11] 



JFIGURE OF THE EARTH. 



211 



the volume of the spheroid, M=- ira 3 (l e) ; hence 



o 



M 





 r 



4 



-- 

 15 



a 



-T. 



r* 



8. We will now proceed to the more general case of a point situated 

 anywhere with respect to the spheroid. 



Let the line OM intersect the surface of the spheroid in L, and 

 describe a sphere with centre O and radius OL. This sphere will in general 



lie partly within and partly without the spheroid ; we shall take the 

 thickness of the shell between them as negative in one case and positive 

 in the other. Let CLA be the meridian through L, and P any point 

 on the surface of the sphere. Join PL with an arc of a great circle, 

 and call the angle PLa between PL and the meridian of the sphere, <f>. 



Also let (X4 = a = c(l+e), OM=r, cosCL = \, cosLP = p,. 

 Then cos CP = cos GL cos LP sin CL sin LP cos (j> 



At the point L the radius vector of the spheroid exceeds c by 



Similarly at P the radius vector of the spheroid exceeds c by 

 ec sin 2 CP = ec ( 1 cos 2 CP}. Hence the thickness between the sphere ca and 



the spheroid is, at P, 



ec(X 2 -cos 2 CP) 



= ec {X 2 (l -fj?} + 2\p. (1 - X 2 )* (1 -^ cos <-(!- X 2 ) (1 - ju 2 ) cos 2 <j>}. 



It is to be remarked that all the elements of the shell for which 

 LP or p remain the same may be considered to be at equal distances 

 from M, and the directions of the attractions make equal angles with 

 MO. Consequently the potential of the shell at M and its attraction 

 in the direction MO will not be affected if we replace any two elements 



272 



