FORCE, A SPACE RATE OF ENERGY 143 



) (5) 



P _p kMm kMm _ kMm f x 

 R R+x~ R \R+x 



Now R is about 3960 miles. If we are concerned only 

 with separations which are small as compared to 3960 

 miles we have R+x approximately equal to R and 

 hence we may obtain from equation (5) an approxi- 

 mate l value. 



p _p _kMmx xgx 



It follows that near the surface of the earth the in- 

 crease in potential energy is directly proportional to 

 the increase in separation. Dividing this increase by 

 the distance x we obtain as the rate 2 at which the 

 p.e. increases with distance, kMm/R 2 . In other words, 

 the p.e. of the system formed by the earth and a mass 

 m increases kMm/R 2 energy units for each unit of 

 length by which the separation increases. 



Now the earth is an oblate sphere with its larger 

 radius at the equator. Raising a given mass a given 

 distance would produce a smaller increase in p.e. 

 at the equator than at the poles since R would be 

 greater at the equator. In general the change in p.e. 

 corresponding to the vertical movement of one gram 

 through one centimeter, will depend upon the region 

 of the earth's surface where this displacement takes 

 place. The gravitational unit of energy, the gram 



1 For example if x is 1 mile the increase in potential energy as 

 given by equation (6) is about one quarter of a per cent larger than 

 as given by (5). 



2 The space rate of change of p.e. is then of the form n ^ lt 



which is the force according to Newton's definition. Cf. foot- 

 note p. 141. 



