- **Classical Physics**
(*http://www.physicsforums.com/forumdisplay.php?f=61*)

- - **Tsiolkovsky's rocket equation question**
(*http://www.physicsforums.com/showthread.php?t=655033*)

tsiolkovsky's rocket equation questionHow would the equation look if instead of knowing the
effective exhaust velocity we knew the force the exhaust was exerting on the rocket. The equation is: [tex] \Delta V = v_e * ln(\dfrac{m_0}{m_1}) [/tex] would [tex] \Delta V [/tex] still be proportional to the log of the initial mass over the final mass? http://en.wikipedia.org/wiki/Tsiolko...ocket_equation |

Re: tsiolkovsky's rocket equation question[itex]\Delta v[/itex] does not depend on thrust. Only on I
_{sp} of the propellant. As long as I_{sp} is some constant, the [itex]\Delta v[/itex] will always be proportional to natural log of the mass ratio. |

Re: tsiolkovsky's rocket equation questionah k. i followed the link to specific impulse and it helped me understand.
http://en.wikipedia.org/wiki/Specific_impulse Quote:
thrust divided by the rate of the flow of mass from the rocket. that makes sense. (and yeah, this is with the specific impulse being constant). |

Re: tsiolkovsky's rocket equation questionQuote:
_{sp}*g_{0}) |

Re: tsiolkovsky's rocket equation questionDepending on definition. For impulse per weight, v = I
_{sp}*g. For impulse per mass, v = I_{sp}. Both conventions are used, mostly, depending on application. For rocket taking off from Earth's surface, I_{sp} per weight is a more directly useful quantity. For rocket accelerating in deep space, you just want the exhaust velocity, so I_{sp} per mass. |

Re: tsiolkovsky's rocket equation questionQuote:
_{sp}*g_{0} = V_{e, effective}. It isn't a matter of per weight or per mass, and g_{0} is strictly a conversion factor in this case. That's simply the definition of specific impulse and effective exhaust velocity. No matter where the rocket is, an I_{sp} of 300 seconds is exactly the same as an effective exhaust velocity of 2940 meters per second.(Note that I use g _{0} rather than g - this is because no matter where you are in the solar system (or elsewhere), g_{0} = 9.8 m/s^{2}, and since it is a conversion factor rather than a variable, it is independent of the local gravity field) |

Re: tsiolkovsky's rocket equation questionQuote:
_{sp}.1) I _{sp} = dp/dw = (dp/dm)/g2) I _{sp} = dp/dm = v_{e}Both are used in the literature and you differentiate by the units. First definition gives you units of inverse seconds. Second definition gives you units of m/s and is identical to exhaust velocity for a conventional rocket. |

Re: tsiolkovsky's rocket equation questionI've never seen the second one called I
_{sp} - everywhere I've seen it used, it was called effective exhaust velocity. If it is called I_{sp} anywhere, it is at least a somewhat nonstandard usage. Also, just because I'm in a somewhat nitpicky mood at the moment, it's not necessarily identical to exhaust velocity. It's identical to exhaust velocity if and only if the nozzle is perfectly expanded (and thus the pressure thrust is zero). Otherwise, there will be a difference between effective exhaust velocity and actual exhaust velocity. |

Re: tsiolkovsky's rocket equation questionQuote:
_{sp} in seconds, Europeans in units of meters/second. American engineers tend to use customary units, where there's a problem with the pound: Is it a unit of mass or a unit of force? This problem doesn't exist in SI units, and since "specific" typically means per mass, European engineers tend to specify I_{sp} in units of meters/second. |

All times are GMT -5. The time now is 01:21 AM. |

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.

© 2014 Physics Forums