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Tricky Integration in Rocket-Momentum Derivation

  1. Oct 11, 2011 #1
    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! :wink:) 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. jcsd
  3. Oct 11, 2011 #2
    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.

    Follow this link for a description of LaTeX:

    https://www.physicsforums.com/blog.php?b=3216 [Broken]
     
    Last edited by a moderator: May 5, 2017
  4. Oct 12, 2011 #3

    Borek

    User Avatar

    Staff: Mentor

    LaTeX description in the blog is incorrect - tags should take form [noparse][tеx][/tеx][/noparse] and [noparse][itеx][/itеx][/noparse].
     
  5. Oct 12, 2011 #4
    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
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