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Law of Gravitation problem

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data
    You weigh 518 N at sidewalk level outside the World Trade Center in New York City. Suppose that you ride from this level to the top of one of its 410 m towers. Ignoring Earth's rotation, how much less would you weigh there (because you are slightly farther from the center of Earth)?


    2. Relevant equations
    F = (GMm)/r2
    Mass of Earth: 5.97x1024 kg
    Radius of Earth: 6378100 m


    3. The attempt at a solution
    518 = ((6.67x10-11)(5.97x1024)(m))/(63781002)
    m = 52.92 kg

    F = (GMm)/(r+h)2
    F' = ((6.67x10-11)(5.97x1024)(52.80))/(63785102)
    F' = 517.94 N

    So my calculated difference is .0575 N, but apparently that's not right. Not sure where I've gone wrong here, any help is much appreciated.
     
    Last edited: Apr 2, 2013
  2. jcsd
  3. Apr 2, 2013 #2

    SammyS

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    Look at [itex]\displaystyle \ \frac{1}{r^2}-\frac{1}{(r+h)^2}
    =\frac{r^2+2rh+h^2-r^2}{r^2(r+h^2)}=\frac{2rh+h^2}{r^2(r+h^2)}\approx\frac{2h}{r^3}\ .[/itex]
     
  4. Apr 2, 2013 #3

    PeterO

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    Gravity is an inverse square relation, so I would add 410 to the Earth radius you had to find the ratio of increase [I get that r is now 1.000064 times the original. Square that 1.000129
    so the inverse says the weight force is reduced by that factor.

    518 / 1.000129 gave me about 0.0666 N less than 518, rather than your .0575 answer.
     
  5. Apr 3, 2013 #4
    Thank you both very much, got the right answer and I actually understand how I got there!
     
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