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My calc prof would be so disappointed

  1. Nov 23, 2009 #1
    I've been out of practice for one quarter now and it's already starting to show.

    I've been thinking about gravity, and I wanted to find a "true" eq for freefall. (one that incorporates the inverse square relationship)

    I instantly hit a wall, because I'm trying to find X in terms of T, from dV/dT in terms of X.

    I can't even think of how to start this problem...

    Any help?
  2. jcsd
  3. Nov 23, 2009 #2
    What's wrong with the "fake" one? y = -(g/2)t^2 + y_0. It has that relationship stuff you wanted. I suppose a more true eq would be if you consider wind resistance in the form of kv^2.

    Then, dv/dt = kv^2 - g, an easy diff eq. Then once you have V, integrate it again.
  4. Nov 23, 2009 #3


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    It would help if you showed more of your work, but I can help get you started. In the event that you've already carried through with my suggestions, just let me (and others) know and we'll provide the help you need! :smile:

    From Newton's Law of Universal Gravitation, we know that

    [tex]F = m\frac{dv}{dt} = -\frac{k}{x^2}[/tex]

    Using the chain rule, we know that [itex]\frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dt}[/itex], which we apply so that

    [tex]F = mv\frac{dv}{dx} = -\frac{k}{x^2}[/tex]

    You should be able to solve that with a straight forward integration (don't forget the constant :wink:). After that, you'll end up with a differential equation of the form [itex]\frac{dx}{dt} = f(x)[/itex], so it might be helpful if you remember that [itex]\frac{dt}{dx} = \frac{1}{dx/dt}[/itex]. By the way, sorry if there are sign errors (I'm doing this off the top of my head). Good luck!
  5. Nov 24, 2009 #4


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    I presume you mean you want a formula for an object falling through a distance large enough that you cannot just assume that the force due to gravity is constant:
    [tex]ma= m\frac{dv}{dt}= -\frac{GMm}{r^2}[/tex]
    We can, of course, immediately cancel the "m"s. Also, since v= dr/dt, we have, by the chain rule, that dv/dt= (dv/dr)(dr/dt)= v dv/dr. That is, we can change that equation in v, r, and t into an equation involving only v and r:
    [tex]v\frac{dv}{dr}= - \frac{GM}{r^2}[/tex]
    which is a "separable differential equation".

    [tex]v dv= -GM \frac{dr}{r^2}[/itex]
    which can be integrated:
    [tex]\frac{1}{2}v^2= \frac{GM}{r}+ C[/itex]

    I doubt your calc prof would be all that disappointed. This is a technique, called "quadrature", for solving non-linear differential equations that is not normally taught in a calculus course. Notice, by the way, that that equation can be rewritten
    [tex]\frac{1}{2}mv^2- \frac{GmM}{r}= mC[/tex]
    and is basically "conservation of energy".

    If you take, as initial condition, that at t= 0, r(0)= R and v(0)= 0, an object falling from height R, we have
    [tex]\frac{1}{2}(0^2)= \frac{GM}{R}+ C[/tex]
    so C= -GM/R and
    [tex]\frac{1}{2}v^2= \frac{GM}{r}-\frac{GM}{R}[/tex]

    Now solve that for v:
    [tex]v^2= \frac{2GM}{r}- \frac{2GM}{R}[/tex]
    [tex]v= \frac{dr}{dt}= -\sqrt{\frac{2GM}{r}- \frac{2GM}{R}}[/tex]
    (I took the negative root because I am taking "+" velocity upward.)

    [tex]\int\frac{dr}{\sqrt{\frac{1}{r}- \frac{1}{R}}}= \sqrt{GM}t[/tex]
    to find r as a function of t.

    Your calc prof would also not be too disappointed if you could not do that integral. It is not integrable in terms of elementary functions. It is a type of "elliptic" integral, so called because it arises in finding the length of an arc of an ellipse. Since the general motion in an "inverse square" gravity field is along an ellipse, that should not be too surprising.
  6. Nov 24, 2009 #5
    That was an amazing post, HallsofIvy. Thank you!
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