Page:Encyclopædia Britannica, Ninth Edition, v. 6.djvu/212

Rh COME T The arc u&quot; - u is equal to the difference of true ano malies, and the true anomaly at the first observation (v) will be obtained from comet discovered at the Observatory of Marseilles by M. Sorrelly on the 6th of December 1874, employing three observations taken at that Observatory on December 7, tan. 4t/ = cotan. l(u&quot; - u ) - ^ or from . . . (IX.) tan. v The perihelion distance (q) = r. cos. 2 ^v = r&quot;. cos. 2 |z&amp;gt; &quot;. The longitude of the perihelion, reckoned on the ecliptic to the node and thence on the orbit, is given by ir= & + u -v. . for direct motion TT = & + v - u. . for retrograde motion As a further verification of the calculations, we have which should give the former value of L We have now only to determine the time of perihelion passage (T), by finding the interval between the first observation and the perihelion (T) from v and q, by means of the equation r= &quot;&quot; tan - tan - (2g)t Tc being the Gaussian constant [8 235581 4] K + P,5228787]tan.|,,.) (XII.) Similarly we may find the interval from the third observation to perihelion by substituting v &quot; for v ; the times thus separately determined should agree, and this agreement will afford a third check upon the accuracy of our work. Thus the whole of the elements of the parabolic orbit are found, and it is always desirable to ascertain how the geocentric place calculated from these elements for the time of the second observation agrees with the position ob served ; the first and third places are necessarily represented. In the computation of a geocentric position from para bolic elements we may proceed thus : Find the interval from perihelion passage to the time for which we require to compute (t T), in days and decimals. Put cotan. 2 / = 3& (2?)- 2 (t - T) cotan. = 3 . / cotan. v then tan. i v 2 cotan. 2 (XIII.) We have thus the true anomaly and radius-vector. Then, if the motion be direct, cos. . cos. (9 - & ) cos. (v + IT - Q&amp;gt; ) } cos. A sin. (0 - &) = sin. (v + ir 8,). cos. i },. (XIV.) sin. &amp;gt; =sin. (v + ir- a). sin. i ) or, if the motion be retrograde, cos. A . cos. (a - 6) = cos. (v - w + & ) ) cos. A. sin. (& - 0)_sin. (v-w+ ft), cos. i &amp;gt;. . . (XV.] sin. A i =sin. (V-IT + &). sin. i } equations which give the heliocentric longitude and latitude (6, A). The geocentric longitude and latitude (a, /8) and the true distance from the earth (A) are then obtained from A . cos. $ . sin. (o - A) = r . cos. A . sin. (0 - A) A. cos. . cos. (a- A) = r cos. A. cos. (0- A) + R ^ (XVI.] A. sin. 8 r. sin. A If the position of the comet as referred to the equator is required, Put tan.. N= sin a Then tan - R - A - (XVII. n. a tan. . Decl. =tan. (N + e ). sin. R.A .6, and 26, viz. : Mean Time at Marseilles. 1874. H. M. s. Dec. 7 6 40 52 16 16 43 31 .. 26 6 18 Comet s Right Ascension. Comet s Declination. Greenwich Mean Times. Right Ascension (arc] Declination. 7-26339 240 6 8 + 36 38 50 16-68190 242 46 27 + 45 37 37 26-24751 246 14 45 + 55 34 12 16 24-52 +36 38 50 16 11 5-82 +45 37 37 16 24 58-97 +55 34 12 Converting the right ascensions into arc, and the times nto decimals of a day, after subtracting 21 m 35&quot; from them, or reduction to the meridian of Greenwich, we have Dec. The obliquity of the ecliptic, from the Nautical Almanac, was 23 27 28&quot;, and hence, by the formulae p. 183, we find the following positions referred to the plane of the ecliptic, and interpolating for the above times from the same ephemeris the corresponding longitudes of the sun and log. radii-vectores of the earth, correcting the sun s longitudes for aberration, and reducing all to mean equinox of 1875 0. .... o // + 55 31 52 3&quot; + 64 35 27 j3 &quot; + 74 17 30 Log. R 9-9933590 R&quot; 9-9929263 R &quot; 9-9926916 The right ascension and declination are thus converted into longitude and latitude for the first observation : a ....225 2 30 ....221 47 16 ...212 44 3 A ...255 32 49 A&quot;.... ....265 7 46 A &quot;.... ....274 52 15 R..A 5 Log. tan. 8 Log. sin. R.A Log. tan. N N t N-6 Log. cos. (N ) Log. tan. R.A Log. cos. N Loo 1, tan. . a 240 6 8 + 36 38 50 Log. sin. a - Log.tan.(N - e) - -9-8497775 -0-3135946 + 9-8715409 - 9-9379771 Log. tan. (3 +0 1633721 + 55 31 52 - 9.9335638 139 21 56 23 27 28 Precession to 1875 Nutation in longi tude with contrary sign to Nautical Almanac + 3-3 ....+ 7-5 115 54 28 - 9-6404058 + 0-2403519 - 9-8807577 - 9-8801731 Correction to longitude. . . . + 10 8 + 0-0005846 Correction . . Long. M.Ea. 1875-0 // 225 2 19 + 11 ...225 2 30 And so for the second and third positions. The interpolation of the sun s longitudes and the log. radii-vectores of the earth from monthly page iii. of the Nautical Almanac requires no illustration. We now form the angles a A&quot;, a&quot; - A&quot;, a &quot; - A&quot;, &amp;lt;fcc., and take out the sines and cosines required; and it is always convenient to have these functions and other of the principal quantities copied in plain figures on a paper separate from the calculations. Thus we have, -A&quot;.. ...319 54 44 -A&quot;.. ...316 39 30 &quot;- A&quot;.. ...307 36 17 A .. 329 29 41 &quot;-A&quot; &quot; ...297 51 48 -A &quot;.. ...310 10 15 &quot;- A .. ...317 11 14 &quot;- A .. ... 19 19 26 &quot; o .. .. 347 41 33 Sine. -9-8088592 -9-8365440 -9-8988564 -9-7055368 -9-9464841 Cosine. +9-9352968 +9 6696554 +9-8096060 +9-8654465 +9.9748170 +9-9899024 As an example of the calculation of the orbit of a comet by Olbers s method, we will compute the elements of the We have t&quot;-t = 9-41851 days; t &quot; -t&quot; = 9-5G561 days;
 * &quot;&quot;- - 18-98413 days.