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Angular Displacement

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Earth's orbit around the sun is nearly circular. The period is 1 yr = 365.25 days. In an elapsed time of 2.0 days, what is earth's angular displacement?

    2. Relevant equations

    Theta=Theta final - Theta inital

    3. The attempt at a solution

    I'm very confused and need help starting it!
     
  2. jcsd
  3. Sep 26, 2009 #2
    It stands to reason that if Earth travels 365 degrees (2Pi radians) in 365.25 days, then Earth travels ___ degrees in 2 days. You should be able to set up an algebraic equation for this one.
     
  4. Sep 26, 2009 #3

    rock.freak667

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    if period is the time taken for one revolution and it is an approximate circle. Then in 1 time period (365.25 days) how many radians does it travel?
     
  5. Sep 26, 2009 #4
    2 pi i think
     
  6. Sep 26, 2009 #5

    rock.freak667

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    Right right good good...


    so if in 365.25 days it rotates 2pi radians

    in 1 day how much will it rotate?
     
  7. Sep 26, 2009 #6
    (2 * pi radians) / 365.25 = 0.0172024238
     
  8. Sep 26, 2009 #7

    rock.freak667

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    good good

    so in 1 day it rotates 2π/365.25 radians.

    So in 2 days how much does it rotate?
     
  9. Sep 26, 2009 #8
    0.0344048476 radians
     
  10. Sep 26, 2009 #9

    rock.freak667

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    so in 2 days it rotates 0.0344048476 radians, isn't this what the question asked for ? (you can convert it degrees depending on what the question wants)
     
  11. Sep 26, 2009 #10
    YES! thank you so much for your help :)
     
  12. Sep 26, 2009 #11
    Is there anyway you can also help me find out the change in the earth's velocity? i know that velocity is change in distance over time. but it keeps telling me that i can' tput my answer in radians/seconds so i dontk now how to do it
     
  13. Sep 26, 2009 #12

    rock.freak667

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    Well you know 2 days it rotates 0.0344048476 radians

    the angular velocity ω is defined as d/dt or

    [tex]\omega = \frac{\theta_2 - \theta_1}{t} = \frac{change \ in \ angular \ displacement}{time}[/tex]


    and you have the change in angular displacement is 0.0344048476 radians.

    So in 2 days what is ω ?
     
  14. Sep 26, 2009 #13
    .0172 radians/ days.. but its not asking for angular velocity so it doesn't want my answers in radians/ days or hours or seconds
     
  15. Sep 26, 2009 #14

    rock.freak667

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    Yes but we need ω to get v. Convert ω to radians/second.

    Now what is the relationship between v, ω and r ? (r is the distance from the center of rotation -> the sun in this case)
     
  16. Sep 26, 2009 #15
    i converted w to radians/ second and got. 1.99 x 10^-7. i have no idea what the relationship is
     
  17. Sep 26, 2009 #16

    rock.freak667

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    So you know the equation v=rω?
     
  18. Sep 26, 2009 #17
    OHH so i would just do v= 6378 km ( 1.99 x 10^-7) and that would give me my answer in km/second?
     
  19. Sep 26, 2009 #18

    rock.freak667

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    Well I don't know the distance between the sun and the Earth, but that is what you would do.

    Also if that is wrong, try using r=radius of sun + distance between the sun and the earth + radius of the earth. Not sure they meant to use the sun and earth as point masses or how they normally are.
     
  20. Sep 26, 2009 #19
    OHH so i just do v= 6378 km ( 1.99 x 10 ^-7) and my answer comes out in km/s?
     
  21. Sep 26, 2009 #20

    rock.freak667

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    uhm yes basically.
     
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