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LX. The Effects of the Diurnal Rotation on the Upper 

 Atmosphere. By C. G. Darwin, B.A* 



AS the earth rotates the forces of viscosity tend to drag 

 the lower atmosphere with it, so that the mass motion 

 of the air in this region has the same angular velocity as 

 that of the solid earth itsalf. Now, at a certain height above 

 the adiabatic region the density becomes so small that there 

 are practically no collisions between the molecules, and here 

 viscosity can therefore no longer have play. A molecule 

 will rise from out of the collision region, perform a portion of 

 an orbit, and then fall back again. Now, during its free path 

 the angular momentum and not the velocity of the molecule 

 is constant, and so it would be expected that at altitudes 

 above the collision region the mean motion of the air would 

 lag behind that of the earth. This paper was undertaken to 

 find the amount of this lag. We shall see, however, that it 

 is entirely counterbalanced by another effect. It is very 

 probable that this may have been perceived by writers on 

 the subject : but I have found no explicit statement of it ; 

 and the non-existence of the lag does not seem to me to be 

 an obvious fact at first sight. 



Suppose that there is a definite layer in the atmosphere, 

 such that below it the gas is moving with the same angular 

 velocity as the earth, and that above it there are no collisions. 

 Let this layer be at a distance r from the centre of the 

 earth. Take spherical polar coordinates at the centre of 

 the earth, fixed in direction. If we neglect for the present 

 any molecules describing permanent orbits outside r , all the 

 gas at any greater height will be due to molecules projected 

 from the sphere r . 



Consider a molecule projected from r 0i X , l with a velocity 

 V at angle to the vertical in an azimuth <p . Let this 

 molecule pass through r, A,, I with velocity V in the direction 

 6 } 6. The equation of energy then gives 



Y 2 -2a-=Y 2 -2a-, 

 ' r ■ r 



g being the attractional part of gravity and a the earth's radius. 

 The equation of angular momentum gives 



?'V sin = r \ sin # , 



and since the orbit is in a plane through the earth's centre, 

 we have by spherical trigonometry 



sin \ sin o =sin \ sin 0. 



* Communicated bv the Author. 



