SATELLITES UPON THE FORM OF SATURN’S RING. 
115 
Terms involving argument 0 5 n : 
a- 0 ,n = 0, c 5 ,„ = 0, 
2~X () k2u sin (2 n<p — r) X u cos r 
3 K A., ~4a 4 ) 2e 5 , 0 ’ 
X„ (in 2 —a ,,„) cos (2 
6 («1,0^5,0 — 4n 4 ) 
Terms involving argument 0 5 , 2 „: 
n 2 ) cos 2- 
«5.2» 
(05.0 -'ll 2 ) 
U (^l.n — W 2 ) sill 2 t 
4 (^i,o^o.o n/) 
4 — 
Y — 
2n 
3uvX n sin (.3 n<p — t) 
8 (9;i 4 -« i , u 0 5 ,o) ’ 
9(n 2 -d,. 0 ) 
i X„ cos l 
(3 71(f) — t) , 1 
(rq, 0 -rr) cos 
2tA„ cos 
( 71(f) — T) 
16(9 
n 4 
^5,u) 
► 
4 an I 
(O.5,o- — -n 2 ) 
(<<-' i.n + n 2 ) («,,„- 
n 2 ) 
(sin 2rA 0 
sin ( 
{n<f> — t) 
8 ku l 
(«i,o0 5 , u -n 4 ) 
Terms involving powers products of the 0’s follow in similar fashion. 
If we summarize the parts specially recpiired, we find 
(01,0— n 2 ) (05,o-^ J ) = 
4/c 2 n 2 + ^- ( 0 , 5 ,o — n 2 ) cos 2t0], 2)1 — ku cos 2t0 2 , 2(t + cos 2 T t 
A, 2,1 
-i («i,o~ n 2 ) cos 2 - 7 - 03,2 «+•■•;• • 
(17) 
and 
2c(rq,A,o-n 4 ) = b l '( 
( 05 ,u — n 2 ) sin 2t 0 1i2 «— icn 2 sin 2t0 2 , 2 „ + «:h" sin 2t0 4 , 2 „ 
— b l («1,0 — W 2 ) sin 2 t 0 . 3 , 2n + • • • ; • • 
(18) 
where, as already stated, 
(«i,o — n 2 ) (0 5 ,u — n2 ) — 4kV. 
It is necessary to examine the expressions just obtained in order to see whether 
the complete integral of equations (6) has been found. 
The integer n is determined so as most nearly to satisfy the relation 
(0i,o-^ 2 ) ( 05 , 0 —w 2 ) = 4 kV, 
when 
© 1 , 0 ’ 05 ,o and k are known. 
The negative value of n will also satisfy this relation. 
On solving equation (17), for each value of n there will be, in general, two values 
of 2r, equal and opposite in sign. So that altogether there are four distinct values 
of 2t obtainable. Each of these with the corresponding value of n will give a 
