Of KEPLER'S PROB LEM. 235 



To this adding nat. fin m — .9034776, we have nat. fin p' zz 

 •9049154, and ?' n 64° 49'. To have a more corred; value of 

 ?, repeat the calculation, 



Log.e'= 3-937S334 



log. fin p' =■ 9.9566250 



2 X log. cof 9' =- 19.2578320 



log. f^ fin(p' cof = <p' = 3. 15 19904, whence nat. fin fr:. 9048966, 

 therefore (p = 64° 48' 33". 



3. To find I, we have tan A = ^ X ^'" ^ X fee 45' j 



cof'P-'" 

 2 



log. e =. 2.9687667 



log. fin <p — 9.9565982 



log. fee 45° = I C.I 505 1 50 



Sum — 19.0758799 

 log. cof ^ = 9-9999994 



log. tan A = 9. 075 8 8 05 = log. tan 6'' 47' 29" 



A 



But fin ^ = tan - X fin 45°, therefore 

 2 2 



log. tan- - 8.7733146 



log. fin 45° = 9.8494850 - 



log. fin - = 8.6227996 = log.fin 2°24'i6".5. Therefore 



2 



^ zz 2°. 24.16". s, and->|/ n: 4''.48'. 33*, fo that /* =: (p — i' — 60°, 



tlie eccentric anomaly required. 



15. It remains now to confider the cafe of the problem, ap- 

 plicable to comets, where the eccentricity is very great, or near- 

 ly equal to unity. Here the general folution may be ufed, and, 

 even without any new fimplification, arifing from the peculiari- 

 ties of the cafe, will bring out an accurate refult witli very little 



troublfe; 



