Consider a velocity change $dV$ for a rocket accelerated by an on-board thruster. Is the total amount of fuel consumed proportional to $dV$, or to $dE$, where $dE$ is the change in kinetic energy, or to some other measure, and if so, which measure?

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    what's with the immediate -1 score? – Andrew Palfreyman Apr 12 at 16:50
  • Some users seem to go through and down-vote questions they just don't like for some reason. I don't get it either. – Tom B. Apr 13 at 6:43
up vote 1 down vote accepted

For this question you should consider the rocket equation:

$$v(t)-v_0 = u \ln \left ( \frac{M_0}{M(t)} \right ) - gt$$

where $v_0$ and $M_0$ are the initial speed and mass, $v(t)$ and $M(t)$ are the speed and mass after a time $t$ and $u$ is the exhaust velocity relative to the rocket body. You can ignore the $-gt$ term if you are, for example, in outer space.

Rearranging for $M(t)$, you get

$$M(t) = M_0 \ e^{-\frac{dv+gt}{u}}$$

After a time $t$, the mass $dm$ of fuel consumed is then

$$dm = M_0 \left ( 1 - e^{-\frac{dv+gt}{u}} \right ) = M \left ( 1 - e^{\frac{dv+gt}{u}} \right )$$

I hope this answers your question.

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