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Tsiolkovsky rocket equation

  1. Sep 20, 2013 #1
    I am trying to understand the Tsiolkovsky rocket equation.
    I am looking at this right now:

    they said that the momentum of the rocket itself is:
    p1 = (m-dm)(v+dv)

    and the momentum of the gas was:
    p2 = dm(v-u)

    Here is the problem- u is relative to the rocket.
    In my opinion p2 should be:
    dm( (v + dv) -u )
    because the speed of the rocket has changed too

    Where is my mistake?

    Thank you,
  2. jcsd
  3. Sep 21, 2013 #2

    Filip Larsen

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    Gold Member

    welcome to PF, Marina.

    The gas is ejected at a time where the speed of the rocket is v, that is, before the rocket gets the small dv increase in speed.
  4. Sep 21, 2013 #3
    Thank you!
    I have been trying to figure out this for hours, and it turned out to be so simple :)
  5. Sep 21, 2013 #4
    Hi Marina

    There is no mistake in your reasoning .You are correct in your expression of the momentum of the ejected mass . The expression in the given link is flawed.

    Fortunately the two expressions ,one given in the link ,the other by you ,give the same final result .

    Hope that helps :)
    Last edited: Sep 21, 2013
  6. Sep 21, 2013 #5
    Thank you for your response
    I cant get the final result with my equasion. Can you please show me how it is done.
    I am trying for about 24 hours now :/
  7. Sep 21, 2013 #6
    I like to do it as a differential equation.

    d(total momentum)/dt = 0
    m = rocket mass, function of time
    v = rocket velocity, function of time
    u = exhast-gas velocity backward from the rocket

    d(rocket momentum)/dt + d(exhaust momentum)/dt = 0
    d(m*v)/dt - (v-u)*(dm/dt) = 0
    The first term is easy
    The second term is (exhaust velocity) * d(exhaust mass)/dt
    Note: d(exhaust mass)/dt = - d(rocket mass)/dt
    (exhaust velocity) is in the direction of the rocket in the outside observer's reference frame
    Relative to the rocket, it is - u
    Relative to the observer, you need to add v, giving v - u

    One can do some easy rearrangement:
    d(m*v)/dt - (v-u)*(dm/dt) = (dm/dt)*v + m*(dv/dt) + (u-v)*(dm/dt) = m*(dv/dt) + u*(dm/dt)

    Thus, we get
    dv/dt = - (u/m)*(dm/dt)

    One can make u a function of time, or add other forces, like gravity or drag.

    But one can easily solve this equation for u a constant:
    v = u*log(m0/m)

    where m0 is the initial mass.
  8. Sep 21, 2013 #7
    Marina ...have you got the desired result ?
  9. Sep 21, 2013 #8
    The extra term will produce a term quadratic in the differentials. Quadratic terms can be safely neglected because they are an "infinitesimal of higher order"
  10. Sep 21, 2013 #9
    Thank you! Now it works for ma all the time.
    You guys are the best!!!!!!
  11. Sep 22, 2013 #10

    Filip Larsen

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    Gold Member

    I'm glad Marina got her problem solved.

    Not sure why you would say that, Tanya?

    The derivation looks quite standard to me. The infinitesimal parcel of propellant dm is consider to leave the rocket with a relative speed of -u at time t when the rocket exactly has speed v, not at time t+dt when the rocket has speed v+dv. This is quite similar to how the mass of the propellant dm is considered to leave the mass of the rocket at time t so that dm is not "present" at time t+dt, that is, the mass of the rocket at time t+dt is m-dm and not m.
  12. Sep 22, 2013 #11
    It's not like that makes any difference anyways since all extra terms that appear are infinitesimal of higher order and can be safely neglected.
  13. Sep 22, 2013 #12
    Hi Filip

    The derivation is quite standard ,but not quite right :rolleyes:

    The momentum of the system(rocket+unspent fuel ) at time t is Mv .At time t+dt , dm mass with relative velocity 'u' has been ejected.The velocity of the rocket changes to v+dv.The velocity of the ejected mass is (v+dv-u).

    The momentum of mass ejected comes in picture at time t+dt not at time t .And the velocity of the rocket at time t+dt is v+dv, not v . Hence the momentum of the ejected mass should be (dm)(v+dv-u).

    If what you say is right ,then how the initial momentum of the system at time t is taken Mv and not (M-dm)v + dm(v-u) ?

    Interestingly,the end result is same in both the cases,one outlined in the link ,the other I have explained :)
    Last edited: Sep 22, 2013
  14. Sep 22, 2013 #13

    Filip Larsen

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    Gold Member

    Before you claim all textbooks on the subject are incorrect on this you may want to consider that time does not play the role you think it does when deriving the rocket equation.

    You are supposed to model what happens at time t by adding up the total momentum "before" and "after" time t as if both "before" and "after" happens exactly at time t. That is, the time between "before" and "after" situation is zero, not dt. And just in case you now wonder where time then enters the picture it does so implicitly from the inclusion of a kinematic quantity (speed v) in the rocket equation and not from an explicit inclusion of a dt (which is then later ignored as a second order effect anyway).

    It would also be weird to model the transfer of propellant mass at time t but the corresponding transfer of propellant impulse at time t+dt. Since impulse relates to force that means that if the rocket is turned on exactly at time t then propellant will be exhausted but both the force and the acceleration will be zero at that point in time?
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