To find the declination of a star. 7. Calculate the moon's right ascension and declination, when its latitude is 5° 0' 7" N., its longitude 64° 54' 1", and the obliquity of the ecliptic 23° 27′ 45′′. 68. Problem. To find the declination of a star. Solution. I. Observe its meridian altitude, and its declination is at once found by one of the equations [404 406.] II. If the star does not set, and both its transits are observed, we have 69. Problem. To find the position of the equinoctial points. Solution. Since the right ascension of all stars is counted from the vernal equinox, and since the two equinoxes are 12 apart, the present problem is the same as to find the right ascension of some one of the stars, which may afterwards serve as a fixed point for determining the right ascension of the other stars. Observe the declination of the sun for several successive noons near the equinox, until two noons are found between which its declination has changed its sign; and observe also the instant of the sun's transit across the meridian on these days, by a clock whose rate of going is known. Then, by supposing the sun's motions in declination and right ascension to be uniform at this time, which they nearly are, the time of the equinox, that is, of the sun's being in the equator, is found by the proportion To find the right ascension of a star. the whole change of declination: either declination = the sideral interval between the transits - 24" the sideral interval between the transits of the equinox and that of the sun at this declination ; (479) and this interval is the difference between the right ascensions of the sun at this declination and the equinox. If the passage of a star had been observed in the same day, the right ascension of the star would have been the interval of sideral time of its passage after that of the vernal equinox. 70. EXAMPLES. 1. If the sun's declination is found at one transit to be 7' 9".5 S., and at the next transit to be 16′ 31′′.1 N.; what is the sun's right ascension at the second transit, if the sideral interval of the transits is 24h 3m 388.21 ? Solution. 79.5+16′ 31.1=23′ 40′′.6=1420′′.6= ar. co. 6.84753 2. If the sun's declination is found at one transit to be 18′ 38.8 S., and at the next transit to be 5' 3.2 N.; what is the sun's right ascension at the second transit, if the sideral interval of the transits is 24h 3m 388.4? Ans. 0h 0m 468.6. 3. If the sun's declination is found at one transit to be 5' 57".9 N., and at the next transit to be 17′ 26′′.3 S.; what is the sun's right ascension at the second transit, if the sideral interval of the transits is 24h 3m 35$.71 ? To find the obliquity of the ecliptic. 71. Problem. To find the obliquity of the ecliptic. Solution. I. Observe the right ascension and declination of the sun, when he is nearly at his greatest declination; that is, when his right ascension is nearly 6 or 18h. If he were observed at exactly his greatest declination, the observed declination would obviously be the required obliquity. But for any other time, the sun's declination and right ascension are the legs of a right triangle, of which the obliquity of the ecliptic is the angle opposite the declination. Hence sin. O's R. A. tang. obliq. (480) tang.'s Dec. Now if we put h the diff. of 's R. A. and R. A. of solstice, sin. (obliq.-'s dec.)=(obliq.-'s dec.) sin. 1" tang.2h sin. (obl.+'s dec.) (483) obl.-'s dec.: =cosec. 1"tan.2 h sin. (obl.+'s dec.) (484) h2 cosec. 1" tan.2 1a sin. (obl.+'s dec.) and the second member of (484) may be regarded as a correction in seconds to be added to the 's dec. to obtain the obliquity, and the obliquity in the second member need only be known approximately. To find the obliquity of the ecliptic. 72. EXAMPLES. 1. The right ascensions and declinations of the sun on several successive days were as follows: June 19, R. A. = 5h 50m 533, Dec. = 23° 26′ 45′′.2 N. To find the obliquity of the ecliptic. Solution. Assume for the obliquity the greatest observed declination, or 23° 27′ 45′′, and the corrections of all the observations may be computed in the same way as that of the first, which is thus found, cosec. 1" tang. 2 1 = 225 tang. 1′′ h9m 78 - 5478 6.43570 2 log. 5.47598 23°26′ 45′′+23° 27′ 45′′ - 46° 54′ 30′′ sin. 9.86348 23 27 44 .9 In the same way the 2d observation gives the 3d observation gives the 4th observation gives 23 27 45 .2 23 27 45.3 23 27 45.3 the 5th observation gives sum 117° 18′45′′.5 The mean = 23° 27′45′′.1 |
