Rocket Losing Mass: Finding Ratio of Max Momentum to Initial Mass

[Karol]

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.

 

[mfb]

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).

 

[haruspex]

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 realized 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.

 

[Karol]

Assume thrust, $F=m\dot v$, is constant.

If the velocity V_r 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 can't 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

 

[mfb]

The expressions for P(t) and its derivative look more complicated than they should.
Can you find P(V)? Maybe that is easier.

 

[haruspex]

$$0=m\cdot dV+dm\cdot V_r\rightarrow \int_0^V dV=-\int_{M_0}^m \frac{dm}{m}\rightarrow V(t)=ln\left( \frac{M_0}{m} \right)=ln\left( \frac{M_0}{M_0-Ct} \right)$$
$$P=m(t)V(t)=(M_0-Ct)ln\left( \frac{M_0}{M_0-Ct} \right)$$
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.

 

[Karol]

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

 

[mfb]

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.

 

[haruspex]

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?

 

[Karol]

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

 

[mfb]

You don't have u^v', you have ln(u^v). 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.

 

[Wily Willy]

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.

 

[Karol]

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

 

[Wily Willy]

Karol, could you post the correct answer?

 

[Karol]

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: $\frac{1}{2},\quad \frac{1}{3},\quad \frac{1}{e},\quad \frac{1}{\pi},\quad \frac{1}{4},\quad$

 

[Wily Willy]

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.

 

[Karol]

You don't have u^v', you have ln(u^v). 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

 

[mfb]

$$V(m)=ln\left( \frac{M_0}{m} \right),\quad P(m)=mV(m)=m\cdot ln\left( \frac{M_0}{m} \right)$$
$$\dot P=ln\left( \frac{M_0}{m} \right)-m\frac{m}{M_0}\cdot \frac{1}{m^2}=ln\left( \frac{M_0}{m} \right)-\frac{1}{M_0}$$
$$\dot P=0 \rightarrow e^{\left( \frac{1}{M_0} \right)}=\frac{M_0}{m} \rightarrow m=\frac{M_0}{\sqrt[M_0]{e}}$$
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.

 

[haruspex]

$$P=m(ln \: M_0-ln\: m), \quad \dot P=ln \: M_0-ln\: m-\frac{1}{m}$$

Better, but this time you dropped a factor m in that step.

 

[Karol]

$$P=m\cdot ln\left( \frac{M_0}{m} \right)=m(ln \: M_0-ln\: m), \; \dot P=ln \: M_0-(ln\: m+1)$$
$$\dot P=0\;\rightarrow\; ln \: M_0-(ln\: m+1)=0\;\rightarrow\; m=e^{(ln\:M_0-1)}=\frac{M_0}{e}$$
That's correct i think, according to @Wily Willy

 

[Karol]

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

 

[haruspex]

$$P=m\cdot ln\left( \frac{M_0}{m} \right)=m(ln \: M_0-ln\: m), \; \dot P=ln \: M_0-(ln\: m+1)$$
$$\dot P=0\;\rightarrow\; ln \: M_0-(ln\: m+1)=0\;\rightarrow\; m=e^{(ln\:M_0-1)}=\frac{M_0}{e}$$
That's correct i think, according to @Wily Willy

That's it.

 

[Karol]

Yes but what about P(V) post #21?

 

[haruspex]

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 $\int m.dV$.

 

[Karol]

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 $\int^V_0 m(V)dV$

 

[Wily Willy]

The first thing that catches my eye is that momentum should always have units of kg(m/s). You are missing that when you start here. It does not effect your solution because it would be canceled by division once you set the derivative equal to zero.

P=m⋅ln(M0m)=m(lnM0−lnm),P˙=lnM0−(lnm+1)

This is right. (Never-minding the velocity.)

P˙=0→lnM0−(lnm+1)=0→m=e(lnM0−1)=M0e

And this is right. Congrats! You know most people consider rocket equations the hardest part of introductory mechanics, so if you can handle this you know you can handle anything else!

