170 Prof. P. Lowell on a Method for Evaluating 



Heat received and Heat retained. — Such, then, would be the 

 mean annual temperature of the planet, were the heat retained 

 as well there as here. I am far from saying that such is the 

 temperature. For the retention is not the same on the two 

 planets, being, on account of its denser air, much better on 

 the Earth. But that such is the amount received is enough 

 to suggest very different ideas as to the climatic warmth 

 from those hitherto entertained. 



Temperature deduced from Heat retained. — To obtain some 

 idea of the heat retained and of the temperature in consequence 

 we may proceed in this way : — 



Let y = the radiant energy received at the surface of the 

 Earth. 

 y 1 = that similarly received on Mars. 

 e = the relative emissivity or the coefficient of radiation 

 from the surface of the Earth, giving the ratio of the loss in 

 twenty-four hours to the amount received in the same time, 

 due to factors other than the transmissibility of the air, which 

 is separately considered. 



e l = the like coefficient for Mars. 



Clouds transmit approximately 20 per cent, of the heat 

 reaching them ; a clear sky at sea-level 50 per cent. Con- 

 sequently as the sky is halt the time cloudy, the mean trans- 

 mission of its air-envelope for the Earth is *35 e. For Mars 

 it is *60 e\. 



To get, then, the mean temperature of the planet in degrees, 

 A' 7 from the heat retained, which is the daily mean receipt 

 less the mean loss, we have the following equation, the mean 

 temperature of the Earth being 519°'4F. abs., 288 0, 5C. above 

 absolute zero. 



_jr = ^(1-^600 

 ^8-5 */y .(1^-85 e) ' 



where ; the first term under the brackets denotes the daily 

 receipt, the second the nightly loss. 



Determination of e. — To find e we have the data that the 

 fall in temperature toward morning on the Earth under a 

 clear night sky is about 18° F. or 10° C, under a cloudy one 

 about 7° F. or 4° C. Taking the average day-temperature 

 from these data at 292° abs. on the centigrade scale or 19° C, 

 and considering that for an average day sky and a clear night 

 the transmission or loss is 



i(-35 + -50> or -425^; 



while for an average day and a cloudy night it is 



i(-35 + '20)e or '275 e; 



