Now, since P exceeds 90° its cosine is negative, and consequently must be taken negatively. Hence PA = 41° 9′ 46′′ + 26° 2′ 53′′′ = 67° 12′ 39′′, and the work indicated by equa. (2) will stand thus:

The sum

50° 5' 47".... 9.8071953 4) 67° 12′ 39′′.... 9.5880938 26° 2′ 53′′....

.. 10.0465177

20, cos AB 73° 56′ 40′′.... 9.4418068

Thus the distance required is 73° 56′ 40′′ of a great circle, or in English miles, reckoning 69 to a degree, it is 5139 nearly.

Example II.

The latitude of St. Helena is 15° 55' S. its longitude 5° 49′ W.; the latitude of the Bermudas 32° 35′ N. longitude 63° 32′ W. Required their distance?

Here since the latitudes are of different kinds, we shall have

PA 90° + L = 90° + 15° 55′ = 105° 55'.

The preceding formulæ applied to these data, will give 73° 26' for the distance required.

Example III.

What is the distance on a great circle between St. Mary's, latitude 37° N., longitude 25° W., and Cape Henry on the same parallel of latitude, but in longitude 76° 23′ W.?

Ans. 40° 31, which is about 36 miles less than the distance measured upon the parallel itself.

Example IV.

Required the distance on a great circle between the island of St. Thomas, latitude 0°, longitude 1o E., and Port St. Julian, in latitude 48° 51′ S., and longitude 65° 10′ W.

Ans, 74° 35'.

Note. In the 3d example the operation becomes shortened because the triangle PAB is isosceles; in the 4th it is shortened because the triangle is quadrantal.

PROBLEM VIII.

30. To determine the elongation of a planet from the sun, at the time when it appears stationary; on the supposition of concentric circular orbits, in one and the same plane.

As thus restricted the problem may be solved by means of the principles of plane trigonometry, combined with those of central forces in dynamics.

Let two concentric circles EeI, PpQ, be drawn, the former to represent the orbit of the earth, the latter that of the planet, the common centre s of both circles being imagined the place of the sun. While the earth moves from E to e in its orbit, let the planet move from P top in its orbit; then, when the planet appears sta tionary the right lines EP, ep, drawn from the earth to the planet will be parallel. In that state of things draw the radii SE, se, SP, sp; from E and P draw tangents to their respective circles to meet each other in T; and join ST. So will the triangles SET, SPT, be right angled at E and P; and the parallel lines EP, ep, will be so near each other, that the intercepted arcs Ee, PP, of the orbits, may be regarded as coinciding with the corresponding portions of their respective tangents.

Let v be the velocity of the earth in its orbit, v' that of the planet. Then, since the arcs Ee, Pp, are described in the same interval of time, they will be to each other as those velocities, that is,

Ee: Pp :: V: v'.

Also, since the arcs Ee, Pp, are confounded with the portions of their respective tangents which meet at T, and the parallels EP, ep, divide TE, TP, proportionally, we have ET: PT :: Ee: Pp; and consequently,

ET: PT :: V: v'.

Put SE a, SP = = b, and the required angle SEP of elongation = E. Then, since SPSE:: sin E: sin EPS, we shall have

a2

Therefore, v : v ́ :: √/ (r2 — — sin2E): √/ (r2 — sin1E).

b2

a2

Whence v2 (y2 — sin 2E) = v'2 (r2 — — sin 2E);

31. A still simpler and more convenient expression for sin E may be obtained by introducing the radii of the respective orbits for the velocities of the bodies. Thus, by the theory of central forces (Gregory's Mechanics, book ii. art. 282) v: v′ :: √b: √a;

Hence (63-ab2) r2 = (b3 — a3) sin 2E,

Supposing r and a to be each unity, we shall have

Here the positive sign obtains in the case of the inferior planets, the negative sign in that of the superior planets.

Example.

The mean distance of Jupiter from the sun being 5.2028 times that of the earth; required the elongation of the planet from the sun at the time of its apparent station?

The logarithmic operation for tan E indicated by equa. (4) is as follows:

From 2 log b

..5.2028.... 1.4324742

Take log (6+1) 6·2028 ....

Diff. +20....

0.7925878

20.6398864

Its half is tan E .. 115° 35'.... 10-3199232

Here, the tangent being negative, it belongs to the second quadrant (chap. iv. art. 5).

Note. By like operations the mean elongations of the other planets at the time of their apparent stations may be found as below.

SCHOLIUM.

32. The inversion of our 4th equa. furnishes in certain cases a convenient expression by which to approximate to the mean distance of a planet from the sun.

For from tan E =

6 + 1'

we find

b = ‡ tan 2e̱ ± tan x (1 + Į tan 2ɛ)*:

an equation which gives the radius of the planet's orbit in terms of that of the earth, from the elongation at the instant when the motion of the planet became direct after having been retrograde.

=

Thus, in the case of Ceres it was found by observation that E at the time of apparent station was 122° 37′ 40"; whence b was found 3-2018. The slight difference between this elongation and the one just given, arises from this,-that the latter was actually observed, the planet moving in an elliptical orbit, while the former is what would be observed if the orbit were circular.

The following is the logarithmic operation, for finding the mean value of b in Ceres:

20 is log tan2 E ..0·60995.... Ī·7852952 Add..

1.

Log (1 + tan2 E) = log .. 1.60995.... 20-2068125