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Gravational field 2

  1. Mar 26, 2007 #1
    Two isolated masses, M1 = 2.20 kg and M2 = 607 kg are initially rest, a distance d = 171 cm apart. Their gravitational attraction is the only force acting. Calculate the time it takes for M1 to move from that distance to 169 cm from M2. Assume that M2 does not move and that the force is constant over that small distance, and equal to that at 170 cm.

    i want to use F=G(m1m2)/r^2=m1a so Gm2/r^2=a

    6.67x10^-11*607/(1.71/2)^2=a

    then when i have a i am going to use x=xo+vot+.5at^2 to find t

    i am not getting the right answer please help me
     
  2. jcsd
  3. Mar 26, 2007 #2

    hage567

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    Actually, I think I will stick with my original advice. Like I said, you shouldn't be dividing the distance by 2. If the force is constant, I would think you take it at the 1.71 m.
     
    Last edited: Mar 26, 2007
  4. Mar 26, 2007 #3
    At a blush. I think the decision to halve the 171 is mistaken
     
  5. Mar 26, 2007 #4
    Obviously you are supposed to consider that the acceleration is constant, or else you would have to deal with some differential equations.
     
  6. Mar 26, 2007 #5
    well the problem tells you to use 170,
    so we have :
    f=6.67E-11*607*2.2/.170^2=?

    Divide by 2.2 to get a, then use kinematics

    PS oops thats meant to be 1.7
     
    Last edited: Mar 27, 2007
  7. Mar 26, 2007 #6
    ok i will try that
     
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