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What is the gravitational force on the satellite?

  1. Jul 27, 2005 #1
    A satellite of mass 1900 kg used in a cellular telephone network is in a circular orbit at a height of 690 km above the surface of the earth.

    What is the gravitational force on the satellite?
    Take the gravitational constant to be G = 6.67×10−11 N*m^2/kg^2, the mass of the Earth to be m_e = 5.97×1024 kg, and the radius of the Earth to be r_e = 6.38×106 m.


    my answer was 1.51*10^4 {\rm N} which was correct, but im confused on the next question: What fraction is this of its weight at the surface of the earth?
    Take the free fall acceleration to be g = 9.80 m/s^2.
     
  2. jcsd
  3. Jul 27, 2005 #2

    mukundpa

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    The value of g changes with height. 690 Km is comparable to the radius of the Earth. Have you considerd this.

    Hint g' = g[Re/(Re+h)]
     
  4. Jul 28, 2005 #3
    so for that equation doi use the answer times the mass of the earth to find the fraction, theyre talking about?
     
  5. Jul 28, 2005 #4

    mukundpa

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    yes
    weight of the settelite is mg,
    hence ratio of the weights will be mg'/mg=g'/g
    where g' is accelerationdue to gravity at height h and g is that at the surface of earth.
     
  6. Jul 28, 2005 #5

    HallsofIvy

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    You needed the mass of the earth to find the gravitational force on the satellite in orbit but you don't to find the gravitational force on the satellite on earth (its weight). Just use F= mg where g= 9.81 m/s2 and m= 1900 kg, as given.

    Divide the two to find the fraction.
     
  7. Jul 29, 2005 #6
    i did what you guys told me but i keep getting wrong. i tried what mukundpa and halls suggested but i ended up wrong.
     
  8. Jul 29, 2005 #7
    wouldn't the second part of the question just be [tex]\frac{r^2_{orbit}}{r^2_{earth}}[/tex]?
     
  9. Jul 29, 2005 #8
    nvm i already solved it. i used the number i got as the wrong numerator
     
  10. Jul 29, 2005 #9

    mukundpa

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    I am extremely sorry for mistype and not looking back seriously
    Actually the derivation is like that

    at the surface of earth acceleration due to gravity is

    g = GM/Re^2 Re is radius of earth

    and at a height h from the surface
    g' = GM/(Re+h)^2

    Hence g' = g [Re/(Re+h)]^2

    Therefore the required fraction is mg'/mg = [Re/(Re+h)]^2
    =[6380/(6380+690)]^2 = 0.814
    is it correct
    sorry again
     
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