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Magnitude of displacement vector

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data
    The Earth moves around the Sun in a circle of radius 1.50 X 10^11 m at approximately constant speed. Taking today's position of the Earth as origin, draw a diagram showing the position vector 3 months, 6 months, 9 months, and 12 months later. Draw the displacement vector between the 0 month and 3 month positions and calculate the magnitude of the displacement vector for this 3-month interval.


    2. Relevant equations
    x= rcos[tex]\phi[/tex]
    y= rsin[tex]\phi[/tex]
    x^2 + y^2 = r^2



    3. The attempt at a solution
    my friend told me that the diagram is supposed to look like the sine function and that the amplitude= 1.5 X 10^11 m, but I still don't understand how to calculate the magnitude of the displacement vector.
     
  2. jcsd
  3. Sep 22, 2009 #2

    kuruman

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    Your friend told you wrong. The Earth goes around the Sun much like the tip of a clock hand goes around the face of the clock. If "today" the tip is at 12 o' clock (0 month position), it will be at the 3 o' clock position 3 months later, the 6 o'clock position 6 months later and so on. Can you draw the displacement vectors and calculate their magnitudes now?
     
  4. Sep 22, 2009 #3
    So do I just take the square root of ((1.50 X 10^11)^2 + (1.50 X 10^11)^2) to get the magnitude of the displacement vector?
     
  5. Sep 23, 2009 #4

    kuruman

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    For 3 months, yes. What about 6 months?
     
  6. Sep 23, 2009 #5
    Between the 0 month and 6 month positions would I take (1.50 X 10^11) + (1.50 X 10^11)?
     
  7. Sep 23, 2009 #6

    kuruman

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