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In applying these formula, the declination is to be regarded as affirmative or negative, according as it is north or south.
When the mean right ascension and declination of a star is known for a given time, and the annual variations have been computed by the above formulæ, its mean right ascension and declination at a subsequent time. not distant from the former more than 20 or 30 years, or, in case of the pole star or other star near the pole, 10 or 15 years, are obtained by adding to the given right ascension and declination, the product of the corresponding annual variation by the number of years, and parts of a year, in the interval. In like manner, the right ascension and declination may be obtained for a prior time, only subtracting the products instead of adding them.
55. To find the aberration of a fixed star in right ascension and deelination for a given time.
Let s, Fig. 60, be the star, E the earth, BQ the equator, P its pole, LC the ecliptic, V the vernal equinox, N the pole of the declination circle PSG, and sNM an arc of a great circle, which, passing through the pole N, must be perpendicular to PSG. Also, let D be the point of the ecliptic towards which the earth is moving at the given time, and sD an are of the great circle, in which the plane sED cuts the celestial sphere. Then (132), the direct effect of aberration is to make the star appear to be at a point s' between s and D, such that ss' 20".36 sin Ds.
Let s'd be perpendicular to PG. Then, as ss' is very small, we may regard the triangle s'ds as rectilineal. We have, also, without sensible error, G's
Gd, Pe Ps, and s'd =es. Now, since G's' Gd Gs
sd, it is evident the effect of aberration on the star s, which is in the first quadrant, is to diminish the right ascension by the quantity G'G, and the declination by sd. Consequently,
Aber. in right ascen.
Put a = 20′′.36, A — VG = star's right ascension, D Gs declination, S = sun's longitude, and a = GVF obliquity of the ecliptic. Then, using small arcs and angles instead of their sines, we have,
es = ePs. sin PeePs. sin Ps G'G cos D;
sin F sin VF, or, sin F
also, es = s'd ss' sin DsF-20′′.36 sin Ds sin DsF a sin Ds sin DsF. Hence, G'G cos D a sin Ds sin DsF.
But, sin DsF: sin F:: sin DF: sin Ds, or, sin Ds sin DsF sin DF; and in the right angled spherical triangle VGF,
a cos A sin D
Then (A), Aber. in right ascen.
G'G =m sin DF Again, in the right angled triangle VGF, we have (App. 49), tang VG cos & tang VF, or, cot VF 90° - Ÿ. p. Then, since VD 180°
a sin A
cos D sin VF
= cos & cot A
a sin A
a sin A cos D cos p
Aber. in right ascen. = m sin (S + ¢) Now, in the small triangle s'ds, we have, sd ss' cos s'sd ss' sin MsD But, sin MSD: sin M:: sin MD: sin Ds, or, sin Ds sin MsD = sin M sin MD; and, sin MV: sin NV:: sin MNV, or, sin GNs: sin M
a sin Ds sin MsD.
cos A sin D
sin NV sin GNs
a sin Ds sin MsD
a cos A sin D
sin ɛ cot D
900. Then, sin MD sin MV = cos 0, and cot MV
= n sin MD
cos & tang A
a sin M sin MD
sin & cot Dcos & sin A
sin (MV + VD) = sin (S + 0), tang 0, or, cot MV tang 0.
*When MV is less that 90°, as in the figure, it is 90° + ◊ 360° that is to be regarded as equal to MV.
Aber. in declin.
n sin (S + 0)
The increase or diminution of an arc by 180°, changes the sign of its sine or cosine, but does not affect its numerical value. It is, therefore, evident that if the value of the arc & be increased or diminished by 180°, the expression for the aber. in right ascen. (F) will still be true; for the signs of both factors m and sin (S + ) being thus changed, there will be no change in the sign of the product. Hence, we may always take affirmative, and not exceeding 180°. Similar observations apply to the
The quantities p, 0, m and n change but little for a number of years, and therefore, when once computed for any star, they serve for a long time in computing the aberration of that star. Table IX contains the values of these quantities for 30 principal fixed stars. The values of m and n are all made affirmative by increasing the values of and by 180°, when requisite.
56. To find formula for the lunar nutation in right ascension and declination.
To obtain these formulæ we have recourse to certain results established by Physical Astronomy. It has been proved that the phenomena of nutation may be explained on the supposition that the pole of the equator, instead of moving strictly in a circle about the pole of the ecliptic (126), moves in a small ellipse about the mean place of the pole, that is, around that point in the circle at which the pole would be if the nutation did not exist, and in a period equal to that of the moon's nodes. The major axis of this ellipse is situated in the solstitial colure, and is to the minor axis in the ratio of the cosine of the obliquity of the ecliptic to the cosine of twice the obliquity. The major axis has been found to be equal to 18".44;* and hence, the minor axis is 13′′.73.
