How Does Taylor's Mechanics Address Rocket Momentum with Changing Mass?

[Bashyboy]

Hello,

I am reading section 3.2, concerning the analyzation of a moving rocket with a changing mass. (I couldn't find a preview of the book in google books, so hopefully someone out there has this textbook.) Here is an except from the book, but be warned that I am adding notes in brackets:

"At time t, the momentum [of the rocket] is P(t) = mv [m and v are the mass and velocity of the rocket at time t]. A short time later at t + dt, the rockets mass is (m + dm), where dm is negative, and its momentum is (m+dm)(v+dv). The fuel ejected in the time dt has mass (-dm) and velocity v - vex [v is the velocity of the rocket as viewed by some stationary person on earth, and vex is rate at which the fuel flows out, relative to the rocket]. Thus, the total momentum (rocket plus the fuel just ejected) at t + dt is
P(t+dt) = (m + dm)(v+dv) - dm(v - vex)."

As one might notice, as I did, they accounted for the momentum of the fuel at time t+dt, which is - dm(v - vex) (there's a negative because the momentum is in the opposite direction); but at time t, they did not, for the expression is P(t) = mv. Where is the momentum term for the fuel at time t?

 

[AlephZero]

Where is the momentum term for the fuel at time t?

Before it burns, the fuel has the same velocity as the rocket, so its momentum is part of the P(t) = mv.

You could write
(m+dm)v - (dm)v
if you really want to separate out "the momentum of the fuel of mass -dm that you are going to burn next" and "the momentum of everything else". But of course (m+dm)v - (dm)v = mv.

 

[Bashyboy]

Isn't the velocity of the fuel at any instant v - vex?

 

[AlephZero]

The fuel has velocity v before it burns, and v - vex after it burns.

At time t, the bit of fuel with mass dm hasn't burned yet. At time t+dt, it has burned.