Three Approximate Solutions of Kepler^s Problem. 205 



THREE APPBOXIMATE SOLUTIONS OF KEPLEKS 



PROBLEM. 



By H. A. Howe, A. M., Assistant at the Cincinnati Observatory. 



Method I. 



Prof. Grunert, in his JSTeue N aeherinigsweise Aufloesung cler Kep- 

 ler''schen Aufgabe, gives the following equation : 



1-f-e cos ^ M sin (E— i Mj-f-F (E) 



1— e ~ sin I M cos (E— ^ M)— F (E)' (1) 



where e, M and E designate respectively the eccentricit}-, the mean 

 anomal}', and the eccentric anomaly, while 



F (E) = i- (E— M— sin (E— M) ). 



Clearing equation (1) of fractions, and reducing it, we have 



tan (E— i M ) = 1+.^ tan AM— , ^ 2_F_(E) 



^ ^ ^ 1— e ^ (1— e) cog ^ M cos (E— | M)' ^ ' 



Let tan (E'— ^ M) = -~^ tan ^ M and a = —^^. Substituting and 

 1 — e " 1 — e * 



transposing in equation (2) we have 



^^^^3^ZlM__ = tan(E'-iM)-tan(E_iM), (3) 



_ sin (E'-E) 



cos(E'— ^M)cos(E— l-M)' ^ ' 



... si„(E.-E)=iL^^:^l|=J^F(E). (5) 



We have also 



, t? ^^'7}!^^ , ^,, = tan ( E'— ^ M ) — tan A M. (6 ) 



cos|-Mcos(E — |M) V 2 / 2 V y 



2e_ 

 -e 



Since tan (E'— 4 M) = 1+? tan A M, if g = ^ 



sin (E'— M) = q sin \ M cos (E'— ^ M). (7) 



But • ' 



sin (E— M) = E— M-fE'— E— 2 F (E'); . (8) 



E— M =5sin-^Mcos(E'— |M)-|-(E— E')-}-2F(E'), (9) 

 =i9cos^Msin(E'— ^M)-j-(E— E')+2F(E'),(10) 



