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Homework Help: Scaling of the vertical projectile problem nondimensionalization

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Restate the vertical projectile problem in a properly scaled form. (suppose x<<R).


    Initial conditions: x(0)=0, dx(0)/dt=Vo

    Find the approximate solution accurate up to order O(1) and O(e), where r is a small dimensionless parameter. (i.e. the solution is given by a function f(e)).

    Hint: Suppose 1/(1-e) is a term involving the small parameter e (i.e. e<<1) in the dimensionless equation, do the taylor expansion 1/(1-e)=1+e+....If we approximate 1(1-e) by 1 in the equation then we will obtain the O(1) solution, if we approximate 1/(1-e) by 1+e in the equation, we will obtain the O(e) solution.

    2. Relevant equations

    3. The attempt at a solution

    I properly nondimensionalized the equation, getting the following:

    d2y/dT^2=-K/(y+1)^2 where K = gR/Vo^2.

    Now, as for the O(1) and O(e) stuff, I am completely baffled as to what the problem is asking for. Can someone please explain to me what exactly I am supposed to be doing?
    Last edited: Jun 6, 2010
  2. jcsd
  3. Jun 6, 2010 #2


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    i think they want you to do a series expansion of 1/(y+1)^2, so you get something like

    [tex] \frac{1}{(y+1)^2} = a + by + cy^2 +... [/tex]
    where a, b & c are constants

    the differential equation then becomes
    [tex] \frac{d^2 y}{dT^2} = -\frac{K}{(y+1)^2} \approx -K(a + by + cy^2 +...) [/tex]

    the O(1) case is equavalent to constant gravity, as it negelcts all term O(y) and above
    [tex] \frac{d^2 y}{dT^2} = -\frac{K}{(y+1)^2} \approx -K(a) + O(y) [/tex]

    the O(y) case includes the first order change, and neglects the terms O(y^2) and above
    [tex] \frac{d^2 y}{dT^2} = -\frac{K}{(y+1)^2} \approx -K(a + by) + O(y^2) [/tex]

    note this is possible because generally y<<1, so the contribution of y^n terms dcreases rapidly with increasing n
  4. Jun 6, 2010 #3
    I actually redid the nondimensionalization part and it was correct, but not properly scaled which explains why I didn't get a term that was <<1. The correctly scaled nondimensional equation is d2Y/dT^2=-1/(1+eY)^2 where e<<1.

    It doesn't really change though, I still don't see where the 1/(1-e) comes into play. When the problem statement says "the solution is given by a function f(e)", do they mean that O is a function of epsilon alone? I'm having trouble seeing what exactly O is - is it a solution Y to the differential equation or simply the right hand side of the differential equation after you taylor expand it?
    Last edited: Jun 6, 2010
  5. Jun 6, 2010 #4


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    i haven't checked with ur derivation of the differntial equation... if ur happy with it, stick with it...

    O actually means something about how one sequence "dominates" another... but i wouldn't bother going into it that far...

    Though not exactly correct, I always just took "O" to mean order.. so if you cut off your sequence at O(1), that means the order of the terms neglected are of the order of x or O(x) ...

    As x<<1, the order of x is << that of 1, simalarly O(x^2)<<O(x) and so on...
  6. Jun 7, 2010 #5
    So O is just the order of the taylor approximation that replaces the right hand side?
    Last edited: Jun 7, 2010
  7. Jun 7, 2010 #6


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    Yep, so in my first post for the O(1) case, you neglect all terms O(y) and above... and so on
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