88 W. A. Norton on Comets. 
2 2 
Integrating, Te ksina.z +C=5. 
If we suppose the initial velocity to be zero, v= 0, when 2=1, : 
and ei E 
2 2 
- & —ksina.r-+C=0, or, C=" keine. 
2 Me | 3 y2 a—r i 
Whence, = pr? G-2) — &sin a (2 —7) =z i pr(*=*)—bsin 
y2 
(2 _ r) _—— ie e . (1) i 
Let Z= greatest distance passed over in the direction NZ, and 
- — i sina—0: — nate 2 
we have Z —~*sina=0; or, Z= 7 (2) 
This value of z is the distance NZ, X being the point where 
the orbit is tangent to the line ZX, perpendicular to NZ. _ Put 
ing a=90°, we get for the distance to which a particle will — 
cede from the nucleus, when emitted in the direction NS, 
But, by equ. (2), Zsina=* =H; also Zsina=ZsinNZV 
=NV; hence NV=H, and the point Z will fall on VT, draws 
through V, at the distance H from N, and perpendicular to NS. 
To find X, the point of tangency, resume equ. (1); and, sinc 
remote from the nucleus r may be neglected in comparison with | 
dz? be 
z, we have a= v= 2(pr—ksina.z)  . . (4) 
1 dz 1 dz 
Whence, SOME Nprcksinat apr (i, 
oneal 
2 opr 
Or, “tedinigy tes AV 2Qpr —2ksin a..z 
4 
ksina salt LT ae 
This formula is quite accurate for determining any portion of . 
the interval of time sought, for the beginning of which z a 
ison with 7, and as the motion is far more rapid in 1 
vicinity of the nucleus than at a distance from it, we may obta? — 
pretty nearly the whole interval of time, from N to X, by UR 
posing z, at the beginning, to be several times r; but, on ©” 
