Positional Astronomy:
Precession

The Earth’s axis is tilted to its orbital plane.
The gravitational pull of the Sun and the Moon on the Earth’s equatorial bulge
tend to pull it back towards the plane of the ecliptic.
Since the Earth is spinning, its axis precesses.
The North Celestial Pole traces out a precessional circle
around the pole of the ecliptic,
and this means that the equinoxes precess backwards around the ecliptic,
at the rate of 50.35 arc-seconds per year
(around 26,000 years for a complete cycle).

Around 2000 years ago,
the Sun was in the constellation of Aries at the spring equinox,
in Cancer at the summer solstice,
in Libra at the autumn equinox,
and in Capricorn at the winter solstice.
Precession means that all of these have changed,
but we still use the old names
(e.g. the First Point of Aries for the vernal equinox),
and the symbols for the vernal and autumnal equinoxes
are the astrological symbols for Aries and Libra.

Exercise1:

The Vernal Equinox occurs nowadays
when the Sun is in the constellation of Pisces.

Pisces covers a section of the ecliptic
from longitude 352° to longitude 28°;
at longitude 28° the ecliptic passes into Aries.

How many years would we have to go back,
to find the Sun at “the First Point of Aries”
at the vernal equinox?

Click here for the answer.

Precession is caused by the Sun and the Moon.
However, the Moon does not orbit exactly in the ecliptic plane,
but at an inclination of about 5° to it.
The Moon’s orbit precesses rapidly,
with the nodes taking 18.6 years to complete one circuit.
The lunar contribution to luni-solar precession
adds a short-period, small-amplitude wobble
to the precessional movement of the North Celestial Pole,;
this wobble is called nutation.

Ignoring nutation,
luni-solar precession simply adds 50.35 arc-seconds per year
to the ecliptic longitude of every star,
leaving the ecliptic latitude unchanged.

This definition assumes the ecliptic itself is unchanging.
In fact, the gravitational pull of the other planets perturbs the Earth’s orbit
and so it gradually changes the plane of the ecliptic.
If the equator were kept fixed,
the movement of the ecliptic would shift the equinoxes forward along the equator
by about 0.13 arc-seconds per year.
This is planetary precession,
which decreases the Right Ascension of every star by 0.13 arc-seconds per year,
leaving the declination unchanged.

Combining luni-solar and planetary precessions gives general precession.
(Lunar nutation and planetary precession also produce slight changes in the obliquity of the ecliptic)

Because of precession,
our framework of Right Ascension and declination is constantly changing.
Consequently, it is necessary to state the equator and equinox
of the coordinate system to which any position is referred.
Certain dates (e.g. 1950.0, 2000.0) are taken as standard epochs,
and used for star catalogues etc.

To point a telescope at an object
on a date other than its catalogue epoch,
it is necessary to correct for precession.

Recall the formulae relating equatorial and ecliptic coordinates:
     sin(δ) = sin(β) cos(ε) + cos(β) sin(ε) sin(λ)
     sin(β) = sin(δ) cos(ε) - cos(δ) sin(ε) sin(α)
     cos(λ) cos(β) = cos(α) cos(δ)

Luni-solar precession affects the ecliptic longitude λ.
The resulting corrections to Right Ascension and declination
can be worked out by spherical trigonometry.
But here we use a different technique.

Consider luni-solar precession first,
recalling that it causes λ to increase at a known, steady rate dλ/dt,
while β and ε remain constant.

To find how the declination δ changes with time t,
take the first equation and differentiate it:
     cos(δ) dδ/dt = cos(β) sin(ε) cos(λ) dλ/dt
To eliminate β and λ from this equation,
use the third equation:
     cos(δ) dδ/dt = cos(α) sin(ε) cos(δ) dλ/dt
i.e.            dδ/dt = cos(α) sin(ε) dλ/dt

To find how the Right Ascension α changes with time,
take the second equation and differentiate it:
     0 = cos(ε) cos(δ) dδ/dt + sin(ε) sin(δ) dδ/dt sin(α) - sin(ε) cos(δ) cos(α) dα/dt
i.e.     sin(ε) cos(δ) cos(α) dα/dt = dδ/dt [ cos(ε) cos(δ) + sin(ε) sin(δ) sin(α) ]
                                           ;           = cos(α) sin(ε) dλ/dt [cos(ε) cos(δ) + sin(ε) sin(δ) sin(α) ]
Cancelling out sin(ε) and cos(α) from both sides gives:
     cos(δ) dα/dt = dλ/dt [ cos(ε) cos(δ) + sin(ε) sin(δ) sin(α) ]
Dividing through by cos(δ) gives:
     dα/dt = [ cos(ε) + sin(ε) sin(α) tan(δ) ] dλ/dt