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
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