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

[erobz]

More comprehensive treatment.　Say rocket is at rest at t=0. Momentum of the system is conserved zero.
0=m(t)v(t) + R \int_0^t ( v(u)+U )dt
differentiating by t,
\frac{d}{dt}[m(t)v(t)]+R [ v(t)+U ]=\frac{d}{dt}[m(t)v(t)]+Rv(t)-F=0
where m(t) is mass of rocket body and fuel in tank. $m(t)=m(0)-Rt$,
$v(t)$ is speed of rocket,
$U$ is relative speed of ejected fuel to rocket. $U<0$ when $v(t)>0$,
$F=-RU$.

So, since you’ve defined the ejecta momentum to be that integral, that treatment works. So it wasn’t all that difficult to correct after all. Thanks for sharing!

P.S. in your top line you’ve wrote $v(u)$ in the integral.

 

[anuttarasammyak]

P.S. in your top line you’ve wrote $v(u)$ in the integral.

Yes, it is in the integral. My typo not dt but du, dummy variable for integration. I corrected it in the post. Thanks.

 

[erobz]

@anuttarasammyak So, even more generally it starts out with:

$$ 0 = m(t)v(t) - \int_{0}^{t} \dot m(u) ( v(u)+U(u))~du $$

I think the momentum of the rocket + unburnt fuel should always be opposite the momentum of the ejecta

but since $\dot m$ and $v + U$ are both negative, we need $m(t)v(t)$ to be positive. Which is leaning me toward the minus sign?

 

[erobz]

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:

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

 

[haruspex]

Obviously the momentum carried by the exhaust hasn't been taken into account. Isn't this obvious from the correct derivation?

To the OP, almost surely not. But your simple explanation quoted above, together with the example in post #2, should do it.

 

[erobz]

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.

 

[haruspex]

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.

 

[erobz]

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!

 

[anuttarasammyak]

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
0 = m(t)v(t) + \int_{0}^{t} -\dot m(u) ( v(u)+U(u))~du
with U(u)<0 or
0 = m(t)v(t) + \int_{0}^{t} -\dot m(u) ( v(u)-U(u))~du
with U(u) >0 for normal thrust.

Thus
\frac{d}{dt} [m(t)v(t)]=\dot m(t) v(t)+F(t)
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
\frac{d\ mv}{dt}=F holds in condition of \dot m(t)=0

 

[erobz]

F(t) = \frac{d}{dt} [ m(t)v(t)]+ \dot m(t) v(t)
The second term in RHS is the correction to OP.

You lost the ejecta velocity relative to the rocket.

 

[anuttarasammyak]

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).

 

[erobz]

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.

 

[vanhees71]

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!