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Homework Help: Weight in Terms of Distance From Earth

  1. Mar 3, 2008 #1

    bfr

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    [SOLVED] Weight in Terms of Distance From Earth

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

    At what height above the earth's surface will an astronaut have a true weight that is one half his weight on earth?

    2. Relevant equations

    F=G(m1*m2)/r^2
    (GM)/(4*pi^2)=(R^3)/t^2

    3. The attempt at a solution
    I did:
    F/2=(G*m1*m2)/r^2
    r=sqrt((2*G*m1*m2)/F)

    My teacher got sqrt(2) times the radius of the earth. Where do I go from there?
     
    Last edited: Mar 3, 2008
  2. jcsd
  3. Mar 3, 2008 #2

    tiny-tim

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    Hi bfr!

    You're confusing yourself by using the letter r for two different distances.

    Hint: call the earth's radius r, and call the radius you're looking for s.

    Then write two equations, one with r and one with s.

    Then … does that help? :smile:
     
  4. Mar 3, 2008 #3
    g=GM/R^2

    so g is directly proportional to 1/R^2
    g'/g = R^2/R'^2
    1/2=R^2/R'^2
    let R = 6400*1000 m
    R'^2 = 2R^2
    R' = sqrt2 R = 9050km
    atitude = 9050-6400 =2651km above the Earth's surface

    of coz , u may realize that
    GM =gRe^2 , where Re is the radius of the Earth
    then
    g'= gRe^2/R'^2
    which is the same.
     
  5. Mar 3, 2008 #4

    tiny-tim

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    Yes! :smile: But a bit long-winded.

    You could just say:

    F =G.m1.m2/r^2
    F´ =G.m1.m2/r´^2
    so F/F´= r´^2/r^2 = (r´/r)^2.

    Do you see the advantage now of being careful to use different letters?! :smile:

    [size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​
     
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