[geocentrism] Fw: (geocentrism) geostationary / geosynchrous sat.
- From: "Philip" <joyphil@xxxxxxxxxxx>
- To: <geocentrism@xxxxxxxxxxxxx>
- Date: Wed, 27 Oct 2004 10:36:09 +1000
The matter of orbital mechanics has captured me. I am not a star gazer, but it
seems as I delve into this subject of astronomy, a question that has always
been in the background is this. Why do we always see the same stars every night
for the full 365 days of the year? Shouldn't the night sky of June be what we
could not see because it was behind the sun of the day sky in December.
If that sounds like a conundrum, put it another way. Right now today, I cannot
see the stars that are in the sky back of the sun. But 6 months from now they
will should be my night sky view. Yet from what I remember, the big dipper is
always in view every night. Am I missing something? This is from the
heliocentric moving earth view point of course.
On the other hand, if thecosmos with all its stars moved around a stationary
earth, and only made an annual inclination deviation up and down for the
seasons, seasons, then yes we would always see the same night time stars.
Again, perhaps I missed something.
If your page is big enough this diagram should indicate what I was getting at .
<<<<night sky---O Earth in June-------O sun-------Dec,O Earth night sky>>>>>
Entirely diagramaticly opposite night skys. Now back to my other project.
I've done some figuring. hold/save the following facts for later and please
correct me if I erred.
From figures given for the mean sizes of our planet in the solar system, the
earth radius is 6,378k which I calculated to give an equatorial circumference
of 265,031k, which computes to a rotational velocity of 1,670 k/h which is
about 600k faster than the speed of sound.
The earth is at a mean distance of 1496 X 10 5th k from the sun
1496X10 5th X 44/7 all over 365X24 equates to a speed around the sun 107,345
k/h
Next, the Geosynchronous sat is at a distance from the centre of the earth of
35,786k + 6,378.14 = 42164k
Orbit = 44/7 X 42164 = 265031k/24
which is equating to 11,043 k/h
That seems slow to me. However, if all this is correct I will apply these to my
diagram. I need to know if anyone can tell me, looking down upon the solar
system from the N aspect which direction we are circling the sun. I am going to
assume from memory or presumption that it is anticlockwise.
Philip.
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