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So in my lectures we derived the rocket equation which is given by:

$$F = m \cdot\frac{dv}{dt} + u\cdot \frac{dm}{dt}$$

Where m is the mass of the rocket, and u is the speed of the ejected fuel.

But if I use

$$F = \frac{dP}{dt} = \frac{d(mv)}{dt}$$

Where $P$ is the momentum I get

$$F = m \cdot\frac{dv}{dt} + v\cdot \frac{dm}{dt}$$

My question is why are these different? Isn't this a contradiction?

Qmechanic
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  • how did you derive the first equation that you gave? – Alex Robinson Feb 27 '18 at 15:03
  • The equation $F=ma$ is to be used for a system of constant mass even though the Mathematics makes it appear that the system can have a variable mass. – Farcher Feb 27 '18 at 15:06
  • @Farcher - It's the equation $F=\frac{dP}{dt}$ that is to be used for a system of constant mass rather than $F=ma$. In the case of a constant mass system, $F=ma$ and $F=\frac{dP}{dt}$ are equivalent. – David Hammen Feb 27 '18 at 15:12
  • https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Most_popular_derivation – Augs Feb 27 '18 at 15:16
  • Depends on what you want to know for your rocket system I guess. And then it depends on how you draw your system boundaries (i.e. with an external thrust force or a constant mass system). – jjack Feb 27 '18 at 16:48

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