750 Prof. J. H. Poynting on Prof. Lowell's Method for 



radiation reflected down again and the radiation downwards 

 of the energy absorbed by the atmosphere. 



This is brought out clearly in the footnote on p. 172, 

 where he uses a formula of Arrhenius, to which I am unable 

 to refer, but which I think he must misinterpret in making 

 it give his result. The inadequacy of his method is well 

 shown by its application to the cloud-covered half of the 

 earth's surface. He finds that this half only receives 02 "of 

 the radiation which the clear sky half receives. The surface 

 temperature under cloud should therefore be only \/0"2 = Q'ti1 

 of that under clear sky. If the latter is 300° A. the former 

 is only about 200° A. Common observation contradicts this 

 flatly, for the difference is at most but a few degrees. 



On another point common observation appears, at any rate 

 at first sight, to contradict Professor Lowell. He assumes 

 that the loss in the visible spectrum radiation in ils passage 

 through the atmosphere is practically all due to reflexion, 

 and he puts it down as about 0*7 of the whole in clear sky. 

 If this were true the reflexion from the sky opposite to 

 the sun would I think be vastly greater than it is. While 

 card-board reflects diffusely about 0*7 of sunlight. But 

 when a piece of white cardboard is exposed normally to 

 the sun's rays it is several times brighter than the cloudless 

 sky. 



The " greenhorse effect " of the atmosphere may perhaps 

 be understood more easily if we first consider the case of a 

 greenhouse with horizontal roof of extent so large compared 

 with its height above the ground that the effect of the edges 

 may be neglected. Let us suppose that it is exposed to a 

 vertical sun, and that the ground under the glass is " black " 

 or a full absorber. We shall neglect the conduction and 

 convection by the air in the greenhouse. 



Let S be the stream of solar radiation incident per sq. cm. 

 per sec. on the glass. Of this let rS be reflected, aS be 

 absorbed, and jfS be transmitted by the glass. Then 

 r + a-M = l. Let the ground send out radiation R per sq. 

 cm. per sec. and of this let rjl be reflected, a 2 R be absorbed, 

 and tiR be transmitted by the glass. Here also i\ 4- «i + t^ = 1. 

 It is to be noted that since the edges are far distant R is 

 incident on each sq. cm. of glass. The glass, then, absorbs 

 aS + axR, and as it is thin it maybe taken as having the 

 same temperature on each side, so that it sends down to the 

 ground ^(aS + «iR), the other half going upwards into space. 

 Equating receipt and expenditure of radiation by the ground, 



R=^S+r I R + i(«S + a 1 R), 



