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Gravitational field strength (ratios)

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data

    A 6.2*10^2-kg satellite above Earth's surface experiences a gravitational field strength of magnitude 4.5-N/kg. Knowing the gravitational field strength at Earth's surface and Earth's radius, how far above Earth's surface is the satellite? (Use ratio and proportion.)


    2. Relevant equations

    Fg = (GMm)/r2
    g = GM/r2
    gE = 9.8-N/Kg
    rE = 6.37*10^6-m


    3. The attempt at a solution

    gs/gE = rs2/rE2
    rs= √((rE2gs)/gE)

    and my answer comes out to be 4.3*10^6-m. But the answer at the back of the book is 3.0*10^6-m, what did I do wrong? Thanks
     
  2. jcsd
  3. Feb 24, 2013 #2

    cepheid

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    The field strength is INVERSELY proportional to the square of the radius. So it should be

    gs/gE = rE2/rs2

    which is the reciprocal of what you had on the right hand side.

    Think about it: for gs ,the rs is on the bottom, and for 1/gE, the rE is on top. It also makes intuitive sense. We know gs is smaller than gE, so the ratio should have the smaller distance on top.
     
  4. Feb 24, 2013 #3

    cepheid

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    Don't forget, also that rs is the distance between the satellite and the centre of its orbit, which is not quite what the problem is asking for.
     
  5. Feb 24, 2013 #4
    ah that make sense, thanks for everything!
     
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