Astronomia Britannica exhibiting the doctrine of the sphere, and theory of the planets decimally by trigonometry, and by tables : fitted for the meridian of London ... / by John Newton ...

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Title: Astronomia Britannica exhibiting the doctrine of the sphere, and theory of the planets decimally by trigonometry, and by tables : fitted for the meridian of London ... / by John Newton ...
Author: Newton, John, 1622-1678.
Publication: London :: Printed for the author by R. and W. Leybourn, and are to be sold by Thomas Piercepoint ...,; 1657.
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Subject terms: Astronomy -- Early works to 1800.; Planetary theory -- Early works to 1800.; Astronomy -- Mathematics -- Early works to 1800.
Link to this Item: http://name.umdl.umich.edu/A52255.0001.001
Cite this Item: "Astronomia Britannica exhibiting the doctrine of the sphere, and theory of the planets decimally by trigonometry, and by tables : fitted for the meridian of London ... / by John Newton ..." In the digital collection Early English Books Online. https://name.umdl.umich.edu/A52255.0001.001. University of Michigan Library Digital Collections. Accessed June 4, 2024.

Pages

CHAP. 15. Of the Motion of Mars.

THere being no other variety in calculating the place of this Pla∣net, then what hath been already shewed, in the motions of Sa∣turn and Jupiter, we will proceed in the same method, and ga∣ther first the middle motions for the former time given, and then shew the Dimensions of his several Orbs, as we shall have occasion for them.

Page 125

An. Christ.	Longit. ♂	Aphel. ♂	Node ♂
Yeares 1500	245. 61611	146. 80916	45. 40250
80	193. 32778	1. 75139	1. 07194
6	68. 23750	. 13139	. 08028
Iuly	111. 10306	. 01250	. 00778
D 16	8. 38500	. 00096	. 00057
H 18	. 39306
P 4564	. 00996
Mean Mot.	627. 07247	148. 70540	046. 56307
Aphel. Subt.	14. 70540	Rest Anom.	••19. 36707

Supposing the Semidiameter of ♂ his Orbe: 100. 000
His eccentricity according to Bullialdus is: 9239
Semidiameter of the Earths Orbe: 656••8
The sine of his angle of Inclination: 3230
And the Arch answering thereto: 1. 85111
And therefore suppose the R. of the Earths Orbe: 100. 000
The eccentricity of Mars will be: 14075
The Semidiameter of his Orbe: 152350
The parts of his greatest Inclination: 4921

The Anomaly 119. 36707 is the angle A M E in the Ellipsis of the 13 Chapter, and therefore the halfe of it is the halfe sum of the angles M E H and M H E 59. 68353

2. The side M E	304700
3. The side M H	28150
Summe	332850 co. ar.	4. 4777515
Differ.	276550	5. 4417726
Tang: ½ sum	59. 68353	10. 2330382
Tang: ½ diff:	54. 86290	10. 1525633
Difference	4. 82063 Angle M E H

Difference doubled 9. 64126 Angle M B H or the Aequa∣tion sought, and to be subtracted from the planets mean Longitude, be∣cause the Anomaly is lesse then 180. viz. 119. 36707.

Meane longitude of Mars: 267. 07247
Aequation subtracted: 9. 64126
Mars his Eccentrick place: 257. 43121

Page 126

To finde his distance from the Sun.

As the sine of M B H	9. 64126 co. ar.	0. 7760404
To the side M H	28150	4. 4494783
So is the sine of M B H	60. 63293	9. 9402403
To the side B H	146473	5. 1657590
From the eccentrick place sub: Node.		119. 36707
Argument of Latitude		128. 06614
Whose complement is K L		51. 93586

To finde the Reduction.

As the sine of 90
To Cosine the great inclin: X K L	1. 85111	9. 9997732
So tang: of K L	51. 93494	10. 1061739
To tang: of X K	51. 92039	10. 1059471
Differ.	. 01455 Reduction.

Because the argument of Latitude is more then 90, the Reduction must be added to the Eccentrick place.

The eccentrick place of Mars: 257. 43121
Reduction adde: . 01455
Eccentrick reduced: 257. 44576

To finde the present inclination.

As Radius
To the greatest inclin: E B	9921	3. 6920533
So the sine of K L	51. 93494	9. 8960878
To the X L	3873	3. 5881411

Which are the parts of inclination agreeing to the common Radius, 152350. But the distance of Mars from the ☉ is to be put for the Radi∣us, and then the parts of inclination will be 3724.

For as common Radius D L	152350 co. ar.	4. 8171576
To ♂ distance B H or A L	146473	5. 1657590
So is X L	3873	3. 5881411
To X L	3724	3. 5710577

To finde the distance of ♂ corrected by Curtation.

As A L	146473 co. ar.	4. 8342410
To Radius		10. 0000000
So is X L	3724	3. 5710577
To the sine of L A X	1. 45703	8. 4052987

Page 127

As Radius
To A L	146473	5. 1657590
So Cosine of L A X	1. 45703	9. 9998596
To A X	146426	5. 1656186

To finde the second inequality of Mars.

We must have given, 1 The angle N A S, which is to be found by subducting the ☉ place from the eccentrick of ♂ reduced, or this from it, so that lesse then 6 signes may remain, this remainer is the Anomaly of the Orbe, and the complement thereof is the angle N A S, or the halfe, is the halfe sum of the opposite angles.

Example.

The eccentrick of ♂ reduced: 257. 44576
The ☉ true place: 154. 07347
Anomaly of the Orbe: 103. 37229
Complement is N A S: 76. 62771
Halfe Anomaly: 51. 68614

These given with the sides N A & S A, I say,

As the greatest side N A	146426 co. ar.	4. 8343814
Is to Radius		10. 0000000
So is the lesser side S A	100895	5. 0038707
To the tang. of	34. 56887	9. 8382521
Adde	45
As Radius
To co-tang.	79. 56887	9. 2650444
So tang. ••/•• sum	51. 68614	10. 1022929
To tang. ½ diff.	13. 11536	9. 3673373
Summe	64. 80150 angle A S N
Differ.	38. 57078 angle A N S

Because the Suns place was subtracted from the eccentrick of ♂ redu∣ced, therefore the angle of elongation A S N 64. 80150 must be added to the ☉ place 154. 07347 and then the place of ♂ will be 218. 87497.

To finde the distance of ♂ from the Earth.

As the sine of A N S	38. 57078 co. ar.	0. 2051769
To the side A S	100895	5. 0038707
So sine of N A S	76. 62771	9. 9880626
To the S N	157438	5. 1971102

Page 128

To finde the latitude of Mars from the Earth.

As the side S X	157438 co. ar.	4. 8028898
To the side X L	3724	3. 5710577
So is Radius		10. 0000000
To tangent of X S L	0. 13560	7. 3739475
Which is the quantity of ♂ his Northern latitude.