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Velocity of the Moon (magnitude and direction)

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data

    1. The problem statement, all variables and given/known data

    The velocity of the Moon relative to the center of the Earth can be approximated by varrowbold(t) = v [−sin (ωt) xhatbold + cos (ωt) yhatbold], where v = 945 m/s and ω = 2.46 multiplied by 10−6 radians/s. (The time required for the Moon to complete one orbit is 29.5 days.) To approximate the instantaneous acceleration of the Moon at t = 0, calculate the magnitude and direction of the average acceleration during the following two time intervals.

    (a) between t = 0 and t = 0.200 days
    ______ m/s
    ______ degrees (counterclockwise from the +x axis)

    (b) between t = 0 and t = 0.0020 days
    _____ m/s
    _____ degrees (counterclockwise from the +x axis)


    3. The attempt at a solution

    When I try to solve for Vx and Vy I get the same result for each one, yielding zero, and if I try another method, i get very minute numbers. None of which appear to be correct. I'm at a loss for how to do this problem
     
  2. jcsd
  3. Jan 19, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi corykowa! Welcome to PF! :smile:

    (on this forum, just use the bold (B) tag for vectors: v = v[−sin (ωt)i + cos (ωt)j] :wink:)
    What formula did you use for average acceleration?

    Show us what you tried, and then we'll know how to help! :smile:
     
  4. Jan 19, 2010 #3
    Hi thanks for the help. I was thinking since the moon is moving around us, that it would be a centripetal acceleration problem? But maybe that's where I was going wrong. So I was using that formula in relation to vectors...

    a = -lim (v/2*theta)2vsin(theta)ey as lim of theta approaches zero = (-v^2/R)ey
     
  5. Jan 20, 2010 #4

    tiny-tim

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    Hi corykowa! :smile:

    (just got up :zzz: …)
    (have a theta: θ and try using the X2 tag just above the Reply box :wink:)

    No, you're misreading the question.

    It's only asking you for the average acceleration.

    (a), over 0.2 days, will give you a figure close to the instantaneous acceleration, and (b), over 0.002 days, will give you a figure even closer.

    The formula you need is average acceleration = (v(t2) - v(t1))/(t2 - t1) :wink:
     
  6. Jan 20, 2010 #5
    Oh thank you much! I just tried that equation and my response "differs from the correct answer by orders of magnitude". I'm not too sure what that means... I feel like i must be off by 10^x in my answer but not too sure how to go about figuring that out?? I got 945 when t=2 and 1.89 when t=.002. Also, I still don't understand how to solve for the direction (in relation to the +x axis). I figure it has something to do with arctan?? Thank you!
     
    Last edited: Jan 20, 2010
  7. Jan 21, 2010 #6

    tiny-tim

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    Hi corykowa! :smile:

    (just got up :zzz: …)
    (try using the X2 tag just above the Reply box :wink:)

    Did you remember to convert the days into seconds? :wink:

    Show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:

    (and y/x is the tangent of the angle from the x-azis)
     
  8. Jan 21, 2010 #7
    I had completely spaced converting days into seconds. However, what I ended up doing was:

    differentiate v(t) = v( -sin(wt) + cos(wt)) which is
    -v(w) (cos(wt) + sin(wt)) you have the acceleration
    v(w) = 2.32 x 10^-3 meter/sec^2
    At t = 0 days, the acceleration is - 2.32 x 10^-3 meters/sec^2

    This appears to be what my homework was asking me as webassign accepted this as correct.

    I'm still in the process of figuring out the direction part though. I'm trying: [tex]\theta[/tex] = artan Vy/Vx

    Thank you for your help!
     
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