Rocket equation

For a body of varying mass $M$ ejecting material at fixed relative velocity, conservation of momentum gives a differential relation between $dM$ and the body's change $dv$. Integrating it produces a logarithm of the mass ratio.

When matter leaves a moving body with zero instantaneous relative velocity, it carries the body's current momentum per unit mass but creates no additional recoil term. A short-interval momentum balance including the newly released material gives $q\dot{v}=F_{\mathrm{ext}}$ for the remaining body's mass $q(t)$. The expression $\frac{d}{dt}(qv)=F_{\mathrm{ext}}$ omits the outgoing momentum flux. Nonzero relative ejection velocity instead produces the thrust term of the rocket equation.

A classical rocket ejecting fuel at rate $\alpha >0$ and relative exhaust speed $u>0$, while collecting stationary dust of mass $\beta dx$ per traveled distance, obeysThese are mass conservation and momentum conservation for the rocket, incoming dust and outgoing gas. The dust-loading term supplies the momentum needed to accelerate incoming stationary material. Put $\lambda =\sqrt{\alpha /(\beta u)}$ and $V=\lambda u$. While the engines operate and $0\le v<V$, an initially stationary rocket satisfiesFor $\lambda >1$, continuing to $V$ would drive total mass to zero. For $0<\lambda <1$, total mass instead grows without bound and $V$ is approached only after infinite time under indefinitely maintained thrust. In both cases a finite fuel supply runs out before that limiting speed. Total stored mass must not be confused with remaining fuel.

For collinear motion without external forces, a rocket ejecting material backwards at constant speed $u$ in its instantaneous rest frame obeys $m{\gamma }^{2}dv+udm=0$. Here $m$ is the rocket's rest mass, including stored internal energy, and $\gamma$ is its Lorentz factor in the chosen inertial frame. The equation follows from momentum conservation, conservation of relativistic energy, and the relativistic velocity-addition formula; expelled rest mass is not simply the decrease in rocket rest mass.

For zero initial velocity, body masses $kM,(1-k)M$ and fuel masses $k{m}_0,(1-k)m_0$, the rocket equation gives the displayed final second-stage speed after burning the first-stage fuel, discarding its body without impulse, and burning the second-stage fuel. It assumes $M>0$ and $0\le k<1$. The speed exceeds the single-stage value when $0<k<1$ and fuel is positive. Its limit as $k\to 1^{-}$ concerns a surviving second stage of vanishing mass; at $k=1$ no such stage exists.

If a vehicle moving at speed $v$ ejects mass forward at relative speed $u>0$, thenAs $dM<0$, the vehicle slows down.