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Earth and Acceleration due to gravity

  1. Jan 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Earth is commonly thought of as a sphere, but this is not true. because of Earth's spin, it closely resembles an oblate spheroid, which is just a fancy name for "shaped like a squashed orange". The effect of this spin is a tidal bulge that forms at the equator so that the equatorial radius of Earth is about 21 km greater than the polar radius of 6370 km. The mass of the earth is 5.979 x 10^24 kg.
    a) Calculate the acceleration due to gravity at the north pole using the definition of gravitational field.
    b) The acceleration due to gravity at the equator is different from that at the North Pole because the radius is 21 km longer and because of the spinning of the earth. use the definition of gravitational field and centripetal acceleration to calculate the acceleration due to gravity at the equator.
    c) A 1.00 m pendulum with a 500 g bob is constructed at the equator. An identical pendulum is constructed at the North Pole. They are set to oscillate at the same time, with identical amplitudes. Which pendulum will make the most swings in a time of exactly 8 minutes? Support your answer.


    2. Relevant equations
    acceleration due to gravity= GM/R^2
    G= 6.67 x 10^-11
    Centripetal acceleration= V^2/r


    3. The attempt at a solution
    a)g=GM/r^2
    = (6.67x10^-11)(5.979x10^24) / (6370000m)
    =9.828 m/s^2
    I think I did this part of the question right, since the answer makes sense to me.

    b)Using the same equation as the first, and adding on 21 km to the radius, I got the answer of 9.764 m/s^2 Now, the question says to also use centripetal acceleration to answer it. I'm not sure how to combine that, so that is where I am stuck.

    c) I didn't want to attempt this until I get the right answer for the first two. I'm not sure what equation I use for this, though?
     
  2. jcsd
  3. Jan 8, 2009 #2

    LowlyPion

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    At the equator there is centrifugal acceleration outward that would be figured as ω²r that would subtract from your gravitational value.
     
  4. Jan 8, 2009 #3

    LowlyPion

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    The period for a pendulum is generally T = 2π(l/g)1/2
     
  5. Jan 9, 2009 #4
    Either I don't recall or the textbook didn't go over the ω²r equation. What exactly does the ω stand for?

    Thanks for the second equation, I will use it as soon as I get the first two questions done.
     
  6. Jan 9, 2009 #5

    LowlyPion

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    dθ/dt = ω = v/r

    v²/r = ω²r

    ω = 2πf
     
  7. Jan 9, 2009 #6
    I'm sorry (I must be dumb) I still don't understand.
     
  8. Jan 9, 2009 #7

    LowlyPion

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    An object at the equator is rotating at the radius of the earth at that point.

    However you want to calculate the acceleration - an acceleration that acts opposite to the gravitational acceleration at that point - then choose one method or another.
     
  9. Jan 9, 2009 #8
    For all practical purposes, the Earth can be thought of as simply "spherical". An increase in radius at the equator of 21 km is a mere deviation from its spherical radius of less than .33%, which yields deviations in calculations that are commonly negligible.
     
  10. Jan 9, 2009 #9

    Redbelly98

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    Another way to calculate this is from the v2/r equation to figure out how much the acceleration is reduced due to the centripetal acceleration. You'll need to figure out what v is at the equator first.
     
  11. Feb 12, 2009 #10
    Hes just confusing you, ω is the angular velocity ω=[tex]\Delta[/tex][tex]\theta[/tex]/[tex]\Delta[/tex]T which can be used to calculate the velocity v=rω, or in this case v²/r = ω²r which simplifies down to ac= ω²r (ac=angular acceleration)

    V=2[tex]\pi[/tex]r/T
    [tex]\Rightarrow[/tex] ac=((2[tex]\pi[/tex]r/T)^2)/r
    [tex]\Rightarrow[/tex] (4[tex]\pi[/tex]^2r^2)/T^2r
    [tex]\Rightarrow[/tex] since ω=[tex]\Delta[/tex][tex]\theta[/tex]/[tex]\Delta[/tex]T
    then ac=ω²r
     
    Last edited: Feb 12, 2009
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