6 Wilde, Moving Force of Terrestrial and Celestial Bodies. 



minute of time, describe in its fall i6 feet i inch. For 

 the versed sine of that arc which the moon, in the space 

 of one minute of time, would by its mean motion describe 

 at the distance of 60 semi-diameters of the earth is nearly 

 16 feet I inch. Wherefore since that force in approaching 

 to the earth in the reciprocal duplicate proportion of the 

 distance, and, upon that account, at the surface of the 

 earth 60 X 60 times greater than at the moon, a body in 

 our regions falling with that force ought, in the space of 

 one minute of time, to describe 60X60X 16 feet i inch. 

 And with this very force we actually find that bodies here 

 upon earth do really descend. And therefore the force 

 by which the moon is retained in its orbit becomes, at the 

 very surface of the earth, equal to the force of gravity 

 which we observe in bodies there. And therefore the 

 force by which the moon is retained in its orbit is that 

 very same force which we commonly call gravity." 



12. I have already demonstrated that the moon's 

 orbital velocity is 38160 miles per minute, and that a 

 body would describe an arc of 4'936 miles of the earth's 

 circumference in one second, during which time it would 

 fall through the versed sine of that arc =193 inches. 

 Now as the versed sines of arcs of the same length are as 



the radii inversely from the centre, we have — ^:~q — 



= 3"2i inches as the versed sine of 4936 miles of the 



moon's orbit. Again, as versed sines are as the squares 



r , ■ , 38-160 miles 



of their arcs, we have f~~"^i — 7'7V x r2i, or nearly 



' 4'936 miles ^ / J J > j 



16 feet I inch, as the versed sine of 38'i6o miles 



through which the moon falls towards the earth during 



one minute of time in accordance with Newton's 



demonstration. 



13. The assumption by Newton of the whole number 



