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Magnitude of acceleration due to gravity

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Satellites are placed in a circular orbit that is 1.20 × 106 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?



    2. Relevant equations
    My book is saying that the relevant equations are:

    W = G((ME m)/r2)

    W = mg


    3. The attempt at a solution
    After looking at this post, I tried the formula g = MG/r^2. (M being the mass of the Earth)

    g = ((6.67 x 10^-11)(5.97 x 10^24))/((1.20 x 10^6)^2)

    This^ yielded 276.53 m/s^2, which is apparently wrong.
     
  2. jcsd
  3. Mar 3, 2014 #2
    g' = (Re^2/r^2)*g

    r = Re + h where Re is the radius of the earth and h is the height above it.

    g = acceleration due to gravity on the earth

    substitute the values given in the equation in that formula and you should get the answer for g'
     
  4. Mar 3, 2014 #3

    DataGG

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    Gold Member

    As Ronaldo said, ##r## is the distance of the object to the CM of earth. That means, as Ronaldo stated: $$r=1,20*10^6+r_{earth}$$
     
  5. Mar 3, 2014 #4
    It looks like you guys are correct, because I got the answer correct and the book says the following...: http://twitpic.com/dx955g

    Thank you!!
     
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