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What happens to the gravitational field strength's magnitude if

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data
    What happens to the gravitational field strength's magnitude if
    a) r decreases by a factor of 4?
    b) r increases by factor of 2

    2. Relevant equations
    i'm not sure which equation they are referring to? it may be g ∝ 1/r^2


    3. The attempt at a solution
    I tried to do this:
    1/ (r-4)^2 and i got 1/(r^2 - 8r + 16)
    the answer at the back of the book for a) says gravitational field strength's magnitude is supposed to become 16 times greater and for b) it's supposed to become 1/4 as great
    please help
     
  2. jcsd
  3. Apr 2, 2013 #2

    PeterO

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    The radius has not decreased by 4, but by a factor of 4.

    So r → r/4 (rather than r → r-4)
     
  4. Apr 2, 2013 #3
    hey thanks, but is it g ∝ 1/(r/4)^2?
     
  5. Apr 2, 2013 #4
    Fg=Gmm/r^2 so if you bring the masses 4 times closer and Gmm remain the same, then Fg' = Gmm/(r/4)^2. It follows that Fg'=Gmm/(r^2/4^2) which is Gmm/r^2/16, thus 16 (Gmm/r^2). Since Gmm/r^2is the original force, the new one is 16 times greater.

    Likewise you can show that increasing r to 2r will make (2r)^2 = 4r^2, so the force will be 4 times smaller.
     
  6. Apr 2, 2013 #5

    PeterO

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    Expand that expression and what do you get? Compare it to the original 1/r2
     
  7. Apr 2, 2013 #6

    PeterO

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    Also a simple approach is: this is an example if an "inverse square law" [some others are intensity of light, intensity of sound, electrical attraction between charges ...]

    The inverse part tells you the change is opposite: reduce the separation - increase the force.
    The square part tells you the size of change. change "r" by a factor of 4 → a "g" change of 42 (which is 16).
     
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