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Math Help - How long will it take to fall?

  1. #16
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    There is a simple formula you can use.
    If you are standing at s_0 and you throw an upward or downward (this makes it positive or negative) with accelaration of gravity g. Then the function that tells you the height after t seconds is:
    S(t)=-\frac{1}{2}gt^2+v_0 t+s_0
    Of course in a vacuum.
    ---
    Topsquark, may I ask you two questions?

    It is true that when an object travels further up its effect from gravity diminishes. So what equation describes this, a cubic?
    If the initial speed is not too great it is an arc of a sinusoid.

    RonL
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  2. #17
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker
    There is a simple formula you can use.
    If you are standing at s_0 and you throw an upward or downward (this makes it positive or negative) with accelaration of gravity g. Then the function that tells you the height after t seconds is:
    S(t)=-\frac{1}{2}gt^2+v_0 t+s_0
    Of course in a vacuum.
    ---
    Topsquark, may I ask you two questions?

    It is true that when an object travels further up its effect from gravity diminishes. So what equation describes this, a cubic?
    Assuming Newtonian gravity (which should suffice) for an object launched from the surface of the Earth we get the equation
    0 = (1/2)mv^2 - (1/2)mv_0^2 - \frac{GmM}{R+y} + \frac{GmM}{R}
    from energy conservation, where v0 is the initial (launch) speed, M is the mass of the Earth, R is the radius of the Earth, G is the Universal Gravitation constant, m is the mass of the object, and y is the height above the Earth. (Note: g = \frac{GM}{R^2}) Or we could use Newton's 2nd:
    \frac{GmM}{(R+y)^2} = ma (Positive downward.)

    Neither equation is trivial to solve for y(t). I can't find my notes on how I did it, but I seem to recall solving the energy equation for y(t) at some point in the past. However, I don't recall getting any great inspiration from it either.

    -Dan

    The more I stare at the energy equation, the more I doubt I solved it. I may have solved a first or second approximation where y is small.
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  3. #18
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick
    Thanks dan, you get a +rep . I have no idea what how to calculate most of the stuff but it's always nice to see what would be involved...
    Thank you, though I can't claim credit...most of it came from one of my textbooks.

    I was thinking that if you really wanted to do the project you could still model the equations even if you didn't have the derivations. The quadratic resistance case is a pretty ugly (oxymoron?) equation but easy enough to program on a fancy calculator, or in Excel if you wanted to make a graph.

    -Dan
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  4. #19
    Grand Panjandrum
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    Quote Originally Posted by topsquark
    Assuming Newtonian gravity (which should suffice) for an object launched from the surface of the Earth we get the equation
    0 = (1/2)mv^2 - (1/2)mv_0^2 - \frac{GmM}{R+y} + \frac{GmM}{R}
    from energy conservation, where v0 is the initial (launch) speed, M is the mass of the Earth, R is the radius of the Earth, G is the Universal Gravitation constant, m is the mass of the object, and y is the height above the Earth. (Note: g = \frac{GM}{R^2}) Or we could use Newton's 2nd:
    \frac{GmM}{(R+y)^2} = ma (Positive downward.)

    Neither equation is trivial to solve for y(t). I can't find my notes on how I did it, but I seem to recall solving the energy equation for y(t) at some point in the past. However, I don't recall getting any great inspiration from it either.

    -Dan

    The more I stare at the energy equation, the more I doubt I solved it. I may have solved a first or second approximation where y is small.
    You shouldn't need to solve it, it is a projection of a conic onto the "up"
    axis, because it is a degenerate inverse square law orbit.

    RonL
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  5. #20
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack
    You shouldn't need to solve it, it is a projection of a conic onto the "up"
    axis, because it is a degenerate inverse square law orbit.

    RonL
    (Chuckles) You know, because it's a projectile motion problem rather than a classical orbital problem I never thought of that... Good call!

