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Consider a rocket in free space of initial mass (including fuel) of $m_0$ using up/ejecting fuel at a constant rate $r$ (mass per time). That means that at time $t$ the rocket has mass $m=m_0-rt$. The velocity of the exhaust is $v_e$.

When firing the rocket imparts $\dot{m}v_e$ impulse per unit time (force) on the exhaust (at first I fought this is where the problem lies but after more considerations it seams legit). This must be equal (and opposite) to the impulse imparted on the rocket during that time and hence the rate of change of momentum.

$$ -\dot{m}v_e = \frac{d}{dt} (mv) = \dot{m}v + m\dot{v}$$ substituting in $m=m_0-rt$ and $\dot{m}=-r$ we get $$ rv_e = -rv + (m_0-rt)\dot{v}$$ applying the reverse chain rule (I know I could have simply not expanded the term for momentum and gotten the same result) and integrating $$ rv_e = \frac{d}{dt} (m_0-rt)v$$ $$ trv_e +c= (m_0-rt)v$$ rearranging, determining the constant (for say $v(t=0)=0$) and substituting $t=t_{final}$ we get that: $$\frac{\Delta v}{v_e}=\frac{tr}{m_0-rt} = \frac{\text{fuel mass}}{\text{vehicle mass}}$$

which is quite different from the actually equation of $$\frac{\Delta v}{v_e} = \ln\left(\frac{m_0}{m_f}\right)$$

I understand the derivation of this equation, what I'm missing is at what point in the derivation did my method go wrong. I have a feeling that stating that the fuel is ejected at a constant mass rate trivialised some calculus and made me miss a term of factor which would give the exponential/logarithm, but thats just a hunch.

stafusa
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Michal
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2 Answers2

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Svavil and garyp are correct when they point to the $\dot{m}$ term as being the problem. Consider the case where the ship is ejecting blocks rather than an exhaust gas. With each block expelled at a speed of $v_e$ and mass $\delta m$ the change in the speed of the ship would be $$(m - \delta m) \delta v = v_e \delta m$$ As the size of the blocks goes to zero the $\delta m$ term disappears, or mathematically $$\lim_{\delta m \to 0} (m - \delta m) = m$$ So the equation becomes $$m dv = v_e dm$$ Which can be rearranged to $$ dv = v_e \frac{dm}{m}$$ From that you can get the solution you quote. One thing to note is that the solution is independent of the rate or changes in the rate the gas is expelled, it only requires that the speed relative to the ship is constant.

John M. Cavallo
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This must be equal (and opposite) to the impulse imparted on the rocket during that time and hence the rate of change of momentum.

$$ -\dot{m}v_e = \frac{d}{dt} (mv) = \dot{m}v + m\dot{v} $$

This is where your train of thought went wrong. The correct equation for a rocket free of external forces is, assuming $v_e$ is magnitude of exhaust's velocity (thus it is not a signed quantity),

$$ -\dot{m}v_e = m\frac{dv}{dt}. $$

This is because change of momentum associated with the rocket (with rate of change $d(mv)/dt$) is not just due to effective force of the expelled gases acting on the nozzle $-\dot{m}v_e$, but also due to loss of momentum that fuel+oxidizer had before they left the rocket.

This lost momentum, per unit time, is $-\dot{m}v$. To get the rate of change of momentum, we thus have to subtract this loss from the effective force:

$$ -\dot{m}v_e + \dot{m} v = \frac{d(mv)}{dt}. $$ See e.g. my answer here for explanation and general derivation of the equation of motion for variable mass systems:

https://physics.stackexchange.com/a/142773/31895

Ján Lalinský
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