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Secular and periodical motions.

CHAPTER VI.

PRECESSION AND NUTATION.

73. The ecliptic is not a fixed, but a moving plane, and its observed position in the year 1750 has been adopted by astronomers as a fixed plane, to which its situation at any other time is referred.

The motion of the ecliptic is shown by the changes in the latitudes of the stars.

74. Celestial motions are generally separated into two portions, secular and periodical.

Secular motions are those portions of the celestial motions which either remain nearly unchanged, or else are subject to a nearly uniform increase or diminution, which lasts for so many ages, that their limits and times of duration have not yet been determined with any accuracy.

Periodical motions, are those whose limits are small, and periods so short, that they have been determined with considerable accuracy.

75. The true position of a heavenly body, or of a celestial plane, is that which it actually has ; its mean

Position of the mean ecliptic.

position is that which it would have if it were freed from the effects of its periodical motions.

The mean position is, consequently, subject to all the secular changes.

76. The mean ecliptic has, from the time of the earliest observations, been approaching the plane of the equator at a little less than the half of a second each year, thus causing a diminution of the obliquity of the ecliptic.

Let NAA' (fig. 41) be the fixed plane of 1750, and NA, the mean ecliptic for the number of years t after 1750. Let A be the vernal equinox of 1750, and AQ the equator. Let

II = NA and a = the angle ANA , ; then, upon the authority of Bessel, the point of intersection N of the ecliptic, which is called the node of the ecliptic, with the fixed plane, has a retrograde motion, by which it approaches A at the annual rate of 5'.18, and if this point could have existed in 1750, its longitude would have been 171° 36' 10", so that I= 171° 36' 10"

- 54.18 t.

(485) Moreover, the angle which the mean ecliptic makes with the fixed plane increases at the annual rate of 0.48892, but this rate of increase is itself decreasing at such a rate, that at the time t this angle is

Il = 0".48892 t - 0",0000030719 t2 (486)

77. Problem. To find the change of the mean latitude of a star, which arises from the motion of the ecliptic.

Change of mean latitude.

Solution. Let

L = the *'s lat. in 1750 8 L = its change of lat. 4= its long. in 1750 — 171° 36' 10" +55.18 t (487)

= its long. referred to the node of the ecliptic 84 = its change of long. from the node; then, if Z (fig. 42) is the pole of the fixed plane, P that of the ecliptic, and B the star; we have

PZ=
=n,

ZB = 90° L, PB = 90° — L-8 L
PZB = 90° + 1, P = 90°. 4 - 81.

Draw ZC perpendicular to PB, and we have, since PZ, PC, and CZ are very small,

PC = PZ cos, P

n sin. (4+84)

or

= a sin, 1

cos. PZ: cos. PC=cos, BZ: cos. BC

or

BZ-BC

PC-PB-BZ=-L=rsin. A

8 L-
-n sin. A

(488) =-(04.48892 t-04.0000030719 + 2) sin. A.

Again, the triangle ZPB gives, by (354),

sin.j(PZB+P): cos.}(PZB-P)=tan.1 :tan.}(PB+PZ)

But

(PZB+P)=90°-181, 1 (PZB-P) = 1+101,

Mean celestial equator.

whence
A = cos. A tang. L

(489) = (0“.48892 t – 0".0000030719 t2) cos. A tang. L.

78. The mean celestial equator has a motion by which its node upon the fixed plane moves from the node of the ecliptic at the annual rate of about 50%, while its inclination to the fixed plane has a very small increase proportioned to the square of the time from 1750.

Thus, if AQ (fig. 41) is the equator of 1750, and A' Q that for the time t, so that A is the vernal equinox of 1750, and A, that for the time t.

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then A' moves from A at the annual rate of 50".340499, and this rate is diminishing so that at the time t

p = 50.340499 t 0".0001217945 ta, (490)

and the value of w in the year 1750 was

w' = 23° 28' 18",

and is increasing at a rate proportioned to the square of the time, so that

w = w + 0".00000984233 +2. (491)

79. Problem. To find the change of the mean obliquity of the ecliptic and that of longitude.

Solution. Let (fig. 41)

NA,Q=W, NA, = y.+ 1;

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Now in calculating the parts of 41 Y ф

and w

" W, which are proportional to the time, we may, since 4 and y, differ but little as well as w and w and since n is small, put

19

1+ (4+4,) = , sin. į (4 — 4,) = 3 (4 ¥,) sin. 1" tang. 2 = 4 tang. 1= t sin. l= 3 (0”.48892) t sin. 1"

} (w tw,) = w', tang. 3 (", — w)= (, — w) sin. 1"

cos. ¿ (4 4,)=1,
which, substituted in (492) and (493), give

4 –4, = 04.48892 t sin. II cotan. w (494)
-w= 04.48892 t cos. II,

(495)
which are thus computed,
09.48892
9.68924

9.68924 171° 36' 10" cos, 9.99532 sin. 9.16446

a)

1

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