What ratio of final to initial mass of a rocket to achieves the highest energy efficiency - the highest ratio of final mass kinetic energy to chemical energy expended? And more generally the relation of efficiency to mass ratio. This is a bit hypothetical because we are not usually interested in efficiency in this sense, but the answer is curious none the less.
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I'm assuming this will be evaluated in the frame where the ship starts at rest? Since you want the ratio of chemical energy expended, then we assume the ship always has 1 unit of fuel. If you take $r$ as the ratio of final to initial mass, then $m_f$ is just $r^{-1}$ and $m_0$ is $r^{-1} - 1$.
Then the final kinetic energy of the ship is (ignoring constants): $$E \propto m_f \space (\Delta v)^2$$ $$E \propto (r^{-1} - 1) \left ( \ln{\frac{r^{-1}}{r^{-1} - 1}}\right )^2$$
That derivative is a bit more than I want to deal with right now, but a calculator says it has a maximum near $r=0.7968$
answered May 30, 2023 at 23:05
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$\begingroup$ I don't follow this because the mass ratio calculated from the expressions is not $r$. In any case im assuming something more usual - the energy expended is proportional to the mass lost $(m_0 - m_f)$, ie for a given chemical propellant, so $$ E_{kin}/E_{chem} \propto m_f (\ln{m_0 / m_f})^2 / (m_0 - m_f) = (\ln{r^{-1}})^2 / (r^{-1} - 1) $$ I get max at $r \propto .25, r^{-1 }\propto 5.0$ $\endgroup$– ddddmmmmCommented May 31, 2023 at 10:06
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$\begingroup$ Apologies. I flipped $r$ and it is the fuel fraction instead of final to initial. So the fraction you asked for is (1 - r) or 0.2032. $\endgroup$ Commented May 31, 2023 at 15:55