— Central Forces. 291 
Art. X.—On Central Forces ; by Prof. Taroport Srnone. 
(Continued from p. 73 of this Volume.) 
Pur Fo=1—ecos.9, then Fnt=1—ecos. nt, F’nt=e sin. nt ; 
| # e® dsin.®nt 
hence by (0) 1-ecos.9=1—e cos. nt-e? igi «. 2 nde 
e* = d* sin.‘nt 
1.2.3. neq ete. but by (5) 1—ecos. o= hence by substi- 
luting this value, and the values of sin.?nt, sin.?nt, &c. (see La 
_ Croix’s Traité de Calcul Différentiel, etc. p- 314,) then taking the 
differentials (as indicated be the sda ine a Pane wad 
there results the equation ~ omit eae e cos. nt —F cos. Qnt =; = a 
a 3nt — 3 cos. nt) ~ 75 “5 —si(4 cos. 4nt. —4.2? cos. 2nt) — ete. 
(1), (see Mec. Cel. Vol. I, p. 179.) It may be well to observe, that (s) 
§ives the solution of Kepler’s Problem, supposing v to be calculated to 
_ terms, including e* only ; but it iseasy to see that the value of v can 
be easily calculated by the method which I have given to terms in- 
volving. any integral positive powers of e which may be desired... If 
€>1, the conic section is an hyperbola, a= its semitransverse - 
€= its focal distance +a, as in the case of the ellipse. In this 
p’dv 
a(e*—1) BAG Bie. es 
Curve p’ =a(e?— 1), T= Le cos. — (2), e'dt=r? de= (77 cos, vy 
—1)#dv 
put. SS (ae then Mi peas oF (3). By (2) 
Ar ieee 
dr ae(e? ~1) sin. vdv ; ave? —1X = 
= and sin. v= 
Cae v)? aif 
rdr Qtte. 2 
we hence (9) becomes ndt 1: pu "a cos. ° 
= an (242) " es 
or r= a | dt=ed je ” » or by inte- 
ae ae (4); 1 thes, ndt= ed tan. ?— cos-5 
gration 1 nt =e tan. eck tan. (a+ = (5), v, 9, and ¢ being reckoned 
rom the perihelion, and P= the semicircumference of a circle rad. 
i 
