 Solution

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.