy needed to lift the fuel. The Space Shuttle, for instance, had a mass ratio of about 16. So in your example, to lift the 10 tonne "rocket" we would need 1600 tonnes of fuel. So there is basically zero savings in having the fuel separate from "the rocket". This is basically what the SS's ET was after all, a "separate rocket" providing the fuel.

So why more? Because two rockets flying in formation have higher drag than one.

There are some interesting takes on this though. One is to launch a low altitude rocket that sprays fuel behind it. Now your main rocket scoops it up as it climbs. Because a rocket is heaviest on the pad, this can actually save a crapload of fuel, and it does not require the two rockets to fly in formation. There are significant practical problems here, obviously, wind not the least of them. Another possibility is to scoop up the air, either as oxidizer to offset liquid oxygen, or simply as working mass as in laser propulsion. So far none of these concepts has reached production.

  • $\begingroup$ Thank you for your answer but you miss the point. You are assuming a specific agency of two rockets that I have not detailed at all, and by mentioning a side ship I only theorized a rocket empty of fuel. The way the fuel is provided is irrelevant here. Now, you write zero or the same or more. Drag apart (forget any effect due to the side ship), the fuel need of the remaining rocket is neither zero nor equal nor superior to a normal rocket. $\endgroup$ – Exocytosis May 8 at 20:08
  • $\begingroup$ It seems my question was not clear that I don't care about the fuel used to lift the accessory ship, so I am editing the question. $\endgroup$ – Exocytosis May 8 at 20:10
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    $\begingroup$ 93 tons of fuel $\endgroup$ – Rick May 8 at 23:09

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