In Portland's OMSI there is a hands-on water bottle rocket station. (https://www.youtube.com/watch?v=cdtmVY76_PQ). The rockets are normal PET bottlers. The visitors fill their bottle with an amount of water and then fill the remaining volume with compressed air at a given pressure.

The challenge is to find the best water-to-air ratio so that the rocket flies highest. Too much water is bad, not only because it makes the rocket heavy. As I explained to my son, the compressed air is also the energy store for this rocket (not the water, since water is almost incompressible). To have less compressed air means to have less energy available.

But then I got stuck because by inversion this means that an "all air" configuration should be best: Most available energy, highest kinetic energy, highest speed of empty bottle. This is obviously wrong. It was clear experimentally that the best ratio is somewhere in the middle. Also it makes intuitive sense that some mass in form of water is needed to produce thrust, since actio = reactio. In order to produce momentum, mass is needed to "push off of".

I'm aware of the fairly complex rocket flight physics. (For example, https://www.ohio.edu/mechanical/programming/rocket/analysis1.html gives an accessible overview.) But because I am not interested in an exact result much of it can be neglected. The basics are fairly simple: Energy stored in the compressed air is transformed into kinetic energy of the expelled water, rocket and earth, plus "losses" through heat from turbulences.

My question is on a more general, abstract level. Momentum or not, we have a given energy in the air which must go somewhere.

Where does the energy go which is stored in the compressed air in a "compressed air only" configuration? It should be more energy than with a partly water-filled bottle; but the rocket's final velocity (and hence kinetic energy) is much lower. Did we produce that much heat? I don't think so. Did we accelerate the earth? No, the "burn phase" was short.

I am missing something. What is it?

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    $\begingroup$ All rockets in the atmosphere will convert all of their internal energy into heat in the end, but that's not what causes the "efficiency problem". The velocity change of a rocket is governed by the rocket equation: $\Delta v = v_{exhaust}\ln {m_0\over m_1}$ and with nothing but air filling at comparably low pressure the mass ratio of a bottle rocket is very small. So even if you manage to increase $v_{exhaust}$ somewhat it falls victim to a much too small mass ratio. $\endgroup$
    – CuriousOne
    Sep 7, 2015 at 13:33
  • $\begingroup$ @CuriousOne Ok. And where does the energy go, if the v-exhaust is not rising adequately? $\endgroup$ Sep 7, 2015 at 13:38
  • $\begingroup$ Like I said, the energy always goes into heat in the end, but rocketry is not an energy but a momentum transfer problem. Unless your propellant mass is a significant fraction of the rocket's mass (in commercial vehicles it's way over 90% of the total mass!) you don't have an efficient rocket. $\endgroup$
    – CuriousOne
    Sep 7, 2015 at 13:43
  • $\begingroup$ Let's run the numbers: a 2l bottle rocket pumped to 50psi (3.5bar) has a propellant mass of something like 2l*3.5*1.5g/l=10.5g. An empty 2l bottle has a mass of 54g, with fins etc. probably a whole lot more. But even so your best case mass ratio is (10.5+54)/54=1.194. ln(1.194)=0.178. OTOH, a water filled rocket can achieve a mass ratio of e.g. (1000+54)/54=19.5 and ln(19.5)=2.97, which is some 16.7 times better than the air filled rocket. $\endgroup$
    – CuriousOne
    Sep 7, 2015 at 13:57
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    $\begingroup$ Having said all of this, it occurred to me that an air bottle rocket would also need a proper nozzle to be effective (that is not necessary for an incompressible exhaust medium like water), even as an air rocket. Indeed, one could make an air-rocket highly efficient by powering an expansion turbine which uses air from the outside to improve the mass ratio considerably. That, of course, would be a VTOL jet with a jet engine and not a rocket, anymore. $\endgroup$
    – CuriousOne
    Sep 7, 2015 at 14:09

1 Answer 1


It was already discussed in the comments that a water rocket needs to push "something" out. It is instructive to do the calculation in a little more detail to see where the "energy" goes. For this I will consider the relative share of energy going to the rocket and the "expelled matter" (gas, or water) as a function of the expelled mass. To simplify things, we will assume that all matter is expelled as a single entity with a certain velocity; in reality you might need to integrate, but any inequality that holds for a small amount of expelled matter will hold for the integral over many such amounts.

I will use upper case symbols for quantities relating to the "rest of the" rocket (mass M, velocity V, momentum P - without the expelled mass) and lower case for the expelled matter(m, v, p). From conservation of momentum, $P = -p$ so $M\cdot V = - m\cdot v$. The energy of the rocket $E_r$ and expelled mass $E_m$ will be respectively:

$$E_{r} = \frac12 M V^2 = \frac{P^2}{2M}\\ E_m = \frac12 m v^2 = \frac{p^2}{2m} = \frac{P^2}{2m}$$

It follows that the ratio of (energy in rocket)/(energy in expelled matter) is

$$\frac{E_r}{E_m} = \frac{m}{M}$$

In other words - the lower the mass of the expelled matter, the greater the relative amount of energy it contains. In the limit of "no water", the little bit of air mass contains virtually all the energy.

  • $\begingroup$ The argument I was looking for. While the momentum must be equally distributed, the energy doesn't have to. And due to the energy growing with the square of the velocity, faster expelled fuel "carries" more of the energy "away". Almost all energy used in a photon drive ends up in the photons, for non-relativistic velocities. $\endgroup$ Sep 7, 2015 at 21:12
  • $\begingroup$ @PeterA.Schneider Regarding photon drives, that equation ends up quite differently if you also take the mass-energy of matter into account (E=mc²). If you do, a photon drive should be much more attractive than a matter based reaction drive in the division of energy between vehicle and reaction-stuff. $\endgroup$
    – JanKanis
    Sep 10, 2019 at 7:28
  • $\begingroup$ @JanKanis Well, you have to take that into account, but how does it make a photon drive more attractive? More attractive than it already is: It is attractive exactly because the reaction mass is minimal. Mass is hard to lift, transport and accelerate; energy, on the other hand, is relatively abundant with nuclear reactors. Yes, it would be better to expel more mass at $c$, but we don't have it. $\endgroup$ Sep 10, 2019 at 8:47
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    $\begingroup$ @PeterA.Schneider If you measure the energy that goes into your 'exhaust' vs energy that goes into your spaceship, then a regular mass expelling rocket puts a lot more energy into the exhaust (kinetic energy + mass energy) than a photon rocket puts into its photon exhaust for the same amount of thrust/momentum. (At least that's what I expect, I haven't done the math.) $\endgroup$
    – JanKanis
    Sep 10, 2019 at 9:01
  • $\begingroup$ @JanKanis For some reason revisiting this question and your answer.-- As to your last remark: True (that if you count the expelled mass as energy according to $E=mc^2$, the overall energy expended with a regular drive is enormous and the energy efficiency many orders of magnitude worse than with a photon drive. But that assumes that the expelled mass could be converted to "energy" (i.e., photons) instead which is currently impossible. $\endgroup$ Dec 18, 2020 at 11:56

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