THE ELEMENTS OF THE ORBIT OF A SATELLITE. 
843 
In the same way it may be shown that {— 1 } and { 1 } are necessarily different. 
Therefore {— 1 } and {— 2 } being terms of the third and fourth orders may be 
dropped. 
It follows from this discussion that, as far as concerns the present problem, 
(e»)cos(2«) = (e»){2}+(e)[{lH-{S}]+(e s )[{0} + {2} + {4}]+(e s )[{lj + {3}]+(e»)[{2}] 
(e)cos(30=(e){3}+(e 2 )Q2} + {4}]+(e»)[{]} + {3}]+(e«){2} 
(e)cos(0 = (e 2 )!lJ+(e=)[{0} + {2}]+(e 3 )[{l} + {3}]+(e‘)f2J 
(e 2 ) cos ( iff) = (e 2 ) {4 j + (e 3 ) {3 ] + (e 1 ) { 2} 
(e 2 )cos(0) = (e 2 ){0} 
And the sum of these expressions is equal to (a). 
We thus get the following rules for the use of the expansion (269) of cos (Jc0-\-(3) 
for the determination of <4> (a) : 
When Jc= 2, omit in © 3 terms in cos (Jc —3), cos (£+3) 
in terms in cos (Jc —4), cos (Jc — 2 ), cos (&-f- 2 ), cos (&+4) 
When Jc=3, omit in © 3 term in cos (£+ 2 ) 
in ©3 terms in cos (Jc —3), cos (&+ 1 ), cos (A'+3) 
all of ©4 
When Jc= 1 , omit in © 2 term in cos (Jc — 2 ) 
in © 3 term in cos (Jc— 3), cos (Jc— 1 ), cos (£+3) 
all of © 4 , 
When Jc= 4, omit in © 2 term in cos (£+ 1 ) 
in ©o term in cos (Jc), cos (£+ 2 ) 
all of © 3 , ©4 
Then following these rules we easily find, 
When Jc= 2 , /5=ct 
cos ( 2 #-f“ a ) = (1 _ 4e 2 +Yfe 4 ') cos ( 2/2 + a) — 2 e(l —-Je 2 ) cos (/2 + a + ui) 
-f2e(l — -/e 3 ) cos (3/2-fa —zn-)-f |e 2 cos (a-(-2CT)d-b^e 2 cos (4/2-f a_ 2 ct) . (270) 
When Jc— 3, /3=a — 
cos (30+a— ct) = (1 — 9e 2 ) cos (3/2-fa—rrr) — 3e(l —- 4 x e 2 ) cos (2/2-fa) 
+ 3e cos (4/2 +a—2nr) + - 8 L e 2 COS (/2 + a + ts-) . (271) 
