Rocket losing mass

  • Thread starter Karol
  • Start date
  • #26
13
0
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!
 
  • #27
13
0
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.
 
  • #28
35,714
12,293
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=Vr. Plugging it back into the equation for m, we get m=M0e-1.


There is a way to get the result V=Vr 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 -Vr relative to the rocket, this happens at V=Vr.
 
  • #29
1,380
22
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
 
  • #30
13
0
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.
 
  • #31
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
##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##.
 
  • #32
1,380
22
##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##
 
  • #33
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
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 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 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.
 
  • #34
1,380
22
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##
 
  • #35
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
$$CV_r\int^{\frac{m_0-u}{C}}_{m_0}\frac{-du}{C\cdot u}$$
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.
 
  • #36
1,380
22
$$CV_r\int^{u}_{m_0}\frac{-du}{C\cdot u}=V_r ln\:\left( \frac{m_0}{u} \right)=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)$$
But what's m(V)? how does it help?
 
Last edited:
  • #37
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
$$CV_r\int^{u}_{m_0}\frac{-du}{C\cdot u}=V_r ln\:\left( \frac{m_0}{u} \right)=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)$$
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.
 
  • #38
1,380
22
$$v=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)=V_r ln\:\left( \frac{m_0}{m} \right)=v(m)\rightarrow m(v)=m_0e^{-\frac{v}{V_r}}$$
$$P(m)=V_r\cdot m\cdot ln\left( \frac{m_0}{m} \right),\;\dot P(m)=V_r\left[ ln\left( \frac{m_0}{m} \right)-1\right]=0\rightarrow m(P_{max})=\frac{m_0}{e}$$
If i search for the velocity at maximum momentum:
$$P(v)=m(v)v=m_0e^{-\frac{v}{V_r}}\cdot v,\;\dot P(v)=0 \rightarrow v=V_r$$
 
  • #39
1,380
22
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.
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?
 
  • #40
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
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.
 
  • #41
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
$$v=V_r ln\:\left( \frac{m_0}{m_0-Ct} \right)=V_r ln\:\left( \frac{m_0}{m} \right)=v(m)\rightarrow m(v)=m_0e^{-\frac{v}{V_r}}$$
$$P(m)=V_r\cdot m\cdot ln\left( \frac{m_0}{m} \right),\;\dot P(m)=V_r\left[ ln\left( \frac{m_0}{m} \right)-1\right]=0\rightarrow m(P_{max})=\frac{m_0}{e}$$
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.
 
  • #42
1,380
22
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?
 
Last edited:
  • #43
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+m_g\frac{d\vec{v}_g}{dt}+\vec{v}_g\frac{d\vec{m}_g}{dt}$$
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 momemtum 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?
 
Last edited:
  • #44
1,380
22
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
 
  • #45
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
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
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.
 
  • #46
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
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.
 
  • #47
1,380
22
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)
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 mr is a function of time too.
 
Last edited:
  • #48
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,144
7,272
$$\frac{d\vec{p}}{dt}=m_r\frac{d\vec{v}_r}{dt}+\vec{v}_r\frac{d\vec{m}_r}{dt}+C(\vec{v_r}-V_r)dt=0$$
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 mr 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.
 
  • #49
1,380
22
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
 
Last edited:

Related Threads on Rocket losing mass

  • Last Post
Replies
4
Views
2K
Replies
5
Views
548
  • Last Post
Replies
5
Views
4K
Replies
3
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
13
Views
5K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
4K
Top