406 Archdeacon Pratt on the Figure of 



of the distance, for the following reason. The Himalayan mass 

 attracts very much as if it were a huge prism running not far 

 from latitude 33° at Q, and not far from east and west. Now 

 the attraction of a long uniform prism upon a point opposite its 

 middle equals the mass of the prism divided by the product of 

 the point's distances from the middle and either of its extremi- 

 ties. If therefore the point be near (compared with the prism's 

 length), the force will nearly vary inversely as the distance ; but 

 if it be more distant, the force will approach much more to the 

 inverse square. I will therefore suppose that below C the force 

 thus varies. For points of the ocean due south of Hindostan, 

 we may suppose the force to act in parallel lines (that is, at right 

 angles to a line east and west through Q), and inversely as the 

 square of the distance from that line. The distance of C from 

 Q = 1061 miles, and the attraction at C = gravity x tan 7". 



m M 



Hence --.^^...q , or force at C, = -^ tan 7", M being the mass of 

 (1061)^ fl^ 



the earth, and a its radius ; 



.-. m=(0-0000024)M. 



If u be the distance of any point of the ocean from the line 

 through Q, 6 its latitude, then u = 2asin^ {S3°—6), and 



— - j "2 <?M is a term which must be added to the potential V in 



the equation of the surface given by fluid equilibrium. This 

 equation is 



const = Y + ~ (o^a^ cos^ d-\ — , 



or const = — (1 — € sin^ 6)-\ , 



r ' u 



r being the radius vector of the ellipse whenm = 0; 



or 1=-(1— esm^^ +^- 

 r ^ 'Mm 



• = «(l-esm^^+^^), 



e being the ellipticity. 



From this, by differentiating, we have 



rde- ^^^^'^^+u sin^l (33^=^- 



Hence if -^ be the angle through which the normal to the ocean- 

 surface is thrown back by the attraction, 



m r/.cosec|(33°-^) 

 "^•^^=2M dd • 



