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Rocket losing mass

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A rocket launches from a space base at 0 velocity and looses mass at constant rate C. what is the ratio between the rocket's mass at maximum momentum to it's initial mass.

    2. Relevant equations
    Newton's second law: ##F=dP=\frac{d(mV)}{dt}##
    Conservation of momentum: ##m_1v_1=m_2v_2##

    3. The attempt at a solution
    To find maximum momentum i differentiate momentum:
    $$F=dP=\frac{d(mV)}{dt}=m\frac{dV}{dt}+V\frac{dm}{dt}=ma+V\cdot C=0$$
    $$m(t)a(t)=-C\cdot V(t),\ m=M_0-C\cdot t\rightarrow (M_0-C\cdot t)a(t)=-CV(t) \rightarrow t=\frac{M_0}{C}+\frac{V(t)}{a(t)}$$
    Both a(t) and V(t) are functions of T and i can't solve for t.
    I am not told anything about the velocity the mass leaves the rocket.
     
  2. jcsd
  3. Jun 16, 2015 #2

    mfb

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    It is v0, you can assume that it is constant (otherwise the problem is not solvable). The value does not matter, as the velocity will cancel out in the calculations later.

    You don't need that. Just assume that there is enough fuel to reach the point of maximal momentum (we know that there is, because the rocket has a constant mass flow rate).
     
  4. Jun 16, 2015 #3

    haruspex

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    Yes, I realised my blunder but didn't manage to delete before you immortalised it.
    Karol, expanding F as ##\dot P = m\dot v + \dot mv## is never a good idea. It would be true of a body in which mass is being created or destroyed without affect its velocity. See section 6 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-momentum-impacts/.
    Assume thrust, ##F=m\dot v##, is constant.
     
  5. Jun 17, 2015 #4
    If the velocity Vr of the expelled gases is constant and the mass per second is also constant then ##\delta m\cdot V_r=F\cdot\delta t## and since ##\delta t## is constant then F, the force exerted on the rocket is constant, is that right?
    But if so i cant take ##m\dot v## as constant since m changes and ##F=\dot P = m\dot v + \dot mv## and there is another member ##\dot mv##.
    But maybe i don't understand F. i think your F in ##F=m\dot v## is the same F in mine: ##F=\dot P## and is the total force applied on the rocket, so how can they be the same? they have 2 different equations, your is shorter than mine.
    Conservation of momentum between the gases and rocket:
    $$0=m\cdot dV+dm\cdot V_r\rightarrow \int_0^V dV=-\int_{M_0}^m \frac{dm}{m}\rightarrow V(t)=ln\left( \frac{M_0}{m} \right)=ln\left( \frac{M_0}{M_0-Ct} \right)$$
    $$P=m(t)V(t)=(M_0-Ct)ln\left( \frac{M_0}{M_0-Ct} \right)$$
    I differentiate P in order to find maximum momentum:
    $$\dot P=-Cln\left( \frac{M_0}{M_0-Ct} \right)+\frac{(M_0-Ct)^2}{M_0}=0$$
    $$ln\left( \frac{M_0}{M_0-Ct} \right)=\frac{(M_0-Ct)^2}{CM_0}$$
    I can't solve
     
  6. Jun 17, 2015 #5

    mfb

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    The expressions for P(t) and its derivative look more complicated than they should.
    Can you find P(V)? Maybe that is easier.
     
  7. Jun 17, 2015 #6

    haruspex

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    You lost a factor Vr, but it doesn't matter - just makes it dimensionally wrong.
    You've differentiated the log function incorrectly. You should not end up with a quadratic.
     
  8. Jun 18, 2015 #7
    How? i know only P=m(t)V(t), how to make P(V)?
    $$0=m\cdot dV+dm\cdot V_r\rightarrow \int_0^V dV=-V_r\int_{M_0}^m \frac{dm}{m}\rightarrow V(t)=ln\left( \frac{M_0}{m} \right)=V_r\cdot ln\left( \frac{M_0}{M_0-Ct} \right)$$
    $$P=m(t)V(t)=V_r(M_0-Ct)ln\left( \frac{M_0}{M_0-Ct} \right)$$
    I don't know where i made the mistake:
    $$P=m\cdot ln\left( \frac{M_0}{m} \right), \quad \dot P=-Cln\left( \frac{M_0}{m} \right)+m\frac{m}{M_0}$$
    I differentiated in a different way:
    $$P=ln\left( \frac{M_0}{m} \right)^m\rightarrow \dot P=m\left( \frac{M_0}{m} \right)^{(m-1)}\cdot\left( \frac{m}{M_0} \right)^m=\frac{m}{M_0}$$
    $$\dot P=0\rightarrow M_0=Ct \rightarrow t=\frac{M_0}{C}$$
    The mass at P'=0 is 0, so it doesn't help
     
    Last edited: Jun 18, 2015
  9. Jun 18, 2015 #8

    mfb

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    The power of m is inside the log, not outside. You mixed that somehow in the other approach.

