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Centripetal acceleration of an orbit to the earth help!

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data

    A geostationary satellite goes around the Earth in 24hr. Thus, it appears motionless in the sky and is a valuable component for telecommunications, including digital television. If such a satellite is in orbit around the Earth at an altitude of 35 800 km above the Earth's surface, what is the module of its centripetal acceleration?

    2. Relevant equations
    I believe I should use
    ac = (4∏2r) / T2

    T= 24 hr = 86400s
    r = 35800 km = 35 800 000 m = 3.58 x 107 m

    3. The attempt at a solution
    I plugged in the values:
    ac = (4∏235 800 000) / 864002
    ac = 0,189

    The answer SHOULD be 0.223 m/s2

  2. jcsd
  3. Nov 14, 2011 #2

    D H

    Staff: Mentor

    I have highlighted your error:

    Do you see the problem?
  4. Nov 14, 2011 #3
    ahaha... isnt altitude the radius?
  5. Nov 14, 2011 #4

    D H

    Staff: Mentor

    No. Altitude is the distance between the satellite and the closest point on the surface of the Earth. Radius is the distance to the center of the Earth.
  6. Nov 14, 2011 #5
    OOOOH so i add the radius with the altitude, and THATS the r value i use! Right?
  7. Nov 14, 2011 #6
    Nvm, I got it, thanks again !
  8. Nov 14, 2011 #7

    D H

    Staff: Mentor

    Very good, and you're welcome.
  9. Nov 19, 2011 #8
    how do you calculate the distance from the center of the Earth for a geosynchronous satellite orbit?
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