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Using ratios to obtain the weight of a person at a specific distance

  1. Apr 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Assuming the person weighs 980 N on the surface of the Earth, how would you use ratios to obtain the weight of a person at 128 000 km above the surface of the Earth.


    2. Relevant equations

    Fg is proportional to 1/d2


    3. The attempt at a solution

    The answer in the key says 0.072 N. I didn't know exactly where to start but by working backwards from the answer I have determined that it is 20 times the distance from the centre of Earth. I know how to solve these questions using ratios if I am given the times the distance from the centre of Earth.

    For example for a previous question: Three times the distance from the centre of Earth, here is how I solve it.

    Fg is proportional to 1/d^2

    1x3 = 3 therefore 980 = 1/3^2

    980 / 9 = 109 N

    This method works everytime when I am given the times the distance from the centre of Earth. But when I am given a specific distance I do not know how to get the answer. Can someone please help? Thanks!
     
  2. jcsd
  3. Apr 1, 2013 #2
    Since [itex]F\propto\frac{1}{d^2}[/itex], You can write [itex]F = k\frac{1}{d^2}[/itex] where k is the constant of proportionality.

    So if [itex]F_1\propto\frac{1}{d_1^2}[/itex] and [itex]F_2\propto\frac{1}{d_2^2}[/itex]

    then [itex]\frac{F_1}{F_2}=\frac{\frac{1}{d_1^2}}{\frac{1}{d_2^2}}[/itex].

    The constants of proportionality cancel.
     
  4. Apr 1, 2013 #3
    The radial distance from the center of the earth to the surface is ??? What is the ratio of 128000 km to this distance?
     
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