ILLUSTRATIVE EXAMPLE 257 



giving the velocity at any point in the form 



Since v = , this equation becomes 

 dt 



(d) 



r/ 



so that t = I dr (e) 



On performing the integration, we can find the time required to describe any 

 portion of the path. Let us first consider the different types of solution. 

 We see from equation (c) that v vanishes when 



so that if F 2 < 2 gra, there is a positive value of r, intermediate between + a 

 and + oo , for which the velocity vanishes. Thus if F 2 < 2 gra, the projectile 



goes to the point at distance and then falls back on to the earth. If 



F 2 > 2 gra, we find that there is no positive value of r for which v vanishes, so 

 that the particle goes to infinity : it escapes from the earth altogether. 



When F 2 = 2 gra, the velocity vanishes at infinity ; thus the particle just 

 escapes from the earth's attraction, but is left with zero velocity. Its kinetic 

 energy of projection is just used up in overcoming the earth's attraction. 



Let us consider first the special case in which F 2 = 2 ga. We find that equa- 

 tion (e) reduces to 



where C' is a new constant of integration. 



If we measure time from the instant of projection, we must have t = when 

 r = a, so that 



and on eliminating (7', 



