= Solution
Write the total remaining rocket <mass> as $\mathcal M=M+m(t)$, with $\dot m=-\alpha$. Over an interval $dt$, an exhaust <mass> $dm_e=\alpha\,dt$ is expelled backwards with inertial <velocity> $v-u$ to first order. Applying <momentum conservation> to the rocket and this exhaust gives
$$
\mathcal M v=(\mathcal M-dm_e)(v+dv)+dm_e(v-u)+o(dt).
$$
Cancellation and division by $dt$ give the <rocket equation>
$$
\boxed{(M+m)\frac{dv}{dt}=\alpha u.}
$$
With $M>0$, $m(t)=m_0-\alpha t$, and $0\le t\le m_0/\alpha$, integration gives
$$
v(t)-v_0=u\log\frac{M+m_0}{M+m(t)}.
$$
At burnout,
$$
\boxed{v_{\rm final}=v_0+u\log\left(1+\frac{m_0}{M}\right).}
$$
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.
Back to article page
