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Netwons second law and rocket equation

  1. Mar 9, 2009 #1
    according to newtons second law:
    [tex]\vec{F}=\frac{d\vec{p}}{dt}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}[/tex](1)
    ie force impelled on a body is equal to the rate of change of momentum
    however when we use calculus to derive rocket equation we get:
    [tex]m\frac{d\vec{v}}{dt}={\vec{v}_{gas\; relative \;to \;rocket}}\frac {dm}{dt}[/tex](2)
    where [tex]\vec{v}[/tex] is the velocity of the rocket
    my problem now is that many text book concludes that the net force on the rocket
    [tex]\vec{F}=\vec{T}=m\frac{d\vec{v}}{dt}={\vec{v}_{gas\; relative \;to \;rocket}}\frac{dm}{dt}[/tex] where T is the THRUST
    but when you apply (2) in (1) would not the force on rocket by gas be
    [tex]\vec{F}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}=({\vec{v}_{gas\; relative \;to \;rocket}+\vec{v})\frac{dm}{dt}[/tex]?????
     
  2. jcsd
  3. Mar 9, 2009 #2
    What's dm?...its 0...mass is not gonna change!
     
  4. Mar 9, 2009 #3
    if you use p=m(t)*v(t) and use product rule in differentiation you will get it dm/dt
     
  5. Mar 9, 2009 #4

    tiny-tim

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    Hi calculus_jy! :smile:
    I don't understand what you're doing here …

    the RHS is the rate of change of momentum of the gas, so what is the LHS supposed to be? :confused:
     
  6. Mar 9, 2009 #5

    arildno

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    There is a LOT of confusion on this issue.

    The major reason for that is that one doesn't clearly distinguish between "material systems" and "geometric systems".

    Now, a "material system" consists of the SAME particles throughout the observation period. Thus, this is the type of system that Newton's 2.law is valid for!

    A "geometric system" is defined as whatever mass particles happen to reside within a specified spatial region throughout the observation period. It does NOT, in general, consist of the same material particles through the observation period, and HENCE, Newton's 2.law is not naively applicable here!

    Because particles may ENTER, and LEAVE that spatial region, we must correct for the flux of momentum over the boundaries of the specified region in order to get a "tweaked" Newton's 2.law.

    I made a thread of this some time ago, where the rocket equation is properly derived:
    https://www.physicsforums.com/showthread.php?t=72176
     
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