Netwons second law and rocket equation

AI Thread Summary
The discussion centers on the application of Newton's second law to the rocket equation, highlighting the confusion regarding momentum and mass change. It emphasizes that the force acting on a rocket is derived from the rate of change of momentum, yet there is debate over how to correctly apply this when considering the gas expelled from the rocket. A distinction is made between "material systems," where the same particles are present, and "geometric systems," which may involve different particles over time, complicating the application of Newton's laws. The conversation suggests that a proper understanding of these systems is crucial for accurately deriving the rocket equation. Overall, clarity in the definitions and application of these concepts is essential for resolving the confusion surrounding the calculations.
calculus_jy
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according to Newtons second law:
\vec{F}=\frac{d\vec{p}}{dt}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}(1)
ie force impelled on a body is equal to the rate of change of momentum
however when we use calculus to derive rocket equation we get:
m\frac{d\vec{v}}{dt}={\vec{v}_{gas\; relative \;to \;rocket}}\frac {dm}{dt}(2)
where \vec{v} is the velocity of the rocket
my problem now is that many textbook concludes that the net force on the rocket
\vec{F}=\vec{T}=m\frac{d\vec{v}}{dt}={\vec{v}_{gas\; relative \;to \;rocket}}\frac{dm}{dt} where T is the THRUST
but when you apply (2) in (1) would not the force on rocket by gas be
\vec{F}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}=({\vec{v}_{gas\; relative \;to \;rocket}+\vec{v})\frac{dm}{dt}?
 
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What's dm?...its 0...mass is not going to change!
 
if you use p=m(t)*v(t) and use product rule in differentiation you will get it dm/dt
 
Hi calculus_jy! :smile:
calculus_jy said:
however when we use calculus to derive rocket equation we get:
m\frac{d\vec{v}}{dt}={\vec{v}_{gas\; relative \;to \;rocket}}\frac {dm}{dt}(2)

I don't understand what you're doing here …

the RHS is the rate of change of momentum of the gas, so what is the LHS supposed to be? :confused:
 
There is a LOT of confusion on this issue.

The major reason for that is that one doesn't clearly distinguish between "material systems" and "geometric systems".

Now, a "material system" consists of the SAME particles throughout the observation period. Thus, this is the type of system that Newton's 2.law is valid for!

A "geometric system" is defined as whatever mass particles happen to reside within a specified spatial region throughout the observation period. It does NOT, in general, consist of the same material particles through the observation period, and HENCE, Newton's 2.law is not naively applicable here!

Because particles may ENTER, and LEAVE that spatial region, we must correct for the flux of momentum over the boundaries of the specified region in order to get a "tweaked" Newton's 2.law.

I made a thread of this some time ago, where the rocket equation is properly derived:
https://www.physicsforums.com/showthread.php?t=72176
 
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