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

[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