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13t is occasioned by the action of the planets, and diminishes the effect of the actions of the sun and moon. This affecting only the ecliptic, does not affect the north polar distance.

The terms depending on the longitude of the moon's node serve for determining the effect of what has been called the lunar nutation. The terms depending on the longitude of the sun serve for determining what has been called the solar nutation. The terms depending on twice the longitude of the moon, and twice the longitude of the sun, are too small to be noticed, except in the nicest researches.

I. Precession in right ascension (A), and north polar distance (NPD).

Taking the angle FNM (the change of longitude in the preceding problem 50", 32 t, t not exceeding a few years, by equation (a).

=

The luni-solar precession in right ascension=

50", 32t. (cos + sin O. sin A. cot NPD) =

46", 18t+20", 04 t. sin A. cot NPD,

from which subtracting 0", 13t, the general precession in right ascension will be had.

By equation (6) the precession in NPD =

-50", 32 t.sin O. cos A-20", 04 t. cos A.

Cor. 1. The precession in right ascension will be subtractive, when sin A. cot NPD is negative and greater than cotO, which can never happen but when cot NPD is greater than cot O. Therefore the right ascension of every star, the polar distance of which is greater than 23° 28′, is always increased by precession.

Cor. 2. Precession affects equally the north polar distance of every star having the same right ascension.

II. Lunar nutation in right ascension and north polar

distance.

Let the angle FNM-17", 30. sin N, N being the longitude

of the moon's node, and the change of obliquity=9, 25. cos N. See equations (e) and (f). Then by equations (a), (d).

Nutation in right ascension =

— 17", 30. sin N. (cos O+ sin O. sin A. cot NPD),

+9", 25. cos N.cos A. cot NPD =

=

-15, 87. sin N-[8". 07. cos (AN)+1", 18. cos (A+N) ]. cot NPD.

By equation (b) and (c),

The nutation in north polar distance =

17", 30. sin N. sin O. cos A-9", 25. cos N. sin A,

– 8′′, 07. sin (A—N) — 1′′, 32. sin (A + N).

III. Solar nutation in right ascension and north polar distance.

Let the angle FNM1", 25. sin 2 S, S being the longitude of the sun, and the change of obliquity=0, 54. cos 2 S, and we obtain by a process similar to that in II.

Solar nutation in right ascension =

-1", 15. sin 2S-[0, 51. cos (A-2 S) + 0, 02. cos (A +2S)]. cot NPD.

Solar nutation in north polar distance =

-0", 51. sin (A-2 S)-0", 02. sin (A + 2S).

The solar nutation has also been called the semi-annual equation, because depending on twice the sun's longitude it goes through its period in half a year.

Prop. XIX. To deduce the effect of the aberration of light O on the right ascension and declination of a star.

Let S (fig. 63) be the star; ED the equator, N its pole; MELH the ecliptic, and MS a great circle perpendicular to the circle of declination NSHD.

Let L be the point of the ecliptic towards which the earth is moving, which is always 90° behind the place of the sun, Take

Sp so that sin Sp: sin SL :: vel. of earth: vel. of light :: sin 20": rad, or which comes to the same,

Sp: 20":: sin SL : rad,

then p is the apparent place of the star as affected by aberration (art. 281 and 283).

Draw pq parallel to the equator, and then Sq= aberration in declination. Also pNq= aberration in right ascension. Let n = 20"1.

1..pq: Sp:: sinp Sq: rad

Sp:n:: sin SL : rad

pNqpq:: rad: sin Nq, or sin NS sufficiently near.

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2..Aberration in decl.= Sq=Sp.cosp Sqn.sin SL. sin MSL

=20". sin M. sin ML.

From the above expressions very convenient practical formulas may be deduced.

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and n. sin M are constant for a given star,

the

aberration in right ascension varies as the sine of LH, and the aberration in declination varies as the sine of LM.

1. Let L= the sun's longitude, then, because the aberration in right ascension (A) varies as sin LH=

sin (EH-(L-90°)) sin (90° — (LEH)) =

=

cos (LEH), we may express it by m. cos (Les K), and supposing m positive, KEH will be the sun's longitude when the aberration is a maximum and positive. K and m may be found in the following manner:

tan K (EH) =

tan ED COS HED

tan A cos O'

When L 90°, the point L is in E.

=

Then m (cos 90-K) = m. sin K = ab. in R. ascens. =

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It only remains to be known to what quadrant K belongs: tan K has the same sign as tan A.

Because m is positive

sin A sin K

must be negative, cos D being

always positive. Therefore A and K must be in opposite semicircles; and, their tangents having the same signs, they must be in opposite quadrants.

2. Again, because the aberration in declination varies as sin ML sin (MEL-90) = cos (180-ME-L), we may express it by m'. cos (LK'), and supposing m' positive, K'— 180°— ME will be the sun's longitude when the aberration is a maximum and positive.

K' and m' are found in the following manner:

tan K'tan ME.

By spherical trianglea PEM

tan ME =

sin PE

cot P. sin E+ cos PE. cos E

Therefore, because cot P-cot SPD. (= decl.)

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Also sin M. sin ME sin PE. sin P= cos A. sin D.

Therefore m' =

=

-20".cos A. sin D
sin K

The quadrant to which

K' belongs is thus determined, the sign of tan K' is known from its value above given. The sign of m' being positive, cos A. sin D and sin K' must have different signs, but knowing the signs of tangent and sine of an arc the quadrant is known.

a Cagnoli Trig. p. 270. Luby's Trig. p. 73. art. 120.

The quantities m, m', K, K' being constant for the same star, render this a very concise method of computation, when the aberrations of the same star for several days are required, or when tables of the aberrations of a given star are required.

Indeed m and m' are not strictly constant on account of the variable velocity of the earth, but the variation is so small that usually it is not considered.

When a single place of a star is required, then general tables are more convenient.a

The three last problems containing the effects of refraction, precession, nutation solar and lunar, and aberration of light, are of constant use in practical astronomy.

Prop. XX. Given the mean anomaly of a planet, to find the true anomaly.

Let the semiaxis major of the planet's orbit = 1, its eccentricitye, m the mean anomaly, the eccentric anomaly, and y the true anomaly.

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then m+d

eccentric anomaly nearly, (art. 233) for which sub

stitute p, and let p+x=2, the eccentric anomaly.

Now by the same article

(fig. 33) ACL (m) = ACI (p+x) + LCI.

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Hence the seconds in LI, or the angle LCI =

SC. sin ICD

sin I"

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a Woodhouse's Astr. p. 466, &c. Conn. des Temps. 1810, p. 422 and 442.

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