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Satellite Question

  1. Jan 17, 2005 #1
    I hate this question:

    I don't know if I'm doing something wrong because it seems so hard. Can i get someones way of doing this question...

    Communications satellites are placed in orbit so that they remain stationary relatice to a specific area on the Earth's surface. They are given the name synchornous satellites because, to maintain such a position, their period as they orbit must be the same as the Earth's.
    What is the height of such a satellite measured from
    a) the centre of the Earth, and
    b) the surface of the Earth
     
  2. jcsd
  3. Jan 17, 2005 #2
    If you hate it, then don't do it....
    read the sticky
     
  4. Jan 17, 2005 #3

    dextercioby

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    They are called geostationary satellites.

    Their trajectory can be a circle.Apply the 2nd law to the satellite in the inertial frame given by the center of the Earth.

    Daniel.
     
  5. Jan 17, 2005 #4
    I'm not given an acceleration though

    2nd law:?

    net f=ma

    ma=M(4pi^2r)/T^2

    is that it?
     
  6. Jan 17, 2005 #5

    dextercioby

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    Okay,now use the fact that the movement is not in a straight line,but on a circle...

    What is the RHS of your equation??

    Daniel.
     
  7. Jan 17, 2005 #6
    man you've got to help me here......Okay I'll tell you what i've done...

    i got (r+x) as my radius.....x being the distance from earth surface to satellite. therefore

    4pi^2(r+x)/T^2=GM/(r+x)^2

    I continued after that but this is probably all wrong please help......:)
     
  8. Jan 17, 2005 #7

    dextercioby

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    Perfect.Your equation is good.Now just plug in the numbers and tell me your result.Total distance (from the Earth's center).

    Daniel.
     
  9. Jan 17, 2005 #8
    LOL that can't be right dext. I mean I get cubes when i cross multiply and it's all messed up......I get stuck later i can't find x

    I got stuck at

    x=7.51(10)^22/(1.2E14+1.9E7x+x^2)

    THATS so wrong ain't it
     
  10. Jan 17, 2005 #9

    dextercioby

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    You needn't solve for "x".Your unknown is not necessarily "x".Think about it. :wink:

    Daniel.
     
  11. Jan 17, 2005 #10
    dexter i need to know r+x which is equal to height but in order to know that I need to solve for x right......then add x with r which i have already......

    I'm sorry but my iq is nothing compared to yours
     
  12. Jan 17, 2005 #11

    dextercioby

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    How about this equation??Can u solve it...?

    [tex] (R+x)^{3}=\frac{T_{rot}^{2}GM_{E}}{4\pi^{2}} [/tex]

    Daniel.

    P.S.How about "thinking out of the box"??? :wink:
     
  13. Jan 17, 2005 #12
    OMG !!!!! HAHAHAHAHAHA DEXTERCIOBY.....I'm so Amazed....seeing that equation was like magic to me.....like how the hell OMG
    you know what I did was the following !!!!! :
    4pi^2(r+x)(r^2+2rx+x^2)=Gmt^2

    and then multiplied each one

    THANK YOU OMNISCIENT ONE
     
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