256 MOTION UNDER A VARIABLE FORCE 



We notice that the right hand of equation (92) is the kinetic 

 energy of the particle. Also, since the force opposing the motion 



of the particle along its path is S, its potential energy is I S ds. 



Thus equation (92) expresses that the sum of the kinetic and 

 potential energies remains constant it is the equation of energy 

 for the motion of the particle. From a knowledge of the total 

 energy at any instant of the particle's motion, we can determine 

 the constant (7, and can then proceed to integrate equation (93), 

 if possible. 



ILLUSTRATIVE EXAMPLE 



Assuming that the value of gravity falls off inversely as the square of the dis- 

 tance from the earth's center, determine the motion of a projectile fired vertically 

 into the air, the diminution of gravity being taken into account. 



Let a be the radius of the earth, and g the value of gravity at the surface. 

 Then, at a distance r from the earth's center, the value of gravity will be 



Since the particle moves along a radius drawn through the center of the 

 earth, we may measure all distances from the earth's center, and the distance 

 r from the earth's center may replace the coordinate s of 204. The value of 



the force S resolved along the path is , so that the equation of motion is 



mga? _ d 2 r 



The equation of energy, as in equation (92), is 



rmga' 2 , 



C I - dr | mv 2 , 



or G H - = \ mv 2 . (a) 



Let us suppose that the particle was projected from the earth's surface with 

 velocity V. Putting r = a in equation (a), the value of v must be V, so that 

 we must have 



and this equation determines the value of C. Eliminating C from equations (a) 

 and (6), we obtain 



