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Centripetal acceleration of a satellite orbiting Earth

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data
    A satellite is orbiting the Earth in a stable orbit of which the radius is twice that of the Earth. Find the ratio of the satellites cepripetal acceleration to g on the Earth's surface.


    2. Relevant equations
    centripetal acceleration = (v^2)/r or (4rpi^2)/T^2


    3. The attempt at a solution
    Well the only thing i can figure is the rafius of the satellite which is 12.8x10^6
    Otherwise I have no idea where to start :(
     
  2. jcsd
  3. Apr 15, 2010 #2

    tiny-tim

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    Hi ride4life! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    As usual, it's good ol' Newtons' second law that's needed, this time together with another of Newton's laws! :tongue2:

    What does that give you? :smile:
     
  4. Apr 15, 2010 #3
    Re: Gravitation

    Ummm...
    Fc=Fg
    mac=Fg
    mac=mg
    ac=g
    ?????
     
    Last edited: Apr 15, 2010
  5. Apr 15, 2010 #4

    tiny-tim

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    ok … now combine that with your original g = v2/r :smile:
     
  6. Apr 15, 2010 #5
    Re: Gravitation

    v2=GM/r
    v2=(6.67x10-11x6x1024)/6.4x106
    v=7907.67m/s
     
  7. Apr 15, 2010 #6
    Re: Gravitation

    ahhh...

    ac=GM/r2
    ac=(6.67x10-11x6x1024)/(2x6.4x106)2
    ac=2.44
     
  8. Apr 15, 2010 #7

    Matterwave

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    Re: Gravitation

    Remember to always include the units or else the answer makes no physical sense.
     
  9. Apr 15, 2010 #8

    tiny-tim

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    Sorry, not following this …

    you're making it very complicated :redface:

    The question only asks for the centripetal acceleration (as a multiple of g on the Earths' surface) …

    you have centripetal acceleration = GM/r2, so … ? :smile:
     
  10. Apr 15, 2010 #9
    Re: Gravitation

    centripetal acceleration = 2.44
    ratio compared to Earth's g which is 9.8 is 2.44/9.8 which is about 1:4
     
  11. Apr 15, 2010 #10

    HallsofIvy

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    Re: Gravitation

    It is only "about" 1:4 because of the over-complicated way you did that. That was tiny-tim's point. It isn't necessary to find the actual accelerations.

    [tex]a= \frac{GM}{r^2}[/tex]
    If r= 2R (R is the radius of the earth) then
    [tex]a= \frac{GM}{(2R)^2}= \frac{GM}{4r^2}= \frac{1}{4}\frac{GM}{r^2}= \frac{1}{4}g[/tex].

    The centripetal acceleration of a satellite at radius twice the radius of the earth is exactly 1/4 g.
     
  12. Apr 15, 2010 #11
    Re: Gravitation

    ahhh...
    i see what you mean, way less complicated than my way :biggrin:
     
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