Meteorites to the Earth's Orbit. 9 



in their orbits very near to the sun. Assuming that the orbits 

 were parabolas we have for all the stones whose perihelion dis- 

 tances were less than one-half, Sin^S < J. If there be drawn 

 circles, AA, AA, 45° from S and from E, then will all the 

 stones whose absolute quits were in the central zone, APP A A A 

 which is bounded by the circles A A, have perihelion dis- 

 tances greater than one-half and less than unity. Of these 

 there are 103 out of a total of 116. If the same orbits are as- 

 sumed to have had semi-major axes equal to 5, then the circles 

 A A would have to be drawn a fraction of one degree farther 

 from S and from E to serve as the limiting curve to orbits 

 whose perihelion distances exceed one-haLf. 



It appears from figure 1 that these 116 stones were, with a 

 few exceptions, following the earth in their orbit about the 

 sun. This could happen from either one or more of three pos- 

 sible causes : 



Firstly, that nearly all the stones in the solar system are 

 moving in direct orbits, very few in retrograde orbits ; — 



Or, secondly, that stones moving in retrograde orbits for 

 some reason, as for example their great relative velocity, 

 may not have been able to pass through the air and to reach 

 the ground in solid form ; — 



Or, thirdly, that stones moving in such retrograde orbits, 

 and coming through the air, may be falling while men sleep, 

 or for some like reason may fail to be found. In other words, 

 the effective cause may work above the air, in the air, or below 

 the air. 



Let us assume, as an hypothesis, that neither of the first two 

 are the true causes. In that case we should have the stones 

 moving in every direction as they cross the earth's orbit. 

 There should be about as many orbits having retrograde mo- 

 tions as direct motions. Hence the absolute quits of all stones 

 coining into and hence, by hypothesis, coming through the air, 

 should be symmetrically distributed in their longitudes relative 

 to the sun. At least there should be as many absolute quits 

 in the G-hemisphere as in the Q-hemisphere (figure 1). Take 

 account now of the earth's motion and locate the relative quits. 

 All these stones whose absolute quits lie outside of the circle 

 TT will have their relative quits in the G-hemisphere. Upon 

 the hypothesis of parabolic orbits and of an equable distribu- 

 tion of the absolute quits over the celestial sphere the number 

 of relative quits in the G-hemispbere should be to those in the 



Q-hemisphere as 1-fcos— : 1— cos— , or as 17:3. The relative 

 4 4 



quits should then be very much more numerous in the G-hem- 

 isphere than in the Q-hemisphere. 



Furthermore, suppose that the heavens visible at a given time 



