- #36
Karol
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$$CV_r\int^{u}_{m_0}\frac{-du}{C\cdot u}=V_r ln\:\left( \frac{m_0}{u} \right)=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)$$
But what's m(V)? how does it help?
But what's m(V)? how does it help?
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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 said:$$CV_r\int^{u}_{m_0}\frac{-du}{C\cdot u}=V_r ln\:\left( \frac{m_0}{u} \right)=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)$$
But what's m(V)? how does it help?
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?haruspex said:I'll write Ptot for total momentum and Pm for the momentum of the rocket mass. And I'll write tf 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.
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.Karol said: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?
Excellent.Karol said:$$v=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)=V_r ln\:\left( \frac{m_0}{m} \right)=v(m)\rightarrow m(v)=m_0e^{-\frac{v}{V_r}}$$
$$P(m)=V_r\cdot m\cdot ln\left( \frac{m_0}{m} \right),\;\dot P(m)=V_r\left[ ln\left( \frac{m_0}{m} \right)-1\right]=0\rightarrow m(P_{max})=\frac{m_0}{e}$$
If i search for the velocity at maximum momentum:
$$P(v)=m(v)v=m_0e^{-\frac{v}{V_r}}\cdot v,\;\dot P(v)=0 \rightarrow v=V_r$$
As I indicated, you have to be very careful using ##\dot p= m\dot v +v \dot m##.Karol said:$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+m_g\frac{d\vec{v}_g}{dt}+\vec{v}_g\frac{d\vec{m}_g}{dt}$$
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 mentioned in my last post, the final equation above should end with a reference to mr, not mg (to match with the change of sign). That makes it the same as I posted, ##\frac{dm_r}{dt}=-C##, vg=vr-V.Karol said: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 dvg=0
haruspex said:As I mentioned in my last post, the final equation above should end with a reference to mr, not mg (to match with the change of sign). That makes it the same as I posted, ##\frac{dm_r}{dt}=-C##, vg=vr-V.
I want to maintain the original intention of the formula:haruspex said:As I mentioned in my last post, the final equation above should end with a reference to mr, not mg (to match with the change of sign)
That dt shouldn't be there at the end.Karol said:$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+C(\vec{v_r}-V_r)dt=0$$
Correcting that to $$m_r\frac{d\vec{v}_r}{dt}+C\cdot V_r=0$$Karol said:$$m_r\frac{d\vec{v}_r}{dt}+C\cdot V_r\,dt=0$$
But i can't solve since mr is a function of time too.
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##?haruspex said: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.