THE ELEMENTS OF THE ORBIT OF A SATELLITE. 
795 
If we form U and T, and solve tlie equation for Dh/, we obtain the same results. 
Again 
Ki 2 +Jr 2 H = -2a / b^ 1 = 2a / b/c 1 y 1 
dt 
■o K \ "t f3^ b _ 11 
-1- a -'1“^ + ^!-^ + ^ 
O *1 + *^ 1 
b l“ b 21 j 
= a '^] 3 (yi+ 2tK i z i) b y ( 204 ) 
Hence the equation for £ is 
—D 4 £=(/c 1 + K 2 )(y 1 + 2£/c 1 z 1 )q-2/c 1 y i -fthe same with 2 for 1 
And by the rules of solution 
A£ =-21- 
cc'bt U 2^(«l 2 -«2 2 ) 
Since 
Hence 
Ki+k 3 +2 k 1 -2 k 1 ( / Ci+k 2 ){^^+^-| +&c. 
+ &c. 
1 - 
K 1 ~ *2. 
Zo 
f l —Kc,y 2 (/Cj— K 2 y 2 
1 /c 2 + « y 1 /q + a 
Cl l 
b (/q— k 2 ) 2 1 b (/q—/c 2 ) 
,4 
/Co + « „ /q + « 
*i=^—Ci, %=-^-c 3 
1 dZ/\ _ , /c 2 + a / 1 dZ 2 \ _ , /q +« 
Af dt ) a a (/Ci — ^ 2 ) 2 ’ \A 2 ' dt ) a a (/c 1 — /Co) 2 
( 206 ) 
If we form U and T, and solve the equation for Df, we obtain the same result. 
Terms depending on the variation of (3. 
The results may be written down by symmetry. 
z and y are symmetrical with £ and y, and therefore unaccented X’s are symmetrical 
with accented ones, and vice-versa; a is symmetrical with (3, and vice-versa . 
The suffixes 1 and 2 remain unaffected by the symmetry. 
