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Homework Help: Satellite orbit question: does mass affect this problem?

  1. Mar 4, 2006 #1
    I think I have an answer for the following problem, but I'm not sure if it (or the assumptions used to get it) are correct:

    Consider a 540 kg synchronous satellite in a circular orbit in the same plane as our equator. Find the height H of the satellite above the Earth's equator.

    What I got:

    msat - mass of satellite = 540 kg
    mearth - mass of earth = 5.98 x 10^24 kg
    r = radius = distance between earth's center and the satellite = r
    G = 6.67 x 10^-11 (N m^2 / kg^2)
    v = velocity of the satellite
    a = acceleration for uniform circular motion
    radius earth = 6380 km

    F = ma
    a = v^2 / r

    G (msat * mearth) / r^2 = msat ( v^2 / r )

    v = (2*pi*r) / T

    T = 1 day = 86,400 seconds

    substituting...and cancelling msat on both sides

    G (mearth / r^2 ) = (2*pi*r)^2 / (r T^2 )


    r^3 = ( G*mearth*T^2 ) / (4*pi^2)

    r^3 = 7.54 x 10^22 meters cubed.

    r = 4.23 X 10^7 meters or 42,300 km

    total radius - earth radius = height above the earth's surface

    42,300 km - 6,380 km = 35,920 km or 35,920,000 meters above the earth's surface
  2. jcsd
  3. Mar 4, 2006 #2
    Mass does not affect it.

    I got the same answer as you. Google also has the same answer.
  4. Mar 5, 2006 #3
    Thank you.
  5. Mar 5, 2006 #4


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    Science Advisor
    Homework Helper

    Since you did ask whether "mass affects this problem," you should know that the equations you set up assume that mass doesn't matter by taking the Earth to be stationary. If you were to do the problem a little more carefully, the Earth and the satellite would revolve about their center of mass.

    You would then find that the mass appearing in the equation you wrote for the period would contain the combined mass of the Earth and satellite (instead of just the Earth's mass) so then, yes, the mass of the satellite contributes to the period.

    Of course, for the problem at hand, the mass of the satellite is so small compared with the mass of the Earth that the difference it makes in calculating the geosynchronous radius is entirely negligible.
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