 

[Wily Willy]

You can get the velocity by going back and differentiating the total momentum of the rocket and the gas. I would suggest not using the notation C for the rate that the fuel is burnt. Just keep in mind that the total momentum is a constant, the velocity of the gas is a constant, and that the change in the mass of the rocket is equal and opposite to the change in the mass of the gas.

 

[mfb]

Okay, as you found the solution we can explore alternatives with more help. Here is the velocity calculation I suggested:

P=m*V

We know from post #4 that $V=V_r \ln \left(\frac{M_0}{m}\right)$, solving this for m gives $m=M_0 \exp\left(\frac{-V}{V_r}\right)$
Therefore,
$$P=M_0 V \exp{-V/V_r}$$
$$\dot{P} = M_0 \exp\left(\frac{-V}{V_r}\right) \left(1-\frac{V}{V_r}\right)$$
It is easy to see that this becomes zero at V=V_r. Plugging it back into the equation for m, we get m=M₀e^-1.There is a way to get the result V=V_r without calculation just based on momentum conservation: the change of momentum of the rocket is given by the "momentum exhaust". This is zero exactly when the exhaust has zero velocity. As it moves with -V_r relative to the rocket, this happens at V=V_r.

 

[Karol]

You can get the velocity by going back and differentiating the total momentum of the rocket and the gas... Just keep in mind that the total momentum is a constant, the velocity of the gas is a constant, and that the change in the mass of the rocket is equal and opposite to the change in the mass of the gas.

The result is as mfb said:
$$m(V)=M_0e^{\left( -\frac{V}{V_r} \right)}\left( 1-\frac{V}{V_r} \right)$$
But how to get to that with your method?
I tried this way:
From a coordinate system moving at the momentary velocity V the change in momentum of the gases and rocket equals zero: $0=m\cdot dV+dm\cdot V_r$
But i can't differentiate it any further, i think mathematically it's not allowed.
I try to take the total momentum in an inertial frame but i don't know which mass to take for the gases: should i sum up all the gases that were expelled from the beginning or the infinitesimal amount? and the gases from different times have different velocities also.
So i am left with the infinitesimal amount:
$$\dot P=\frac{d(mV+dmV_r)}{dt}$$
But i can't differentiate that either

 

[Wily Willy]

My way is really the same as mfb's. I was trying to be "less direct" in my advice and keep in line with the forum's rules.

Anyway, I was just saying that determining a function for momentum with velocity as the only variable might be easier if you just started from the beginning with careful notation of the variables involved.

 

[haruspex]

$0=m⋅dV+dm⋅V_r$
But i can't differentiate it any further, i think mathematically it's not allowed.

Well, no, you can't differentiate it because in full it is $0=dP=m⋅dV+dm⋅V_r$. You can divide through by dt to give $0=\dot P=m⋅\dot V+\dot m⋅V_r$, whence $(m_0-Ct)\dot V=C V_r$.

 

[Karol]

$0=dP=m⋅dV+dm⋅V_r$. You can divide through by dt to give $0=\dot P=m⋅\dot V+\dot m⋅V_r$, whence $(m_0-Ct)\dot V=C V_r$.

But $\dot m=-C$ and $\dot m V_r=-C\cdot V_r\rightarrow 0=\dot P=m⋅\dot V+\dot m⋅V_r=C\cdot V_r-C\cdot V_r=0$

 

[haruspex]

But $\dot m=-C$ and $\dot m V_r=-C\cdot V_r\rightarrow 0=\dot P=m⋅\dot V+\dot m⋅V_r=C\cdot V_r-C\cdot V_r=0$

I'm finding the notation confusing. I'll write P_tot for total momentum and P_m for the momentum of the rocket mass. And I'll write t_f for the time at which the rocket momentum is maximised.
$P_m=mv$, $\dot P_{tot}=0=m\dot v -Cv_r$, $\dot P_m(t_f)=0=\dot m(t_f)v(t_f)+m(t_f)\dot v(t_f)=-Cv(t_f)+ Cv_r$.
So $v(t_f)=v_r$, which should not surprise. It says the rocket's momentum is maximised when the exhaust being emitted at that time has zero momentum.
But this doesn't get us to the answer. To get an answer we need to solve a differential equation. There may be an easier way, but one way is to go back to the basic rocket equation $\dot P_{tot}=0=m\dot v -Cv_r=(m_0-Ct)\dot v -Cv_r$. Rearrange and integrate wrt t.

