Of KEPLER'S PROBLEM, 235 



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

 •9049154, and <?' rz 64 ° 49'. To have a more correct value of 

 P, repeat the calculation, 



I'Og.e*- 3-9375334 



log. fin <t>' = 9.9566250 



2 X log. cof ¥ =~ 19.2578320 



log. e 2 fin <p' cof 2 <p' — 3.1519904, whence nat. fin <p — .9048966, 

 therefore <p = 64 ° 48' 33". 



3. To find h we have tan A z= e X fin ^ X fee 45 * 



cof^=-^ 

 2 



log. * = 2.9687667 

 log. fin <p — 9.9565982 



log. fee 45 = 10.1505150 



Sum — 19.0758799 



, ,.0) -r— m 



log. cof 2—-— = 9-9999994 



log. tan A = 9. 0758 8 05 — log. tan 6° 47' 29" 



But fin * - tan A x fin o therefore 

 2 2 



A 



log. tan- - 8.7733146 

 log. fin 45 - 9.8494850 



log. fin ' = 8.6227996 = log.fm 2 6 24'i6".5. Therefore 



^ = 2 °. 24. 1 6' / .5, and^ = 4°-48'. 33^ fo that /* ~ <p — <J» = 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 Amplification, arifing from the peculiari- 

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



trouble. 



