60 KINEMATICS. [120. 



surface, P Q the initial position of the moving point at the time 

 / = o, P its position at the time /; let OP^R, OP = s Q , OP = s; 

 and let g be the acceleration at P v j the acceleration at P. 

 Then, according to Newton's law, j \g=R^ : s 2 . This relation 

 determines the value of JJL in (10), which becomes 





the minus sign indicating that the acceleration tends to dimin- 

 ish the distances counted from O as origin. 



The integration can now be performed as in Art. 117. Mul- 

 tiplying by ds/dt and integrating, we find |V =gR*/s + C. 

 If the initial velocity be zero, we have v = o for S = S Q \ hence 



c= - 2 > and 





Here again the minus sign is selected after extracting the 

 square root, since the velocity v is directed in the sense opposite 

 to that of the distance s. 



Substituting ds/dt for v, separating the variables v and s, and 

 integrating, we find 



(.7) 



120. Exercises. 



(1) Find the velocity with which the body arrives at the surface of 

 the earth if it be dropped from a height equal to the earth's radius, and 

 determine the time of falling through this height. 



(2) Interpret equation (17) geometrically. 



(3) Show that formula (16) reduces to v = V^p (Art. 113) when 

 s = R and ^ s = h is small in comparison with R. 



