Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie

Ring around the world

Options
  • 17-06-2012 9:20pm
    #1
    charles_92688
    Registered Users Posts: 52 ✭✭


    Assume the earth is a perfect sphere of 25000 miles.
    You have a string (of infinitesimal thickness) encircling the earth. The length of the string is three feet longer than the circumference of the earth or in other words it is 25000 miles + 3 feet long.

    Without using a calculator, how high above the ground is the string above the ground, equidistant, with the shape of the string a perfect circle around the earth?

    Part B: Assume instead of the earth circumference being 25000 miles, it is 25000^25000^25000 light years in circumference and you have a string that is three feet longer. How high is the string above the ground equidistantly?

    Part C: Assume the circumference is one inch (about like a marble maybe), and you have a 37 inch string, how high above the surface of the "marble" is the string?


Comments

  • Options
    #2
    3DataModem
    Registered Users Posts: 3,973 ✭✭✭3DataModem


    pi over 6 feet... Or about 7 inches in both cases.

    Or is it a trick question?


  • Options
    #3
    Eathrin
    Registered Users Posts: 8,076 ✭✭✭Eathrin


    Is it 3/2pi feet?


  • Options
    #4
    charles_92688
    Registered Users Posts: 52 ✭✭charles_92688


    3DataModem wrote: »
    pi over 6 feet... Or about 7 inches in both cases.

    Or is it a trick question?

    Yes, (about 5.8 inches) but still that is mind boggling. That delta r is independent of C. No matter how big the sphere is, an extra three feet of string requires about 6 inches of delta radius. Mind boggling.


  • Options
    #5
    charles_92688
    Registered Users Posts: 52 ✭✭charles_92688


    Eathrin wrote: »
    Is it 3/2pi feet?

    Yep. I find this very interesting. It is independent of C.


  • Options
    #6
    Eathrin
    Registered Users Posts: 8,076 ✭✭✭Eathrin


    Yeah I got the same for B, very interesting.


  • Advertisement
  • Options
    #7
    Eathrin
    Registered Users Posts: 8,076 ✭✭✭Eathrin


    For C I have 18/pi inches


  • Options
    #8
    charles_92688
    Registered Users Posts: 52 ✭✭charles_92688


    Eathrin wrote: »
    For C I have 18/pi inches

    You mean for delta R you get 18/pi, right? (Not C, we gave you C.)


  • Options
    #9
    charles_92688
    Registered Users Posts: 52 ✭✭charles_92688


    Eathrin wrote: »
    Yeah I got the same for B, very interesting.

    It is indeed interesting. I asked this (25000 mile) to a math prof once and he replied off the cuff, "An infinistesimal delta r" which "seems" right but no matter what C is, the delta r is always about 5.8 inches.

    Simply, 2 * pi* r = C
    2 * pi * dr = dC

    Our dC is always 36 so our dr is always 36/(2*pi) ~ 5.8 inches.


  • Options
    #10
    Eathrin
    Registered Users Posts: 8,076 ✭✭✭Eathrin


    I meant part C, I didn't realise my answers were the same, doh!
    I just did it using simple algebra, rather than calculus.


Advertisement