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Freefall from a great height, changing g

  1. Feb 21, 2008 #1
    This isn't a homework problem exactly - I'm attempting to teach myself introductory physics and there's a question that's bugging me. The kinematics equations that I've seen so far for falling bodies all assume that little g is a constant, i.e. the distance of the fall of the object from its initial height above the earth is small enough that delta g is zero. My question is how the equation would change for freefall where the starting height was far enough that little g was changing during the fall?

    2. Relevant equations

    y(t) = 1/2gt^2, g = G*M1*M2/r^2

    3. The attempt at a solution

    My best guess is that instead of using little g as a constant, i'd integrate F = G*M1*M2/r^2 over the distance traveled and somehow plug that into the equation instead, but my knowledge of both calculus and physics is too shaky to know if I'm on the right track. Can anyone help?

    Many thanks in advance.
     
  2. jcsd
  3. Feb 21, 2008 #2

    HallsofIvy

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    Yes, that's (almost) exactly right- the gravitational force is given by F= -GM1M2/r2 (don't forget the "-". We are measuring r upward but the force is downward). Since we still have "force= mass*acceleration", if M2[/sup] is the mass of the falling object, then acceleration = -GM1/r2. Acceleration is, of course, the derivative of speed with respect to time so we have to integrate, not with respect to the distance but with respect to time. Since the distance does appear in the formula, we have the "differential equation"
    [tex]\frac{dv}{dt}= -\frac{GM_1}{r^2}[/tex]

    One thing that makes that complicated is that it is the derivative of v on the left side but we have only r on the right. Since v= dr/dt, we have to use the chain rule to write dv/dt= (dr/dt)(dv/dr)= vdv/dr. Now the equation is
    [tex] v\frac{dv}{dr}= -\frac{GM_1}{r^2}[/tex]

    That is what is called a "separable equation" since we can "separate" the variables v and r:
    [tex]v dv= -\frac{GM_1}{r^2}dr[/tex]
    and then integrate both sides:
    [tex]\frac{1}{2}v^2= \frac{GM_1}{r}+ C[/tex]
    Giving the initial height and speed we could solve for the constant C.

    Once we've done that, since v= dr/dt, we have
    [tex]dr/dt= \sqrt{2\frac{GM_1}{r}+ C}[/tex]

    While that is also a "separable" equation and can be written as
    [tex]\frac{dr}{\sqrt{2\frac{GM_1}{r}+ C}}= dt[/itex]
    the left side does not have an "elementary" integral. It gives what is called an "elliptic integral" (obviously related to the elliptic orbits of planets) though special cases can be easy to solve.
     
  4. Feb 21, 2008 #3
    Ah, that helps a lot. Thank you!
     
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