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Old May 13th 09, 08:33 PM posted to sci.astro.amateur
Dave Typinski[_3_]
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Default Celestial sphere image?

Dr J R Stockton wrote:

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?


I'm glad you brought that up, JR, because I was originally reticent to
post that message for that very reason.

So I did some looking, and lo and behold, it turned out that it isn't
circular resoning after all.

Yes, you calculate Earth's mass using satellite orbit data, but you
also use big G calculated in an entirely different fashion.

First, calculate big G by measuring the acceleration between test
objects in a torsion balance. This is the key step that breaks the
circle.

Second, measure the orbit altitude and orbit period of a satellite.

Third, using your new big G number, calculate the Earth mass required
to produce the measured orbit altitude and period.

Gundlach and Merkowitz at UWash did just that a while ago.

http://asd.gsfc.nasa.gov/Stephen.Merkowitz/G/Big_G.html

Their refinement of big G resulted in an uncertainty of 8x10^16 metric
tons in Earth's mass. That's three orders of magnitude smaller than
the accuracy required to show the difference between solar and
sidereal orbit periods.

After figuring that out, I felt somewhat more confortable posting my
explanation. ;-)
--
Dave