Positional Astronomy:
Geocentric or diurnal parallax

Refraction affects the apparent altitude of a star.
But there are other phenomena that affect its apparent position, too.
One of these is parallax.
Refraction decreases the zenith angle, but parallax increases it.

Our observations are made from the surface of the Earth, not its centre.
This is irrelevant when observing distant objects such as stars.
But for closer objects (e.g. within the Solar System), a correction must be made.
This is geocentric parallax, or diurnal (daily) parallax
(since it varies daily as the Earth spins around its axis).

To an observer at O, the zenith angle of object S appears to be z'.
Its true zenith angle, as seen from the centre of the Earth C, is z, which is smaller.
Parallax is greatest for an observer at O₁,
where the object appears to be on the horizon.
We define the angle of parallax p by p = z'-z.

If a is the Earth's radius, and r is the geocentric distance to the object,
then the plane triangle OCS gives:
     sin(p) / a = sin(180°-z') / r = sin(z') / r
that is,
     sin(p) = (a / r) sin(z')

Parallax is greatest at O₁, where z'=90°.
The parallax here is called the horizontal parallax,
designated by P = 90°-z, where
     sin(P) = a / r.
For small angles, we may take P = a / r,
where P is measured in radians.

In the general case, we may replace the term (a/r) by sin(P), and write
     sin(p) = sin(P) sin(z')
or, since angles of parallax are generally small,
     p = P sin(z')

Exercise1:

A minor planet (asteroid) passes very near the Earth,
at a distance of 200,000 km.
What will be its horizontal parallax?
(Take the Earth to be a sphere of radius 6378 km.)

At St.Andrews (latitude +56°20'),
the minor planet is observed to cross the meridian
at an apparent altitude of +35°.
What does its declination appear to be?

What is its true declination, after correcting for geocentric parallax?

Click here for the answer.

Apart from occasional near-earth asteroids,
the Moon is the nearest natural object,
with average P around 57 arc-minutes.
So for calculating times of moonrise and moonset,
we must use an altitude of
0° - 16' [semi-diameter] - 34' [refraction] + 57' [horizontal parallax]
     = +7'.

Exercise2:

The Moon is at declination -14°.
What will be its hour angle at moonrise
(when the top edge of the Moon first appears over the horizon),
at a latitude of +56°20'?

Click here for the answer.

Allowing for lunar parallax is essential
when predicting occultations of stars by the Moon
(and, of course, solar eclipses).

Exercise3:

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.)

The semi-diameter of the Moon’s disc is 16 arc-minutes.
Will observers at St.Andrews see the Moon occult Aldebaran?

Click here for the answer.

The Earth is not actually spherical.
For more accurate calculations, we use the geoid:
a spheroidal solid which closely approximates the Earth's true shape.
For any particular latitude, this gives corrected values
for geocentric distance a and geocentric latitude.