IN PHYSICAL ASTRONOMY. 
379 
If n{l+2r 0 } = n andn 2 =ii a = ajl+yr 0 j 
If 2 e is the coefficient of sin (n (1 + &) t + £ — w) in the expression for the 
longitude, 
e(l +/) =e(l + k — r 0 ) 
y = 1 — y r 0 + e 1 1 + k — y r 0 | cos (n (1 + k) t + s — & j 
+ ej t cos (n (1 + k,) t + e - 
1 m j a 3 
6 paf 
, , m i a h 
b3 > 0 + '6]^ K ' 
+ d 1 - - iffr** } » ( 1 » (' - $$*») ‘ + - d 
+ e/, COS (1 + & ( ) t + 6 — 
~ = 1 + y r 0 — e 1 1 + & — y r 0 | cos (n (1 + k) t + a — ^ 
— e i f i cos (1 + Jc t ) t + g — 
= l + ^L bi0 - ™< a L b 3l 
6 1 u,a t 3 ’ 6/xa, 2 
-el 1 + ^Lb i0 - 5 m ‘ a lb 3l )cos( n(l -351 b S} )t+B-*\ 
l 6 pa, 3 3,0 12j/,a ( 2 3,1 J \ V 4/xa, 2 7 / 
— e ; f cos (1 + &,) t + s — 
If a < a ; as before, and 
= 1 + r l0 + e, (1 +/') cos (1 + *') * + e , — + e// cos ^n, (1 + */) * + 
we find 
m f 1 , a , I , m I , 5a, J 
r -° - 7T IT &3 *° 2T/ 3 ’ 1 / k ‘- J t U 3 ’ 0 ” 45, 3>l I 
//{(l + V)"(l-3r, 0 )-l} = ^.t s . ! 
If n , { I + 2rJ = n, and n,* = «, = *,{ 1 + y »"»} 
