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Question of latitude of Earth and measured field strength

  1. Dec 26, 2016 #1
    1. The problem statement, all variables and given/known data
    My book said:
    As the Earth is rotating about its axis, objects on the earth's surface are performing circular motion. Therefore, objects on the earth are accelerating and they need a centripetal force. The weight contributes a small portion to the centripetal force needed. Therefore, the measured g is smaller. The decrease in g is largest at the equator. At higher latitude, the radius of the circle is smaller while w is the same. Therefore, the centripetal force needed is smaller. The measured g is larger at higher latitude.

    2. Relevant equations


    3. The attempt at a solution
    However from my calculation, let the latitude be θ, W=weight, N=normal force, R=radius of the earth
    Wcosθ - Ncosθ = mw^2 Rcosθ
    W-N=mw^2(R)..........(1)
    Wsinθ=Nsinθ
    W=N......(2)
    Which is totally the same as considering an object on the equator.
    Why am I wrong, please help
     
  2. jcsd
  3. Dec 26, 2016 #2

    cnh1995

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    https://www.physicsforums.com/posts/5643868/
    This recent thread might help.
     
  4. Dec 26, 2016 #3

    CWatters

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    That last sentence is badly worded in my opinion.

    Gravity provides ALL of the centripetal force needed and a lot more. Your apparent weight is the force left over after some has been used to provide the needed centripetal force.
     
  5. Dec 26, 2016 #4
    I know it. The book actually means that a small portion of weight is used to contribute all the centripetal force. It is because the centripetal force is the net force by W - N. However by my calculations, I don't think the apparent weight ( Normal force ) will decrease when the latitude change (except for the poles) .
     
  6. Dec 26, 2016 #5

    haruspex

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    W and N cannot be in the same straight line.

    By the way, there is another reason N is greater at the poles than at the equator.
     
  7. Dec 26, 2016 #6
    You have applied a cos theta on every term. Why? Have you drawn a picture and figured out which forces point which way and how they vary with latitude?
     
  8. Dec 26, 2016 #7
    Q.png
     
  9. Dec 26, 2016 #8

    haruspex

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    Since the body is static, the centripetal force is the resultant of all applied forces. If W and N are in the same straight line, how can their resultant lie in a plane of latitude? Which of these forces might not be as you have shown it? (Hint: is the Earth really spherical?)
     
  10. Dec 26, 2016 #9
  11. Dec 26, 2016 #10

    haruspex

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  12. Dec 26, 2016 #11
    Q.png
     
  13. Dec 26, 2016 #12

    haruspex

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  14. Dec 26, 2016 #13
    W cosa - Rcosb = mw^2 r.....(1)
    Wsina = Rsinb .....(2) where a<b
    So by (2),
    W/R = sin(b) / sin(a)
    I don't know how to analyze it, does sinb/sina increases when a increases?
     
  15. Dec 26, 2016 #14

    haruspex

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    I assume R is the same as N.
    Can you derive a single equation that does not mention angle b?
     
  16. Dec 26, 2016 #15
    That is a bit long....
    sinb = Wsina / R
    cosb = root (1 - (Wsina / R)^2)
    Put this into (1)
    Wcosa - R root (1 - (Wsina / R)^2) =mw^2(r)
    Wcosa - √[ R^2 - W^2 sin^2(a) ] = mw^2(r)
    However R and r are still variables?..
     
  17. Dec 26, 2016 #16

    haruspex

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    OK, but R is what you are trying to find, so rearrange it as R=...
    You can determine r.

    It might be better to go back to using N instead of R so that you can use R for earth's radius.
     
  18. Dec 26, 2016 #17
    I get something like
    R^2 = W^2 - 2Wcosa(m w^2 r) + m^2 w^4 r^2
    when I did some research r is something like √[ A^2 cos^2(a) + B^2 sin^2(a) ], A should be larger than B in the sphere I drew, so can I say
    r = √[ (A^2-B^2) cos^2(a) + 1 ], where A^2 - B^2 is positive and r's value depends on cos^2 (a) ?

    However, the second term and the last term seems to be contradictory
     
  19. Dec 26, 2016 #18

    haruspex

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    For the purposes of estimating r in this question, you can treat the Earth as spherical, so just Earth radius cos (latitude).
    And yes, you can neglect the m2 term.
     
  20. Dec 26, 2016 #19
    R^2 = W^2 - 2Wcosa(m w^2 r) + m^2 w^4 r^2
    = W^2 - 2Wcosa(m w^2 R cosa) + m^2 w^4 R^2 cos^2 (a)
    = W^2 + cos^2 (a) m w^2 R ( m w^2 R - 2W )
    Therefore, R increases with latitude (a), thank you so much
     
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