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Moon in Orbit

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose the moon were held in its orbit not by gravity but by the tension in a massless cable. You are given that the period of the moon's orbit is T = 27.3 days, the mean distance from the earth to the moon is R = 3.85 x 10^8 m, and the mass of the moon is M = 7.35 x 10^22 kg.

    What would the tension in the cable be?


    2. Relevant equations
    Centripetal Acceleration = v^2/R


    3. The attempt at a solution
    Tension = Ma
    a = v^2/R
    v = 2pi/T

    First I concerted the 27.3 days to seconds which is T = 2.358*10^6. Which means that v = 2.6638*10^-6. Then I plugged that in for a, so a = 1.843*10^-20. So then the Tension should be 7.35*10^22 * 1.843*10^-20 which is 1354.67N. But it is saying that is the wrong answer. What did I do wrong?
     
    Last edited: Sep 13, 2008
  2. jcsd
  3. Sep 13, 2008 #2

    hage567

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    Homework Helper

    Take another look at your equation for velocity. It isn't right.
     
  4. Sep 13, 2008 #3
    thanks.. i figured it out :)

    i calculated the angular velocity and got w = 2.66*10^-6 and then multiplied that by R to get the acceleration.
     
  5. Sep 13, 2008 #4

    hage567

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    No, I mean it doesn't make sense. Velocity is m/s right? Where is the "m" in your equation?! If the moon is moving in a circle, what distance does it go in one revolution?
     
  6. Sep 13, 2008 #5
    Velocity is the "directional displacement" covered over a period of time. Considering that the direction in this case is perpendicular to the radius vector at all points, you'd have to look a bit closer at v = 2*pi/T...is that really the "directional displacement" (distance in this case) over time?...you're missing a variable.
     
  7. Sep 13, 2008 #6

    tony873004

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    Since the tension is replacing gravity, it must be equal to gravitational force. So you could do F=GMm/d^2 for gravitational force.
     
  8. Sep 13, 2008 #7
    You technically could do this if the values they gave you are correct. But it might be best to stick simply to what you were doing earlier. You just need to take into account the right expression for the tangential velocity.
     
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