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Free fall due to gravity

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data

    I was given a problem to solve for the speed of a body falling under gravity [equation (1)] where g is acceleration due to gravity, which was easy enough.. but then i thought i would extend it to the case where g is non-constant, and so arrived at equation (2), (where where z is the height above earth [z'=dz/dt and z=dv/dt and z^-2 means z to power -2], and M is the mass of the earth and G is the gravitational constant)

    2. Relevant equations

    (1) : dv/dt = - g - kv

    (2) : z'' + kz' + GMz^-2 = 0

    3. The attempt at a solution

    I believe this is a non-linear second order DE?? i attempted to solve by setting

    z'' + kz' = 0

    and solving the complimentary equation, which was OK, but when i came to solve for the particular integral

    z'' + kz' = -GMz^-2

    i ran into problems, as after substituting in the D and Q operators (http://silmaril.math.sci.qut.edu.au/~gustafso/mab112/topic12/ [Broken]), i could not use the First Shift Theorem, as the RHS is not in an exponential form...

    Any ideas anyone?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 16, 2009 #2

    Tom Mattson

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    Yes, but rather than get into that let's look at your first order equation for [itex]v[/itex].

    Do the following.

    1.) Show that [itex]\frac{dv}{dt}=v\frac{dv}{dz}[/itex].
    2.) Insert [itex]g=\frac{GM}{z^2}[/itex] into the equation.
    3.) Find an integrating factor that makes this equation exact (it can be done).
    4.) Solve.
     
  4. Feb 17, 2009 #3

    HallsofIvy

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    There is no "complementary equation" nor is there a "particular integral". Those are both concepts in linear differential equations where the non-homogeneous part is a function of the independent variable only.

     
    Last edited by a moderator: May 4, 2017
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