Rocket acceleration problem: confused about Newton's 2nd Law

In summary, the first approach is incorrect because the mass of the object cannot change. The second approach is correct, but requires treating the ejecta as an external force.
  • #36
haruspex said:
To the OP, almost surely not. But your simple explanation quoted above, together with the example in post #2, should do it.
I feel like what was figured by the group out bridges the gap. Something original was learned (at least for me)...I feel like we broke through the barrier of "don't use ##\sum F = \frac{dp_{sys}}{dt}## because funny things happen" with a sound example. That's got to be worth something.
 
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  • #37
erobz said:
I feel like what was figured by the group out bridges the gap. Something original was learned (at least for me)...I feel like we broke through the barrier of "don't use ##\sum F = \frac{dp_{sys}}{dt}## because funny things happen" with a sound example. That's got to be worth something.
Sure, but I feel it too often happens that post #1 shows two methods and asks what's wrong with the failing one, then most of the responses ignore that and just offer ways that work.
By all means show better ways, but first off, answer the actual question the OP is asking.
 
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  • #38
We'll, we have now addressed how to approach the OP's method (1) in a consistent way (IMHO). Typically, that's the approach that is completely discarded, and post #3 approach is usually the one that is interjected (as you say) which resides somewhere in between their first and second approaches.

Either way, this one didn't get too out of control!
 
  • #39
erobz said:
but since m˙ and v+U are both negative, we need m(t)v(t) to be positive. Which is leaning me toward the minus sign?
I would make
[tex]0 = m(t)v(t) + \int_{0}^{t} -\dot m(u) ( v(u)+U(u))~du[/tex]
with U(u)<0 or
[tex]0 = m(t)v(t) + \int_{0}^{t} -\dot m(u) ( v(u)-U(u))~du[/tex]
with U(u) >0 for normal thrust.

Thus
[tex]\frac{d}{dt} [m(t)v(t)]=\dot m(t) v(t)+F(t)[/tex]
RHS first term : Minus. Loss of momentum which fuel to be burnt has hold in motion of rocket.
RHS second term : Plus. Gain of momentum by burning the fuel
The first term is correction to OP trial.

The familiar formula of
[tex]\frac{d\ mv}{dt}=F[/tex] holds in condition of [tex]\dot m(t)=0[/tex]
 
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  • #40
anuttarasammyak said:
[tex]F(t) = \frac{d}{dt} [ m(t)v(t)]+ \dot m(t) v(t)[/tex]
The second term in RHS is the correction to OP.
You lost the ejecta velocity relative to the rocket.
 
  • #41
erobz said:
You lost the ejecta velocity relative to the rocket.
I corrected sign in the post recalling R>0. What you point out is mentioned as F(t).
 
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  • #42
anuttarasammyak said:
I corrected sign in the post recalling R>0. What you point out is mentioned as F(t).
I always get all jumbled up on mundane sign issues. Sorry if you feel I’m bringing you into my nightmare! If you don’t keep track of whether or not you accounted for the negative rate it gets confusing. So here… it’s accounted for, going forward all values are positive.
 
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  • #43
Why do you add more confusing "solutions" to this thread? Concerning sign doubts of velocity components, just keep in mind that velocities are vectors and that we deal with a 1D situation using the corresponding velocity components wrt. a fixed basis vector in that direction!
 

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