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Straightforward and Rigorous Rocket Thrust Derivation

  1. Oct 13, 2014 #1
    I have noticed many of the thrust derivations in textbooks I have seen do not do a straightforward derivation of rocket thrust. The all seem to use the same trick with infinitesimals in a sort of binomial form. For reference:

    Taylor, "Classical Mechanics" Pg. 85.

    I am working on a rigorous and straightforward way of doing this sticking to the assumptions of Newton's 2nd law. That is -- every velocity and acceleration referenced to the lab frame. Then converting over to our known frame for the exhaust as the result is ##F_{thr}=-\dot{m}\mathbf{v}_{er}##. Anyone have a reference using ##\dot{\mathbf{p}}=\sum{\mathbf{F}}## and these assumptions in a straightforward way?

  2. jcsd
  3. Oct 13, 2014 #2
    I'm not totally sure what you mean but check out Mechanics and Thermodynamics of Propulsion by Hill and Peterson Chp 10.
  4. Oct 13, 2014 #3
    Interesting text. I will have to look into that more. It doesn't have the typical derivation I don't like.

  5. Oct 13, 2014 #4

    Philip Wood

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    For me, the most convincing (lab frame) derivation uses momentum conservation. Assume that the rocket is going in a straight line in the x-direction. Work with scalar components of velocities in the x-direction. Let that of the rocket at time t be [itex]v_r[/itex]. Let that of the exhaust gases be -v relative to the rocket. Then the velocity of the jet gases in the lab frame is [itex]v_r - v[/itex]. If the mass of the rocket and contained gases at time t is m, and the mass of jet gases emitted per unit time is [itex]\mu[/itex], then equating system's (rocket + gases) at time [itex](t + \Delta t)[/itex] to its momentum at time t,
    [tex](m- \mu \Delta t)(v_r + \Delta v_r) + \mu \Delta t (v_r - v) = mv_r[/tex]
    Multiplying out, tidying and dividing through by [itex]\Delta t[/itex] gives
    [tex]m \frac{\Delta v_r}{\Delta t} - \mu \Delta v_r - \mu v = 0.[/tex]
    In the limit as [itex](\Delta t)[/itex] approaches zero, so does [itex]\Delta v_r[/itex], so the equation becomes
    [tex]m \frac {dv_r}{dt} = \mu v[/tex] in which v is a constant and [itex]\mu = - \frac{dm}{dt}[/itex].
    This is a bit clumsy, but I think it's convincing.
  6. Oct 13, 2014 #5
    That is the derivation that is in the textbooks that I have, and I don't really like it. It feels ad hoc to me, check the reference that the other gentleman posted for a more thouroughgoing derivation. I have not had a chance to go through it carefully yet, but it looks good. No tricks to get the right answer that is already assumed.

  7. Oct 13, 2014 #6

    Philip Wood

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    Didn't think I'd used any tricks or assumed the right answer, but I can't quarrel with your right to choose whichever derivation you like best!
  8. Oct 13, 2014 #7
    Not necessarily tricks, but if I chug ahead in a straight forward fashion like I do for most other Newton problems I get stuck, and I am assuming there is a reason why the textbooks don't go about it in a straight forward way either.

  9. Oct 13, 2014 #8
    Here is my profs solution to this derivation. I believe he makes a mistake by assuming the change of velocity of the fuel, is from the rest frame to ##v_{fg}## the velocity of the fuel with respect to ground. This is true in a static test, but as the rocket itself picks up forward velocity, that is a bad assumption. So I see his solution as ad hoc. He tries to solve this problem in a straight forward way. Here it is:

    "a. Let’s start out with Newton’s second law for a single particle but without assuming that mass is time independent, along with p = mv. This problem can be done in one dimension by choosing your coordinate to be the radius from the center of the earth to the rocket, so let’s represent the vectors as scalars, using signs to indicate direction. The + direction is away from the center of the earth. Let’s also, for now, write everything in the reference frame of a person standing on the ground next to the rocket.

