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Deriving formula for surface velocity of a satellite in orbit.

  1. Oct 27, 2013 #1
    I may be being stupid here... But I'm trying to derive such an equation to no avail.

    Maybe such a formula requires more than one variable - but I would like to easily display this graphically.
    To start out the derivation I stated that the Surface Velocity = (2pi/T)-(v/r) where v/r=angular velocity of satellite in orbit and 2pi/T=angular velocity of a planet.

    Can anyone help?
     
    Last edited: Oct 27, 2013
  2. jcsd
  3. Oct 27, 2013 #2
    Think I got it (in terms of time, not velocity):

    2pi*sqrt((radius of planet+orbital altitude)^(3)/(GM))-rotationalperiodofplanet
     
    Last edited: Oct 27, 2013
  4. Oct 27, 2013 #3

    gneill

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    Staff: Mentor

    You can work out the angular velocities separately to begin with. You stated the one for the planet as ##2\pi/T##. Call that ##\omega_p##, and make sure that T is the sidereal period for the planet. For the satellite you again made a true statement, namely that the angular velocity is given by ##\omega_s = v/r##, with r the radius of the orbit and v the orbital speed.

    Assuming a circular orbit the orbital speed will be ##v = \sqrt{GM/r}##, where M is the mass of the planet.

    Assuming that the satellite's orbit is in the plane of the equator, the difference in angular velocities is then:
    $$\omega_p - \omega_s = \Delta \omega = \frac{2\pi}{T} - \frac{1}{r}\sqrt{\frac{GM}{r}}$$
    You can move the 1/r into the square root if you want.

    With Δω in hand you can work out the surface speed of the sub-satellite point easily enough.
     
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