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99.9999% sure the book is wrong about this circumference problem

  1. Nov 30, 2012 #1

    ISX

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    1. The problem statement, all variables and given/known data
    The circumference of a ball is 29.6 inches. Given that the radius of Earth is about 6400 km, how many balls would it take to circle around the equator with the balls touching one another?


    2. Relevant equations
    (d)(pi) = circumference
    1 inch = 0.0254 meters
    1 km = 1000 meters

    3. The attempt at a solution
    Ball diameter = ((29.6)(0.0254))/3.14 = 0.239 m
    Earth circumference = (6400)(2)(3.14)(1000) = 40,192,000 m

    # of balls circumnavigating the equator = 40,192,000/0.239 = 168,167,364

    In the back of the book it says 26,700,000 balls (it only uses 3 significant digits) and I can get that if I divide the radius of the earth by the ball diameter: ((6400)(1000))/0.239 = 26,700,000

    I don't see why they are using the radius when it says around the earth, not halfway through it. Am I right here? Didn't know if there was some sort of magic I was missing.
     
  2. jcsd
  3. Nov 30, 2012 #2

    Ibix

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    I agree with you, although you should lose a few significant figures.
     
  4. Nov 30, 2012 #3

    berkeman

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    Staff: Mentor

    It does look like the book's answer is wrong. But I believe that your answer is slightly wrong as well. Can you think of a small error term that you did not take into account in your calculation?
     
  5. Nov 30, 2012 #4

    ISX

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    Yes the ball radius makes the earth's diameter bigger since the ball isnt flat, I didn't think that would hardly make a difference when they always round it giving it a huge tolerance of error.
     
  6. Nov 30, 2012 #5

    Dick

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    I'll agree with that. There's other small approximation errors as well. Like the amount of circumference a ball occupies isn't exactly the same as the diameter of the ball. The 'exact' way to do it would be to find the angle subtended by a ball as viewed from the center of the earth and divide it into 2*pi. Just for fun I did the number the ISX way and then did it that way and looked at the difference. It's pretty close to pi balls. Which you can account for with that radius correction. So you don't even need a 'huge tolerance of error'. The given answer would be right to within a few balls if a more accurate version of pi were used, assuming all of the other numbers were exact.
     
    Last edited: Nov 30, 2012
  7. Nov 30, 2012 #6

    LCKurtz

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    Using your given numbers I get 26,742,651 so I think your text is correct.
     
  8. Nov 30, 2012 #7

    Dick

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    Can you show your work? ISX is 99.9999% sure. Ibix agreed. Berkeman agreed. I did it two different ways and I agree. I'm really interested.
     
    Last edited: Nov 30, 2012
  9. Nov 30, 2012 #8

    ISX

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    Care to explain how?
     
  10. Nov 30, 2012 #9

    LCKurtz

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    Now you guys have me worried -- hold on, I'll check...
     
  11. Nov 30, 2012 #10

    LCKurtz

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    Yup. I must have made the same mistake the book did. I used 29.6 as the radius instead of the circumference. :redface:
     
  12. Nov 30, 2012 #11

    ISX

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    Had me worried there lol.
     
  13. Nov 30, 2012 #12

    Dick

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    I was just thinking you had responded to the initial post without reading anything of what the responses were further down the line. That can be kind of annoying.
     
    Last edited: Nov 30, 2012
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