Timeline for Why does this formula say rocket efficiency depends on velocity?
Current License: CC BY-SA 4.0
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| Apr 10 at 16:39 | history | edited | eshaya | CC BY-SA 4.0 |
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| Apr 10 at 15:51 | history | edited | eshaya | CC BY-SA 4.0 |
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| Apr 6 at 22:23 | comment | added | eshaya | @ErinAnne - The exhaust velocity is varying all the time in the minds of engine designers. | |
| Apr 6 at 2:27 | comment | added | eshaya | @OrganicMarble The rocket equation tells you that if $v_e >> v$ then the fuel does not need to be more than the load. Ion drive engines now have $v_e > 90$ km/s which means for a rocket built in space, say in geosynchronous orbit, one could have fuel tanks that are less than the load and travel around the solar system. Of course, one would need fuel to slow down as well. | |
| Apr 6 at 1:02 | comment | added | Organic Marble | Is there any rocket where "the mass of the fuel doesn't greatly exceed the mass of the load"? | |
| Apr 5 at 18:59 | history | edited | eshaya | CC BY-SA 4.0 |
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| Apr 5 at 18:52 | comment | added | Erin Anne | yes, both of those facts are reflected in my comment | |
| Apr 5 at 18:46 | comment | added | Erin Anne | I'm struggling to see how the derivative operation $dE/dv_e$ makes sense. Exhaust velocity ideally doesn't vary in the frame in which these equations are derived. That seems like it should matter, but it's admittedly been a long time since I did rigorous calculus | |
| Apr 5 at 18:40 | history | edited | eshaya | CC BY-SA 4.0 |
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| S Apr 5 at 18:31 | review | First answers | |||
| Apr 8 at 11:31 | |||||
| S Apr 5 at 18:31 | history | answered | eshaya | CC BY-SA 4.0 |