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Centripetal Acceleration and Satellite Question

  1. Sep 19, 2010 #1
    1. The astronaut orbiting the Earth in Figure P4.30 is preparing to dock with a Westar VI satellite. The satellite is in a circular orbit 600 km above the Earth's surface, where the free-fall acceleration is 8.21 m/s2. The radius of the Earth is 6400 km.
    a) Determine the speed of the satellite. m/s
    b) Determine the time required to complete one orbit around the Earth. min

    2. Relevant equations[/b]
    centripetal acceleration= v squared/ radius
    distance*speed= time

    3. The attempt at a solution[/b]
    To solve for the speed I used 8.21 as the centripetal acceleration and made that equal to v squared/7000 and then solved for v.

    8.21=v squared/7000
    v= 239 m/s

    I then took 14000*3.14 for the circumference of the orbit and multiplied that by 239 m/s for the total time to orbit and divided that by 60 to put the answer in minutes. I got 2919.94 minutes.

    However, both answers are wrong. Can I not use the free fall acceleration for centripetal acceleration?
  2. jcsd
  3. Sep 19, 2010 #2


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    Welcome to PF,

    Are you trying to claim that distance*speed = time? You might want to rethink that one...
  4. Sep 19, 2010 #3


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    Yes, you can, and you should, because gravity is what is providing the centripetal force that keeps the satellite moving in a circle. You might have a computation error. Can you post your calculations for part a?
  5. Sep 19, 2010 #4
    for part a I used the equation centripetal acceleration= v^2/ radius

    8.21= v^2/(6400+600)
    v= 239.73

    what is wrong with this calculation?

    Concerning finding the time. I completely forgot that t=distance/speed.
  6. Sep 19, 2010 #5
    I just realized my problem. I did not do the simple unit conversions from km to m. Everything worked out fine. Thanks for all of your help!
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