KNOWLEDGE 



[Oct. 13, 1882. 



the orbit has been entirely changed) as to leave scarcely 

 room for any doubt that the coiuet has como back again 

 lont; before it was expvctod — how soon to rt-turn yet again, 

 and how soon to be liiially absorbed by the sun, it were at 

 present somewhat rash to say. But in another paper I 

 have given some evidence which makes it look very much 

 as though the end of the comet were not many monthi otl". 



In the next numlnT I sliall consider at length how that 

 destruction of the comet, wluch now seems certain, is 

 likely to be brought about, and witli what probable effects 

 on the sun aiul on this eartk 



(To he coutlnutd.) 



WHEN WILL THE COMET RETURN ? 



THE following, besides dealing with the comet, shows 

 how the motion of a body moving in a very eccentric 

 orbit in a given period may be conveniently dealt with. 



I have shown in my " Geometry of Cycloids," that the 

 common cycloid may be used for measuring the motion of 

 a body in a very eccentric orbit around the sun— at least, 

 this follows directly from what is illustrated there in tlie 

 plate facing p. 209, where the curves for all orders of 

 motion, from motion in a circle to motion in a straight line, 

 under gravity, are given together. We see in that picture 

 a certain curve for an eccentricity of nine-tenths, and this 

 curve lies very close to the common cycloid representing 

 the time-curve for motion in a straight lino (eccentricity 

 unity). How much nearer (in fact, not distinguishable 

 from the cycloid), would be the curve for an eccentricity of 

 ■Ci'Jl'>0 such as the comet of If'lS had! Supposing the 

 period of the comet reduced to one year, the eccentricity, 

 assuming the perihelion distance 500,000 miles, would be 



92,.500 , Q00 -500,00 920 184 

 92,-500,000 "925-185 



Or the distance of focus from perihelion less than one- 

 eight«-enth part of that shown in the plate above mentioned. 

 Thus tlie time-curve would lie only at one-eighteenth part 

 of the distance from tlie cycloid at which the curve for 

 eccentricity 94-10 is placed, which yet lies so near the 

 cycloid that, if drawn by itself, the eye could not recognise 

 any difference from the true cycloidal figure. 



We may, then, for the purpose of a rough calculation 

 such as I am about to make, take the cycloid for our time- 

 curve after the manner described in Section \\L of my 

 treatise on the Cycloid (the section explaining my " Gra- 

 phical Use of the Cycloid and its Companion to determine 

 the Motion of Planets and Comets "). 



The particular problem I wi.sli now to deal with is this : — 



Supposing the comet of 181.'}, which had been before 

 moving in a period of more than 100 years, moved after 

 1>^1.3 in a period of .37 years, and after 1880 in a period 

 of less than .1 years, so as to be back at perihelion on 

 S<'pt^-mlj<-r 17th la.st, we might fairly infer it would not 

 !«• at all unreasonable to infer that its next circuit 

 would \xi accomplihhr-d in a year. Let us see, then, in 

 what time, roughly, it would accomplish any distance from 

 the sun on its elongated orbit 



Lrrt A Pa represent the orbit of a comet, C its centre, 

 AC = earth'g dutmce, or 92,500,000, S the sun (AS = 

 aljout 500,000 milen), AQO a half cycloid liaving a A for 

 axis. Then, as I have iihown in rny " Cycloidal Geometry," 

 if MPQN is drawn parallel to the base AD, the time 

 in which the comet moves 'from A to P is to the time from 

 A to a as QN* is to MN. 



Now, instead of drawing a cycloid aQD, with special 

 care, to give us the times wo want, suppose wo make the 

 calculations suggested by this construction. They are very 

 easy. 



Draw the circle a 11 A cutting MP^N in K. Then we 

 know from the properties of the cycloid that 

 tjN=arcAR-MR 

 Suppose now AM=a third of AC ; that being about the 

 distance traversed by the new comet from S, as observed 

 on Sept. 24. 



Tlien CM --= I CA = -GGGGG . . . if CA = 1 ; then from a 

 table of natural cosines we find that arc AK, whose cosine 

 is 'GGG ... is one of 48° 12' ; and (this we take out at the 

 same time) the sine of this arc is -745476. Turning next 

 to a taV)le of lengths of circular arcs, wo find that the arc 

 AR of 48° 12' has a length of -841249. Hence 

 QX = ■841249--74547G = -095773 

 And AD = semi-circumference = ."i -14 1593. 



Thus the time from A to P = fpny^jT; of half a year 

 95773x182-025 

 = 3141593 *^*y«- 

 log. 95773 = 4-9812431 

 loc. 182-625 = 2-2G15G02 



7-2428033 

 log. 3141592 = G-4971499 



0-745G534 = log. 5-5674. 

 So that on this supposition as to the period, 5 days 13 

 hours 37 minutes would have been required to carry the 

 comet over the distance actually traversed between Sept. 17 

 and Sept. 24, or in 7 days. 



Let us try an or))it of diflerent dimensions ; — 

 Suppose Aa, the major axis, to be only one-half the 

 earth's, so that by Kepler's Third Law the period, instead 



of being one year would be 1 year -4- 2-, or l-^ys (we 

 need not work out this value — as we shall obviously want 

 the logarithm of the number of days in this new period, 

 and this will be obtained by subtracting the logarithm of 

 v/K from the logarithm of 182G25 already in use). 



If now Ato represent about a third of the earth's mean 

 distance on this new scale, where AC represents half 



the earth's distance, then obviously Aot=^~^AC; and 



2^=^ = -3333 Hence Ar is an arc of 70'' 32': 



AC 3 



mi = -942835 (using table of natural sines as before) ; and 



arc rA= 1-23 1039. Thus 



yn=I -231039- -942835=28820'! 

 or time 



288204 1 



from A to /)= 



3141593 



v/8 



^f 3G5- 



days. 



I 



