THE ELEMENTS OF THE ORBIT OF A SATELLITE. 
825 
From this we see that j will always decrease as £ increases at a rate which tends to 
become infinite when \=1 ; and i increases as £ increases so long as \ is less than '5, 
but decreases for values of X between "5 and unity at a rate which tends to become 
infinite when X= 1. If we consider the subject retrospectively, £ decreases,^' increases, 
and i decreases, except for values of X between ’5 and unity. 
This continued increase (in retrospect) of the inclination of the lunar orbit to the 
invariable plane is certainly not in accordance with what was to be expected, if the 
moon once formed a part of the earth. For if we continued to trace the changes back¬ 
wards to the initial condition in which (as shown in “ Precession ”) the two bodies 
move round one another as parts of a rigid body, we should find the lunar orbit inclined 
at a considerable angle to the equator ; and it is hard to see how a portion detached from 
the primeval planet could ever have revolved in such an orbit. 
These considerations led me to consider whether some other hypothesis than that of 
infinitely small viscosity of the earth might not modify the above results. I therefore 
determined to go over the same solution again, but with the hypothesis of very large 
instead of very small viscosity of the planet. 
This investigation is given in the next section, but I shall not retraverse the ground 
covered by the integration of the first method, but shall merely take up the problem 
at the point where it was commenced in the present section. 
§ 20. Secular changes in the proper planes of the earth and moon when the 
viscosity is large. 
Let p = 2 gaw /19u, where v is the coefficient of viscosity of the earth. 
Then by the theory of viscous tides 
tan 2f x = 
2 («,-/ 2 ) 
tan 2f= —, tan g L =- 
• 2/2 
tan p'= 
o i 
( 261 ) 
If the viscosity be very large p is very small, and the angles —2f 1; \tt —2f, 
\tt —g 1; \tt— g are small, so that their cosines are approximately unity and their sines 
approximately equal to their tangents. Hence 
.. p • p 
sm4i, = — sin 4t=—, 
n — Sl n 
2 u . 2p 
sm 2 g i = -—L--, sm 2g— — 
61 n — 2/2 b n 
Then introducing X — fl/n, we have 
sin 4f_ sin 2g 1 _2(1 — X) 
sin/fi ’ sin 1— 2X 5 
sin 2g 
sin 4fi 
= 2(1 X) 
(262) 
Introducing the transformations (262) into (251), we have 
