Of KEPLER'S PROBLEM. 257 



Then, to find a value of x, from the equation, x^ -}- a x zz b : 

 fince l) is ftnall in comparifon of a, it is manifeft that x mud 

 be a very fmall fradlion ; and, confequently, x-' inconfiderable in 



refpedl of a x: therefore x =: - = .05 nearly : and, having cor- 



a 



re<fled this value by the common method, we fliall find, 



A' iz: fin (p n .0547 = fin 3° 8', and fo(p-=z i'j6° $z'. 



As the firfl; approximation which we are now computing, cannot 



be exa<fl, even to, the nearefl minute, it would be fruitlefs to pufl^i 



the calculation to a great degree of accuracy. For the fame 



reafon, I here ufe the proportion in the general rule, becaufe it 



requires but one operation for finding 1^, viz. cof : 



cof ^ : : cof ^ : cof il/. 



Hence ■^ is found =3° i', and therefore p '=■ <p — ■^ = 

 173° 51'; this is the firft approximation to the eccentric ano- 

 maly reckoned from the aphelion, and is too fniall. 



adly, To compute the fecond approximation, we have, 

 />,, m — p 



2 fin 2° c8' , I .00 



e = ^X- --^X^p^^,; whence -= 1.06878; 



-^{m—p) 



I I = .06878 = a; -^— J- =: .0038749 = k Therefore, 



«5 -^ a X — b: but we know, that a near value of x is .0547 t 

 and, having corrected this value, we Ihall find, * rr fin (p = 

 .0540430 ; therefore, fin <p = ;o540430 = fin 3° 5' 52'', and (p = 



176° 54 8". 



As ^|/ is here a fmall angle, we muft ufe the method of Art. 1 2. 



to find it with the requifite exadlnefs : this gives tan A = 



, X — ^^^ — X fee 45 % and fin 2r — tan - X fin 45 " ; there- 



col 



2 



fore log. tan A = 8.868948 1, and 4> = 2" 59' 32'', 



Vol. v.— P. II. Hh wherefore 



