162 Prof. G. H. Darwin. [Dec. 13 y 



Now w(3— w) is a maximum when it is equal to (f) 2 , and therefore 

 the square root can never become imaginary. 



From the sign of the last term in the equation for (3, it is clear that 

 one of the values of /3 is negative. Hence to avoid infinite ellipticity 

 at the centre, the c corresponding to the negative root must be zero. 

 Hence the solution of Clairaut's equation (2) is 

 e =*a«-f+V[(!) a -»(3-»)]. 

 1 d 



The surface value of -(ea 2 ) is clearly n—\-\-*/ {(f ) 2 — n(S— n)}.. 



t da 



Thus from (3) we have 



-=K»-i)+ivW-»(3-»)}- 



t 



Then substituting for w from (9) in (5), 



3(5—%)' 



And since M=|r7r/>=f7r — ?— , we have for the precessional constant 



C— A 5— n /t> i . 



-rr-=q 



C d— n 



Now let us collect these results, and express them in terms of h 

 instead of n. The solution is 



At7T o 



And the mass inside of any radius a is — a 3k . 



oh 



4-6* 

 W Z-3k—l 



k(2-Sk) 



4-6* 



dp . 



div a dh(l-Jc) 



And when &=§ , p = 27rlogiv, w^-=27r, w=a~ l . 



dw 



The length of the modulus at the surface is 1/3(1— &) of the- 

 planet's radius. 



e=e^" 3 * +3 v[(f) 2 -m-*)]. 



-^i-^+iy{(f) 2 -Ki-^}. 



! ^ - . fMa 2 . 



2 + 3& 

 A 2 + 3& 



3 , (t-W. 



