Rocket losing mass

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Homework Statement


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.

Homework Equations


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

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.
 

Answers and Replies

  • #2
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I am not told anything about the velocity the mass leaves the rocket.
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.

There seems to be missing information. You need to know how much of the initial mass is fuel.
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).
 
  • #3
haruspex
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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).
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.
 
  • #4
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Assume thrust, ##F=m\dot v##, is constant.
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
 
  • #5
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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.
 
  • #6
haruspex
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$$0=m\cdot dV+dm\cdot V_r\rightarrow \int_0^V dV=-\int_{M_0}^m \frac{dm}{m}\rightarrow V(t)=ln\left( \frac{M_0}{m} \right)=ln\left( \frac{M_0}{M_0-Ct} \right)$$
$$P=m(t)V(t)=(M_0-Ct)ln\left( \frac{M_0}{M_0-Ct} \right)$$
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
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.
 
  • #7
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Can you find P(V)? Maybe that is easier.
How? i know only P=m(t)V(t), how to make P(V)?
$$0=m\cdot dV+dm\cdot V_r\rightarrow \int_0^V dV=-V_r\int_{M_0}^m \frac{dm}{m}\rightarrow V(t)=ln\left( \frac{M_0}{m} \right)=V_r\cdot ln\left( \frac{M_0}{M_0-Ct} \right)$$
$$P=m(t)V(t)=V_r(M_0-Ct)ln\left( \frac{M_0}{M_0-Ct} \right)$$
You've differentiated the log function incorrectly. You should not end up with a quadratic.
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
 
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  • #8
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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.
 
  • #9
haruspex
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I don't know where i made the mistake:
$$P=m\cdot ln\left( \frac{M_0}{m} \right), \quad \dot P=-Cln\left( \frac{M_0}{m} \right)+m\frac{m}{M_0}$$
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?
 
  • #10
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The power of m is inside the log, not outside. You mixed that somehow in the other approach.
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
 
  • #11
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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.
 
  • #12
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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.
 
  • #13
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You know V(m), you can convert this to m(V). Then P(V)=m(V)*V, and that expression has a nice derivative.
$$V(m)=ln\left( \frac{M_0}{m} \right),\quad P(m)=mV(m)=m\cdot ln\left( \frac{M_0}{m} \right)$$
$$\dot P=ln\left( \frac{M_0}{m} \right)-m\frac{m}{M_0}\cdot \frac{1}{m^2}=ln\left( \frac{M_0}{m} \right)-\frac{1}{M_0}$$
$$\dot P=0 \rightarrow e^{\left( \frac{1}{M_0} \right)}=\frac{M_0}{m} \rightarrow m=\frac{M_0}{\sqrt[M_0]{e}}$$
And again it's not correct
 
  • #14
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Karol, could you post the correct answer?
 
  • #15
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You are differentiating ln(M/m) incorrectly. Write it as ln(M)-ln(m), then try
$$P=m\cdot ln\left( \frac{M_0}{m} \right)=m(ln \: M_0-ln\: m), \quad \dot P=ln \: M_0-ln\: m-\frac{1}{m}$$
$$\dot P=0\rightarrow ln \: M_0=ln\: m-\frac{1}{m}$$
I can't solve.
Karol, could you post the correct answer?
The possible answers are: ##\frac{1}{2},\quad \frac{1}{3},\quad \frac{1}{e},\quad \frac{1}{\pi},\quad \frac{1}{4},\quad ##
 
  • #16
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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:
 
  • #17
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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
$$P=ln\left( \frac{M_0}{m} \right)^m,\quad \dot P=\frac{m^m}{M_0^m}\cdot m \left( \frac{M_0}{m} \right)^{m-1}\cdot \frac{-M_0}{m^2}=1$$
Incorrect
 
  • #18
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$$V(m)=ln\left( \frac{M_0}{m} \right),\quad P(m)=mV(m)=m\cdot ln\left( \frac{M_0}{m} \right)$$
$$\dot P=ln\left( \frac{M_0}{m} \right)-m\frac{m}{M_0}\cdot \frac{1}{m^2}=ln\left( \frac{M_0}{m} \right)-\frac{1}{M_0}$$
$$\dot P=0 \rightarrow e^{\left( \frac{1}{M_0} \right)}=\frac{M_0}{m} \rightarrow m=\frac{M_0}{\sqrt[M_0]{e}}$$
And again it's not correct
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.
 
  • #19
haruspex
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$$P=m(ln \: M_0-ln\: m), \quad \dot P=ln \: M_0-ln\: m-\frac{1}{m}$$
Better, but this time you dropped a factor m in that step.
 
  • #20
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$$P=m\cdot ln\left( \frac{M_0}{m} \right)=m(ln \: M_0-ln\: m), \; \dot P=ln \: M_0-(ln\: m+1)$$
$$\dot P=0\;\rightarrow\; ln \: M_0-(ln\: m+1)=0\;\rightarrow\; m=e^{(ln\:M_0-1)}=\frac{M_0}{e}$$
That's correct i think, according to @Wily Willy
 
  • #21
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Can you find P(V)? Maybe that is easier.
$$0=m\cdot dV+dm\cdot V_r\;\rightarrow m\int^V_0 dV=V_r\int^{M_0}_m dm\;\rightarrow mV=V_r(M_0-m) \;\rightarrow m=\frac{M_0 V_r}{V_r+V}$$
$$P(V)=m(V)V=\frac{V\cdot M_0 V_r}{V_r+V},\; \dot P=\frac{V_r}{(V_r+V)^2},\;\dot P=0$$
It can't be
 
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  • #22
haruspex
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$$P=m\cdot ln\left( \frac{M_0}{m} \right)=m(ln \: M_0-ln\: m), \; \dot P=ln \: M_0-(ln\: m+1)$$
$$\dot P=0\;\rightarrow\; ln \: M_0-(ln\: m+1)=0\;\rightarrow\; m=e^{(ln\:M_0-1)}=\frac{M_0}{e}$$
That's correct i think, according to @Wily Willy
That's it.
 
  • #23
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Yes but what about P(V) post #21?
 
  • #24
haruspex
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Yes but what about P(V) post #21?
The step in the first right arrow in post #21 looks wrong to me. m is a variable, you can't take it outside ##\int m.dV##.
 
  • #25
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The step in the first right arrow in post #21 looks wrong to me. m is a variable, you can't take it outside ##\int m.dV##.
So how to get P(V)? i can't integrate ##\int^V_0 m(V)dV##
 

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