1 16 On the Formation of the Tails of Comets. 



of the tail ; m— moment of inertia of any portion of the tail, 



and m'= moment of inertia of the whole tail ; then, m=fx 2 xdx 



x 3 

 =-q--fc. Integrating between the limits #=0, and x=l, we 



m' I 3 I 3 



have -^-=o, and m'=2' q-, (2). As we are ignorant of the di- 

 mensions of the head at the time of the perihelion passage, for 

 which the calculation is to be made, we will neglect its moment 

 of inertia with respect to its centre, which on any admissible 

 supposition is but a very small fraction of the moment of inertia 

 of the tail, and will, so far as it has any effect, only make v, the 

 velocity of translation, the greater. 21= mass of the tail ; and 

 1000 x 21+21= 1001x2/= whole mass. Put b = distance of 

 centre of gravity of whole mass from centre of nucleus, and a = 

 distance of same point from the centre of gravity of tail. (b = 

 24,975 miles = 25,000 miles (nearly); a=24,975,025 miles.) 

 Also let M= moment of inertia of whole mass — neglecting as 



I 3 

 above. Then, by the principles of mechanics, M=2'or-{-2JX 



I 3 

 2- ~-+2lXa 2 +(1000 x2l)b 2 



a 2 +(1000x2Z)6 2 :and k 2 = — - — 



I 2 

 2+a 2 +l000b 2 



1001x2/ 

 (3). Making the calculation we obtain k 2 = 



1001 



831,876,980,528. 



Next, to find w, I take the formula for a parabolic orbit, t 



D f N/- 



(tang.£w+£ tang. 3 &u), (4) ; in which t= time elapsed 



since the instant of the comet's arrival at its perihelion, or a cer- 

 tain interval before; m= anomaly corresponding to the time t ; 

 D= perihelion distance ; and m= intensity of sun's attraction at 

 the unit of distance. Taking 1= radius of earth's orbit, and one 

 day for the unit of time, m=0.00030; D = 0.005263 (=500,000 

 miles): and making u= 90°, t= one hour (very nearly). But, 

 as there is some uncertainty in the perihelion distance, we will 

 take £=1£ hours, or 2^=3 hours. During this interval of three 

 hours, in which the comet passed from 90° on one side of the 

 perihelion to 90° on the other side, the average angular velocity 

 of revolution was 0.000291 of a mile per second. Equation (4) 



