- #1

- 7

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Taylor first writes down the change in the momentum of the rocket as dP = P(t + dt) - P(t) = (m + dm)(v + dv) - mv = mdv + vdm, where he neglected the small term dmdv. So far, so good.

Next he writes down the change in momentum of the expelled fuel. He defines the velocity v

_{ex}of the expelled fuel, so that the velocity of the fuel relative to the ground or some other fixed point would be v - v

_{ex}. Then he claims that the change in momentum of the expelled fuel would be -dm(v - v

_{ex}) which is negative because the momentum of the fuel is opposite the momentum of the rocket.

Now my question: Shouldn't the change in momentum of the expelled fuel also depend on dv? And how could I develop the expression for dP of the fuel systematically, similar to how Taylor did it for the rocket? In a time Δt the mass of the expelled fuel would be Δm, and as Δt goes to zero, Δm would also go to zero. This would mean that it is not possible to determine a momentum of the expelled fuel at a specific instant t, and that only a change in momentum can be expressed. If this is true, then I cannot develop the expression for dP of the fuel as P(t + dt) - P(t). But how else can I do it? Taylor does not convince me because he does not mention what happens to dv. What exactly is going on here?