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The basic question is how the speed of a rocket change. Since you are always expelling mass at a fixed velocity, with a given momentum, you must also gain that momentum yourself, so that it is conserved. Now assume that by some reason, even though you are expelling mass, your mass doesn't change. You are throwing mass with a fixed velocity, so your own ...

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You asked for an intuitive answer. A rocket accelerated by burning fuel and expelling the combustion products at high velocity. Conservation of momentum says that if you expel the same mass faster, you will get greater acceleration. This gives rise to the proportionality with $v_{cx}$. As for the logarithmic part: if you imagine two rockets of mass $m$ ...

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The equation is basically telling you that the change of velocity of the rocket, $v-v_0$, is proportional to the expulsion velocity of the fuel $v_{ex}$. But it is not quite equal, since as the rocket is burning fuel, it is getting lighter. This is why you also take into account the rate of change of the mass, $\ln(m_0/m)$. So, how do we derive this? ...

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My guess is that the exhaust is not perfectly in line with the balloon's centre of mass. In other words, the exhaust is slightly off-centre. That will cause the thrust of the escaping gas to push the nozzle slightly to the side, causing the balloon to rotate sightly as it travels forward. If this deflection stays constant you will get a perfect circle. If ...

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