SOI

Astronomy.

if*?

2i. Having thus obtain’d the diurnal Are#* '^pmultiply this by any Number of Days and D ^

-- i’luuiuci U1 J_yyy5

Parts of a Day, it will give the Area P R QS, ot ^

Anomaly, for the given Time.for "January 23 D. 6 H. 11

Tims let it be M u

.<1

Then from the Time of Perihelion,Subduct the given Time,

D.

jFeb. 19Jan. 23

s-

8

6

M*

ll

The Difference will be 27

Wherefore to Log. of diurnal Area 8.718 = 0-^^Add the Log. of the given Time 27,0833= 1 - 48 ^

The mean Anomaly required, = 236,1 = 20' J ^

22. Having therefore the Area P R QS = 23 ^ ?I ’can find A Q_= x, from the Equation x> -j" 3 x 'for if when the Quadrantal Area P S R is 100,

S R = x — I, then ’tis plain, x* %x ~ I + 312 a in that -Cafe. Therefore when the mean

11 1 '

vveP

,u*

-

r6 'i

is but -ris Part of this, we have ^ 3 x — cl .

0,04; which will be a constant Multiplier for r c ^u 5any given Anomaly to fit it for the Equation-©,04 X 236,1 = 9,444 = 3* in the prest* 1 jvC*

which resolved according to the usual Methodsx 1,65 nearly.

23. Then by. the Nature of the ParabolaAP- 1 . ’ 65 * 1 ’ 6 - 5 ^ 1,3612. Also AP + P S '" S

fas

tkt 1,8612, the Distance of the Comet from th® ‘ 3 . L> ‘ {Jthe given Time. But to express this Distance I -g a ftl*fame Parts as the Sun’s mean Distance from thecontains 1,00000, we must consider that the P erl W jfiDistance P S — 0,22206; whence S R = ‘'’’tad} 01Wherefore fay, As 1 : 0,44412 :: 1,8612 :the Expression required24. In the Right-an b .the Sides, we find the An

24. In the Right-angled Triangle Q_Ale Sides, we find the Angle QS'A — 62° z6? ' , e 0 C 'the obtuse Angle P S Q_— 117 0 33I', fh^^

iiocentric Distance of the -Comet from the I ei j ^