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?
$\begingroup$
$\endgroup$
2
asked Apr 12, 2018 at 16:47
-
1$\begingroup$ what's with the immediate -1 score? $\endgroup$– Andrew PalfreymanCommented Apr 12, 2018 at 16:50
-
$\begingroup$ Some users seem to go through and down-vote questions they just don't like for some reason. I don't get it either. $\endgroup$– Tom B.Commented Apr 13, 2018 at 6:43
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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.