 

[Karol]

one way is to go back to the basic rocket equation $\dot P_{tot}=0=m\dot v -Cv_r=(m_0-Ct)\dot v -Cv_r$. Rearrange and integrate wrt t.

$$(m_0-Ct)\dot v -CV_r\;\rightarrow \dot v=\frac{CV_r}{m_0-Ct}\;\rightarrow \int\dot v=\int\frac{CV_r}{m_0-Ct}$$
$$v=CV_r\int^t_0\frac{dt}{m_0-Ct},\;m_0-Ct\triangleq u,\; dt=-\frac{du}{C}, \; t=0\rightarrow u=m_0$$
$$CV_r\int^{\frac{m_0-u}{C}}_{m_0}\frac{-du}{C\cdot u}=-V_r\left[ln\:u\right]^{\frac{m_0-u}{C}}_{m_0}=V_r ln\left( \frac{Cm_0}{m_0-u} \right)=V_r ln\left( \frac{m_0}{t} \right)$$
It's incorrect and doesn't lead me to the answer of what's m(v). also i am not sure i am allowed to make the step: $\int\dot v=v$

 

[haruspex]

$$CV_r\int^{\frac{m_0-u}{C}}_{m_0}\frac{-du}{C\cdot u}$$
It's incorrect and doesn't lead me to the answer of what's m(v). also i am not sure i am allowed to make the step: $\int\dot v=v$

the upper bound on the integral is wrong. It contains two errors. One could have been found by checking the dimensional consistency.

 

[Karol]

$$CV_r\int^{u}_{m_0}\frac{-du}{C\cdot u}=V_r ln\:\left( \frac{m_0}{u} \right)=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)$$
But what's m(V)? how does it help?

 

[haruspex]

$$CV_r\int^{u}_{m_0}\frac{-du}{C\cdot u}=V_r ln\:\left( \frac{m_0}{u} \right)=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)$$
But what's m(V)? how does it help?

The above expression is what you have found as being equal to v, no? So you have v as a function of t. You can easily convert that to v as a function of m, and hence get mv as a function of m. Then see what m maximises mv.

 

[Karol]

$$v=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)=V_r ln\:\left( \frac{m_0}{m} \right)=v(m)\rightarrow m(v)=m_0e^{-\frac{v}{V_r}}$$
$$P(m)=V_r\cdot m\cdot ln\left( \frac{m_0}{m} \right),\;\dot P(m)=V_r\left[ ln\left( \frac{m_0}{m} \right)-1\right]=0\rightarrow m(P_{max})=\frac{m_0}{e}$$
If i search for the velocity at maximum momentum:
$$P(v)=m(v)v=m_0e^{-\frac{v}{V_r}}\cdot v,\;\dot P(v)=0 \rightarrow v=V_r$$

 

[Karol]

I'll write P_tot for total momentum and P_m for the momentum of the rocket mass. And I'll write t_f for the time at which the rocket momentum is maximised.
$P_m=mv$, $\dot P_{tot}=0=m\dot v -Cv_r$, $\dot P_m(t_f)=0=\dot m(t_f)v(t_f)+m(t_f)\dot v(t_f)=-Cv(t_f)+ Cv_r$.
So $v(t_f)=v_r$, which should not surprise. It says the rocket's momentum is maximized when the exhaust being emitted at that time has zero momentum.

