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Universal law of gravitation

  1. Jun 11, 2007 #1
    1. The problem statement, all variables and given/known data
    At what height above the earth's surface is the acceleration due to gravity 10% of that at sea level?

    2. Relevant equations
    F= Gm1m2/r(squared)


    3. The attempt at a solution
    I think some how one of the masses is moved over so that F/m=a and I think .10 is multiplied times something like (r squared) or (r + x)squared.
     
  2. jcsd
  3. Jun 11, 2007 #2

    G01

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    As you said, F/m=a. When at the surface of the Earth at sea level:

    [tex]\frac{F}{m}=\frac{GMm}{m(r+x)^2}=\frac{GM}{x^2}[/tex]

    where x= radius of Earth, and since at the sea level r would be zero.

    Now what is 10% of this? After you find this, can you set up an equation and solve for the distance from Earth? If you need more help feel free to ask. Good Luck!
     
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