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Gravitational Forces

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data
    The moon is 3.9 × 105 km from Earth's center and 1.5 × 108 km from the sun's center. If the masses of the moon, Earth, and sun are 7.3 × 1022 kg, 6.0 × 1024 kg, and 2.0 × 1030 kg, respectively, find the ratio of the gravitational forces exerted by Earth and the sun on the moon. (Use G = 6.670 × 10-11 Nm2/kg2.)


    2. Relevant equations
    F=G(m1m2)/d²
    Where F is the force, G is gravity constant, m1 and m2 are masses, and d is distance.


    3. The attempt at a solution
    F=G(mEarthmMoon)/d²
    F=6.67×10-11((6×1024)(7.3×1022))/3.9×108²
    F=192074950690335305719.92110453649
    F=1.9e+20N

    F=G(mSunmMoon)/d²
    F=6.67×10-11((2×1030)(7.3×1022))/1.5×1011²
    F=432808888888888888888.88888888889
    F=4.3e+20N

    1.9e+20N / 4.3e+20 = 2.2
    and I tried its inverse, 0.44

    Neither of these are correct.
    I am probably missing something from the question.
    What is wrong about this and where?
     
    Last edited: Apr 1, 2009
  2. jcsd
  3. Apr 1, 2009 #2

    LowlyPion

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    First of all lose the precision. It only makes it difficult to see what's going on.

    Second of all since they want a ratio. Take it right upfront and recognize that you can express it more simply as:

    Fe/Fs = Me*ds2/Ms*de2

    Makes calculation a little easier.
     
  4. Apr 1, 2009 #3
    Me*ds2/Ms*de2
    (6×1024*1.5×1011²)/(2×1030*3.9×108²)
    0.44

    Imputing this as an answer results in an incorrect.

    I'm not really sure what the question is asking for, since what I did seems to be correct, as far as I can tell.
     
  5. Apr 1, 2009 #4

    LowlyPion

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    Earth to Sun Ratio = (6.0 x 1024 kg) * (1.5 × 108 km)2 / 2.0 × 1030 kg * (3.9 × 105 km)2 = .4438

    Sun to Earth ratio = 2.253

    The numbers look right.
     
  6. Apr 2, 2009 #5

    tiny-tim

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    Hi 312213! :wink:

    Well, that method should work …

    but why are you putting G and mmoon into the figures?

    This is a dimensions problem, and all you need is the ratios.

    Try again! :smile:
     
  7. Apr 2, 2009 #6
    Well this was just the first way I took it. As ratios, the numbers are the same.

    Fe/Fs = Me*ds2/Ms*de2

    Me*ds2/Ms*de2
    (6×1024*1.5×1011²)/(2×1030*3.9×108²)
    0.44

    Still 0.44.
    I think the question is looking at something else. Or wrong altogether.
     
  8. Apr 2, 2009 #7

    berkeman

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    Staff: Mentor

    Merged two duplicate threads into this one...
     
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