RE: Newbie GEO question

From: Barker, Jefferson (JBARKER) (
Date: Thu Oct 11 2001 - 17:47:36 EDT

  • Next message: Bram Dorreman: "Re: Newbie GEO question"

    Jonathan Wojack has the right idea but his accuracy is off somewhat.
    The speed of a satellite in orbit is related to it's altitude.  The higher
    the orbit, the slower the speed around the Earth  relative to the center of
    the Earth.  Earth satellites never escape the Earth's gravity.  In fact,
    Earth's gravity is the only force keeping them in a curved orbit around the
    Earth.  A satellite in orbit never has a speed of zero because then it would
    fall directly toward the center of the Earth.  Any satellite must have
    tangential velocity (lateral velocity) so that the two bodies won't hit each
    other.  Even the Moon, a Earth orbiting satellite, must have lateral
    velocity around the Earth.  Satellites in geostationary orbit must keep
    moving or they'll crash into the Earth.
    For a satellite to appear stationary in the sky to a ground observer two
    conditions must be met.
    	The period of rotation of the satellite must equal the period of
    rotation of the Earth.
    	The plane of rotation of the satellite must coincide with the plane
    of rotation of the Earth (i.e., the satellite must remain over the Equator
    at all times.
    Then, assume the following:
    	Satellite's mass is negligible compared to Earth's mass
    	The orbit is perfectly circular (they never are)
    	The mass of the Earth is homogeneous (it isn't)
    	The radius of the Earth's surface is the same in all directions
    (Earth is not a perfect sphere)
    	Perturbations (fluctuations)caused by the solar wind, the
    gravitational attraction from the Moon, Sun and other planets are very
    	The Earth takes exactly 24 hrs to rotate on it's axis (it doesn't)
    With these assumptions in effect, Wojack's logic then can be applied.
    	Earth's radius to the surface at the Equator = 3,963.19556 miles
    	Period of rotation = 24 hrs.
    	Altitude of a geostationary orbit is about 22,300 mi above the
    surface of the Earth
    Then: v = circumference of orbit /period of rotation 
    	v = 2 pi x (radius of Earth + alt of sat)/period
    	v = 2 x 3.1416 x (3,963 mi + 22,300 mi)/24 hr
    	v = 163,671.01 mi /24 hr
    	v = 6,819.625 mi/hr = 10002.117 ft/sec = 1.89434 mi/sec
    Look's like Jonathan Wojack was just about right.
    That's a lot faster than any rifle bullet.  A satellite may be in a
    geostationary orbit, but it must move fast to keep up with the Earth's
    rotation.  Otherwise the satellite will appear to the ground observer to
    move across the sky.
    Jeff Barker
    Sr. Space Instructor
    US Army Command and General Staff College
    Fort Leavenworth, Kansas
    -----Original Message-----
    From: Jonathan T Wojack []
    Sent: Thursday, October 11, 2001 3:06 PM
    Subject: Re: Newbie GEO question
    > I'm new to the list and have a question regarding GEO sat's.
    > At what speeds do they travel?
    > I realize that relative to the earth at 22,000 miles they travel 0, 
    > but
    > let's say bring their orbit down to LEO, how fast would they be 
    > traveling?
    > 17,500?
    > I was able to observe one in my homemade 10" telescope, very cool. 
    > Stars
    > move, Satellite doesn't.
    I'll theorize:
    If we assume that GEO sats orbit at 22,000 miles (I think it's closer to
    25,000 miles, but I'm not sure), then I theorize that we can find their
    speed (relative to what I don't know).  I think we can just extend the
    radius of the Earth 22,000 on paper to discover the satellite's speed:
    Diameter = Pi*d
    Diamteter = Pi*(22,000+22,000+8,000[Diamter of Earth])
    Diameter = Pi*(52,000)
    Diameter = 163,362 miles.
    Now, this theoretical distance is traveled in approximately 24 hours:
    Speed = s[Distance]/t
    Speed = 163,362/24
    Speed = 6,806 miles per hour
    This translates into 1.89 miles per second, which translates into the
    perferred 3 kilometers per second.
    [This was all a big guess.  I hope it's right!   : - )]
    Jonathan T. Wojack       
    39.706d N   75.683d W      
    4 hours behind UT (-4)
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