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Relativistic rockets

  1. Oct 20, 2007 #1
    Hello there

    I've been looking at this page

    http://en.wikipedia.org/wiki/Relativistic_rocket

    at the relativistic version of Tsiolkovsky rocket equation, but something puzzles me


    With chemical rockets, the mass that is being converted into energy is so tiny that almost
    all fuel mass can be considered to end up as propellant, but when we are talking about relativistic speeds, the energy requirements are so high that a major portion of fuel would end up being converted to energy.

    What puzzles me is how to incorporate this into this equation.

    For example, if 80% of the rocket mass was fuel, and there was, say a fusion process, and the product was propellant with kinetic energy equal to matter that was converted into energy how would you incorporate this into the equation? Because now, not all of the fuel mass is propellant, and yet there was indeed a loss in ships mass (which was converted into kinetic energy)
    Any ideas?

    Perhaps my assumptions about the equation are wrong.
    I assumed that the mass ratios are just reaction mass vs. total mass, and have nothing to do with that portion of mass that is converted into energy
     
  2. jcsd
  3. Oct 20, 2007 #2

    pervect

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    The wikipedia page you cite talks about this in some detail - have you looked at it closely?

    What's important is (as the page says) the amount of momentum you get when you utilize one unit (say 1kg) of fuel. This is known as "specific impulse".
     
  4. Oct 20, 2007 #3
    Now this is embarrassing

    I did actually read the whole page, and entered everything into matlab, to get some nice curves out, and then started getting strange results, actually I was getting out complex numbers, lol, so I thought maybe I didn't understand the equation right, which is why I started this thread.

    Either way, it works fine now after I double checked it

    thanks

    I hope this thread doesn't end up as useless
     
  5. Oct 20, 2007 #4
    But just so this thread doesn't go to waste, I'm going to ask, this this equation still hold if your "exhaust" is in form of radiation?
    In other words, can a self-fueled laser be a classical rocket.
    I know radiation has momentum, but I'm just wondering if there is a catch
     
  6. Oct 20, 2007 #5

    pervect

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    The specific impulse of a photon / laser exhaust is 1 in geometric (relativistic) units. In meters/second, it's just 'c'. You might also see this as 3*10^8 netwon seconds / kg.

    Sometimes you see specific impulse in seconds as the Wikipedia article mentions, you have to multiply the figure in seconds by the acceleration of gravity in m/s^2, i.e by 9.8 m/s^2, to convert this to the correct units of velocity. Thus the specific impulse of a photon drive would be about 30 million seconds (354 days, about 1 year).
     
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