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Circular motion of a satellitte

  1. Apr 6, 2005 #1
    A satelitte is kept in a circular orbit 300km above the surface of the earth by the force of gravity. At this altitude the acceleration due to gravity is only 8.9m/s^2. the radius of the earth is 6.4 x 10^6m.

    A.) Calculate the period of the satellite?
    B.) Calculate the speed of the satellite?

    For A. i got 5451s
    and for B. i got 7722m/s

    can anyone tell me if thats what they got too?


    CAlculate the frequency of with which the earth would have to rotate so that an object on the surface of the earth at the equator would just become "weightless"(all of the gravitational force on it would be necessary to keep the object in its "orbit" as the earth rotated.)

    So for this one, my final answer , my frequency was 1.97 x 10^-4.,..and by the way, what would be the units for frequency here?
    for this question, my centripetal acceleration , i used 9.8m/s^2..
    thanks for anyones help in advance!
  2. jcsd
  3. Apr 6, 2005 #2
    for a) and b), yes i got the same. i'm not sure how to do the 2nd one though.
  4. Apr 6, 2005 #3
    I have the same answers for A and B. I am not sure what you mean in the second part of your question, do you mean what angular velocity would the earth require to counteract the gravational pull of the eath at the equator (i.e. force centripital = force gravity)?
  5. Apr 6, 2005 #4


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    Gravity is always a centripetal force.However,it needn't be equal to[itex] m\omega^{2}R_{eq} [/itex].When it dooes,that "weightless" part is due to the annihilation of gravity's pull and the centrifugal force experinced by a body in circular movement.

  6. Apr 6, 2005 #5
    yeah, the second question is confusing..im really not sure what it means, that's exactly what the question is though..
    what i did was Ac = V^2/r Ac being 9.8m/s^2 and r is 6.4 x10^6.. I used that to find the velocity which is 7919m/s.
    Then i used the formula v = 2(pi)r/T to solve for T, for T i got 5077. to find frequency, it's 1 divided by the period(T) so i got 1.97 x 10^-4
    and im not sure wht the units for frequency is...
  7. Apr 6, 2005 #6


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    And one more thing

    [tex] g_{eq}\simeq 9.78 m s^{-2} [/tex]
    [tex] R_{eq}\simeq 6378 Km [/tex]

  8. Apr 6, 2005 #7


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    U found [itex] \nu [/itex],which is frequency of the rotation movement.

    The method i proposed delivered [itex] \omega[/itex]...
    It's angular velocity,equivalently,angular frequency.It's the famous [itex] \omega [/itex] ...

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