Particle on the Surface of a Smooth Rotating Globe. 465 



and period of rotation of the smooth level globe are those 

 of the earth. Thus we have 



V=10m./s.=864km./day . . (101) 

 a>R = 4xl0 4 km./day .... (1(>2) 



-^=•0216 and °^=46. . . (10*3) 



Case I. a. — \ = 0. The particle is projected along the 

 equator from west to east and continues to move along 

 the equator. 



Case I. b. — When \ Q is small, fi is large and h is small. 

 The elliptic functions may be represented with sufficient 

 accuracy by circular functions, as in the analogous case of 

 the pendulum. 



Here / /2Vo> \ 



x=\ cosU/ -j£- .n 



(j> = Yt/R. 



The period = 27ta / 7 r Yf - =4*8 days. 



V 2 Voj 



The wave-length =4160 km. 



Case I. c. — As X is increased, the angle at which the 

 equator is crossed is increased. If the equator is crossed at 



right angles a= - and therefore fi=l, & = 1/ V2. 



The tables give E = 1*351, K = 1*854, and hence the 

 formulas (7*6), (7'32), (7*5) show that the period 

 = 136 hours = 5*7 days, the wave-length = 2240 km., and 

 the amplitude RA : =935 km. 



Case I. d. — If \ is increased still further, the equator is 

 crossed in the retrograde direction. The distance between 

 successive crossings is reduced. The distance vanishes when 

 the relation between the elliptic integrals is E = ^K. The 

 track is now a figure of eight. The solution of the equation 

 E = ^K is & = sin65°-35=-909. This gives us /x = *459, 

 RX =1200 km., and the equator is crossed at 130°*7. 

 The period (found by writing K = 2*321 in (7*6)) is 

 7*1 days or 170 hours *. 



* The solution of E = ^K is given by Hess (I. c. p. 27) and Love 

 (I. c. p. 53) as &=sin 64 0, o'5. Comparison with Legendre's Tables 

 {Fonctions Elliptiques, table viii. p. 290) shows that the correct solution 

 is that given above. 



