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Homework Help: Centripetal forces and satellites

  1. Aug 22, 2011 #1
    Hi im kind of new to the forum nice to be here! here is my question.

    1. The problem statement, all variables and given/known data

    A satellite 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 8.9 m/s. The radius of the earth is 6.4*10^6m

    a) calculate the period of the satellite

    b) calculate the speed of the satellite


    2. Relevant equations

    Ac=(4(Pi^2)(r))/(T^2)

    Where ac is the acceleration
    T is the period
    r is the radius

    V=(2(pi)r)/(T)

    3. The attempt at a solution

    a)

    Using the first equation listed

    T= (2(pi))(r/ac)^(1/2)
    r=(6.4*10^6)+(300km)(1000m/1km)
    ac=8.9m/s

    T=5451.6 seconds

    b) Using our known T,
    V=(2Pi)(6700000)/(5451.6)
    V=7722 m/s

    Thanks in advance, I feel I may not have the concept down well enough and this may not be the way to do it.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 22, 2011 #2

    PhanthomJay

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    Gold Member

    That looks right to me. Welcome to PF as a poster!!
     
  4. Aug 23, 2011 #3
    ** r = radius
    r = RE + h = (6.4*10^6m + 300,000m) = 6.7x10^6

    **C = Circumference
    C=2(PI)r

    T=period
    T=C/v

    **for circular motion there are 2 components of acceleration, radial acceleration and tangential acceleration. the radial (centripetal acceleration) is shown:

    Ar=(mv2)/r

    **because gravity from the Earth is causing the satellite to undergo centripetal acceleration:

    mg = (mv2)/r

    **the masses of the satellite cancel to show the speed of the satellite:

    (rg)1/2=v

    v = 7722 m/s

    ** the period
    T = 2(PI)r/v = 5448s

    Based on what I got, I would say you did a good job!
     
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