Conceptual Problem with Thrust

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The discussion centers on the conceptual understanding of thrust in rocket dynamics, specifically addressing the equation m(dv/dt) = -u(dm/dt). The user questions why the thrust is not represented as m(dv/dt) + v(dm/dt) due to the changing mass of the rocket. The response clarifies that while the mass loss affects acceleration, it does not alter the force equation, which remains F = dp/dt. The correct formulation incorporates both mass loss and thrust, leading to dp/dt = v(dm/dt) + F, confirming that m(dv/dt) = F.

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  • Knowledge of momentum conservation in variable mass systems
  • Basic calculus for differentiating equations
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Hi everyone, as you can tell from the title, I'm having some trouble understanding thrust. Here's what my textbook says:

"m(dv/dt) = -u(dm/dt) [where u is the speed of the exhaust relative to the rocket and dm/dt is negative]

Now dv/dt is the acceleration of the rocket, so the left side of this equation (mass times acceleration) equals the net force, or thrust on the rocket,

F = -u(dm/dt)"

But I thought F = dp/dt. So, why wouldn't the thrust, or net force, on the rocket be m(dv/dt) + v(dm/dt) since the mass of the rocket varies with time?

Thanks in advance for any help. (Sorry if I posted in the wrong section.)
 
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Safebox said:
Why wouldn't the thrust, or net force, on the rocket be m(dv/dt) + v(dm/dt) since the mass of the rocket varies with time?
The acceleration is affected by the decreasing mass of the rocket, but not the force.
 
Equation F=dp/dt assumes you are not loosing mass. If you are loosing mass, you are also loosing momentum along with it. If the rocket is moving at velocity v, an element of mass dm has momentum vdm. If you are loosing/gaining that mass at rate dm/dt, the gain/loss of momentum due to mass gain/loss is v(dm/dt).

So the correct equation is: dp/dt = v(dm/dt) + F. If you now expand dp/dt the way you did, you get: v(dm/dt) + m(dv/dt) = v(dm/dt) + F, or m(dv/dt) = F.

Good catch by the way. I remember this threw me off while trying to convert the rocket formula to relativistic case a few years back. This is actually an important step that's hand-waved away in most derivations of rocket formula.
 

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