    -Dan
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  6. #21
    Grand Panjandrum
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    Quote Originally Posted by topsquark
    (Chuckles) You know, because it's a projectile motion problem rather than a classical orbital problem I never thought of that... Good call!

    -Dan
    Yes I think its quite neat

    RonL
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  7. #22
    Forum Admin topsquark's Avatar
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    First approximation to the energy equation

    0 = (1/2)mv^2 - (1/2)mv_0^2 - \frac{GmM}{R+y} + \frac{GmM}{R}

    Note that this is the first integral of the Newton's 2nd equation in the previous post where this appears. (5 posts ago)

    Solve this equation for v:
    v = \pm \sqrt{\frac{2GM}{R+y}+\left ( v_0^2-\frac{2GM}{R} \right ) }

    v is positive for an object moving upward and negative for moving downward, so we can split the solution into two pieces for upward and downward motion.

    For simplicity, take a = 2GM (NOT the acceleration!) and b = v_0^2-\frac{2GM}{R}. This equation is then
    v = \pm \frac{dy}{dt} = \pm \sqrt{\frac{a}{R+y}+b }

    Take the first approximation to this equation where y is considered to be small. (ie. keep y to first order, y << R.) This equation becomes an approximate linear first order equation:
    \frac{dy}{dt} \pm \frac{a}{2R^2 \sqrt{a+\frac{b}{R}}} y =  \pm \sqrt{\frac{a}{R}+b}

    Letting c = \frac{a}{2R^2 \sqrt{a+\frac{b}{R}}} and d = \sqrt{\frac{a}{R}+b} we see that
    \frac{dy}{dt} \pm cy = \pm d
    has the solution
    (*) y(t) = Ae^{Bt}+D

    Consider now the upward motion solution. (Use the "+" signs.) Inserting this into equation * we get that B = -c and D = \frac{d}{c}, so
    y(t) = Ae^{-ct} + \frac{d}{c}.

    Now v = \frac{dy}{dt}(0) = v_0. Thus A = - \frac{v_0}{c}.

    y(t) = - \frac{v_0}{c}e^{-ct} + \frac{d}{c}.

    Back-subbing to get to the original variables:
    y(t) = - \frac{2R^2v_0 \sqrt{a+\frac{b}{R}}}{a} exp \left [ -\frac{a}{2R^2 \sqrt{a+\frac{b}{R}}}t \right ] + \frac{2R^2 \sqrt{\frac{a}{R}+b} \sqrt{a+\frac{b}{R}}}{a}

    y(t) = - \frac{R^2v_0 \sqrt{2GM+\frac{v_0^2-\frac{2GM}{R}}{R}}}{GM} exp \left [ -\frac{GM}{R^2 \sqrt{2GM+\frac{v_0^2-\frac{2GM}{R}}{R}}}t \right ] + \frac{R^2v_0 \sqrt{2GM+\frac{v_0^2-\frac{2GM}{R}}{R}}}{GM}

    y(t) = \frac{Rv_0 \sqrt{2GMR^2+Rv_0^2-2GM}}{GM}  \left ( 1 - exp \left [ -\frac{GM}{R \sqrt{2GMR^2+Rv_0^2-2GM}}t \right ] \right )

    To get the rest of the motion, solve y(T) for maximum height, then use the solution to the equation \frac{dy}{dt} - \frac{a}{2R^2 \sqrt{a+\frac{b}{R}}} y = - \sqrt{\frac{a}{R}+b} with the initial condition \frac{dy}{dt}(T) = 0. (Or simply consider that the equation for y(t) for the whole motion will be symmetric about t = T.)

    As I mentioned previously, it doesn't give me any great inspirations, though it's a nice exercise in applied differential equations.

    -Dan

    I was just thinking. I did this analysis a few years ago and it never occured to me until now that the solution needs to be forced to be continuous about the time t = T. That could add some interesting complications in solving for the downward motion equation. I'm not going to bother with it. Consider this as a problem "for the interested student."
    Last edited by topsquark; August 29th 2006 at 12:25 PM. Reason: Addendum
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