ON THE LOSS OF ENEEGY BY FRICTION OF THE TIDES. 75 



where «S is the density at the distance r from the center, a is the earth's 

 radius, and 



wi = 141° 40' 28" = 2.4727 G = 4.426 



With the Laplacian law of density, then 



^ = I ^^' / / r: (^)' '^" "^ (^) '°' ^^^^^^ (^) 



= |^a. 



«/ 



= — ;ra^— J(3m^— 6) sin m— (m^— 6m) cos m] 

 With the same law of density the mass of the earth (M) is 



M =Ga^ III f — j sin w— cos fd^d^df — j 



= 47rGa9 



pa 



' / — sinm — d(~) 

 /a o \aj 



= 47ra'— ,[sin m—m cos m] 

 m^ 



Consequently, by the division of these two expressions, 



2 r (3m^— 6) sin w— (m^ — 6w) cos ^ "JTir 2 

 ~3m^L sin m—m cos m J 



Substituting in this the numerical value of m, there results 



/ = .335 Ma} 

 and therefore 



T = .168MaV 



Taking the radius of the earth at 3,958 miles, its mean density at 5.5, and 

 the sidereal day as 86,164 seconds, it becomes 



T = 159 X 10" foot-pounds 



At the rate of loss due to tidal friction as calculated above, this amount 

 of energy would last 43,900x10" years. The day would be lengthened 

 by 1 second in about 500,000 years. 



