Re: The 3 minute 56 second illusion

oriel36 <kelleher.gerald@xxxxxxxxx> wrote:

On Jun 10, 8:43 pm, Dave Typinski <möb...@xxxxxxxxxxxxx> wrote:

86400 seconds per solar day divided by 1.00271793 orbits per solar day
equals 86,166 seconds per orbit.  Close enough.
Do you think that the people who launch and operate satellites know
what they're doing?

http://www.ncgia.ucsb.edu/education/curricula/giscc/units/u014/tables/table02.html

69.17 miles at the Equator corresponds to both 4 minutes of clock time
and 1 degree of geographical separation leading to the maximum
Equatorial circumference of 24901.2 miles in 24 hours and 360 degrees.

I do not know how you are going to get the maximum Equatorial
circumference from the 23 hour 56 minute 04 second value

Because you're failing to account for the orbital component of
rotation. You wish to use the Earth-Sun line as a reference, but you
fail to acknowledge that this line itself rotates through roughly one
degree in one day.

In order to obtain accurate answers, one must set up a calculation
using a reference frame appropriate to a given problem. Given how
you've set up the calculation, you are using an incorrect frame of
reference. Consequentially, you are obtaining faulty results.

As you've set it up, the proper frame of reference is the fixed stars.

Zenith to zenith, 360°, in 86,164 seconds.

Relative to the center of the Earth, a point on the equator moves 68
miles /more/ that the equatorial circumference in 86,400 seconds
because the Sun's apparent position against the stars moves while the
Earth rotates.

Solar noon to solar noon, 360°, in 86,400 seconds.

Using the fixed stars as a reference:
One sidereal day = 86,164 seconds = 360°
One solar day = 86,400 seconds = 361°

Using the Earth-Sun line as a reference:
One sidereal day = 86,164 seconds = 359°
One solar say = 86,400 seconds = 360°

It's just math. It works.

and I really do not care

That's unfortunate and apparent.
--
Dave
.

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