280

ASTRONOMY.

tance the space between the two parallels is altogether insen-sible.

(84.) If the Earth be at G, and Mars at I, he is in opposition ;ret it situated as in the figure, he is not seen exactly opposite to theSun , but a little below the ecliptic at K, or, he has south latitude.If Mars be at R when the Earth is at G, he is in conjunction. Ifthe Earth be at E, and Mars in the node, and in opposition at F,then will the centre of the Earth’s shadow be directed to Mars , andwere not the Earth much less than the Sun, Mars would be enve-loped in the shadow, or eclipsed. Since this never happens, it shewsthat the Earth’s shadow ends in a point before it reaches Mars , andthat the Earth from that planet would then appear as a black spoton the Sun ’s disk.

(85.) Let the Earth be at N, and the Moon at M, then the Moon will appear from the Sun at O, and from the Earth at L, where theline SL parallel to N d meets the starry heavens. In both casesshe would have north latitude; because she is nearly in the middlebetween her nodes, her geocentric latitude would be about 5° 17',being the angle subtended at the Earth by a perpendicular from theMoon to the plane of the ecliptic, but the heliocentric latitudewould be less than 1 j', or less than the 395th part of the geo-centric latitude, because the perpendicular would subtend anangle at the Sun so much less : when the Moon is at t she is inopposition, or at the full, and appears a little above the ecliptic a P.Much after the same manner may many other circumstances relat-ing to the motions of the planets be elucidated.

( 86 .) The true heliocentric longitude and latitude of a superior orinferior planet being known, its distance from the Sun , and also theSun’s true place and distance from the Earth at the same time, thegeocentric longitude and latitude may be determined in the fol-lowing manner: First find the commutation, by subtracting the

true longitude of the Sun , + 20 " bn account of the aberration oflight, from the ecliptic heliocentric longitude of the planet; and theremainder, if under six signs, or w hat the remainder wants of 12signs, if above six, will be the commutation. Next take the dif-ference between the logarithm of the distance of the Sun from theEarth, and that of the planet’s distance from the Sun , by sub-tracting the least from the greatest, and adding 10 to the index ofthe remainder ; then seeking in the logarithms of the tangents theangle corresponding to the difference, we obtain a quantity whichwill always exceed 45°; and 45° being subtracted from it, the re-mainder will be an angle of use in finding the elongation of theplanet. The logarithm of the tangent of this last quantity addedto the logarithm of the tangent of half the commutation, givts thelogarithm-tangent of an angle, which added to half the commuta-tion, in the case of the superior planets, or subtracted for an inferiorplanet, the sum or difference will be the elongation of the planetfrom the Sun, as seen from the Earth . Then to find the geocentriclongitude, add the elongation to the Sun ’s true place, it the com-mutation be less than six signs ; but subtract it, if greater than sixsigns ; and the sum or difference will be the true place of theplanet, as seen from the Earth . To find the geocentric latitude,use the following proportion : the sign of the commutation is tothe sine of the elongation, as the tangent of the heliocentric lati-tude is to the tangent of the geocentric latitude. Obs. The geo-centric latitude is always of the same name with the heliocentric.The following example will be sufficient to illustrate the precedingrules:

Examp. Required the geocentric longitude and latitude of Mer-cury, on April 1 , 1813, at 13' 21 ",4 past noon, meantime atParis , or at 4' 1”,4 past twelve noon, equal time at Greenwich.

Calculation of the Sun’s true Place and Distance from the Earth, at the proposed Time , from De La Lande’s Tables, in the Third

Edition of his Astronomy.

Epoch’s of Q’s long. 1813.

( April 1 ,

Ilis motion for.x 13 minutes,

( 21",4

O’s mean place at proposed time.

Equation of ©’s orbit...

O’s place equaled.

Secular equation.

Equation arising from.4 If.

?

First equation of $ .

Second equation of $ .

Nutation.

O’s true place.

O’s mean long.

•Arg. 1.

O’s Apogee. | ))

Arg. 2.

U

Arg. 3.?

Arg. 4.cf

Arg. 5.$

Arg. 6 .Nut.

s. o '

9 9 45 12,1

2 29 41 38,0

32,00,9

s. 0 ' "

3 9 42 31

15,5

937

82

453

228

257

156

720

117

967

16

606

13

0 9 27 23,0+ 1 55 25,9

3 9 42 46,5

0 9 27 23,0

19

681

413

837

983

619

O’s mean anomaly,from ©, (Arg. O’s

ID +13

fr° m ] 5 + 5

— 1

0 11 22 48,9— 2,2 ■

+ 0,7

— 4,4

+ 7,1

+ 3,5

— 0,3

— 11,5

8 29 44 36,5Log. of O’s distmean anomaly,).Secular variation

Equation arising

0 11 22 41,8

True log. of O from

0,000105

Calculation of the Heliocentric I.ongitude and Latitude oj Mercury, jor the given time from the same Tables.

Epoch of $’s mean longitude, 1800 .

Mean long, of $

g’s Aphelion.

5’s Si

s. 0

3 18 10 38

11 20 36 23

0 12 24 23

2 13

3,7

s. 0 ' "

8 14 20 51

0 0 12 11

14

S 0 ' M

T 15 56 48

0 0 9 23

11

I-Iis motton for - 1 .

(21",4 .

3 21 13 40,7

+ 17 23 44,3

8 14 33 16

3 21 13 40,7

1 16 6 22

4 8 37 13,5

4 8 37 25,0

- 11,5

7 6 40 24,7

_ Y '

2 22 30 51,5

__ J

True place of in his orbit ...

$’s mean Anomaly

Y

g from Q

4 8 37 13,5— 3 15,4

Reduction to the ecliptic .....

^’s heliocentric latitude

6° 56'22” N.

4 8 33 58,1