A SIMPLE METHOD OE EIXDIKG THE l^ADII 

 VECTORES OE A COMET'S ORBIT. 



P.v THOMAS ASH1-: ('Ri;(;\N. 



To illustrate this method I hn\(' taken Comet 

 1908, c, the positions of which are those given hy 

 Professor Kobold in the Asfronomische Nachrichten 

 for Berlin mean midnight on the 2n(l, 4th and 5th 

 October, 1908, the corresponding Greenwich mean 

 time being 11 hours 6 minutes 25 seconds. 

 According to the Eii^jUsIi Meclujiiic, Vol. 

 LXXXN'III. No. 2274, page 282, these positions are 

 subject to a correction of 90 seconds to add to the 

 Right .Ascension and l'-5 to subtract from the 

 Dcrlination ; thev then are as follows : — 



2nd. 4th. 6th. 



Comet's R.A. ... J13' 51' 0" 308° 10' 30" 303" 34' 15" 

 Do. Dec. N.... 70° 4' 48" 67° 26' 42" 64° 30' 42" 



The corresponding places of the sun being — 



2nd. 4th. 6th. 



Snn's K.A. ... ISS 34' 21" 190° Zi' 26" 192° 12' 32" 



Do. Lont;. ... 189 20' 20" 19riS'34" 193° 16' 52" 



Find the earth's orbital motion between the 

 obserxations. that is, from the 2nd to the 4th and 

 from tile 4th to the 6th, and resolve each separately 

 into two components, one a motion almost directh' 

 away from the comet, the other a motion in the same 

 direction as that in which the comet, whose motion 

 is retrograde, is moving. Since the comet is at a 

 finite distance, its positions are dependent upon our 

 point of viexN', that is, upon the place of the earth in 

 her orbit. The Declinations must therefore be 

 reduced to the mean Equinoctial and the Right 

 Ascensions corrected for the angular difference 

 between the directions of the first point of Aries as 

 seen from the sun and from the earth, at the instant 

 of observation. Also, with the components of the 

 earth's orbital motion we must reduce the places of 

 the two last observations to that of the first. The 

 corrected and reduced places of the comet then 

 are : — 



2nd. 4th. 6th. 



R.A 305° 16' 39" 295= 21' 16" 286° 28' 30" 



Dec 71° 13' 48" 68' 51' 24" 66° lO' 42" 



And these referred to the plane of the Mcliptic 

 give— 



2nd. 4th. 6th. 



Geoc. Long. ... 286= 44' 32" 282° 58' 42" 279' >' 50" 

 Geoc. Lat. ... 49° 49' 3Z" 46 32' 47" 43° 14' 42" 



From the foregoing data we next find the \alue of 

 r at the first observation, for the angle P = 97"24' 12" 

 and of the two sides, PS' = 40= 10' 28" and PS = 90°', 

 whence SS' is found to equal 1-57937 in parts of 

 radius, and this is the value of r at iiiiif distance. 

 Now, of the three bodies with which this problem 

 deals, viz., sun, earth and comet, since the sun may 

 be considered as stationai"}-, it is evident that if we 

 eliminate the effects of the earth's orbital motion in 



the iiUer\al between iIk' (ibser\alions from the angle 

 P, we will have tlu; values of P' and of P", and a 

 siinple propf)rtion will give the values of r' and r". 

 assuming for the present that the cornet's motion is 

 uniform, therefore — 



97°-403 : 96''-0694 :: 1-57937 : 1-55745 r' 



Also— 



y6°-0694 : 94°-59526 :: 1-55745 : 1-533841 r" 



These are the values of r' and r". determined from 

 that of r on the supposition that the comet's motion 

 is uniform. This, we know, is not the case. The 

 comet's motion is increasing ; therefore r' and r" will 

 both be too great by a quantity equal to the second 

 difference of r, r' and r", which must be subtracted 

 from the values just found. We have next to find 

 ■■ c," or the length of the arc of the orbit which the 

 comet appears to describe between the first and last 

 observations. Using the same formula, we have ; 

 P = 7"41'42". PS' = 40° 10' 28", PS = 46° 45' 18", 

 which gives S'S in parts of radius as -142097, to 

 which we must add -00227 the second difference 

 alread\- found, and we obtain finally : — 



r = 1-57937. r"^ 1-53157. c = -144367 

 all at unit liistcjncc. w ith w hich to satisfy the equation. 



3651 

 ^'=-12^ 



2 t = 8 days also 



(r + r" + cli! - (r + r ' - c)i 



365-25 

 37-6992 



9 ■ 6885 ; therefore. 



8 



^^r-^. = -8257 which is the number that we have 

 9 -6880 



to obtain by trial values of r, r" and c, in order to 

 satisfy the equation. Using figures to the fourth 

 decimal place only, I find that 1 •0537 gives -8255, 

 which satisfies the equation within -0002 ; therefore 

 the radii vectores of the comet's orbit on the dates 

 computed are : — 



1 -6641 

 1 - 6642 



1-6389 1-6137 

 1-63SS 1-6136 



Professor Kobold givc« 

 To find the corresponding distances of the comet 

 from the earth, or the values of D, D' and D". In 

 the triangle CSE, formed by the comet, sun and 

 earth, we have foimd the side CS, which equals r, 

 and SE = R, is known from the Greenwich date of 

 the observation, we have therefore to find the angle 

 at E. Taking the comet's R.A. at the first observa- 

 tion, add to it the angle T SE and reduce the 

 observed declination to the apparent equinoctial of 

 the date, we then have 



corrected R.A. = 322° 25' 21"; rednced Dec. = 71° 49' 3". 

 and these referred to the plane of the Ecliptic give 



