ioo-i02] Stability 101 



Equation (279) becomes 



and this may be regarded as an equation to determine r. Let its solution be 

 supposed to be 



e s h+ ........................ (282). 



Then, on carrying out the integration with respect to r in equation (280) 

 we obtain 



= ^ /Y(a 2 



dS. 



Clearly the term (a 2 x 2 + 6 2 y 2 ) (5ef) vanishes on integration, so that k 2 may 

 be put in the form k<? + A& 2 , where 



^ jj(a 2 x 2 + 6 2 y 2 ) (^ + 2/ 2 ) cZ5. 



The values of /, # are readily found from the condition that the solution 

 (282) shall satisfy equation (281). Carrying out the necessary computations, 

 we obtain 



& 2 = 0-844105, 



so that the moment of inertia is given by 



.................. (283). 



The Stability Criterion 

 102. Collecting results, we have now found for the pear-shaped figure 



^-= 0-14200 (1 -H 0-052270 2 ) (284), 



k* = 0-8441 (1 - 0'09378e 2 ) (285), 



whence it is readily found that the moment of momentum M is connected 

 with the moment of momentum M of the critical Jacobian ellipsoid by the 

 relation 



M = M (1-0-06765O (286). 



Thus it appears that M < M , the moment of momentum decreases as 

 we pass along the pear-shaped series. This series is accordingly proved to 

 be unstable. 



