ASTRONOMY. In the problem it is required to find when the trace whose equation is (6) is a parabola. In this case cos. L-sin.2 ß-sin.2 D+2 sin. B. sin. L(1+sin. ß. sin. L) = 0 844. Let D= the declination of the sun (it 23° 28' nearly,) when in either tropic, 90° - H the hour angle from midnight at which the sun rises at one place, and H + h, that at which the sun rises at the other place, and let the required latitudes of these places be called L, L. Then from the two right-angled spherical A whose legs are D, H; D, H + h and whose opposite to D are 90° L, 90° - tan. D = sin. H x tan. (90° L) cot. D x {cos. h sin. H + sin. h. cos. H} √(1 tan. D tan. L) } which gives the latitude of one place, when that of the other is known. 845. D' Let A, A' be the right ascensions of the stars, D, their co-declinations, and the required distance, measured on a great circle of the sphere. Then we have a spherical ▲ whose sides are D, D' and x, and the angle between D and D' is A - A'. Consequently cos. (A A)= COS. x cos. D. cos. D' sin. D. sin. D' and cos. x = sin. D. sin. D' x cos. (A~A') + cos. D x cos. D' which gives x. 846. To avoid unnecessary prolixity in this subject, let us once for all give a general view of what is termed by some writers the doctrine of the sphere. In the spherical ▲ whose sides are the codeclination D, the colatitude of the place L, the zenith distance Z, and two of whose are the hour 4 from noon H, and azimuth a; if any three of these quantities be given, the other two may be found by the rules and formulæ of trigonometry. we thence deduce cos. Z= cos.D. cos. L±sin.Lcos. {sin. 'D-sin. L(1+cos.α)}(5) and to find H we then have 1 sin. L sin.2 & sin. Z sin. H = x sin. a sin. D (6) Case 4. Given L, Z, H, to find D, a. This case in like manner gives {sin. cos.Zcos.L±sin.Lsin.H√ { sin. Z—sin. 'L(1+cos.H)}..(7) 1. sin. L sin. œ α cos.D Also sin. H = sin. a a 2 (10) cos.L cos.D.cos.Z±sin.Zcos.Hx{sin. D-sin. 'Z(1+cos.H)}(12) 1 sin. Z sin. H cos.L= 1-sin.2 Z sin. a cos.D.cos.Z±sin.Zcos.a {sin.'D-sin. Z(1+cos.a)}..(14) ASTRONOMY. 847. In Tycho Brahe's System of the World, the sun is supposed to move round the earth at rest, and the planets to move round the sun. Now, generally speaking, a planet will appear retrograde, in any system, when moving in that part of its actual path which is convex to the earth. (Woodhouse, p. 558.) Hence if that path be any where convex to the earth, the planet will have retrogradation. Let P be the periodic time of the Sun round the earth, p that of Venus round the sun, and R, r the distances of the sun from the earth, and of Venus from the Sun : then at the inferior conjunction, the actual orbit of Venus will be convex or concave to the Earth, according as (see 467,) Rp2 - rP2 is negative or positive. But if R = 1, we know from the tables that r = .723332. Also P 365.25 and p224.7. These quantities being substituted in the above formula, give a negative result. Consequently, Venus will appear retrograde in the Tychonic system. 848. The difference of the meridian altitudes of the two stars is the difference of their declinations. Call this given difference d, and let d' denote the difference of the altitudes at one o'clock, and a the given difference of the azimuths at that time, and 90 L the required latitude. Also let z, z' denote the zenith distances. Then from the spherical A whose sides are z, z' and d, we have But z' .. cos. z' cos. d COS. Z COS. z sin. z. sin. z' cos. (z + d') and we get by substitution and the arithmetic of sines cos. 2z tan. d'. sin. 2z = A, VOL. II. A = sin.22zA2+ 2A tan. d' sin. 2z+tan.' d' sin.2 2z. 2 R |