THE ELEMENTS OF THE ORBIT OF A SATELLITE. 
7G5 
to it at the point where the axis of z' meets the sphere, and project on this plane the 
poles of the lunar orbit and of the earth. We here in fact map the motion of the two 
poles on a tangent plane to the celestial sphere. Let x , y he a pair of axes in this 
plane parallel to our previous x, y ; and let x', y be the coordinates of the pole of the 
lunar orbit, and %, y be the coordinates of the earth’s pole. Then 
x—j sin N, y — —j cos N; g'=—i sin xp, y' = i cos xp . . . . (123) 
Let x, y, £ y be the coordinates of these same points referred to another pair of 
rectangular axes in this plane, inclined at an angle <p to the axes x', y'. 
Then 
x— x' cos <f>-{-y' sin (p , g' cos <p-\-y sin (p 
y— — x sin (p~\~y' cos <f> , 77 = —sin <£+ 7 /cos <p 
From (123) and (118) we have therefore 
L x sin(/c 1 ^ + m 1 — <p)-\-L 2 sin (/c 3 £-bm 3 —<£)^l 
x — 
y = —L x cos(K l t-{-m 1 — (f)) — L 2 cos (/c 2 ^ + m 2 — <p) i 
£= 
v= 
— L x sin (k^-)- up — (f>) — L 2 sin (/c 3 ^+ m 3 — <p) 
L x cos(«: 1 ^ + m 1 — <p>)-\-L 2 cos (K:o^+m 3 — <p) 
Now suppose the new axes to rotate with an angular velocity k 2 , and that 
(f)=K 2 t+m 2 . 
Then 
X—L x sin [ (k : — /fo) t — b nq — m 3 j 
V +4 = - L x cos [ (k x - k 3 ) t + m 1 — m 3 ] I 
£= — L x sin [ ( k x -K. 2 )t-\- nq — m 2 ] 
7] — L 2 ~L x cos [(«■] — /o^d-nq —m 3 ] 
(124) 
These four equations represent that each pole describes a circle, relatively to the 
rotating axes, with a negative angular velocity (because k x — k . 2 is negative). The 
centres of the circles are on the axis of y. The ratio 
distance of centre of terrestrial circle Z 3 ' + « b 
distance of centre of lunar circle — Z 2 a /c 2 + /3 
(125) 
the distances being measured from the pole of the ecliptic. And the ratio 
MDCCCLXXX. 5 F 
