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Rocket Motion

  1. Nov 26, 2004 #1
    Im having this problem with a rocket equation. Ill state the problem then show what i've done

    Let [tex]M[/tex] = mass of rocket and fuel

    [tex]M_f[/tex] = mass of fuel

    [tex]M_0[/tex] = rockets total initial mass (including fuel)

    (this is given as [tex]10^5 kg[/tex])

    [tex]V_R[/tex] = rockets velocity

    A [tex]10^5 kg[/tex] rocket has a total weight-to-payload ratio of 10 to 1. i.e [tex]M_f = 9M_0[/tex]

    The rocket starts from rest, pointing vertically. Its exhaust gases are expelled at a rate [tex]k (=450 kg/s)[/tex] with a constant velocity [tex]v_0(= 2,600 m/s)[/tex]. Assume the rocket experiences a constant gravitational force with [tex]g = 9.8 m/s^2[/tex]


    Work out the answers to the following questions using the above symbols together with [tex]k, v_0[/tex] and [tex]g[/tex]. Put in numerical values only at the end.

    (1) What is the burn-out time?

    (2) What is the rockets final velocity, assuming it rises vertically during the boost phase?

    (3) How high is the rocket at the end of its boost phase?


    Ok, i've done (1) and am trying to do (2). Im trying to set up the differential equation of the rockets motion.

    I said that at [tex]t = 0[/tex]

    Total momentum = 0

    then at [tex]t = dt[/tex]

    [tex]dp_{tot} = (M - dM_f)(dV_R) - (v_0dM_f)[/tex]

    neglecting 2-nd order terms

    [tex]\displaystyle{\frac{dp_{tot}}{dt} = M\frac{dV_R}{dt} - v_0\frac{dM_f}{dt}}[/tex]

    Then as the only external force on the rocket is gravity

    [tex]\displaystyle{-(M - dM_f)g = M\frac{dv_R}{dt} - v_0\frac{dM_f}{dt}}[/tex]

    I also know that [tex]\displaystyle{\frac{dM_f}{dt} = -k[/tex]

    so

    [tex]\displaystyle{-(M - dM_f)g = M\frac{dv_R}{dt} + v_0k}[/tex]

    Is this at all right so far?
     
  2. jcsd
  3. Nov 26, 2004 #2

    Tide

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    That looks good to me! Of course, you will drop the dM terms as it is infinitesimal.
     
  4. Nov 27, 2004 #3
    Thankyou very much for your help Tide, I was a bit unsure about that [tex]dM_f[/tex] bit.

    I continued on but have got a bit confused again!

    I know that the time for the burn out is [tex]200[/tex] from the first part, so I did this

    [tex]\displaystyle{-Mg = M\frac{dv_R}{dt} + v_0k}[/tex]

    [tex]\displaystyle{\frac{v_0k}{M} + g = -\frac{dv_R}{dt}}[/tex]

    Kind of wandering on I did this

    [tex]\displaystyle{\int^{200}_0 \left(\frac{v_0k}{M} + g\right) dt = -\int^{v_f}_0 dv_R}[/tex]

    This is worrying me though because i know that [tex]M[/tex] is not constant...

    What have I done wrong and what is my next step?

    Thankyou in advance
     
  5. Nov 27, 2004 #4

    Tide

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    M is not constant but you know the rate at which mass is expelled: [itex]M = M_0 - kt[/itex]
     
  6. Nov 28, 2004 #5
    Silly me! Thankyou!
     
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