Let ELF, Fig. 61, be the ecliptic, N its pole, NLM the solstitial colure, EMF the mean equator, P the mean place of the pole, AbCd the ellipse in which the pole is assumed to move, and ABCD a circle about the centre P. Then, according to the investigation in Physical Astronomy, if the arc ABO be made equal to the longitude of the moon's node, and Oc be
* This is the value given by Struve, in No. 426 of the Astr. Nach., as recently obtained from a series of observations made by him at Dorpat.
drawn perpendicular to NL, the point p, in which it meets the ellipse, will be the true place of the pole, and the great circle E'G'F', described about the pole p, will be the true equator. Let s be the situation of a star, and let pr be drawn perpendicular to the declination circle PG, and Em perpendicular to E'G'F'. Then, observing that p stands here in place of P', Fig. 20, we have, as for the annual variations (App. 54),
E'm + gG′
Nut. in right ascen. =
prpP sin pPs
F'm E'E COS ε
gGpr tang G's pr tang DpP sin pPs tang D
Pr pP cos pPs
Now in the triangle pPN, we have,
sin PN: pP :: sin PpN: pNP, or LL', or E’E.
But, as Npc may be regarded as a right angle, or Np parallel to Nc, we have, PpN APp. Hence, as PN,
sin ɛ : pP : : sin APp : E'E
Therefore, E'm E'E cos pP sin APp cot & As pPs APp + APs APp + MG APP + EG +A 90°, and as sin (A
cos A, and cos (A
90) sin A, we have (App. 13 and 14), sin pPs sin APp sin A cos pPs cos APp sin A
We have, also,
pP sin APp sin &
a sin N: pc
Hence (D and E),
9G' pP sin APp sin A tang D-pP cos APp cos A tang D (G) PrpP cos APp sin A + pP sin APp cos A. .
Now, in the small right angled triangles OcP and pcP, we have,
a sin N,
Oc pc=pP sin APp. But, by a property of the ellipse,
DP : dP : : Oc: pc;
a : b
cos APp cos A
sin APp cos A.
arc ABO = longitude of moon's node ;
N = APO.
b sin N.
b sin N.
OP cos APO
a cos N.
Substituting the values of pP sin APp and pP cos APp in (F, G and H), we have,
E'm + gG cos N.
Then, tang o'
and, E'm + gG' or (A),
b (cote + sin A tang D) sin N — a cos A tang D
b cos A sin N + a sin A
- b (cot & + sin A tang D)
b cos A
a sin A
a cos A tang D =
a cos A tang D
b(cot & + sin A tang D)'
m' (sin N cos q' + cos N sin q′) nut. in right ascen. m' sin (N + p′).
n' cos o'
n' sin o'.
= m2 cos p'
m' sin p'.
tang A, n' =
a cos A tang D
a sin A
Pr=n' (sin N cos 0' + cos N sin e') = n′ sin (N + 0), nut. in declin. n' sin (N+0')
(R) The observations in the last Article relative to p, 0, m and n, apply equally here to ', e', m' and n'.
To find the solar nutation in right ascension and declination. It has been found that the solar nutation may be explained by assuming the pole of the equator to describe a small ellipse about its mean place, in like manner as for the lunar nutation. For the solar nutation, we have, Fig. 61, PC0′′.555, Pd = 0".500, and the arc ABO twice the sun's longitude. Hence, if S be the sun's longitude, the formulæ for the solar nutation in right ascension and declination will be the same as for the lunar, except that, instead of the values of a and b in the last Article, we shall have, a = 0′′.545 and, b = 0′′.5, and, instead of N, we shall have, 28.
Note. As these values of a and b are about of their values for the lunar nutations, we may obtain approximate values of the solar nutations, by computing as for the lunar, only using, in (N and R), 2S instead of N ̧ and taking of the results.
57. Given the eccentricity of the orbit of a planet and the mean anomaly, to find the true anomaly.
Let the semi-ellipse PDA, Fig. 62, represent one-half the orbit, C the centre, S the place of the sun in one focus, D the place of the planet in its orbit at any time, and F the place at which it would have been at that time if its angular motion had been uniform. On the diameter AP let