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Drilling a hole through the center of earth to the other side, and dropping an object

  1. Sep 3, 2005 #1
    Hi.
    I'm having trouble solving this situation.
    Imagine that a hole is drilled through the center of the Earth to the other side. An object of mass m at a distance r from the center of the Earth is pulled toward the center of the Earth only by the mass within the sphere of radius r. Write Newton's Law of gravitation for an object at the distance r from the center of the Earth, and show that the force on it is of HOoke's law form, F=-kr, where the effective force constant is k= (4/3)pi(density)Gm
    And show that a sack of mail dropped into the hole will execute simple harmonic motion if it moves without friction. When will it arrive at the other side of the earth.

    Ok, so far, I think I got the first part, where I used density= Mass/Volume
    and volume of a sphere is (4/3)pi(r^3), and I isolated the M.
    I replaced the M value with (4/3)pi(r^3)(density) in the law of gravitation formula, and then by grouping some terms together I get F= Kr
    BUT I HAVE NO CLUE as to where the negative is coming from.
    And also, I have no idea where to start proving that dropping a sack of mail into the hole will be SHM, and When will it arrive at the other side of earth.
    I tried using x(t)=Acos(wt) and making A=r of earth
    w (angular frequency)= [4(pi^2)r]/T BUT now I'm getting myself even more confused. A hint would be really appreciated!!! I really need help here and I need a clue as to where I can start finding the amount of time, etc...
    Thank you very much for your time.
     
  2. jcsd
  3. Sep 3, 2005 #2

    Fermat

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    The negative comes from the gravtational force, F, being directed downwards and the distance, r, being measured upwards. Consider them vectors.

    Defn: SHM is defined by x''/x = constant.

    You have the force acting on the mail - now use Newton's 2nd law to achieve the defn of SHM.
     
  4. Sep 3, 2005 #3
    Thank you for responding, but what does x''/x mean?
    if you don't mind, can u tell me what x represents?
     
  5. Sep 3, 2005 #4

    Fermat

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    Sorry about that.

    x is the displacement.
    x'' is the acceleration.

    You may recognise it better as,

    [tex]\frac{\ddot x}{x} = -\omega^2\ \mbox{where} \ \omega = \sqrt{\frac{k}{m}[/tex]
     
  6. Sep 3, 2005 #5
    oh i get it!! but just one more thing, I'm still having trouble finding the time it takes for the mail sack to reach the other side of the earth through the hole....was i going in the right direction in the first post, or is there something i missed?
     
  7. Sep 3, 2005 #6

    Astronuc

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    The period T in a simple harmonic oscillator = 1/[itex]f[/itex]

    [itex]\omega\,=\,2\pi\,f[/itex]
     
  8. Sep 3, 2005 #7
    It's soo simple!! I get it now!!! Thank you soo much everyone!! You have all been a great help to me! Really appreciate it!
     
  9. Sep 3, 2005 #8
    oh no, i'm in trouble again, sorry to be bothering y'all...
    it's just that, well, ok
    first
    I did w=2(pi)f
    and w= sqrt(k/m)
    and the question stated that k=(4/3)pi(density)G(m)
    so I replace that in the w= sqrt(k/m) where the m cancels out. then I am just left with sqrt((4/3)pi(density)G) = 2pi/T
    and I solve for T but i get 2667.27 minutes, whereas the answer is supposed to be after 42.1 min!! I really need some assistance again, sorry!
    THank you very much for your time.
     
  10. Sep 3, 2005 #9

    StatusX

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    I noticed that, from what you said in your first post, it seems like you first took r to mean the radius of earth in your volume calcuation and then changed it to mean the objects distance from the center when you plugged it into the gravity formula. You may know this and just have skipped this step in your post, but keep in mind that the only reason this works is because the contribution from entire shell outside of where you are, (ie, the shell R<r<R_e, where R is your distance from the center and R_e is the radius of earth) cancels out, and so the force you feel is as if you were on the surface of a planet of radius R with the same density as earth. I just want to make sure you know this because naively plugging into the GMm/r^2 equation gives the right answer, but only by a lucky coincidence. The right way to do it would be to integrate or use Gauss's law.
     
    Last edited: Sep 3, 2005
  11. Sep 3, 2005 #10
    Really?! I wasn't aware of that mistake!! Thank you so much. But how can i go about integrating to get F=-kr then? and what about the amount of time it takes for the sack of mail to reach the other side?
     
  12. Sep 3, 2005 #11

    StatusX

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    Do you know Gauss' law yet? Since the earth has spherical symmetry, put the gaussian surface as a sphere centered at the center of the earth and extending out to where you want to calculate the force. You get a*A=4*pi*G*M, where a is gravitational gravitational acceleration, A is the area of the sphere, G is the gravitational constant, and M is the mass enclosed by this sphere. Plugging in 4*pi*r^2 for the area and 4/3*pi*r^3*(density) for M gives back the formula you found. If you haven't done Gauss' law yet, you'll need to set up an integral. This is more complicated, so I'll wait to see if you need help with it before I explain it.
     
  13. Sep 3, 2005 #12

    Astronuc

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    The bag falls to the center of the earth in time T, the it continues to the other side in time T. This is of course without energy loss due to friction or wind resistance. And of course, it would not be practical to build a tunnel to the center of the earth which contains an iron core, most of which is molten, and has a pressure estimated at 350 GPa.
     
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