Three Aioproximate Solutions of Kepler's Problem. 207 



Method II. 



Substituting x sin (E— M) for F (E) in equation (3) and reducing, 

 we obtain 



tan (E'— ^ M) —tan (E— |M) = (tan (E— iM)— tan | M) ax. (11) 

 A further reduction gives 



(l+«a;) tan (E— i M) = tan (E'— ^ M) (1+ ^ a?:). (12) 



c- 2e 



Since i,= —, 



tan (E— ^ M) = tan (E'— ^ M) (l—p ,-4^)*. (13) 



Equation ( 13 ) may be solved by successive approximations, ;>■ 

 being the variable; it is preferable, however, to proceed as follows: 



Neglecting the term {l-\-ax) and substituting — sin^ (E' — M) for ax 



we have 



tan (E— i M) = tan (E'— i M) (1— || sin'^ (E'— M) ). (14) 



Since 1 — z is approximately equal to — j — where 2 is a small quan- 

 tity, equation (14) may be written 



tan (E— I M) = tan (E— iM) 



l-|_«|sin^(E'-M) ^^^^ 



Employing the previous example for an illustration, we have the 

 following values of our constants; log tan (45°-[-ai) =^ 0-2175146; log 



« = 0-42326; log -=0-40452, which is the function of Zech's sub- 

 traction table, corresponding to the argument log tan ( 45°-f-"' ) • 



12 



ap 



los — = 1-06044 



^ M 164° 52' 13"- 83 

 log tan ^ M 9-4319633W 



log tan (E'— I M) 9-6494779??, 



E'— ^ M 155° 57' 22" 



E'— M —8 54 52 



The derivation of this equation was suggested by Prof. Stone. 



