Precession

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So far, this series of pages has considered

how we
assign coordinates to any point in the sky,

and the various
physical effects that may alter its apparent position.

But there
is a more profound problem with the way we determine
coordinates,

relative to the celestial equator and the
ecliptic,

since these are not permanently fixed.

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.

**Exercise:**

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

So if Δλ is the change in λ in a given
time interval Δt,

the corresponding changes in α and δ are

Δα = Δλ [
cos(ε) + sin(ε) sin(α) tan(δ) ]

Δδ = Δλ
cos(α) sin(ε)

This is the effect of luni-solar precession.

We
also have to add in the planetary precession,

which decreases the
RA by a quantity *a*, during the same time interval.

The combination is general precession:

Δα = δλ [
cos(ε) + sin(ε) sin(α) tan(δ) ] - *a*

Δδ = Δλ
cos(α) sin(ε)

To make this easier to calculate in practice,

we
introduce two new variables, m and n:

m = Δλ cos(ε)
- *a
*n = Δλ
sin(ε)

These quantities m and n are almost constant;

they
are given each year in the *Astronomical Almanac*.

The
values for 2000 are approximately:

m
= 3.075 seconds of time per year

n
= 1.336 seconds of time per year

=
20.043 arc-seconds per year

We can now write:

Δα
= m + n sin(α) tan(δ)

Δδ = n
cos(α)

which means that,

if you know the equatorial
coordinates of an object at one date,

you can calculate what they
should be at another date,

as long as the interval is not too
great (20 years or so).

If the object is a star whose proper
motion is known,

then that should be corrected for as well.

**Exercise:**

The coordinates of the Galactic
North Pole are given officially as

α
= 12h49m00s, δ = +27°24'00",

relative to the equator and equinox of 1950.0.

What should they be,

relative
to the equator and equinox of 2000.0?

(For this calculation, take the
values of m and n for the year 1975:

m = 3.074s per year;

n =
1.337s per year = 20.049" per year.)

Click here for the answer.

Alternatively, the *Astronomical Almanac* lists
**Besselian Day Numbers** throughout the year.

Take a star’s
equatorial coordinates from a catalogue,

and compute various
constants from these,

as instructed in the *Astronomical
Almanac*.

Combine these with the Day Numbers for a given date,

to produce the apparent position of the star,

corrected for
precession, nutation and aberration.

Previous section:
Aberration

Next section: Calendars

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