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 Moon’s 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 Moon’s apparent declination will be
+16°31' - 37' = +15°54'.

The semi-diameter of the Moon’s 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".