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Altitude for Specific Weight

  1. Nov 10, 2008 #1
    How far above the surface of Earth would you weigh 69.7 percent of your surface weight?

    This is the work I've done but something isn't right and I'm not getting the right answer:

    69.7(9.8)=6.67x10^-11(5.98x10^24)/r^2 = 683.06 = 3.98x10^14/r^2

    r= 763329.61m (Incorrect)

    someone please help me??
     
  2. jcsd
  3. Nov 10, 2008 #2
    [tex]\vec{w}_0=m\vec{g}[/tex] and [tex]\vec{w}=0.697m\vec{g}[/tex]. Mass remains constant so what's your [tex]\Delta\vec{g}[/tex] and how does this relate with your system?
     
  4. Nov 10, 2008 #3
    Not really sure what g you're referring to, but I can give you the Gravitational Constant and the mass of earth:

    Gravitational Constant= 6.67x10-11
    Mass of Earth= 5.98x1024
     
  5. Nov 11, 2008 #4
    There you go. And the radius of the Earth; that'll give you your [tex]\vec{g}[/tex] at positions near the surface of the Earth. You'll want to compare this with those radii which make your weight--your mass times the gravitational force acting upon you--0.697 what it is at the surface.
     
  6. Nov 11, 2008 #5
    First of all, you multlplied g (=9.81 m/s^2) by 69.7: but what you need is the radius where the acceleration is 0.697 times g.

    Second, once you have found r, you have to subtract the radius of Earth: the question asks "how far from the surface of Earth".
     
  7. Nov 11, 2008 #6
    And if I may add a third hint: you don't have to know G or M (mass of Earth) to answer this question: those quantities drop out. The ratio of weights at different heights only depends on the ratio of r's.
     
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