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lowea001

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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 [itex]M_{0}[/itex] (excluding fuel) originally has fuel of mass [itex]m_{0}[/itex] which is being ejected at constant velocity [itex]u[/itex] downward, relative to the rocket at a constant rate [itex]\frac{dm}{dt}[/itex]. By conservation of momentum, the rocket acquires a velocity [itex]v[/itex] upward. Applying momentum conservation in the frame of an observer we have:

[tex]\frac{d(M(t)v)}{dt} = M(t)\frac{dv}{dt} + v\frac{dM(t)}{dt} = (v - u) \frac{dm}{dt}[/tex] ***the expression after the last equals sign is what I do not get. ***

Since [itex]dM=dm[/itex] we have:

[tex]M(t)\frac{dv}{dt} = -\frac{dM(t)}{dt}u[/tex]

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 [itex]F = v\frac{dm}{dt}[/itex] still work if v is not constant, but changing with time? Shouldn't the right hand side of the line in question read [itex]\frac{dm(v-u)}{dt} = \frac{dm}{dt}v + \frac{dv}{dt}m - \frac{dm}{dt}u[/itex] 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.

A rocket of mass [itex]M_{0}[/itex] (excluding fuel) originally has fuel of mass [itex]m_{0}[/itex] which is being ejected at constant velocity [itex]u[/itex] downward, relative to the rocket at a constant rate [itex]\frac{dm}{dt}[/itex]. By conservation of momentum, the rocket acquires a velocity [itex]v[/itex] upward. Applying momentum conservation in the frame of an observer we have:

[tex]\frac{d(M(t)v)}{dt} = M(t)\frac{dv}{dt} + v\frac{dM(t)}{dt} = (v - u) \frac{dm}{dt}[/tex] ***the expression after the last equals sign is what I do not get. ***

Since [itex]dM=dm[/itex] we have:

[tex]M(t)\frac{dv}{dt} = -\frac{dM(t)}{dt}u[/tex]

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 [itex]F = v\frac{dm}{dt}[/itex] still work if v is not constant, but changing with time? Shouldn't the right hand side of the line in question read [itex]\frac{dm(v-u)}{dt} = \frac{dm}{dt}v + \frac{dv}{dt}m - \frac{dm}{dt}u[/itex] 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.

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