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An object falling through a drilled hole through the earth!

  1. Nov 25, 2003 #1
    A hole is drilled through the earth so an object can be dropped into the hole, its greatest velocity is at the center of earth and decelerates on its way back on the other side. Then continues indefinitely provided there is no air drag.

    How would one find the velocity at any given point, I have DE next semester, but we did some first order ones, so I would think it is something like:

    V^2 = 2A?Y*

    *Ball dropped from rest so Vo is zero

    A is the acceleration caused from the earth, but it would constantly be changing, wouldnt that be an unknown derivative?

    V^2 = 2?R(dg/dr)

    I attempted to solve it, I got:

    V = sqrt((Gm/R)/ln(2R))

    which doesnt make sense....
     
  2. jcsd
  3. Nov 25, 2003 #2

    krab

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    Are you allowed to make the approximation that the earth is a sphere? Then use the fact that the force at a given radius is as if all the mass that is within that radius is concentrated at the centre. Can you assume that the earth has uniform mass density? Then the acceleration at radius r is -rg/R, where R is the earth's radius. This gives you the same equation as that of a spring, or simple pendulum (i.e. "simple harmonic motion"), and you probably already know how to solve it.
     
  4. Nov 25, 2003 #3
    If all you care about is the velocity at a given point, use conservation of energy:

    [tex]E = K + U = \frac{1}{2}mv^2 - GMm/r[/tex]

    (Total energy E is constant.)
     
  5. Nov 25, 2003 #4
    The simplest solution is the most elusive. Thanks :)
     
  6. Nov 25, 2003 #5

    enigma

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    You need to take into account that the M term from GMm/r only applies to the mass 'beneath' you.
     
  7. Nov 26, 2003 #6
    So if one enters a shaft from Europe say, then what effects a body 'Falling' that has nothing below its feet?

    If there were an hypothetical 'Gap' in the Earth from say United Kingdom>>> Sidney Australia, and if there were two test subjects located at the Surface points A (UK) and B (Aussie) who coincide their leaps, does the aussie emerge at point A (UK)and the limey at point B (Aussie)?..or is there another factor due to the situation, and both test subjects come to a rest at the mid-point in shaft, eventually ending up orbiting each other due to the Radial Gravitational effects of the surrounding Earth?
     
  8. Nov 26, 2003 #7

    enigma

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    Sorry, I should have made myself more clear.

    By putting 'beneath' in quotes, I meant that:

    If you build an imaginary cut beneath you, so that you're 'standing' on a smaller sub-sphere of the Earth, that smaller sub-sphere is the part whos gravity affects you. The remaining hollowed-out sphere 'above' you sums to zero force when you integrate the gravitational force over the entire thing.

    (or at least, so I've been told. That's one integration I have never actually calculated, to be honest)
     
  9. Nov 26, 2003 #8

    enigma

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    As long as you assume that the Earth isn't rotating, they'll end up at the other side. Their maximum speed in the center will be smaller than it would be if the force acting on them throughout the descent was the same as the force acting on them at the surface.

    If the Earth is rotating, then coriolis forces would cause our intreppid bungee jumpers to wind up eating hot magma not very far down.
     
  10. Nov 26, 2003 #9

    jcsd

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    OK, if we assume a few things:

    The Earth is spherical and of uniform density

    Friction, drag, the corolis force (this can be ignored by placing the hole through the poles) are ignored.

    We know:

    [tex]a = \frac{GM}{r^2}[/tex]

    Though I won't prove it, as enigma says for a given r you only need consider the mass of the sphere with the radius r, which as I've assumed a spherical Earth with uniform density is given by:

    [tex]M = \frac{4}{3}\pi r^2 \rho[/tex]

    if we substitute this into the first equation we get:

    [tex]a = \frac{4}{3} \pi \rho Gr[/tex]

    As the other terms are constant it is clear that the accelration is directly proportional to the displacement from the Earth's centre which means that we should immediately recognise that the object dropped down Matt's hole has simple harmonic motion, whose accelration and velocity is given by the following equations:
    [tex]v = \omega \sqrt{r_0^2-r^2}[/tex]

    [tex]a = -\omega^2 r[/tex]

    [tex]\inline{r_0}[/tex] is the amplitude ort maximum displacemnt which in this case is equal the radius of the Earth and we can see that [tex]\inline{\omega}[/tex], the angular frequency is equal to [tex]\inline{-\sqrt{(\frac{4}{3} \pi \rho Gr})}[/tex]

    we can now get the equation for the velocity:

    [tex]v = -\sqrt{(\frac{4}{3} \pi \rho Gr)(r_0^2 - r^2)}[/tex]

    As a final note I'll say that obviusly drag and frction can't be ignored and would actually make this into a dampened simple harmonic oscillator, another factor that can't be ignored is the fact that the Earth is not of uniform density and a better approximation is to consider the mass uniformly distrubuted inside the lower mantle and the core and ignore the differing gravity as you move through the upper mantle and crust.
     
    Last edited: Nov 26, 2003
  11. Nov 26, 2003 #10

    krab

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    jcsd: Your formulas are more complicated than need be. Your factor
    [tex]\inline{{4\over3}\pi\rho Gr}[/tex] is simply [tex]\inline{{GM_0r\over r_0^3}}[/tex] where [tex]\inline{M_0}[/tex] is the earth's mass. But we already know what this acceleration is at [tex]\inline{r=r_0}[/tex]; it is [tex]\inline{g}[/tex]. Hence

    [tex]a={4\over 3}\pi\rho Gr=g{r\over r_0}[/tex]

    and this considerably simplifies the calculation. I gave this expression in my first post. Did you miss it?

    So frequency is

    [tex]\omega=\sqrt{g/r_0}[/tex],

    and the final formula for velocity is

    [tex]v = -\sqrt{\frac{g}{r_0}(r_0^2 - r^2)}[/tex].

    (BTW, you have an extra factor of [tex]\inline{r}[/tex] under your square root.)
     
    Last edited: Nov 26, 2003
  12. Nov 26, 2003 #11

    chroot

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    [moderator butting in]
    Good to see you guys using both the display-mode and inline-mode tex so well! Please let me know if you have any comments, complaints, or suggestions about the tex system.
    [/moderater butting in]

    - Warren
     
  13. Nov 26, 2003 #12

    jcsd

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    krab: I only noticed yor post after I'd posted. I derived the equations as I was writing out my post, I thought that they would probably simplify, but I admit I was too lazy to do it.

    The r exponent was a typo, I had to edit the post about four times to get rid of all the typing errors in the latex code, though the inline text worked just fine and it looks better than I thought it would.
     
  14. Nov 26, 2003 #13

    krab

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    Yea. Thanks Warren, the \inline thing works nicely. I found out about it by comparing jcsd's post to mine. I've now edited my post to make use of it.
     
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