Tsiolkovsky's rocket equation question

  1. How 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?

  2. jcsd
  3. K^2

    K^2 2,470
    Science Advisor

    [itex]\Delta v[/itex] does not depend on thrust. Only on Isp of the propellant. As long as Isp is some constant, the [itex]\Delta v[/itex] will always be proportional to natural log of the mass ratio.
  4. ah k. i followed the link to specific impulse and it helped me understand.

    and it says the effective exhaust velocity is also the 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).
    Last edited: Nov 26, 2012
  5. Actually, your parenthetical note at the end there is somewhat irrelevant - specific impulse is directly proportional to effective exhaust velocity. The two terms are just two different representations of the same value (specifically, effective exhaust velocity = Isp*g0)
  6. K^2

    K^2 2,470
    Science Advisor

    Depending on definition. For impulse per weight, v = Isp*g. For impulse per mass, v = Isp. Both conventions are used, mostly, depending on application. For rocket taking off from Earth's surface, Isp per weight is a more directly useful quantity. For rocket accelerating in deep space, you just want the exhaust velocity, so Isp per mass.
  7. Nope. Isp*g0 = Ve, effective. It isn't a matter of per weight or per mass, and g0 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 Isp of 300 seconds is exactly the same as an effective exhaust velocity of 2940 meters per second.

    (Note that I use g0 rather than g - this is because no matter where you are in the solar system (or elsewhere), g0 = 9.8 m/s2, and since it is a conversion factor rather than a variable, it is independent of the local gravity field)
  8. K^2

    K^2 2,470
    Science Advisor

    There are two alternative definitions of Isp.

    1) Isp = dp/dw = (dp/dm)/g
    2) Isp = dp/dm = ve

    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.
  9. I've never seen the second one called Isp - everywhere I've seen it used, it was called effective exhaust velocity. If it is called Isp 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.
  10. D H

    Staff: Mentor

    Not necessarily. Americans tend to specify Isp 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 Isp in units of meters/second.
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