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Period of pendulum moved to Jupiter's moon Io

  1. Dec 3, 2016 #1
    1. The problem statement, all variables and given/known data

    you are taking your pendulum clock with you to a visit of the jupiter moon Io(radious 3643.2Km, mass 8.94X10^22 kg. calculate the duration of a full Oscillation. On the surface this oscillation time was 1s
    2. Relevant equations
    T=2*π√l/g



    3. The attempt at a solution
    T1/T2=√(g2/g1)
     
    Last edited: Dec 3, 2016
  2. jcsd
  3. Dec 3, 2016 #2

    haruspex

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    What are D1 and D2, and how do you get the second equation from the first?
     
  4. Dec 3, 2016 #3
    D means g . If you divideT1/T2=2*π√l1/g / 2*π√l2/g. the you get the equation
     
    Last edited: Dec 3, 2016
  5. Dec 3, 2016 #4

    gneill

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    Moderator's note: I've changed the title of this thread to be more specific and descriptive of the actual problem.
     
  6. Dec 3, 2016 #5

    NascentOxygen

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    What is l (lower-case L)? What is g?
     
  7. Dec 3, 2016 #6
    L is the length of the pendulum and g is the Gravitational costant
     
  8. Dec 3, 2016 #7

    NascentOxygen

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    It's one pendulum clock that gets transported to Io, so how can there be two different pendulum lengths?

    Usually G is the gravitational constant symbol. What numerical value are you using for lower-case g here?
     
  9. Dec 3, 2016 #8
    yes, G differs from Earth to Jupiter's moon Io. For earth, it is 9.81m/s^2... and Io moon is not mentioned. so we need to use gravitational formula to solve it
     
  10. Dec 3, 2016 #9

    NascentOxygen

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    I have clarified the problem statement here, to present it as I believe would have been intended.
     
  11. Dec 3, 2016 #10
    i mean that.
     
  12. Dec 3, 2016 #11

    NascentOxygen

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    So you have a final equation for TIo that involves the mass of Io?
     
  13. Dec 3, 2016 #12
    No, I do not have the equation. I am confused.
     
  14. Dec 3, 2016 #13

    NascentOxygen

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    Where's the confusion?
     
  15. Dec 3, 2016 #14
    3. The attempt at a solution
    T1/T2=√(g2/g1)

    How I can go ahead?
     
  16. Dec 3, 2016 #15

    NascentOxygen

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    You'll need to show the working you followed in deriving that equation, by starting with something that you know to be right.
     
  17. Dec 3, 2016 #16
    If you divideT1/T2=2*π√l/g1 / 2*π√l/g2. the you get the equation l= radious of the Pendulum and g is the gravitational constant.
     
  18. Dec 3, 2016 #17

    NascentOxygen

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    Okay, so you need formulae or equations for g1 and g2 . What formula can you use for this?
     
  19. Dec 3, 2016 #18
    g=GM/r^2
     
  20. Dec 3, 2016 #19

    NascentOxygen

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    Go ahead and see whether you can now finish this.
     
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