Tricky Integration in Rocket-Momentum Derivation

1. Oct 11, 2011

mar2194

Hi,

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=-dmvexhaust

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

dv=vexhaust(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-vinitial=vexhaustln(minitial/m)

Why is the natural log on the right hand side ln(minitial/m)? When I did the integration I got:
v+vinitial=vexhaustln(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?).

2. Oct 11, 2011

Parmenides

mar2194,

First recall that a property of the natural logarithm is --> ln(a/b) = ln(a) - ln(b) and also when you integrate, say, f(x) = dv, (of which F(x) = v) you will have limits to compute for a definite integral. The integral of dv with limits A to B is v(B) - v(A).

Also think of what's being explained in a conceptual sense. The rocket's final mass will be less than its initial mass. This is because it burns fuel as it increases in speed (the fuel will a part of the rocket's mass). So in other words, the mass of the rocket decreases and its velocity increases. Use these notions to evaluate your limits of m and v and see what you get.

https://www.physicsforums.com/blog.php?b=3216 [Broken]

Last edited by a moderator: May 5, 2017
3. Oct 12, 2011

Staff: Mentor

LaTeX description in the blog is incorrect - tags should take form [noparse][tеx][/tеx][/noparse] and [noparse][itеx][/itеx][/noparse].

4. Oct 12, 2011

mar2194

Thanks! So the integration is from m to m initial, correct? That gives me the result the textbook has. Thanks a bunch! I knew it was simple, but it's literally been 2 years since I've taken Calculus.
~mar2194~

Last edited by a moderator: May 5, 2017