Re: CSIRO Satellite (Correction)

From: Richard Clark (rclark@LPL.Arizona.EDU)
Date: Wed Feb 16 2000 - 23:40:18 PST

  • Next message: Brierley David: "92-86C observed"

    Or, using density instead of mass...
    An orbit immediately above the surface of a sphere of density rho has
    a period of:
    
    P = sqrt( 3 PI / (G rho) )
    
    Radius has dropped out.
    
    P = 375844 / sqrt(rho) if everything is in SI. (+/- any mistakes I may
    have made :-)
    
    A 15 minute period implies a density of at least 174000 kg/m^3. Something
    like 30 times the mean density of Earth.
    
    Hmmm, What's the density of uranium?
    
    A Molnyia at perigee would be leisurely by comparison.
    
    Richard Clark
    rclark@lpl.arizona.edu
    
    
    On Wed, 16 Feb 2000, Sue Worden wrote:
    > If I didn't make a mistake in my arithmetic, a satellite in a
    > circular orbit with period 15 minutes would be at a radius of
    > about 2000 km -- somewhere in the outer core, I believe, but
    > my geology is quite rusty...
    > 
    > Yep, I'd call that "low altitude", by golly! ;-)
    > 
    > --Sue Worden (worden@fc.net)
    > 
    > -----------------------------------------------------------------
    > Unsubscribe from SeeSat-L by sending a message with 'unsubscribe'
    > in the SUBJECT to SeeSat-L-request@lists.satellite.eu.org
    > http://www2.satellite.eu.org/seesat/seesatindex.html
    > 
    > 
    
    -----------------------------------------------------------------
    Unsubscribe from SeeSat-L by sending a message with 'unsubscribe'
    in the SUBJECT to SeeSat-L-request@lists.satellite.eu.org
    http://www2.satellite.eu.org/seesat/seesatindex.html
    



    This archive was generated by hypermail 2b29 : Wed Feb 16 2000 - 23:41:20 PST