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Satellite Motion and Electric Fields

  1. Nov 5, 2005 #1
    The period T of an earth satellite is related to the radius R of its orbit by the equation T^2 = A R^3 where A is a constant. The moon may be assumed to move in a circular orbit of radius RM about earth. The period of the moon's orbit is 28 days. The radius of the orbit of a geostationary satellite about earth is Rg. The orbital radii are related by the expression: RM = K Rg
    Find the value of K.
    I found K to be 3.04 but am not sure if i have done it correctly. Can anyone help?

    Also, a constant potential difference is applied beteen two conducting plates creating a uniform electric field. A very small negatively charged sphere is introduced between the plates. It is found that the values of the weight of the sphere, the charge on it and the electric field between the plates are such as to cause it to remain stationary. At time t=0 the upper plate starts to move with uniform velocity towards the lower plate. The potential difference is kept constant, and the plates remain parallel throughout the motion.
    Sketch a graph to show the variation of the electric field strength E between the plates with time t. As E = voltage/distance, I drew a straight line through the origin. Is this correct?
    Describe what happens to the charged sphere while the upper plate is moving.
    I have said that the charged sphere would accelerate towards the positive plate as the electric field strength would be increasing, because the electric force no longer balances with the charge and the weight of the sphere. Don't think this is correct though. Can anyone suggest how I would answer this?
    Last edited: Nov 6, 2005
  2. jcsd
  3. Nov 6, 2005 #2
    Can anyone help me out?
  4. Nov 6, 2005 #3


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    I don't know how to do this, but I took a guess.

    I assumed that something moving in an orbit equal to the earth's Radius would have a period, T, equal to 24 hrs, or 1 day. Is that correct ?

    Assuming that's correct then ...

    Tg² = ARg³
    1 = ARg³
    A = 1/Rg³


    T = (R/Rg)³

    when R = Rm,

    T² = (Rm/Rg)³
    28² = (K)³
    K = 9.2208

    But that value is the square of 3.04, your value.
    Did you square the 28 when you did your calculation ?
  5. Nov 6, 2005 #4
    That isn't necessarily true. However, "geostationary" means stationary with respect to the Earth, so a geostationary satellite will have an orbital period of 24 hours, whatever its orbital radius.
  6. Nov 6, 2005 #5


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    Thanks for the update :smile:
  7. Nov 6, 2005 #6


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    Well, 23 hours 56 minutes and... 4 seconds is it? But close enough!
  8. Nov 6, 2005 #7


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    Kepler's Law has the distance ratio cubed = the period ratio squared.
    your period ratio is 28, so the distance ratio should be cuberoot(784).

    The Electric Field at the start of the motion is NOT zero, but some E.
    The strength does increase linearly at first, doubling as the distance is halved.
    as time goes by, to 3/4 of the time till the plates touch, E = 4 E_o .
    By the time the plates would touch the E-field would be infinite.

    Yea, the Electric Force (which must be up) would become stronger than
    the (downward) gravity Force, so the ball accelerates upward (toward + plate)
  9. Nov 7, 2005 #8
    thank you all for your help
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