المبادرة و ترنح محور الأرض

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positional

المبادرة و ترنح محور الأرض

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.

تمرين:

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?

الحل:

The equinoxes regress along the ecliptic at 50.35" per year.
So the " 0° " point moves westwards against the constellations.


It has moved 28° into Pisces, at a rate of 50.35" per year.
Divide 28° by 50.35" to get: 2002 years ago.

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.

تمرين:

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.)

الحل:

The formulae are: Δα = m + n sin(α) tan(δ) Δδ = n cos(α)

Substitute the coordinates given in the question: α1950 = 12h49m00s = 192.25°, δ1950 = +27°24'00" = 27.40°,
to get Δα = 2.927s per year, or +146.348s in 50 years. Δδ = -19.59" per year, or -979.63" in 50 years.

Add these to the 1950 values of α and δ, to get α2000 = 12h51m26s δ2000 = +27°07'40"

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.