From the story. Riley gave Homer an engineering book, Principles of Guided Missile Design, which the group found very useful. Hickam's account lists mass ratios among the ideas they had to learn, and the mass ratio is the heart of this lesson. (More in October Sky.)
A rocket's mass keeps changing: each bit of exhaust it throws back makes it lighter, so the same push speeds it up more and more. To add all those small pushes, use calculus.
In a short time $dt$ the rocket (mass $m$) throws out mass $dm_e$ at speed $v_e$ backward relative to itself. Momentum is conserved: $m\,dv = v_e\,dm_e$. The rocket's mass drops by that amount, $dm = -dm_e$, so $$dv = -v_e\frac{dm}{m}.$$ Integrate from the full mass $m_0$ to the final mass $m_f$: $$\Delta v = -v_e\int_{m_0}^{m_f}\frac{dm}{m} = v_e\ln\frac{m_0}{m_f}.$$
Key idea
Tsiolkovsky's rocket equation: $\role[change in speed]{velocity}{\Delta v} = \role[exhaust speed]{velocity}{v_e} \ln\dfrac{\role[starting mass]{mass}{m_0}}{\role[final mass]{mass}{m_f}}$. The mass ratio $m_0/m_f$ sits inside a logarithm, so doubling it adds only $v_e\ln 2 \approx 0.69\,v_e$. Thrust is $T = \dot m\,v_e$. Flying straight up, gravity takes away about $g\,t_b$: $\Delta v = v_e\ln(m_0/m_f) - g\,t_b$.
Worked example: Reaching orbit speed
A rocket's exhaust speed is $3{,}000\ \text{m/s}$. What mass ratio gives $\Delta v = 7{,}800\ \text{m/s}$ (ignoring gravity and drag)?
- $m_0/m_f = e^{7800/3000} = e^{2.6} \approx 13.5$.
- So about $1 - 1/13.5 \approx 93\%$ of the launch mass must be propellant. That is why big rockets drop empty stages.
Answer: $\approx 13.5$
Why it works
Each kilogram of propellant must first be lifted and pushed by the propellant burned before it. That compounding is what the logarithm describes.
Common mistake
Putting the propellant mass, or $m_f/m_0$, inside the logarithm. With the ratio upside down the logarithm is negative.