    For convenience let’s write m instead of m(t). Newton’s second law is then ##F=\frac{dp}{dt}= \frac{d(mv)}{dt}=\frac{dm}{dt}v+m\frac{dv}{dt}##. (18)

    Rearranging this equation gives ##m\frac{dv}{dt}=F−\frac{dm}{dt}v=−v_{RG} \frac{dm}{dt}+F##. (19)

    Here ##v_{RG}## is the velocity of the rocket with respect to the ground.
    The key to this derivation is evaluating the net force on the rocket. There are two

    forces: the rocket’s weight pulls it down, and the expelled fuel exerts an upward force on the rocket. The magnitude of the rocket’s current weight is mg. A bit of expelled fuel of mass ##−dm## carries momentum of magnitude ##dp_{f} = −dmv_{FG}## in time ##dt##, so the rocket must pick up momentum ##dp = dmv_{FG}## in time ##dt## by Newton’s third law. Thus the fuel exerts a force on the rocket (using Newton’s second law) of ##\frac{dp}{dt} = dmv_{FG}##.

    Here the signs are getting a bit tricky. The mass of the fuel is positive, but dm is the
    change in the rocket’s mass, which is negative. ##v_{FG}## points downward and ##dm## is negative,
    so ##\frac{dm}{dt}v_{FG}## points upward, as it must to oppose gravity, which points downward. Thus

    ##m\frac{dv}{dt} = (−v_{RG} + v_{FG})\frac{dm}{dt} − mg##. (20)

    To change reference frames in Newtonian physics, remember that velocities transform according to


    ##v_{RG}+v_{GF} =v_{RF} =v_{FR}##. (21)

    ##m\frac{dv}{dt}=(−v_{RG} −v_{GF} )\frac{dm}{dt}−mg=v_{FR}\frac{dm}{dt}−mg##. (22)

    Now, Goldstein interchanges “velocity” and “magnitude of velocity.” The velocity of the fuel as seen by the rocket has magnitude ##v^{\prime}## and direction downward, so the rocket equation should become

    ##m\frac{dv}{dt} = −v^{\prime}\frac{dm}{dt} − mg##. (23)

    You should be able to convince yourself that the signs as shown are correct: if ##v^{\prime}## is positive, we know that dm/dt is negative, but the force of the escaping fuel must be positive to oppose the force of gravity; hence the minus sign in the first term."

    Again, the main problem I have with this is that he is assuming the fuel's initial velocity before the impulse is zero. I say it should be wrt the fuel tank as mass is conserved in combustion, and that original mass came from fuel/oxidizer tank which is already moving after lift off.

  10. Oct 13, 2014 #9

    Philip Wood

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    Don't see what's not straightforward about my application of the P of C of M! It goes right back to first principles. Arguably it's more fundamental than a method based on forces.

    The danger with applying Newton's second law is carelessness over what one is applying it to. It is, in my opinion, invalid to try to apply it to a body of changing mass, because that would be applying it to a succession of different bodies. That's not to say you can't use N2L for rocket motion, just that you have to be careful about defining the body to which you apply it.

    I'll look at your prof's derivation if I have time.
  11. Oct 13, 2014 #10
    I agree with your statement whole heartedly. When mass is changing, it really gets wild. Hard to keep everything straight in your mind's eye. I wonder if it is doable in Hamiltonian mechanics. I am not sure how one would tree the changing mass there either.

    I believe what you are saying is that what you defined as the system is changing as time goes on, because part of the system is being ejected out of what you originally assigned to be the system. Therefore, N2L does not apply unless you have a rigorous way of accounting for that.

