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Thrust question

  1. Mar 25, 2005 #1

    tony873004

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    At launch, the space shuttle’s engines develop 3.0 * 10^7 N of thrust. If the speed of the exhaust gasses is 15 km/s, what mass of gas is exhausted per second?

    [tex]
    F=ma\Rightarrow m=\frac{F}{a}
    [/tex]

    [tex]
    kg=\frac{kg\cdot m/s^2}{m/s^2}
    [/tex]

    [tex]
    kg=\frac{kg\cdot m/s^{\rlap{--} {2}}}{m/s^{\rlap{--} {2}}}\Rightarrow
    m=\frac{3.0\cdot 10^7kg\cdot \rlap{--} {m}/\rlap{--} {s}}{15(\rlap{--}
    {k}\rlap{--} {m}/\rlap{--} {s})\ast 1000\rlap{--} {m}/\rlap{--} {k}\rlap{--}
    {m}}
    [/tex]

    [tex]
    m=2000kg
    [/tex]

    I'm not sure if I'm allowed to cancel out the squared part of s squared, and still associate the 3.0e7 N with units that no longer represent it. Is my answer even correct?
     
  2. jcsd
  3. Mar 25, 2005 #2

    dextercioby

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    [tex] F=\frac{\Delta p}{\Delta t} =v\frac{\Delta m}{\Delta t} [/tex]


    That "Delta t" is obviously 1sec,since we're asked the flow in one second...

    Therefore the mass exhausted in one second is:

    [tex] \Delta m=\frac{F}{\frac{v}{\Delta t}}=... [/tex]

    Daniel.

    P.S.That force corresponds to a loss in momentum in unit time.
     
    Last edited: Mar 25, 2005
  4. Mar 25, 2005 #3

    tony873004

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    [tex] m=\frac{F}{\frac{\Delta v}{\Delta t}}=... [/tex]
    so...

    [tex]m=F\frac{\Delta t}{\Delta v}[/tex]

    [tex]m=3*10^7kg(m/s^2)\frac{1s}{15000m/s}[/tex]

    [tex]m=2000 kg[/tex]

    Amazing I came up with the right answer anyway, because I think I could have been called for illegal use of units. Your way is more elegant. It make sense since this question is in the momentum chapter. Delta P in your example is change in momentum, right?

    Thanks, Dex. You've bailed me out again :smile:
     
  5. Mar 25, 2005 #4

    dextercioby

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    Sorry,it's the other way around with the variation of momentum.It's due to the mass flow,and not the velocity variation.I'll edit the post...

    Daniel.
     
    Last edited: Mar 25, 2005
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