Where did $\dot P_{tot}=0=m\dot v -Cv_r$ come from? i read in a book that we move at the instantaneous velocity v of the rocket, in a non-inertial frame, then during a short period of time (dt) the gases acquire momentum $dmV_r$ (dm is the decrease in mass of the rocket so it's negative), and the rocket acquires momentum: $(m-dm)dv=m\cdot dv-dmdv\approx m\cdot dv$. do you have an other explanation?
$$dP_{tot}=m\cdot dv-dm\cdot V_r,\;\frac{dP_{tot}}{dt}=\dot P_{tot}=m\dot v-\dot mV_r=m\dot v-Cv_r$$
But $\dot P_m(t_f)=0=\dot m(t_f)v(t_f)+m(t_f)\dot v(t_f)=-Cv(t_f)+ Cv_r$ is a precise differentiation, it's not an approximation like the above, so how come the result $v(P_{max})=V_r$ is true?

 

[haruspex]

Where did $\dot P_{tot}=0=m\dot v -Cv_r$ come from? i read in a book that we move at the instantaneous velocity v of the rocket, in a non-inertial frame, then during a short period of time (dt) the gases acquire momentum $dmV_r$ (dm is the decrease in mass of the rocket so it's negative), and the rocket acquires momentum: $(m-dm)dv=m\cdot dv-dmdv\approx m\cdot dv$. do you have an other explanation?
$$dP_{tot}=m\cdot dv-dm\cdot V_r,\;\frac{dP_{tot}}{dt}=\dot P_{tot}=m\dot v-\dot mV_r=m\dot v-Cv_r$$
But $\dot P_m(t_f)=0=\dot m(t_f)v(t_f)+m(t_f)\dot v(t_f)=-Cv(t_f)+ Cv_r$ is a precise differentiation, it's not an approximation like the above, so how come the result $v(P_{max})=V_r$ is true?

Differentiation involves taking a limit as the 'd' terms become arbitrarily small. In m.dm - dm.dv, the term with two d's becomes arbirarily insignificant. It leads to an exact derivative, not an approximation.

 

[haruspex]

$$v=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)=V_r ln\:\left( \frac{m_0}{m} \right)=v(m)\rightarrow m(v)=m_0e^{-\frac{v}{V_r}}$$
$$P(m)=V_r\cdot m\cdot ln\left( \frac{m_0}{m} \right),\;\dot P(m)=V_r\left[ ln\left( \frac{m_0}{m} \right)-1\right]=0\rightarrow m(P_{max})=\frac{m_0}{e}$$
If i search for the velocity at maximum momentum:
$$P(v)=m(v)v=m_0e^{-\frac{v}{V_r}}\cdot v,\;\dot P(v)=0 \rightarrow v=V_r$$

Excellent.

 

[Karol]

Wily Willy posted a solution, i want to ask something about that.
He said:
The momentum of the system is made of the momentum of the rocket and the momentum of its gas exhaust: $\vec{p}=m_r\vec{v}_r+m_g\vec{v}_g$
Differentiation with respect to time gives:
$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+m_g\frac{d\vec{v}_g}{dt}+\vec{v}_g\frac{d\vec{m}_g}{dt}$$
The total momentum of the system is constant. The change in mass of the gas is equal and opposite to the change in the mass of the rocket. Also, the velocity of the exhaust gas does not change:
$$0=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+0-\vec{v}_g\frac{d\vec{m}_g}{dt}$$
Why doesn't the velocity of the gas change? i know that when the gas has left the nozzle it remains at $v_r$, but what confuses me is whether we use an inertial frame or not, because in respect to the laboratory, the inertial frame, every second the gas has a different velocity (in respect to the rocket it has a constant, $v_r$, velocity).
And from when do we start counting when we consider $\frac{d\vec{m}_g}{dt}$ and $m_g$? do we weigh all the gas that was shot from the beginning or do we look at a short interval of time?

 

[haruspex]

$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+m_g\frac{d\vec{v}_g}{dt}+\vec{v}_g\frac{d\vec{m}_g}{dt}$$
The total momentum of the system is constant. The change in mass of the gas is equal and opposite to the change in the mass of the rocket. Also, the velocity of the exhaust gas does not change:
$$0=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+0-\vec{v}_g\frac{d\vec{m}_g}{dt}$$
Why doesn't the velocity of the gas change? i know that when the gas has left the nozzle it remains at $v_r$, but what confuses me is whether we use an inertial frame or not, because in respect to the laboratory, the inertial frame, every second the gas has a different velocity (in respect to the rocket it has a constant, $v_r$, velocity).
And from when do we start counting when we consider $\frac{d\vec{m}_g}{dt}$ and $m_g$? do we weigh all the gas that was shot from the beginning or do we look at a short interval of time?