  12. Oct 13, 2014 #11

    Philip Wood

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    Try this…

    Define body A as the rocket and its contents at time [itex](t + \Delta t)[/itex].
    Define body B as the jet gases which leave during the time [itex]\Delta t[/itex].
    The change in momentum of body B during time [itex]\Delta t[/itex] is [itex]-\mu \Delta t \times v[/itex].
    So, the force on body B, in the limit as [itex]\Delta t[/itex] approaches zero is [itex]-\mu v[/itex].
    So the force on body A is [itex]\mu v[/itex].
    So applying N2L to body A:
    [tex]\mu v = m \frac{dv_r}{dt}.[/tex]
    You may object that m is the mass of the rocket at time [itex](t + \Delta t)[/itex], not at time t. But as [itex]\Delta t[/itex] approaches zero the distinction becomes immaterial.

    If it's the disappearance of terms as [itex]\Delta t[/itex] approaches zero that you don't like, I'd counter that all proofs rely on doing this; it's just that some more sophisticated derivations don't show it happening!
  13. Oct 13, 2014 #12
    My objection would be that ##v## would be in the Earth frame. That is an assumption of N2L. The correct derivaton would have ##v## in the rocket's frame for the final result. My prof made that assumption correctly. I think he made a subtle error elsewhere per my 2nd to last post.

  14. Oct 13, 2014 #13

    Philip Wood

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    I don't understand your objection. -v is the velocity of the jet gases relative to the rocket. So their rate of change of momentum is [itex]-\mu v[/itex]. In the Earth's frame their initial velocity (while sitting in the rocket) is vr and their final velocity (having been ejected) is (vr - v). So their change in velocity is -v. That's also the change in velocity in the rocket's frame (-v - 0 = -v).

    I didn't quite follow your post number 12, but I'd guess you'd want to work in the Earth's frame of reference (assumed inertial). Both my derivations are done in this frame.

    I'd be very interested to learn exactly where you think the lack of rigour lies in my derivations of post 4 and post 11. Post 4 especially, as it goes right back to basics.

    I started looking at your teacher's derivation, but baulked as soon as I saw
    For reasons I explained in my post 9, I don't think you can apply N2L to a body of varying mass in this way, as in Newtonian physics, different mass means a different body. You'll find support for this view in other places. There are cases where the equation I've just quoted will give you the right answer, but that doesn't mean that the argument is valid.
    Last edited: Oct 14, 2014
  15. Oct 14, 2014 #14
    Thanks, Phill. After thinking it through -- I like your method of deriving rocket thrust. I don't think that the N2L approach would work because you would have to change what you call the system in the middle of the derivation, and I don't think Newton's laws allow that. Because of that, I think my professor's solution is wrong. We have a midterm today and I sent him this email as a clarification of my position on his derivation:

    Dr. #####, in reviewing for the midterm I have a concern about any rocket derivation problem that could possibly be on the test. In thinking about this problem very carefully and seeking consultation from others it was pointed out to me that deriving thrust from Newton's second law can be sketchy. It has to do with how one defines the system.

    1. If you define the system initially as the rocket + its fuel, then you can't change this fact at some later time. That would be arbitrary, and inconsistant with the assumptions of Newton's laws (as I understand them).

    My warrant for this point is that you can't change what masses are in and out of the system later once you have made your initial choice. In the chemical reaction that propels the rocket the mass is conserved even it has left the rocket it is still part of the system just like the parts of an exploding shell leave its center of mass unaltered by the explosion.

    2. If mass is conserved, then the left term of ##\dot{p}=\sum{F}## just becomes ma, and therefore ##ma_{cm}=0## because there are no particles are outside of the defined system. This is not a very interesting result.

    Therefore, I am not sure if Newton's 2nd law with it's strict assumptions is the best tool for this derivation. If one lets go of these assumptions and reasons through the problem carefuly, it can be done. Taylor just uses the conservation of momentum and a thought experiment. No product rule on ##\dot{p}##.

    So if asked on an exam to derive thrust, can I use Taylor's method as it does not run into this logical problem?

    Chris ###### "
  16. Oct 14, 2014 #15
    A month or so ago I came up with a derivation of the Tsiolkovsky rocket equation using forces rather than momentum. I don't like momentum. As an upshot, you also get the velocity in terms of time, and the position in terms of time.

