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you are right, it is your formula a), as there is no thrust of the flowing water against the bucket, although the official NASA formula for the rocket chamber thrust implies that there is. Imagine that you instantly open the bucket floor, would the bucket jump to equalize he momentum of the flushed water? If I'm not mistaken, this problem is related to the ...


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It's true that the natural speed limit is the speed of light. So the minimum time required is 1400 yrs. However, it is imposible to reach that speed. We can't even accelerate subatomic particles to that speed. You asked for the fastest rocket in Earth. The actual record of speed is the Helios II spacecraft (obtained from here) which traveled at 70.22 km/s. ...


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Oh yes, particularly if we ever resurrect Project Orion, which works by tossing nukes out the back of the spacecraft and setting them off. https://en.wikipedia.org/wiki/Project_Orion_(nuclear_propulsion) It does, of course, take a pretty good buffer plate to absorb the momentum impulse. Using Tsar Bomba https://en.wikipedia.org/wiki/Tsar_Bomba as an upper ...


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If you use a highly simplified model without an atmosphere, it could slow the rotation of the Earth. It would be like riding a ferris-wheel and throwing a piece of popcorn straight ahead as you go around. AKA negligible. Physically speaking the torque created by the rocket, assuming its bolted down well enough, would be the force it exerts times the radius ...


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The Saturn V threw roughly $m=3\times 10^6{\rm kg}$ out its hinder end at a speed of about $v=3{\rm km\,s^{-2}}$. The angular momentum of this mass thrown tangentially to the ground about Earth's center is then $R_\oplus\,m\,v$, where $R_\oplus$ is the Earth's radius. Assuming the Earth to be uniformly dense for a rough figure, its mass moment of inertia ...


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If you the rate of change of mass is $\dot{m} = \mathrm{d}m/\mathrm{d}t$ and we're talking about passing into the future direction ($\mathrm{d}t > 0$), then $\mathrm{d}m$ is necessarily negative. Thus, $m\mapsto m+\mathrm{d}m$ represents a decrease of the rocket mass. The opposite convention would require that $\dot{m} = -\mathrm{d}m/\mathrm{d}t$ works ...



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