    You know V(m), you can convert this to m(V). Then P(V)=m(V)*V, and that expression has a nice derivative.
     
  10. Jun 18, 2015 #9

    haruspex

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    You are differentiating ln(M/m) incorrectly. Write it as ln(M)-ln(m), then try. What's the derivative of ln(x) wrt x?
     
  11. Jun 18, 2015 #10
    It's inside:
    $$P=ln\left( \frac{M_0}{m} \right)^m \equiv ln\left[ \left( \frac{M_0}{m} \right)^m \right]$$
    I take again the derivative:
    according to ##\left( u^v \right)'=vu^{v-1}\dot u+u^vln(u)\dot v##
    $$\left[ \left( \frac{M_0}{m} \right)^m \right]'=...=-\frac{M_0^m}{m^m}\left( 1+Cln\left( \frac{M_0}{m} \right) \right)$$
    $$\dot P=-\frac{m^m}{M_0^m}\cdot \frac{M_0^m}{m^m}\left( 1+Cln\left( \frac{M_0}{m} \right) \right)=-1-Cln\left( \frac{M_0}{m} \right)$$
    $$\dot P=0\rightarrow ln\left( \frac{M_0}{m} \right)=-\frac{1}{C} \rightarrow t=\frac{M_0 \left( 1-\sqrt[C]{e} \right)}{C}$$
    The mass at time t of maximum momentum:
    $$m=M_0-Ct=M_0 \sqrt[C]{e}$$
    The ratio to the initial mass M0 is therefor ##\sqrt[C]{e}## but that's incorrect
     
  12. Jun 18, 2015 #11

    mfb

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    You don't have uv', you have ln(uv). Your outer derivative would be the log, then you get the exponent as inner derivative, and then the fraction as another step. There is no point in making it that complicated.
     
  13. Jun 18, 2015 #12
    Honestly, I never really learned this topic, and I know that most of my classmates did not. So I decided to do some reading and make an attempt at solving this problem in detail. Please take a look at my solution and let me know what you think.
     
  14. Jun 18, 2015 #13
    $$V(m)=ln\left( \frac{M_0}{m} \right),\quad P(m)=mV(m)=m\cdot ln\left( \frac{M_0}{m} \right)$$
    $$\dot P=ln\left( \frac{M_0}{m} \right)-m\frac{m}{M_0}\cdot \frac{1}{m^2}=ln\left( \frac{M_0}{m} \right)-\frac{1}{M_0}$$
    $$\dot P=0 \rightarrow e^{\left( \frac{1}{M_0} \right)}=\frac{M_0}{m} \rightarrow m=\frac{M_0}{\sqrt[M_0]{e}}$$
    And again it's not correct
     
  15. Jun 18, 2015 #14
    Karol, could you post the correct answer?
     
  16. Jun 18, 2015 #15
    $$P=m\cdot ln\left( \frac{M_0}{m} \right)=m(ln \: M_0-ln\: m), \quad \dot P=ln \: M_0-ln\: m-\frac{1}{m}$$
    $$\dot P=0\rightarrow ln \: M_0=ln\: m-\frac{1}{m}$$
    I can't solve.
    The possible answers are: ##\frac{1}{2},\quad \frac{1}{3},\quad \frac{1}{e},\quad \frac{1}{\pi},\quad \frac{1}{4},\quad ##
     
  17. Jun 18, 2015 #16
    I came up with 1/e if you would like to take a look at my work. I'm somewhat confident I got there the right way. :smile:
     
  18. Jun 18, 2015 #17
    $$P=ln\left( \frac{M_0}{m} \right)^m,\quad \dot P=\frac{m^m}{M_0^m}\cdot m \left( \frac{M_0}{m} \right)^{m-1}\cdot \frac{-M_0}{m^2}=1$$
    Incorrect
     
  19. Jun 18, 2015 #18

    mfb

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    That's not what I suggested (also applies to post 17 which was not there at the time I wrote this post). Also, you got the derivative wrong again, see the mismatching units. You can follow haruspex' advice if you want to keep the mass instead of the velocity, or use the velocity and get rid of the mass as I suggested. Using the same wrong derivative over and over again does not help.

    @Wily Willy: Please do not post full solutions, this is against the forum rules.
     
  20. Jun 18, 2015 #19

    haruspex

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    Better, but this time you dropped a factor m in that step.
     
  21. Jun 19, 2015 #20
    $$P=m\cdot ln\left( \frac{M_0}{m} \right)=m(ln \: M_0-ln\: m), \; \dot P=ln \: M_0-(ln\: m+1)$$
    $$\dot P=0\;\rightarrow\; ln \: M_0-(ln\: m+1)=0\;\rightarrow\; m=e^{(ln\:M_0-1)}=\frac{M_0}{e}$$
    That's correct i think, according to @Wily Willy
     
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