Astronomy

f 62

. k s(L

required, let fall the Perpendicular S F ; the Ar®® C1is equal to the Sector ACQ_and the Triangle 1 L, <Jthat is, equal to ; QC X A Q_+ IQC X SF 5 g &because f Q_C is a constant Quantity, the Area r* ^will be proportional to A Q_+ S F. Hence if o0 r-the Arch Q_N — 8 F, we have the Arch A N P r ^;tional to the Time or mean Anomaly of the Pwhich we can easily find by having the true A* 10

given.

,¥

55* For Example; in the Orbit of Mars ’’ftQjC : S C 152369 14100 ; and because the h e 0of an Arch equal to Radius is 57^,2957 8, fay,- gf

0.0 = 152369^5 - f/jij

SC= 14100

As the RadiusIs to the Eccentricity

So is the Length of the Arch

To the Length of an Arch. B, 5 ^, 302 =©' 7

Then fay, ' . 0 (P

As Radius S C 90°oo=t 0, -

Is to the Sine S F of the Angle 7 -

SCF=ACQ_, which suppose j 3° 00So is the Length of the Arch B = 5 o ,302= 0, 7y_^'

To that of the Arch QN=SF=2 , ,65i:= 0, 4 i 0

56. Therefore A Q_+ Q_N = 30 6 + 2%®5 * qV39' 3". Thus from the eccentric Anomaly ATgain the mean Anomaly AQ_-f- QN == A N,. ftiStis proportional to the Time; and the reverse 0 . ftpviz. from the mean Anomaly A N given, to heccentric Anomaly A C Q_, L to be done bythod of Infinite Series, as follows. Let the Arch™ . a ji<>the Sine of the Arch AN be = e, the G o-sin.ea^/’^d,the Eccentricity S C — g. The Sine of the Ar c ^ tois equal to the Sine of the Arch AN —N Qj> C -5 s d>'the Sine of the Arch A N— y, which Sine is

pressed by a Conv. Series, e —--— + 7^3

et‘ r _ . . <r’/igt» tS

———, as Dr. Kail has strewn in h>s *■

1.2.3.4,

mtry.

57 '