Temperature and Molecular Attraction. 101 



force acting for the same time on two different masses 

 produce in the masses the same amount of kinetic energy ? 

 would therefore seem to be a perfectly proper question. But 

 on analysis it will be seen that the question cannot be 

 directly answered from the usual standpoint. For/s = E and 

 ft = mv, t denoting time. When using the usual nomen- 

 clature we can therefore only speak of the force between the 

 bodies as acting through a certain distance when producing 

 energy, and can consider it as acting for a certain time only 

 with reference to the momentum produced. 



As two bodies come together from an infinite distance 

 under their mutual attraction, kinetic energy is not developed 

 equally in the two bodies. For the force acting on the two 

 bodies is the same. Now f=ma, where a is the acceleration 

 produced, and ft = mat. Since a = v/t, we have ft = mtvjt—mv. 

 Jj\itft is the same for each body. Therefore 



m 1 v, = m 3 v s (7) 



Now the energy of mass ni\ when moving with velocity v t 

 is 1/27W!?;! 2 , and similarly for the energy of mass ra 2 we havy 

 l/2m 2 v 2 2 , and 



l^mpi* + 1 2m 2 iy = - — . 

 Substituting for v 2 its value from equation 7, we have 



and therefore 



t?__im 2_ w 2 km l m 2 kin l m 2 2 1 

 Mi + Wg * nii-\-7n 2 s 



-^ i/o 9 m i l:m Y m 2 hn^m^ 1 



lh q =l 2m 2 vJ = ; = ■ • - 



' m 1 + m 2 s ?//]-(- m 2 s 



(») 



It should be noticed that we have answered the question 

 above asked only by introducing an experimental fact, namely, 

 that masses mutually determine in each other accelerations 

 inversely proportional to the masses accelerated. From 

 equation 8, E 1 w 1 = E 2 m 2 , a result that follows at once from 

 equation 7. 



9. It is convenient to consider the orbit of the smaller 

 body with reference to the centre of mass of the larger body. 

 Considering the most general case, that is where the body 



