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In which case is the period larger? ( Above earth or below?)

  1. Oct 28, 2016 #1
    1. The problem statement, all variables and given/known data
    A pendulum clock is adjusted so that it keeps excellent time on the ground. The clock is brought to a mine of depth h below the ground and then raised a height h above the ground to see the differences. In which case is the error larger?

    2. Relevant equations
    T = 2π √(L/g)

    L = length of the pendulum
    g = gravity on earths surface
    T = period

    g = GMm/Re2

    G = gravitational constant
    Re = earths radius
    M = mass of earth
    m = mass of pendulum (negligible)



    3. The attempt at a solution
    g1 = GMm/(Re - h)
    g2 = GMm/(Re+h)

    g1 = GMm/(Re - h) is for the pendulum underground. Plugging into the period equation gives T = 2π (Re - h) √(L/GMm)

    g2 = GMm/(Re+h) is for the pendulum above ground. Plugging into the period equation gives T = 2π (Re + h) √(L/GMm)

    It's obvious that the pendulum will tick slower below ground and faster above - but what about the error? I think that the errors would be the same. The question seems to imply that they are not though. I think there is a factor that I am not taking into account?
     
  2. jcsd
  3. Oct 28, 2016 #2
    pl. show a work out
    if it was obvious then the problem might not have been posed.
    well i do not have the answer and can not work out as its for you....homework thing...
     
  4. Oct 28, 2016 #3
    I showed my work above. :P
     
  5. Oct 28, 2016 #4
    I'm not asking for the answer, I just want to know if I'm on the right track or if there is something I am missing here.
     
  6. Oct 28, 2016 #5
    suppose your clock is slower then the time for a full period will be larger
    and if your clock is faster then time for full period will be smaller , so ask again when the clock has more elapsed time for full period
    above or below
     
  7. Oct 28, 2016 #6

    haruspex

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    Yes.
    Both of those are wrong, even dimensionally wrong.
    The first is, in a sense, doubly wrong. The gravitational attraction of a uniform sphere of radius R turns out to be the same as for a point mass, of the same mass, only at radii > R. Inside the sphere it is different. Do you have a formula for that?
     
  8. Oct 28, 2016 #7
    Okay - how does this look?
     

    Attached Files:

  9. Oct 28, 2016 #8

    haruspex

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    You are still using the wrong formula for the gravitational attraction when inside the sphere.
    Here's a clue. Inside a unform spherical shell, the shell exerts no gravitational field. So at distance r<Re, what part of the Earth is responsible for the gravitational field? What is the mass of that part? What is the resulting gravitational acceleration?
     
  10. Oct 28, 2016 #9
    This is the resulting gravitational acceleration.
     

    Attached Files:

  11. Oct 28, 2016 #10
    I believe this is how it changes when you are underground. the g without the subscript is earths (9.8 m/s2)
     

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  12. Oct 28, 2016 #11
    If ##h## is the distance from the surface to the particle, then yes.

    Side note: Why don't you learn basic latex/mathJax to format maths then, you do need post a picture every time. It will also help others to understand what you mean.
     
  13. Oct 28, 2016 #12
    I didn't know that was a thing. Now that I'm in the physics major I'll likely be posting here often so I will look into that. Thank you for the resource
     
  14. Oct 29, 2016 #13
    I used percent error and found that it was greater for the pendulum above earth.
     
    Last edited: Oct 29, 2016
  15. Oct 29, 2016 #14
    Method
     

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  16. Oct 29, 2016 #15

    PeroK

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    It's difficult to see what you've done there. To help you with latex, you are looking for something like:

    ##g_a(h) = g(1 \pm f_a(\frac{h}{R}))##

    ##g_b(h) = g(1 \pm f_b(\frac{h}{R}))##

    Where ##g_a(h)## is the gravity a height ##h## above the Earth and ##g_b(h)## is the gravity a depth ##h## below the Earth's surface. ##f_a## and ##f_b## are functions you need to find and compare. ##R## is the radius of the Earth.
     
  17. Oct 29, 2016 #16
    Okay, but why is it plus or minus? I know below is the expression with a minus sign, and I've gotten those results but I figured that since for above ground the mass isn't changing, then the expression for how gravity changes would just be the typical ga = GM/(R+h)2 divided by g = GM/R2. Either way, I've tried it both ways and am getting that the error is larger above earth but with different percent errors for each function. I've been using h = 30m to compare.
     
  18. Oct 29, 2016 #17

    PeroK

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    That may be correct for ##h = 30m##, but what about other values of ##h##.

    It was ##\pm## so I didn't give too much away! Did you get a formula for gravity a distance ##h## above ground, compared to surface gravity?
     
  19. Oct 29, 2016 #18
    Yes that is what I did. In my most previous post with pictures you can see the equations I used although I know it's hard to read. I'll try other values for h. Thanks for the help - I think I am in the home stretch, lol.
     
  20. Oct 29, 2016 #19

    PeroK

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    That's why I posted some latex, so you could hopefully just copy that to get you started.
     
  21. Oct 29, 2016 #20
    What is latex?
     
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