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I Rocket equations of motion w/ drag and gravity

  1. Feb 3, 2017 #1
    I have seen many examples of the EOM for a rocket derived for the following cases:
    • No gravity, No drag
    • Gravity, No drag
    • No gravity, linear drag (b*v where b is a constant)

    I have never seen
    • Gravity, linear drag
    • Gravity, quadratic drag

    I took John Taylor's two examples of linear drag and gravity (without drag), and tried to combine them (since quadratic drag would add even more complexity). A couple basic assumptions at first:
    The mass of the rocket is described as:
    Where [itex]m_o[/itex] is the initial mass of the fueled up rocket, and [itex]-k[/itex] is the rate at which mass leaves the rocket. Also, [itex]u[/itex] is the velocity of the exhaust (which is taken as a constant), and [itex]v[/itex] is the velocity of the rocket itself.

    For linear drag + gravity, the resultant equation of motion would be:

    [tex]\frac{m dv}{dt}=-\frac{u dm}{dt}-mg-bv[/tex]

    At this point, for the case without gravity, you can do separation of variables. However, that doesn't seem possible here:
    (I've substituted -k in for dm/dt)
    So, putting in m->m(t):

    Any help on solving this analytically would be greatly appreciated!
  2. jcsd
  3. Feb 3, 2017 #2


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    Just doing this in my head, so it might not work...
    Looks like you can get it into the form y'=Ay/x+B. Then try substituting y=ux.
  4. Feb 5, 2017 #3
    I finally found a method in Boyce & DiPrima's diff eq book:
    If you can get the diff eq to look like:
    Then you can solve it using an integrating factor, [itex]\mu (t)[/itex]:
    [tex]\mu(t)=exp\int p(t)dt[/tex]
    In my case, [itex]t_o=0[/itex] which helped simplify it. Anyway, I solved and got a somewhat ugly but completely reasonable formula!

    Which gives the velocity of the rocket anytime before burnout. I can integrate and get height, and then use conservation of energy (with quadratic drag this time) to calculate final height. Sweet.
  5. Feb 5, 2017 #4
    Extra cool:

    I used mathematica to take the limit of that equation as [itex]b\rightarrow0[/itex] and got exactly the same result as John Taylor's 3.11 which solves for [itex]v(t)[/itex] for the case of gravity but no drag.

    Mind = blown. So happy this worked.

    (also, same for [itex]g\rightarrow0[/itex], recovers the same as 3.14)
    Last edited: Feb 5, 2017
  6. Feb 5, 2017 #5


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    The method I outlined works too. Did you try it?
    Writing T=(m0/k)-t and V=(ku/b)-v we have dv=-dV, dt=-dT:
    Now putting V=WT
    W' T + W = (b/k)W - g
    Writing α=b/k-1
    W' T = αW - g
  7. Feb 6, 2017 #6
    I did not try that method, since I couldn't get it into the form you proposed:

    The best I can do with that initial equation is:

    Does your method work for that form as well?
  8. Feb 6, 2017 #7


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    My post #5 lays it out in detail.
  9. Feb 6, 2017 #8
    Wow that is really clever. At first I didn't get what you were saying but yeah that works! I would have never thought of that on my own...
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