From these data G ~ 1.4 x 10" 4 dyne/cm^. With M « 10 14 gr 

 sec" 1 , D = 3000 m and a total width of 2.22 x KH Km, it follows 

 per unit area of the cross section that 



„r -1 -2n 10 14 gr sec" 1 



M[gr sec *cm ] = s-r 5 = 1.5 , 



2.22 x 3.0 x 10 X:5 cr 

 and, with a ~ 1.7°, according to Table 1, for cp = 55°, 



tanp - 34 - 43 . 

 This result can only serve to check the o rder of magnitude of 

 a, G, and M, since tanp is given as a small difference between 

 two larger quantities which are known only approximately. There- 

 fore, p can not be determined from these data. However, since the 

 deviation of the true mass transport from the direction of the 



34 



