signs, and the sum will be the aberration in latitude. Considering the true geocentric latitude as affirmative or negative, according as it is north or south, add to it the aberration in latitude, and the result will be the apparent latitude. For the Horizontal Parallax and Semidiameter. Add together the constant logarithm 0.93337, the log. sine of the true geocentric latitude, the arithmetical complement of the log. sine of the heliocentric latitude, and the arithmetical complement of the logarithm of Mercury's radius vector, rejecting ten from the index of the sum, and the result will be the logarithm of the horizontal parallax, in seconds. To the constant logarithm 9.57584 add the logarithm of the horizontal parallax, and the sum, rejecting ten from the index, will be the logarithm of the semidiameter in seconds. Note 1. The true geocentric longitude and latitude and the horizontal parallax of the planet Venus, may be found in the same manner. The constant logarithm 9.98302 added to the logarithm of the horizontal parallax, and ten rejected from the index of the sum, will give the logarithm of the semidiameter. 2. The geocentric longitude and latitude of a superior planet may also be found in the same manner, except that, in finding the arc B, one ten must be retained in the index of the log. tangent, and the sum of C and A must be taken for the elongation, instead of their difference. 3. At the time of conjunction, the angles of commutation and elongation are each nothing, and, consequently, the geocentric latitude and the parallax cannot be found by the rule. In this case, add the log. cosine of the heliocentric latitude to the logarithm of the planet's radius vector, rejecting ten from the index of the sum, and the result will be the logarithm of the curtate distance of the planet. Take the difference between the curtate distance and the earth's radius vector, or the sum of the two, according as the conjunction is inferior or superior, and add together the arithmetical complement of the logarithm of this difference or sum, the logarithm of the planet's radius vector, and the log. sine of the heliocentric latitude, rejecting ten from the index of the sum, and the result will be the log. tangent of the true geocentric latitude. Also, add the logarithm of the above mentioned difference or sum to the logarithm 0.93337, and the sum will be the logarithm of the parallax. EXAM. 1. Required the geocentric longitude and latitude, the horizontal parallax, and the semidiameter of Mercury at the time given in first example of the last problem. Mercury's true longitude Elongation 17° 0' gives, tab. LXXIX., part I. 16 59 53 E 100 13 27 19" Aber. in long. Hence the apparent geocentric longitude is in this case the same as the true. 2. Required the apparent geocentric longitude and latitude, the horizontal parallax and the semidiameter of Mercury at the time given in the second example of last problem. Ans. App. geoc. long. 230° 59' 53", app. geoc. lat. 0° 10′ 42′′ N.; hor. par. 6.0"; and semidiam. 2.3′′. PROBLEM XXV. To Calculate a Transit of Mercury. 1. For Greenwich mean noon of the day on which the transit occurs, find the sun's longitude, hourly motion, the apparent obliquity of the ecliptics and Mercury's heliocentric longitude. To Mercury's mean anomaly, add 10′ 14′′, the mean hourly motion in anomaly, and the sum will be the mean anomaly, an hour after noon. With this anomaly take out again the equation of the centre, and adding 10′ 14′′ to it, subtract from the sum the equation of the centre at noon. The remainder will be Mercury's hourly motion in longitude, nearly. To the sun's longitude, add 180° 0′ 20′′, and the sum will be the earth's longitude. Then, as the difference of the hourly motions of Mercury and the Sun: the difference of their longitudes ::1 hour: an interval of time. When the earth's longitude is greater than Mercury's, add this interval to the mean noon at Greenwich, but when it is less subtract the interval, and the sum or remainder will be the approximate time of conjunction in longitude. 2. Let T be the approximate time of conjunction, taken to the nearest whole hour, and t an interval of two or three hours. For the times T — t, and Tt, find the sun's longitude, radius vector and declination, and Mercury's heliocentric longitude, latitude, and the radius vector, and thence the apparent geocentric longitude and latitude. Take half the sum of the sun's mean anomalies at the times T t and Tt, and it will be the mean anomaly at the time T. In like manner find the radius vectors of the earth and Mercury at the time T. With these find the sun's semidiameter and Mercury's equatorial horizontal parallax and semidiameter. Add together the semidiameters of the sun and Mercury, and, expressing the sum in seconds, call it . To the constant logarithm k. 7.95071 add the logarithm of the sun's semidiameter in seconds, and the sum, rejecting ten from the index, will be the logarithm of the sun's horizontal parallax. the times T Take half the sum of the sun's longitude, at T +t, and it will be his longitude, at the time T. the sun's tabular mean longitude, at the time T. the equation of the equinoxes in right ascension, found from table XXX., with the argument N, at the time T t, and the result will be the sun's mean longitude from the true equinox. With the sun's longitude at the time T, and the obliquity of the ecliptic, find his right ascension. Subtract the right ascension from the corrected mean longitude, and the remainder, converted into time, would be the equation of time. = 3. Let L Mercury's geocentric longitude, a her geocentric latitude, her horizontal parallax, L sun's longitude, A' his right ascension, D' his declination, negative when south, and apparent ɛ == obliquity of the ecliptic. Then taking the values of the quantities at the times T―t and T+t, respectively, find, for each time, the values of p and q from the following formulæ : = Log. a log. (LL) + log. cos+ Ar. Co. log. cos D', log. clog. a + log. tang+log. cos L', log. d = log. b + log. tang. ε + log. cos. L', Tt, and, dividing the result by the number of the hours in 2t, the tient will be p'. Do the same with the values of q, to obtain q'. quo 4. Take T' and T", to represent the times Tt, and T + t, respectively, and let h the difference of the semidiameters of the sun and Mercury. Then, using the values of p and q, at the time T', find N, d, F, t, t, and t", by the following formula; observing that the arc N is to be negative, and will, therefore, be between 0° and 90° when its cotangent is affirmative, but between 90° and 180°, when the cotan gent is negative. Log. cot. N= log. + Ar. Co. log. p'; log. d = log. cot N+ log. p; log. cos F= log. sin N + log. (d-q) + Ar. Co. log. h; The arc F to be affirmative and less than 180°; log. tlog. cos (NF) + log. h + Ar. Co. log. p'; log. 'log. cos (N — F) + log. h + Ar. Co. log. p'; To find auxiliary quantities for computing the effect of parallax on the times of ingress and egress. 5. Let H' be the hour angle at Greenwich, and find its value for the time of first contact, by adding the equation of time to the mean time, and converting the interval between the resulting apparent time and noon, into degrees. And using the value of D' at the time T', find, log. G log. A 3.5563 + log. (π-7') + log. sin N-+- Ar. Co. log. sin F + Ar. Co. log. p'; log. Glog. cos (N + F); log. B log. Glog. sin (NF) log. Blog. cos D'; log. C log. tang M*=log. B+Ar. Co. log. A-+-log. sin D'; log. Dlog.A+Ar.Co.log.cos M; |
