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I was going through a fairly simple proof about rockets/momentum, taking into account the changing mass of the rocket as well as the velocity of the exhausted fuel. I was getting towards the end of it having felt like I understood it well, until I got to this point (this is from a Junior year course in Mechanics btw):

After assuming that there are no external forces on the rocket (so that we can say dP=0):

mdv=-dmv_{exhaust}

Okay, this step makes perfect sense. Then the proof goes on to say (after separating variables):

dv=v_{exhaust}(dm/m) Okay, again, this makes perfect sense.

But after this there is an integration that I don't understand (it's been a while since I've taken Calculus... so I'm not sure what I missed). The author then integrates both sides and obtains:

v-v_{initial}=v_{exhaust}ln(m_{initial}/m)

Why is the natural log on the right hand side ln(m_{initial}/m)? When I did the integration I got:

v+v_{initial}=v_{exhaust}ln(m).

Similar results, but clearly a big difference.

I didn't post this in the homework section because it's not homework (I'm on medical leave from university and am brushing up on material by going through the text before I have to return and take my final exam in Mechanics... yay! ) and I felt it was more of a mathematical question about integration techniques/reasoning. There might be something conceptual that I'm not following. I also haven't posted here in a while but probably will use these boards a lot in the next few months as I prepare to return to school. I see other posts have used LaTeX but I don't know how to implement it if anyone wants to chime in on how to (i.e. what is the tag I have to put around my LaTeX code?).

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# Tricky Integration in Rocket-Momentum Derivation

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