Assuming I have a spacecraft which is $30,000\,\mathrm{kg}$ (roughly the size of the Apollo spacecraft).

If I take a comet and (theoretically) electrolize it perfectly to hydrogen and oxygen. I know have $2.2\times10^{14}\,\mathrm{kg}$ of $\mathrm{H_2}$ and $\mathrm{O_2}$.

Now I burn it. Assuming perfect energy consumption (all released energy goes to propulsion) how fast will my spacecraft get?

On one hand you have the $F=\frac{GM_1M_2}{R^2}$ which decreases due to both $R$ growing and $M_1$ (the mass of rocket+propulsion) decreasing.

How fast will this spacecraft get?

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    $\begingroup$ Why would You do such a nonsensy thing? If You have some source of electric energy in Your spaceship, You would convert in to exhaust gas acceleration as dirctly as possible. (Plasma or ionic drive) If You need to store the energy for some time (e. g. because the source is some radioactive material You can't switch off) You might do that ba electrolysis of water, but of course You would reconvert it to electric energy in a fuel cell. Learn about efficency of energy conversion, especially learn what entropy is. $\endgroup$ – Georg Apr 23 '13 at 14:17
  • $\begingroup$ Hi Spacefan, and welcome to Physics Stack Exchange! This is a site for conceptual questions about physics, not a place to get people to do physics problems for you. If you can edit your question to ask about the specific physics concept that is giving you trouble, I'll be happy to reopen it. See our FAQ and homework policy for more information. $\endgroup$ – David Z Apr 23 '13 at 23:50

The enthalpy of combustion of hydrogen is 286kJ/mol i.e. 286kJ per 18g of water produced. So if the original mass of water in your comet is M (in kg), and you manage to split it completely into hydrogen and oxygen, then the energy released when you react the hydrogen and oxygen is given by:

$$ E = 2.86 \times 10^5 \frac{M}{0.018} \approx 1.6 \times 10^7 M $$

so assuming your figure of $2.2\times10^{14}$kg is the mass of the original comet the amount of energy released would be about $3.5 \times 10^{21}$J.

If you have an isolated mass, far away from any gravitational fields, then you can work out what velocity change corresponds to this kinetic energy, and doing this gives a velocity of about $4.8 \times 10^8$m/s. Obviously this is faster than light, so relativistic effects have to be included. Although this is straightforward, I'll content myself with saying the end velocity would be a large fraction of the speed of light.

But since you mention Newton's equation for gravity I assume you're asking what happens if you start inside a gravity well, e.g. from the surface of the Earth. The simple answer is that with such a large amount of energy available the escape velocity of the Solar System is insignificant and makes effectively no difference to your final velocity. You'd need to start from near the event horizon of a black hole for it to have any serious effect on your final speed.

  • $\begingroup$ The effiency of burning some fuel in a rocket is never 100 %. Second law is not only for steam engines. $\endgroup$ – Georg Apr 23 '13 at 14:10
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    $\begingroup$ @Georg: true, but the OP specifically asked: Assuming perfect energy consumption (all released energy goes to propulsion) how fast will my spacecraft get? $\endgroup$ – John Rennie Apr 23 '13 at 14:13

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