Geocentric

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 O_{1},

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 O_{1}, 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')**

Exercise:

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'.

**Exercise:**

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).

**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.)

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.

Previous section:
Sunrise,
sunset and twilight

Next section: Annual
parallax

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