68 Centrat Forces. 
his B, for posta - (see his fig.) ... AO: SO:: R° ; eR° whichagrees 
R 
with bis proportion for finding B; also his L=—- (R= his radius 5) 
1 
“Lt L—R cos. ¢’: ae —Reos. 9/221 + 1 —e cos. 9’; and by 
(16) 1—e cos. 9 } 1:2: nt? —9’°-+eR° sin. 9’ : 2°, which agrees 
with his proportion for finding E, .*. his E=x°. By adding 2° to 
g’° and repeating the operation with the corrected value of 9°; 
the second part of his process will be obtained, and so on; ob- 
serving that the successive corrections are to be applied according — 
to their algebraic signs; and that the sign of ecos.¢/ in the de- 
nominator of (16) must be changed into ++ when 9° is between 90° 
and 270°, also the sign of eR°sin.o’ must be changed into — when 
o’° is between 180° and 360°, because cos. o’ is negative in the former 
of these cases, and sin. 9’ in the latter ; when o is found, » is easily 
found also by (7). By changing the sign of e cos. v in the denominator 
See 
of (2) it becomes r=7 a —— ” (17); v being counted from the 
aphelion. Put 2a—r=r’, then 7’= the distance of the particle, 
which is supposed to describe the ellipse, from the other focus; let 
v= the angle made by 7’ and a distance of that focus from the 
a(1— 
nearer vertex, then 7’/= (18); and because r, 7’ make 
1+ecos. a2 
equal angles with the tangent at the place of the particle rdv=r/dv’, 
ne ae . ig sy a+2ae cos. v’ + ae* 
orr?dv=c'dt=rr'dv’ (19). By (18) r=2a-r= item 
and cdt=V gp’ Xdi=V ag(l—e*) xdt; by substituting these 
values, and that of 7 as given by (18), in int 19) and reducing, I have 
(1+-2¢ cos. v/ +e?) XW 1 —e? xdv’, g e 
(1+ e cos. v’)? avt Xdi=ndt; or do’= 
1+ -2e cos. v’ +e? cos.? ae 
ndt X ( 1 4-2¢ cos. v +e? ] x(1—e*) 7, or since cos.?0/= 
1+cos. 2v’ 
2 
by rejecting quantities of the order e4, e°, &c. I have 
e? 
dv! = (5 cos. 2v’ +-e* (cos. v —cos. v/ cos. 20’)| Xndt; by known 
cos. 3v’+-cos. v’ 
formule cos. v’ cos. 20 = ue eee hence and by neglect . 