As I indicated, you have to be very careful using $\dot p= m\dot v +v \dot m$.
In the change of momentum of the gas, it will be much safer to think about it from first principles. There's no change in momentum of the gas previously exhausted, so we just have to add the momentum of the gas exhausted in time dt. This will have speed $v_r-V$ in the inertial frame, where V is the constant exhaust speed relative to the rocket. So momentum gain of the gas will be $dm (v_r-V)=C dt(v_r-V)$.
In your second equation above, I assume the $m_g$ should be $m_r$. And why are you showing masses as vectors?

 

[Karol]

Wily Willy solved correctly and i want to use that.
So i can assume he intended that the equations:
$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{dm_r}{dt}+m_g\frac{d\vec{v}_g}{dt}+\vec{v}_g\frac{dm_g}{dt}$$
$$0=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{dm_r}{dt}+0-\vec{v}_g\frac{dm_g}{dt}$$
Are for a short time (dt) since then dv_g=0

 

[haruspex]

Wily Willy solved correctly and i want to use that.
So i can assume he intended that the equations:
$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{dm_r}{dt}+m_g\frac{d\vec{v}_g}{dt}+\vec{v}_g\frac{dm_g}{dt}$$
$$0=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{dm_r}{dt}+0-\vec{v}_g\frac{dm_g}{dt}$$
Are for a short time (dt) since then dv_g=0

As I mentioned in my last post, the final equation above should end with a reference to m_r, not m_g (to match with the change of sign). That makes it the same as I posted, $\frac{dm_r}{dt}=-C$, v_g=v_r-V.

 

[haruspex]

As I mentioned in my last post, the final equation above should end with a reference to m_r, not m_g (to match with the change of sign). That makes it the same as I posted, $\frac{dm_r}{dt}=-C$, v_g=v_r-V.

 

[Karol]

As I mentioned in my last post, the final equation above should end with a reference to m_r, not m_g (to match with the change of sign)

I want to maintain the original intention of the formula:
$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+m_g\frac{d\vec{v}_g}{dt}+\vec{v}_g\frac{d\vec{m}_g}{dt}$$
And the last two members are $\frac{d}{dt}(m_gv_g)$
$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+C(\vec{v_r}-V_r)dt=0$$
$$m_r\frac{d\vec{v}_r}{dt}-C\cdot v+C(\vec{v_r}-V_r)dt=0$$
$$m_r\frac{d\vec{v}_r}{dt}+C\cdot V_r\,dt=0$$
But i can't solve since m_r is a function of time too.

 

[haruspex]

$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+C(\vec{v_r}-V_r)dt=0$$

That dt shouldn't be there at the end.

$$m_r\frac{d\vec{v}_r}{dt}+C\cdot V_r\,dt=0$$
But i can't solve since m_r is a function of time too.

Correcting that to $$m_r\frac{d\vec{v}_r}{dt}+C\cdot V_r=0$$
Yes, m is a function of t, but you know exactly how it varies with t, so you can substitute for m and solve.

 

[Karol]

I solved
$$m_r\frac{d\vec{v}_r}{dt}+C\cdot V_r=0$$
But i don't understand why is:
$$m_g\frac{d\vec{v}_g}{dt}=0$$

There's no change in momentum of the gas previously exhausted, so we just have to add the momentum of the gas exhausted in time dt.

does $m_g\frac{d\vec{v}_g}{dt}$ refer to all the gases expelled from the beginning? if so why is $\frac{d\vec{v}_g}{dt}=0$?
Edit: i try to explain: the gases have different velocities but each one doesn't change with time. but what confuses me is that $v_g=v_r-V_r$ means the momentary velocity, only at the last moment