74 THE TIDAL PROBLEM. 



which may also be written 



„ ^ 62.3 X 0.007409 X 264000 

 ' 512X216X143 



From the reduction of this expression is obtained 



/ {hr+Tsr+^2.256[i+j^]'}d. 



£'^ = 1.124 foot-pounds per second. 



SECTION 



Making the same assumptions with respect to this section that were 

 made with respect to section "b", we find in the formula v = -- 



a 



«-i«°+546 « = «««» 

 so that 



792,000 + Z 



v = 



9000 X 5280 



As before, we take p= 1. The length of the section is 1,280 miles, or 1,280 X 

 5,280 feet. If now we put Z = 5280Xl280x, the expression for the lost 

 energy is 



_ 62.3 X 0.007409 X 5280 X 1 280 

 "" 900^X64 



which reduces to 



- / {(15 + 128x)« + 172.8(15-|-128x)2}da; 



jBg= 142.484 foot-pounds per second 



Combining these results, we have 



^„ = 37.939 E^ = IA24: E, = 142.484 Total, 181.547 



The total work done in the entire strip is therefore 181.547 foot-pounds 

 per second. If the work done on the opposite shore of the basin be the 

 same, the total work done per linear foot of basin is 363.094 foot-pounds 

 per second. In a basin 60,000 miles long this would amount to 11,503 X 10^ 

 foot-pounds per second, or 36,300 X 10" foot-pounds per year of 365i days. 

 The kinetic energy of the earth due to its rotation is given by the 

 formula rp u 2 



where T is the kinetic energy, / is the moment of inertia of the earth, and 

 (o is its angular velocity. The moment of inertia depends upon the law of 

 density of the earth's interior which is not known. We will probably be 

 not far astray in using the Laplacian law of density, i.e., 



r* 

 sin m— 



S = G ^ 



r 



a 



* See Tisserand, M6c. 061., 2, p. 234. 



