Three Approximate Solutions of Kepler's Problem. 



209 



Method III. 



From equation (7) we deduce 



.„, , ^,. sm (L'— M) 



cos (E — i M) = \ — nvr • 



^ - g sm ^ M 



Multiplying both members by \-^ and reducing, we have 



q sin M 



Substituting in (5), we obtain 



Assuming 

 and 



sin(E'_E) = 2^M?:=^)F(E). 

 ^ ^ e sm M ^ ^ 



sin (E'— E) = (E'— E)" sin 1" 



(Ifi 



2 a sin (E'-M )^ cos (E'-J M) ^^^ 



a sin M ^ ' cos A M ^ ' ^ 



(IS) 



(19) 



F (E) = j2 sin' (E'— M), 



equation (18) becomes 



, E)-- sin^(E'-M) 

 ^ 6 e sin M sin 1" 



This equation may be employed in connection with 



tan (F/— l-M) = tan (450-|-w) tan 4 M, 

 or with 



sin (E'— M) = g sin -i M cos (E'— i M). 



Using it with the former equation, the solution of the previous 

 example is effected as follows : log tan ( 45°-f-<^ ) = 0-2175146; 



loo- 



cosec 1" 

 ITe" 



= 5-14655. 



M 



log tan ^ M 

 log tan (E'— I M) 

 W—\ M 

 E'— M 

 log sin (E'— M) 

 log sin* (E'— M) 

 log cosec M 

 log approx. (E' — E)" 

 approx. (E' — E) 

 true (E'— E) 



164° 52' 13"-83 

 9-4319633^i 

 9-6494779?^ 



155° 57'-21"-98 



—8 54 52 

 9-19022>i 

 6-76088 

 0•29765?^ 

 2-20508yi 

 —2' 40"-36 

 —2 39 -71 



