Timeline for Relativistic Rocket Equation
Current License: CC BY-SA 4.0
6 events
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| Sep 28, 2020 at 15:22 | comment | added | Quietghost | @ikrase If the power is beamed to the rocket, then for the relativistic case the concept breaks down. There are 2 effects, namely the momentum transfer from beaming energy (similar to solar sails) and the additional energy that can be added to the fuel that invalidate the momentum balance principle that makes this a rocket. But if you think about it, why try to add this beamed energy to the fuel when you could just reflect it using a mirror for the momentum advantage anyway and not take the conversion losses? (Additional directional control possibly) | |
| Sep 21, 2020 at 9:51 | comment | added | ikrase | If the rocket is powered by beamed power, can't the energy be more than the rest mass of the fuel? | |
| Apr 9, 2020 at 16:54 | history | edited | Quietghost | CC BY-SA 4.0 |
Clarified the meaning of the acceleration term in the first part
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| Apr 9, 2020 at 16:52 | comment | added | Quietghost | Absolutely, I can clarify this. In the rocket reference frame you would feel $a$, but in the inertial reference frame $a$ would be different | |
| Apr 9, 2020 at 0:51 | comment | added | uhoh |
+1 Kudos for having the tenacity to continue to pursue this and get it right! I'm a little uncomfortable with seeing "$\Delta v = a \Delta t$ is larger than 𝑐. There is nothing wrong with this" without at least a caveat that acceleration $a$ is funky here. The whole point of this answer is to explain that using $ma = F$ instead of $dp/dt = F$ was the mistake that led to the confusion. Is it possible to clarify that your $a$ isn't really a meaningful acceleration?
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| Apr 8, 2020 at 17:17 | history | answered | Quietghost | CC BY-SA 4.0 |