# I Derivation of Rocket Equation Using Relative Velocity

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1. Feb 7, 2017

### lowea001

Based on my current understanding of the problem I do not see this following derivation as valid, although this is what was given in my course notes. Although this particular example is from an undergraduate physics course this is not a homework problem: I'm confused about the underlying concept. I will highlight below the specific step I do not understand. Here is the given derivation of the rocket equation:

A rocket of mass $M_{0}$ (excluding fuel) originally has fuel of mass $m_{0}$ which is being ejected at constant velocity $u$ downward, relative to the rocket at a constant rate $\frac{dm}{dt}$. By conservation of momentum, the rocket acquires a velocity $v$ upward. Applying momentum conservation in the frame of an observer we have:
$$\frac{d(M(t)v)}{dt} = M(t)\frac{dv}{dt} + v\frac{dM(t)}{dt} = (v - u) \frac{dm}{dt}$$ ***the expression after the last equals sign is what I do not get. ***
Since $dM=dm$ we have:
$$M(t)\frac{dv}{dt} = -\frac{dM(t)}{dt}u$$
which is the rocket equation.
Okay, so most of that makes sense (kind of), but the relative velocity part seems kind of questionable (or hand-wavy at the very least) since the entire premise behind being able to treat a fluid's change in momentum being only due to its rate of change of mass is that the velocity is constant, which it is clearly not in the case of v - u. In other words, why does the force from a fluid equation $F = v\frac{dm}{dt}$ still work if v is not constant, but changing with time? Shouldn't the right hand side of the line in question read $\frac{dm(v-u)}{dt} = \frac{dm}{dt}v + \frac{dv}{dt}m - \frac{dm}{dt}u$ since v isn't constant? Sorry, I tried to make this as succinct as possible. If someone could point me in the general direction and help me understand how I am misunderstanding this idea I would be very grateful. Thank you.

Last edited: Feb 7, 2017
2. Feb 7, 2017

### haruspex

The first step, d(mv)/dt= mdv/dt+vdm/dt, is problematic. In effect, it treats the mass m(t) of a closed system as something that can vary. This is of course nonsense. Ending up with the right rocket equation appears to be via some sleight of hand, and I think it is that part that had you foxed.