Exercise:
Aldebaran
is at Right Ascension 4h36m, declination +16°31'.
At a
particular instant, the geocentric coordinates of the Moon
are
also Right Ascension 4h36m, declination +16°31'.
Local
Sidereal Time at St.Andrews (latitude +56°20') is 4h36m.
What
will be the apparent declination of the Moon, after correction for
parallax?
(Take the horizontal parallax of the Moon as 57
arc-minutes.)
Both
objects are on the meridian.
So the Moons altitude a =
(90°-φ) + δ
= 50.18°
so its true zenith angle z = (90°-a) = 39.82°
The
shift due to parallax is p = P sin(z'), where P = 57',
and z' is
the apparent zenith angle.
However, we only know z, the true
zenith angle.
For a first approximation, take z' = z = 39.82°
Then
p = 57' sin(39.82°) = 36.5' = 0.61°.
This would make
apparent zenith angle z' = 39.82° + 0.61° =
40.43°.
Re-calculate p = 57' sin(40.43°) = 37.0' = 0.62°.
No need to re-calculate again.
The
apparent altitude will be 37' lower,
due to geocentric parallax,
and so the apparent declination will also be 37' lower.
So
the Moons apparent declination will be
+16°31' - 37' =
+15°54'.
The
semi-diameter of the Moons disc is 16 arc-minutes.
Will
observers at St.Andrews see the Moon occult Aldebaran?
Aldebaran
is still at Right Ascension 4h36m, declination +16°31'
(unaffected by geocentric parallax).
The
top edge of the Moon will appear to be at declination
+15°54'
+ 16' = +16°10'
So
the top edge of the Moon will appear to pass 21' below
Aldebaran:
there will be no occultation.
Back to "geocentric parallax".