# Deriving force from momentum using d(mv)/dt

In summary, the equation for momentum accumulation in the ejecta from a rocket is:$$\sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt}+ \frac{dM_f}{dt} v_{e/r}$$
Homework Statement
How did we Derive force from momentum
Relevant Equations
F=ma
P=mv
F=dp/dt
How did the d(mv)/dt become the other two?
Can someone explain how do we derive for new formulas in physics?

Product rule of differentiation

MatinSAR
malawi_glenn said:
Product rule of differentiation
malawi_glenn said:
Product rule of differentiation
Thank you I just missed it.

You need to be very careful using this ##dp=m.dv+v.dm## form.
Consider e.g. a water tank mass m speed v leaking water at rate ##\rho##. If we blindly apply that formula we get ##\dot p= m\dot v+v\dot m=ma-v\rho##. Since there appear to be no external forces we might conclude ##ma=v\rho##, so the tank is accelerating.
The flaw is that the leaking water carries momentum away with it, which should be represented as an external force ##-v\rho##. Plugging that in gives ##a=0##.

PeroK, malawi_glenn, BvU and 1 other person
haruspex said:
You need to be very careful using this ##dp=m.dv+v.dm## form.
Consider e.g. a water tank mass m speed v leaking water at rate ##\rho##. If we blindly apply that formula we get ##\dot p= m\dot v+v\dot m=ma-v\rho##. Since there appear to be no external forces we might conclude ##ma=v\rho##, so the tank is accelerating.
The flaw is that the leaking water carries momentum away with it, which should be represented as an external force ##-v\rho##. Plugging that in gives ##a=0##.

So why do rockets work? I do not understand the point you are trying to make.

PeroK and MatinSAR
Frabjous said:
So why do rockets work? I do not understand the point you are trying to make.
You can solve rocketry questions using conservation of momentum quite easily without using that equation explicitly. In time ##\delta t## a mass of fuel ##\mu \delta t## is ejected at speed ##u## relative to the craft of remaining mass ##m##. ##m\delta v=u\mu\delta t##. This leads to ##v=u\ln(\frac{m_0}{m-\mu t})## as desired.

You can argue that I am effectively using the equation, but quoting the equation and then trying to apply it has pitfalls, such as the one I outlined. In the rocketry context, what exactly is the system which m represents?

PeroK and MatinSAR
@Frabjous
erobz said:
Just for clarity, correcting post #11 using the proper way to capture the accumulation of momentum in the ejecta from @anuttarasammyak:

$$\sum F_{ext} = \frac{d}{dt} \left( p_{\rm{rocket}} + p_{\rm{fuel}} + p_{\rm{ejecta}} \right)$$

Mass of the rocket is constant ##M_r##

The mass of fuel lost from the rocket is gained by the ejecta ## \frac{dM_e}{dt} = -\frac{dM_f}{dt}##

The velocity of the ejecta in the rest frame is ##v_{e/O} = v_r - v_{e/r}## and is constant after being ejected, but the momentum of the ejecta is accumulating, hence:

$$\sum F_{ext} = \frac{d}{dt} \left( M_r v_r + M_f v_r + \int \frac{dM_e}{dt}( v_r - v_{e/r}) ~dt \right)$$

Applying the derivative:

$$\sum F_{ext} = \overbrace{\left( M_r\frac{dv_r}{dt} \right)}^{\rm {rocket}} +\overbrace{\left(M_f\frac{dv_r}{dt} + \frac{dM_f}{dt}v_r\right)}^{\rm{fuel}}+ \overbrace{ \frac{dM_e}{dt}( v_r - v_{e/r})}^{\rm{ejecta}}$$

$$\sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt}+ \cancel{ \left( \frac{dM_f}{dt}v_r + \frac{dM_e}{dt}v_r \right)}^0 -\frac{dM_e}{dt} v_{e/r}$$

The Rocket Equation:

$$\sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt} + \frac{dM_f}{dt} v_{e/r}$$

The issue comes up often about taking ##F = \frac{dp}{dt} = \frac{d}{dt} \big( mv \big)##.

We (several posters) took the time not too long ago to figure out how to apply it for "The Rocket Equation". I think this is what @haruspex means by "you have to be careful". It wasn't immediately obvious how to get it to work out.

The whole exchange:

PeroK, Frabjous and MatinSAR

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