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Centripetal acceleration of the satellite?

  1. Oct 11, 2013 #1
    1. The problem statement, all variables and given/known data
    An Earth satellite moves in a circular orbit 790 km above the Earth's surface. The period of the motion is 100.5 min.
    (a) What is the speed of the satellite?
    (b) What is the magnitude of the centripetal acceleration of the satellite?

    2. Relevant equations
    a= v2 / r

    T = (2 Pi r)/ V

    3. The attempt at a solution

    Converting km to m: 790km*(103m / 1km) = 7.90*105m
    Converting min to seconds: 100.5min*(60seconds/1min) = 6.03*103seconds

    Now that I have the conversions, I will solve for the unknown variables in the above formulas:
    I think the radius is 7.90*105m and period is 6.03*103seconds

    (a) Speed of satellite:
    v=(2 pi (7.90*105m)) / (6.03*103s)

    Correct answer: 7460 m/s

    because the speed is incorrect, the error follows through to part (b) when substituting in speed.
    How is it that it's so far off?
    Thanks :)
  2. jcsd
  3. Oct 11, 2013 #2


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    Homework Helper

    r should be the radius of the orbit, centred at the centre of Earth. The height of the satellite above the surface of Earth is 790 km, how far is it from the centre of Earth?

  4. Oct 11, 2013 #3
    I had to look it up, but 6371 km is from the surface of Earth to its centre. So I will try now with the new radius of (6371km+790km) * (10^3m/1km) = 7.16*10^6metres.
    ANS: 7460.63 = 7461m/s

    Thanks for your help!
  5. Oct 11, 2013 #4


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    Homework Helper

    You are welcome. :smile:

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