Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 12A (Dynamics and Relativity)

Write the total remaining rocket mass as $\mathcal{M}=M+m(t)$, with $\dot{m}=-\alpha$. Over an interval $dt$, an exhaust mass $d{m}_e=\alpha dt$ is expelled backwards with inertial velocity $v-u$ to first order. Applying momentum conservation to the rocket and this exhaust givesCancellation and division by $dt$ give the rocket equationWith $M>0$, $m(t)=m_0-\alpha t$, and $0\le t\le m_0/\alpha$, integration givesAt burnout,The logarithm depends on the mass ratio, not on the burn rate. This follows from momentum conservation of the complete rocket–exhaust system; applying a constant-mass equation directly to the rocket alone would omit the momentum carried away.

For $0\le k<1$, the first burn reduces the total mass from $M+m_0$ to $M+(1-k)m_0$. The rocket equation therefore givesDetachment exerts negligible impulse, so the surviving second stage keeps velocity $v_1$. Its initial and final masses are $(1-k)(M+m_0)$ and $(1-k)M$, respectively. A second application of the rocket equation givesFor the single-stage comparison from rest, $v_{\mathrm{single}}=u\log (1+m_0/M)$. HenceEquality holds at $k=0$; for positive fuel and exhaust speed it is strict when $0<k<1$. This two-stage rocket with proportional fuel and body masses gains speed by discarding inert body mass before the last burn.

As $k\to 1^{-}$ the expression tends to $2u\log (1+m_0/M)$, but the surviving second-stage mass also tends to zero. At exactly $k=1$ there is no second stage, so its final speed is undefined and that limiting formula cannot be assigned to it. The first stage alone reaches the single-stage burnout speed before detachment. If $m_0=0$, neither design gains speed.