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So here is my problem - as part of my job I present some science demonstrations to children and one of the tricks I regularly use is the bicarb/acetic acid rocket. I thought the other day that a really fun fact to give them to go along with it would be the amount of bicarb and vinegar you would need to, say, escape the Earth's atmosphere. I'm a bit rusty but I think it's less trivial than it sounds. I think you need the rocket equation, which means you need the exhaust velocity, which is in turn quite complex to calculate. Does anyone have any simpler approaches which might work?

A few nuggets which might help:

1 mole of bicarb produces 291 cc of CO2. Escape velocity is ~11000 m/s

Any answers will be very appreciated!

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I think you're going to run into the "Tyranny of the rocket equation" pretty hard here with a 'fuel' having such little thrust. –  Niel de Beaudrap Apr 25 '13 at 12:59
    
Quite. I have a new found respect for the phrase "it's not rocket science". Hard to appreciate until you actually have a go how challenging some of this is. –  Dave Cunnah Apr 25 '13 at 13:01
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The right way to do this problem is to use the rocket equation, as you guessed. The exhaust velocity depends on such details as the nozzle shape, but I doubt the usual equation will work here since that assumes supersonic exhaust. Short of finding a relevant calculation or, better - measuring the exhaust velocity, you can estimate it using the energy produced in the reaction and assuming some efficiency of conversion to kinetic energy of the exhaust gas. –  Michael Brown Apr 26 '13 at 14:20
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Isn't it possible to ignore the rocket equation and obtain a lower bound by simply assuming that all the free energy released from the reaction of bicarbonate and vinegar could turn into kinetic energy propelling the spacecraft+fuel straight upwards? Then it would be a matter of equating the chemical free energy of the fuel to the gravitational potential energy (with respect to infinity, or maybe 100 km altitude) of the spacecraft+fuel on the surface of the Earth. There would still be an exponential dependence on mass, but I don't see how exhaust velocity would have to factor in. –  Nicolau Saker Neto Apr 26 '13 at 20:35

2 Answers 2

rocket fuel is often characterised by its Specific Impulse. Loosely this is a measure of force applied to rocket per mass of fuel. I'm afriad I don't know what this would be for Bicarb...

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Yes, this is the problem I kept running into - these kind of properties are not cataloged for kitchen condiments! Thanks for looking anyway. I think there must be some way of defining the chemical potential in the bicarb and equating that to the energy released, but I can't see the woods for the trees at the moment! –  Dave Cunnah Apr 25 '13 at 12:25

When you say "reach space" do you mean a) get above most of the atmosphere, b) achieve orbit velocity, or c) achieve earth escape velocity? Either way, if you simplify the problem by ignoring air friction, it's only a matter of achieving a certain velocity.

Robert H. Goddard showed that, if you built your rocket a certain way, you could achieve any desired velocity. (I'm very much simplifying and paraphrasing.)

Suppose you wish to accelerate 1 mass unit to velocity $v$. Assume you do this by making a rocket consisting of $M$ mass units, which is all fuel except for the 1 mass unit left at the end. Then the total mass you need is $M = e^{v/c}$, where $c$ is the velocity of the exhaust.

That means, for example, if you double $v$ you have to square $M$, but if you can double the exhaust velocity $c$, you can reduce $M$ to its square root. That's why exhaust velocity is so important.

This does require that the fuel container not only be as light as possible, but it should fall away as the fuel burns. That's why space rockets have multiple stages.

ADDED: Some common exhaust velocities, taken from Goddard's paper, are:
- Common black-powder rocket: 29 m/s
- "Coston" ship rocket: 31 m/s
He found that by carefully building a rocket chamber and exhaust nozzle, he could get up to exhaust velocities of over 2,400 m/s using special gunpowder. That's a factor of 80, meaning M could be reduced by a power of 80.

The specific impulse of $H_2$ is up around 5000 seconds more or less, times $g$ of $10ms^{-2}$ or 50000 m/s, 20 times faster than the gunpowder, so an additional power of 20 reduction.

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