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Gravitational Interactions » Truck and Car

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data

    i need help with this problem (Online HW)

    If the force of attraction between a 1035 kg sports car and a 10680 kg cement truck is 3.930×10-6 N, how far are they apart?

    2. Relevant equations

    F = Gm_1m_2/r^2



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 23, 2011 #2

    Doc Al

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    Staff: Mentor

    Where are you stuck? Rearrange that formula to solve for the distance r.
     
  4. Nov 23, 2011 #3
    r^2 = Gm_1m_2/F

    We know m_1 = 1035, m_2 = 10680, F = 3.930 x 10^-6 and G is 6.67300 × 10^-11 m^3 kg^-1 s^-2.

    So

    r^2 = (6.67300 × 10^-11)(1035)(10680) / (3.930 x 10^-6)

    = [(6.67300 )(1035)(10680) / (3.930)] x [10^-11/10^-6]
    = [(6.67300 )(1035)(10680) / (3.930)] x 10^-5
    = 18768958.6 x 10^-5
    = 1876895.86 x 10^-6
    = 1.87689586 x 10^-12

    so how we find the distance?
     
  5. Nov 23, 2011 #4

    Doc Al

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    Staff: Mentor

    OK.
    You are messing up the exponents. When you move the decimal to the left one place you must add +1 to the exponent. You subtracted instead. (Example: 456.7 = 4.567 x 10^2)
    Once you correctly find r^2, just take the square root.
     
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