    Assumptions regarding the design and properties of the rocket:
    The only force acting upon the rocket is that of its thrust.
    The force due to thrust is constant. ## F_T ## (Total force also, by a coincidence)
    The amount of fuel expended in a second is also constant ## \dot m ##

    The mass at any point in time is ## m_t = m_i - \dot m t ##

    ## F_T = ma ##
    ## F_T = (m_i - \dot mt)a ##

    Acceleration in terms of time:
    ## a(t) = \frac{F_T}{m_i-\dot mt} ##

    Finding velocity is now an initial value problem. I can provide those details if someone asks.
    Integrate and solve for initial velocity to find velocity in terms of time:
    ## v(t) = \frac{F_T}{\dot m}\ln{(\frac{m_i}{m_i-\dot mt})}+v_i ##

    You can do the same for position, but not necessary to derive Tsiolkovsky:
    ## x(t) = \frac{F_T}{\dot m}(t + (t-\frac{m_i}{\dot m})\ln{(\frac{m_i}{m_i-\dot mt})}+v_it+x_i ##

    The time the rocket will be in the air is the ratio of how much fuel it has (m_w), and how much fuel per second it is using (m_dot)
    ## t_{empty} = \frac{m_w}{\dot m} ##

    Now we can substitute t_empty into v(t) to find the velocity at which the rocket runs out of fuel.
    ## v_f = \frac{F_T}{\dot m}\ln{(\frac{m_i}{m_i-m_w})}+v_i ##

    Using the definition of specific impulse, realizing m_i-m_w is the final mass, and assuming v(0) = 0, you can change it to the usual form:
    ## \Delta v = I_{sp}F_T\ln{(\frac{m_i}{m_f}} ##

    In retrospect, this is probably useless if you're trying to prove your ability to derive it via momentum on a test. I prefer this way, though - it's more intuitive for me.

    If anyone has any tips for explaining complex things well, or for formatting long posts, I'd be happy to hear them.

    For clarity:
    Variables involved:
    ##F_T##, Force due to thrust (N)
    ##m##, Total initial mass of rocket (kg)
    ##m_w##, "Wet" mass of rocket, or total initial mass of fuel. (kg)
    ##\dot m##, Fuel "flux", or how much fuel the rocket uses in a second. (kg/s)
    ##a, v, x##, acceleration, velocity, position. All in one dimension. (m, m/s, m/s^2)
    ##t##, Amount of time elapsed (s)
    ##t_empty##, Point where the rocket has run out of fuel (s)
    Last edited: Oct 14, 2014
  17. Oct 14, 2014 #16

    Philip Wood

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    It's well known that the Law of Conservation of momentum can be deduced from Newton's second and third laws (N2L and N3L). But it's a very respectable law in its own right. [Arguably it has a sounder experimental basis than N2L which, to test experimentally, needs an independent method of measuring force.]

    I'd argue that using momentum conservation for the rocket problem is conceptually simpler than bringing in the concept of force which in any case, by using N2L and N3L, drops out leaving only a relation between masses and velocities and time. Why bring in force in the first place?

    Having said that, I believe that the simple derivation I presented in post 11 uses N2L and N3L correctly in the Earth frame.
    Last edited: Oct 14, 2014
  18. Oct 14, 2014 #17


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    Force is just the rate of momentum transfer. So you are implicitly using concept of momentum, even if you avoid using the symbol for it explicitly.
  19. Oct 14, 2014 #18
    Philip: There's many ways to skin a cat. I approach physics and math problems as if I'm trying to simulate it on a computer, because I understand it best that way. Also, I was sleep during the lecture on momentum! :)

    I realize it's immature to make a unilateral decision to shun an aspect of physics. I'm mostly being facetious.

    AT: Can you build on that? Maybe in a PM so we don't hijack the thread. It would be nice to take this opportunity to learn that momentum is actually elegant and simple. (... I doubt it, though...)
  20. Oct 14, 2014 #19


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