cx F: cx F: FR: F'R'

or cc: R:R'.

Hence if P, P' (Fig. 98) denote the points of projection, S, S' the centres of force similarly situated, that is making equal 2, with the directions of projection or tangents PR, R'R' of the orbits PQW, P'Q'W', and PV, P'V' the chords of curvature, we have

PV P'V' PS: P'S'

From S, S' draw SQ, S'Q' making equal angles with SP, S'P', and meeting the circles of curvature in Q, Q' and join PQ, P'Q and produce SQ, S'Q' to R, R'. Then

< PQV (= <YPS = ¿Y'P'S') = <P′Q'V'. Hence putting PSQ (= <P'S'Q) = a

QV SV :: sin. a: sin.

and PV QV :: sin. : sin. (a + B+ 0)

.. PV: SV :: sin. a. sin. ß: sin. 0. sin. (a + B + 0).

similarly,

PV' S'V'

sin. a. sin. B: sin. . sin. (a + B + 0)

and PV SV :: P'V': S'V'

..sin x sin. (a + B + 0) = sin. '. sin. (a + B + 0')

But generally cos. (P — Q) — cos. (P + Q) = 2 sin. P sin. Q.

.. cos. (a + B)

cos. (a+B+20').

cos. (x + B + 20) = cos. (a + B)

.. cos. (a + B + 20) = cos. (∞ + B + 2 0′) and a + B + 20 = a + ß + 20′

or 00'

that is

And by Lemma 7th of Principia, the arcs and chords are ultimately in a ratio of equality.

.. arc PQ: arc P'Q' :: PS : P'S'

also QS Q'S' :: PS: P'S'

(1)

Again if Qq, Q'q' be the next elemental arcs subtending equal 4QSq, Q'S'q'; and yQr, y'Q'r' the tangents or directions of motion

at Q, Q'; then since the velocities at Q, Q' are as

1 1

or as

LyQS (= LyQP + ZPQS = <PVQ + <PQS = ZPV'Q' +<P′Q'S' = <y'Q'P' + <P'Q'S') = Zy'Q'S', or the directions of motions are similarly situated with respect to the distances SQ, S'Q, we have, by what has already been demonstrated,

arc Qq arc Q'q' :: QS : Q'S' :: PS: P'S'

and qS: q'S: QS: Q'S' :: PS: P'S'

}

and so on through the whole extent of the orbits. Hence, by the composition of ratios we have, supposing PSW = <P′S'W'.

arc PW arc P'W' :: PS: P'S'

and SW: S'W' :: PS: P'S'

.(2)

But curves referred to a centre are similar, when the radiivectores containing equal angles are always proportional.

Therefore the orbit PW is similar to the orbit P'W'.

Q. E. D.

535.

Since the velocity is finite, the law required must be

the velocity being 0 when goo.

Now the velocity in a circle at the distance is

g

Now in the circle whose radius is a, the Periodic Time is (440)

the radius of the circle is found from

v2 = 2μ . ƒ — p3dp = 1⁄2 (r^ — ç1), which gives

=0. Now at the centre let the force become repulsive,

Hence the whole velocity acquired by falling and rising is

V = v + v = √ 2μ × ro.

But the velocity in the circle, the force being the same is

.. V: V' :: √ 2:1 and not :: 1: 1.

The mistake in the enunciation probably arose from adding v2 and v2, instead of v and v'.

538.

μ

μ

If and be the absolute forces, and t, t' the Pe riodic Times, according as the centre of force is in the centre or circumference of the equal circles, then by 428, we have

75

:. p = R × sin. ( ~ + 0) = R cos. § . . . . . . . . (b)

the equation to the spiral.