Celestial sphere image?

In sci.astro.amateur message
, Tue, 12 May 2009 18:27:52, Dave Typinski
posted:

A proof closer to home is to examine the relationship of satellite
orbit altitude and orbit period.

Geostationary satellites live at an orbit radius of 42,165 km, plus or
minus a couple kilometers. Given Earth's mass and that orbit radius
and Newton's laws of motion and gravitation, they orbit in one
sidereal day plus or minus a few seconds.

To do a full orbit in one solar day, they'd have to be either a)
higher by about 80 km or b) the Earth would have to be lighter by
about 3x10^19 metric tons.

Since they aren't higher and Earth isn't lighter, they don't orbit in
one solar day, but in a sidereal day. Since they're geostationary,
the Earth itself must complete one rotatation in one sidereal day, not
in one solar day. QED.

But where does the sufficiently-exact figure for the Earth's mass come
from? Does it not come from measurements of orbit times and sizes,
making your argument circular?

--
(c) John Stockton, near London.
Web URL:http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, & links.
Correct = 4-line sig. separator as above, a line precisely "-- " (SoRFC1036)
Do not Mail News to me. Before a reply, quote with "" or " " (SoRFC1036)