Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 4 / 11B (Dynamics and Relativity)

Let $dm$ be the gas mass ejected in a short interval, and let $dD$ be the stationary dust mass acquired. The rocket mass change is $dM=dD-dm$. Before the interval the momentum of the rocket and this dust is $Mv$, because the incoming dust is at rest. Afterwards it isto first order; changing the gas ejection speed by $dv$ affects only second-order terms. Momentum conservation, with no external force, therefore givesDividing by the time interval produces the rocket equation with stationary mass accretion:The incoming material's zero initial momentum is essential: accelerating it to the rocket's speed consumes part of the thrust.

Without dust collection, $dM=-dm$. The rocket equation reduces toIntegrating from $v=0$, $M=M_0$ givesThis is the usual rocket equation with constant relative exhaust speed, valid while the mass remains positive and the prescribed ejection continues.

Now $\dot{m}=\beta$ and $\dot{M}=\alpha -\beta$, soFor $\alpha >0$, set $w=u\beta -\alpha v$. Then $\dot{w}=-\alpha w/M$ and $w(0)=u\beta$. For $\alpha \ne \beta$, integration givesConsequentlyThis solution of the rocket equation with stationary mass accretion covers increasing or decreasing rocket mass, as long as $M>0$. For equal-rate rocket burning and accretion, in the case $\alpha =\beta >0$, the mass is constant and the linear equation instead givesIt is also the continuous equal-rate limit of the preceding expression when written in terms of $t$.

For the no-accretion limit, expand the power as an exponential:Substitution givesrecovering (a). At a fixed time with $M_0-\beta t>0$, the limiting mass is $M=M_0-\beta t$, so the same calculation recovers the time-dependent no-dust solution.