(no subject)
- From: Regner Trampedach <art@xxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Mon, 24 Dec 2007 01:50:22 +1100
Allen,
You have asked for vectors - I'll let the Christmas spirit prevail
and give you some vectors:
I simplify a bit by describing The Earth's orbit around the Sun as a circle
- not an ellipse. That is an approximation, but a good one for this purpose
- it will have no effect on the argument at hand.
The position vector of the centre of the Earth is then given by:
/ xc \ / COS(t/year*2*pi) \
rc = | yc | = | SIN(t/year*2*pi) | (1)
\ zc / \ 0 /
where we have put then Sun at (0,0,0), and the ecliptic plane (plane of the
orbit) is the x-/y-plane and the z-axis is perpendicular to that plane and goes
through the centre of the Sun.
The first equality can also be written rc=(xc,yc,zc) , but the column
vector notation is more compact.
t is time, measured in the same unit as year (e.g., seconds, hours, days)
and the factor of 2*pi is there because the sine and the cosine go through one
full oscillation when the argument changes by 2*pi - Equation (1) therefore
goes through one cycle (orbit) in one year.
Equation (1), above, describes the translational orbit of the Earth around
the Sun. It describes how the position of the centre of the Earth changes
with time. Only position! - it says nothing about orientation.
The position vector of a particular point on the surface of Earth, at
the equator, is given by:
/ xe \ / COS(t/year*2*pi) \ / COS(t/day*2*pi) \
re = | ye | = | SIN(t/year*2*pi) | + | COS(23.5°)*SIN(t/day*2*pi) | (2)
\ ze / \ 0 / \ SIN(23.5°)*SIN(t/day*2*pi) /
where the 23.5° is the tilt of the equatorial plane compared to the ecliptic.
The consequence of that tilt is that the point on the equator will go above
and then below the ecliptic during one day, as seen in the z-component of
the second term. The day is here the sidereal day.
The orientation of the Earth is described by the vector going from the
centre of the Earth, to our chosen point on the equator, i.e., (2) - (1)
[equation (2) minus equation (1)]:
/ xo \ / COS(t/day*2*pi) \
ro = | yo | = | COS(23.5°)*SIN(t/day*2*pi) | (3)
\ zo / \ SIN(23.5°)*SIN(t/day*2*pi) /
This describes the orientation of Earth with respect to the stars.
It changes in a cyclic manner, repeating itself every sidereal day.
There is no reference to anything happening on a yearly cycle.
It should be obvious, that both the rotation described by Eq. (3) and
the translational orbital motion described by (1) are continuous motions
that happen all the time - e.g., the Earth does not jump forward every
time a sidereal day has been completed, instead the Earth moves ahead in
its orbit all the while it also spins on it's celestial axis.
Regner
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