or x-m2 x- · x2 + 2 m x2 is a max.

Hence by the principles of the diff. calc.

1 — m2 — 3 x2 + 4 mx=0,

This equation gives

2m

x = = = = √ 3 + m3. The upper sign can only be used, be

3

cause 3+ m2 is always greater than 2 m.

By this value of x the angle STV will be found about 40° for Venus (art. 110), and the point V of greatest brightness lies between inferior conjunction and greatest elongation. If the distance of Venus from the sun were = the greatest brightness would √5

1

be at the greatest elongation, for then x=

2.

√5

Prop. XI. To find when a planet appears stationary,

Let S (fig. 56) represent the sun; T and P the earth and planet respectively, their orbits being supposed circular.

When P and T are stationary with respect to each other, the line TP moves parallel to itself, and the angles T and P only vary by the apparent motions of the sun, as seen from the earth and planet: which motions are equal to the angular motions of the earth and planet as seen from the sun.

Now the angular motions of the planets about the sun are to each other inversely as their periodic times; and by Kepler's law, the squares of the periodic times are as the cubes of the distances;

hence the squares of the angular velocities of the planets are inversely as the cubes of their distances.

Let t=T, p=P, r=ST, and r'=SP, t'= a small variation in T, and p a corresponding small variation in P.

(1) By trig. r sin t=r' sinp,

(2) and r sin(t+t')=r' sin (p+P).
By equat. (2)

rsint.cost+r cost. sin t'r' sinp. cos p' + 'cosp. sinp'.

sub

Since and are to be supposed indefinitely small, we may stitute = cos t' and t' sin t', &c. Therefore r sin trt'. cost= .sinpr.p' cosp.

=

(3) Hence by equat. (1) rt. cost=r.p' cosp.

But as was shewn above, however small t and p' are

:

By squaring equat. (3), and substituting from this proportion

r.cos't=r.cos'p,

or (1-sint) = (1-sin'p). Hence, by equat. (1),

a

This solution, it is evident, answers both for a superior and an inferior planet.

Problems in which Approximations are used.

In many of the more useful investigations in astronomy, it is sufficient to make use of approximate solutions, as, for instance, in those for finding the effects of the precession of the equinoxes in right ascension and declination; for the effects of the aberration

For a different solution of this Prop. see Luby's Trig. chap. 4, part II.

of light in right ascension and declination; for the effects of parallax in latitude and longitude; and for a great variety of other problems, of which instances are to be found in every part of astronomy. These solutions, although only approximate, admit of all the accuracy that is necessary, and in general are obtained with much greater facility than exact solutions could be. In these solutions it often happens that we substitute the sine instead of the arc, radius instead of the cosine, the tangent instead of the arc, and conversely. It is therefore of some importance to know the limits of the differences of these quantities. Let aan arc, s= its sine, c its cosine, and t= its tangent, radius being unity: then

From whence by the trigonometrical tables it will easily be found that the difference between the arch of 1o.46′ and its sine is only 1", and therefore in many cases one may be safely substituted for the other. The difference between the tangent 53′ and the arc is only 1".

It is often of use in these problems to reduce a small arc, sine,

a These expressions are easily proved by the diff. calc., and some writers have proved them by principles purely trigonometrical. See Luby's Trig. chap. 1. part II.

&c. expressed in decimals of the radius (unity) to seconds, and the contrary. Which may be done as follows, let a arc in decimals : of the radius, a= the seconds in the arc, then as the sine of 1" and : the arc of 1" are nearly equal, we have

Prop. XII. To find how much the time of rising of the sun or a star is advanced by refraction.

Let RS (fig. 57) represent part of the sun's parallel of declination, R the true place of the sun, when it appears rising at D. Then the time of rising is advanced by the angle SPR, the measure of which is the arc of the equator HL; and also the arc DR = the horizontal refraction.

The small triangle SRD may be considered as a plane triangle and

RD: SR:: sin RSD = cos PSQ : rad.

SR: HL:: rad parallel : rad equat :: sin SP; rad.

therefore RD: HL:: cos PSQ X sin SP: rad',

known, and consequently HL, which, divided by 15, gives the time required.

A somewhat more simple solution may be deduced from the above.

See remarks on this problem in Lalande's Astr. 3d edition, vol. 3, art. 4028. Cagnoli Trig. p. 368.

Cor. 1. If RD be taken equal to the diameter of the sun, the time the sun takes in rising will be had.

Cor. 2. If RD be taken equal to the difference between the horizontal parallax of the moon and the horizontal refraction, the time will be had of the retardation of the moon in rising.

Prop. XIII. Given the error in altitude, or in zenith dis- O tance, to find the error in time.

Let rZ (fig. 58) be the observed zenith distance, and Zs the true zenith distance, rPs P the polar distance. Join rs, and draw rn a portion of a parallel to the horizon. Then sn = error in zenith distance, and r Ps the error in the hour angle.

snsr sin srn sin ZrP: rad.

sr: rPs: sin r P: rad.

hence sn: rPs:: sinr ZP X sin r P: rad'.

Otherwise thus:-By spher. trig. art. (106), p. (67).

Hence

cos r Z=sin ZP. sin r P. cos ZPr+ cos ZP. cos r P.

sin rZ.d.rZ= sin ZP. sin r P. sin ZPr.d. ZPr

sin ZP. sin rZ. sin rZP.d. ZPr.

Consequently