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The Greenwich mean sidereal time is defined by the hour angle
between the meridian of Greenwich and mean equinox of date at
UT1:
[A. B6]:
 |
(18) |
in seconds of a day of 86400s UT1,
where
is the time difference in Julian centuries of Universal Time (UT1) from J2000.0.
From this the hour angle in degree
at any instant of time
(Julian days from J2000.0)
can be calculated by
 |
(19) |
For the precision needed in this paper we may neglect the difference between
and
, such
that (Meeus, 2000):
 |
(20) |
Markus Fraenz
2017-03-13