0
$\begingroup$

I think I understand how the equation arises when there is no external force. I'm following the Wikipedia article on this, there is a nice sketch:

enter image description here

At time $t$, the total momentum of the system is $p_i = (m + \Delta m) v$. At time $t + \Delta t$, the total momentum is $p_f = m (v + \Delta v) + (v-v_e) \Delta m$. If the external force is zero, then $p_i = p_f$ and thus $$ (m + \Delta m) v = m (v+\Delta v) + (v - v_e ) \Delta m $$ $$ 0 = m \Delta v - v_e \Delta m $$

and the rest of the derivation follows (here $\Delta m$ is replaced by $- \mathrm{d} m$, because while the $\mathrm{d} m$ corresponds to the rocket's change in mass, $\Delta m$ was the mass expelled, thus $\Delta m = - \mathrm{d} m$) and we get $$ \mathrm{d} v = - \frac{\mathrm{d} m}{m} v_e \quad \implies \quad v_f - v_i = v_e \log(m_i/m_f) $$

So far so good.

Now I'd like to extend this to a situation, when there is an external force on the rocket. We should go back to the equation $$ 0 = m \dot{v} + v_e \dot{m} $$

The left-hand side is zero, because we didn't consider any external forces. I'd be tempted to just write $F_{\text{ext.}} = m \dot{v} + v_e \dot{m}$, however, I have a hard time replacing the left-hand side directly with $F_{\text{ext.}}$; the right-hand side was obtained as a change of momentum of the system rocket + gas, however, I only consider an external force on the rocket - let's say, the rocket is additionally pulled by a rope - the rope stops acting on the parcel of gas $\Delta m$ as soon as it leaves the rocket! Thus, we would somehow need to consider the rocket alone, and for that, the above derivation might not be sufficient. Even if we consider a force like gravity, that acts uniformly on everything, I have a hard time applying it to the equation above, because the parts of the system (rocket + expelled parcel of gas) are moving with respect to each other, and in those cases, we need to break the system further apart, until each part consists only of particles moving together at the same rate. (and by the way, practically, if $F_{\text{ext.}}$ is gravity, do we write $F_{\text{ext.}} = m (t) g$ (only mass of the rocket) or $m_0 g$ (the initial mass, i.e. mass of the rocket + all the gas that was expelled?)

How should I proceed in such case?

$\endgroup$
2

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.