THE ELEMENTS OF THE ORBIT OF A SATELLITE. 
827 
Integrating as in the last section, from £=1 to '88, we have 
log,, tan -|-J = log,, tan -^J 0 + '0238 
log,, tan 11 = log,, tan ^I 0 — '0895 
Taking I 0 =6°, J 0 =17°, we have 1=15° 34', J = G° 9'. 
These values correspond to I,= 1° 15', J,= 3° 37'. 
Again integrating from £=1 to '76, we have 
log,, tan lJ=log; tan |J 0 — '0461 
log,, tan \1 =log^ tan — '2552 
These give J=5° 44', 1=13° 13', which correspond to 1=2° 33', ^—8° 46'. 
The integration will now be continued over another period, as in the last section. 
The following are the results of the computations. 
Table XII. 
1- 
'96 
•92 
•88 
l 0 g(r=G)+io= 
9-65092 
9-64491 
9-62783 
9-59299 
log (A = D)-f-10 = 
9'84629 
9-86040 
9-87686 
9-89622 
Table XIII. 
£ 
1- 
•96 
*92 
•88 
G l y K l -\- /Icn^K* — K 1 ) = 
— •06781 
-•07617 
-•07802 
— •07323 
G(#c a + a)/kn(K. z — k 1 ) = 
•23026 
•21018 
•19033 
•16832 
D(K 1 +a)/kn(K 2 —K 1 ) = 
— "10634 
— •12511 
— •13843 
— •14720 
D(/Co + a)/A:w(«:o — Kl) = 
•36106 
•34521 
•33771 
•33835 
(bG—aD )/Jcn(K 2 —K l ) = 
— -13815 
— •16352 
-*19057 
— •35054 
Substituting these values in the differential equations (250), we have the following 
equidistant values :— 
