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244

E'G'F'. Then, observing that p stands here in place of P', Fig. 20, we have, as for the annual variations, (App. 54),

Nut. in right ascen. =

Nut. in declin.

pr = pP sin pPs

Em+gG

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E'm E'E cos

=

(A)

(B)

(C)

(D)

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(E)

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

sin

=

Therefore,

APp. Hence, as PN = 1,

pP: sin APp: E'E

E'm E'E cos

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As pPs APP + APs

=

pp sin APp

sin

= pP sin APp cot

APP + MG APP + EG — EM =

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90)

=

+ A - 90°, and as sin (A
A, we have, (App. 13 and 14),
sin pPs
8 = sin APP sin A

=

(F) APP = sin

- cos A, and cos (A — 90) =

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cos pPs = cos APp sin A+ sin APp cos A.

Hence, (D and E,)

gG'

=

pP sin APp sin A tang D - pP cos APp cos A tang D
pP cos APp sin A + pP sin APp cos A.

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Now in the small right angled triangles OcP and pcP, we have,

Ос OP sin APO = a sin (360° — N)

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Substituting the values of pP sin APp and pP cos APP in (F, G and H),

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and,

or (B),

Pr
nut. in declin.

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n' (sin N cos' + cos N sin 0) = n' sin (N + 6',)

The observations in the last equally here to p', 6', m' and n'.

(R) Article relative to p, 0, m and n, apply

56. 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, PC 0′′.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, 2S.

=

=

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 the semicircle AGP be described, and let CL be drawn parallel to SF, and GDH perpendicular to AP. Then, the angle PCL PSF, is the mean anomaly and PSD is the true anomaly. The angle PCG is called the eccentric anomaly.

=

Assuming AC, the mean distance of the planet, to be a unit, put,

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T= time of describing semi-ellipse PDA.

t = time of describing arc PD.

By the property of the ellipse,

area PGA

or,

area PGA

area PRA :: AC CR:: area PGS: area PDS,
area PGS :: area PRA: area PDS.

But, by Kepler's second law (153),

area PRA area PDS:: T: t. Hence,

area PGA area PGS :: T:t:: 180°: PCL :: area PGA : sect. PCL.

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Or, for the arc GL, or angle GCL, in seconds, we have (App. 51. cor.),

angle GCL ew sin x

Also, since PCG= PCL + GCL, we have,

x = m + ew sin x

(B)

(C)

For most of the planets the value of e is less than 0.1, and it is not for any of them more than about 0.25. Hence, it is evident (B), that the angle GCL is generally quite small and that it is never large. The sector GCL differs, therefore, but very little from the triangle GCL, and we have (A), triang. GCL triang. GCS, nearly. Consequently, SL is nearly parallel to CG, and the angle PSL is nearly equal to PCG, the eccentric anomaly. Put,

=

2d

= 180°

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diff. of angles CSL and CLS pangle PSL = eccentric anomaly, nearly, x = p + z.

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Now, substituting p + z instead of x in formulæ (C) and observing that as z must be very small we may regard cos z = 1, and w sin z = z, in seconds, we have,

p + z = m + ew sin (p + z) = m + ew sin p + ez cos p, very nearly. m + ew sin p—P, very nearly.

Hence,

z=

1. e cos p

And, since x = p + z, we have,

(F)

m+ ea sin p—P,
x = p +
1 e cos p

, very nearly,

(G)

+ m

If x is desired with still greater accuracy, it may be obtained by taking p equal to the value of x found from formulæ (G), and recomputing with this value. This repetition is, however, seldom if ever necessary.

Now, as AC the ellipse,

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=

1, SC e, and PSD = u, we have by the property of

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9 1+ e cos u

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CS CG cos PCG- CS

(1 — e3) cos u

1 + e cos u

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1+e(1+e) cos x

1—e+ (1 − e) cos x

e cos u

cos x―e.

COS X -e

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1 + e

=

tang2x.

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1+ e

Hence,

tangu = tangx✓

1 e

(H)

Having found the value of p from the expressions (D and E), we find x from (G), and then the true anomaly u, from (H).

58. To determine the height of a lunar mountain.

Let ABO, Fig. 68, be the enlightened hemisphere of the moon, E the situation of the earth, ES' the direction of the sun from the earth, and SM a solar ray, touching the moon in O, which will be one of the points in the curve separating the enlightened, from the dark part of the moon. Also, let M be the summit of a mountain, situated near to O, and just sufficiently elevated to receive the sun's light on its top. To an observer at E, the summit M, of the mountain, will appear as a bright spot on the dark part of the moon. Let the angle MEO be measured with a micrometer, and let the angle of elongation CES', be found from the positions of the sun and moon,

at the time.

We may, without material error regard ES' as parallel to MS, EC as equal to EO, and the angle MES' as equal to CES'. Consequently, sin OME = sin MES' = sin CES'. Hence, from the triangle EMO, we

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Then, since and C are both small, we have (97), OC = EC sin♪

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Observing that Ma is the height of the mountain, we have, by substituting the value of tang C, (B),

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When the moon's apparent diameter was 32° 5.2 and her elongation CES' = 125° 8', the angle MEO for one of the mountains was found by Dr. Herschell to be 40".625. Hence, taking the moon's diameter = 2160 miles, we easily obtain from the preceding formula, the height of the mountain = 1.45 miles.

INTERPOLATION.

59. Interpolation is a method by which, from several given consecutive values of a quantity, corresponding to given values of another quantity on which it depends or with which it is connected, an intermediate value of the first may be found, corresponding to a given intermediate value of the second. Thus, if the values of an astronomical quantity x, whose variation is not very irregular, be given for the times T, T + 1hr., T + 2hrs, &c., its value for an intermediate time, as for instance, for the time T + 1hr, may be found by interpolation.

60. Let the values of the quantity x, at the times T, T + 1, T + 2, &c., be a, a', a", &c. If the first of these values be subtracted from the second, the second from the third, and so on, the remainders are called first differences. If the first of these be subtracted from the second, the second from the third, and so on, the remainders are called second differences. In like manner we obtain third differences, fourth differences, &c. Let the values of x, at the times T, T + 1, &c., and the successive orders of differences be arranged as in the following table; in which the first differences are denoted, by sa, sa', &c, the second by ̧3a, s'a', &c., and the others, in a similar manner.

A

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