because the plane of this ellipsis is not the same with that of the earth's orbit, it follows, that the former will continually change; so that neither the inclination of the two planes towards each other, nor the line in which they intersect, will remain for any length of time unaltered. 285. The various forces by which the motion of the moon is disturbed, and the changes which take place in its orbit, may be investigated in the following manner. See plate IX. fig. 13. Let CABD be the moon's orbit, T the earth, S the sun, P the moon; make SK ST; and let SK SL:: SP2: SK. Then if SK or ST represent the sun's force at T, SL will represent his force at P. Draw LM parallel to PT; divide the force and MS acting parallel to TS. But the force LM, and the part TM disturb the moon's motion.The force LM in its mean quantity is equal to PT, and by so much the force of the earth is increased. Also TM in its mean quantity is equal 3cs to 3PK, acting in a direction PN parallel and equal rad (r): PR178 725r/ to TS; and the force M T draws the moon out of her orbit. Let P p be the periodical times of the earth and moon; then the sun's centripetal force at T (ST): the earth's centripetal force ST PT at P:: : ; therefore the earth's centripetal PP' force at P = PP PTX PP PP ditional force PT :: And this is to the ad- PT: PP: pp. That is, the force by which the moon is retained in her orbit is to the increase of centripetal force by the sun's action:: PP: pp: 178-725: 1.— 3css 178-725rr force = ) 3cs 178.725r And :: S.R (r): PQ = 178.725 = C. If A sine of twice the moon's distance from the quadratures; then the force QR, accelerating or retarding the moon's motion in its orbit, is 3c 178.725 Let z = A = = S.QPR or KPT cos. PTK; then Therefore the increase of the moon's centripetal rad (r): RP (178-725r. 286. Also force PT : force 3PK or PL:: PT: 3P K. Therefore, ex æquo, the force by which the moon is retained in her orbit: disturbing force PL or TM:: PT × 178·725 : 3 PK. Therefore 3PK the disturbing force TM- PT x 178-725 earth's centripetal force on the moon 3a the sine of the moon's distance from the quadratures earth's centripetal force, divided by 178.725 × radius. Let C, c, be the centripetal forces of the sun and earth, s sine of the moon's distance from the quadrature, radius = r. Then the additional force (PT)=178'725' And the disturbing force (TM) 3sc 178.725r Produce TP, and make PR = PL, or TM, and draw RQ perpendicular to TQ. Then QR is the force that accelerates the moon, and PQ is the diminution of its centripetal force. For the force PR is divided into two forces, PQ, and QR, of which PQ, acting towards Q, diminishes the moon's centripetal force; and QR being parallel to the tangent at P, accelerates the moon at P. 287. There are therefore four points in the moon's orbit, each 35° 16' from the quadratures, where the moon's disturbing force makes no alteration in the earth's central force. For it is greatest when A is greatest, that is, when 2 CP is ninety degrees, or C P = 45°. The disturbing force TM, in the syziges A and B, is 2 PT. And therefore the earth's force upon the moon in the syziges, is twice as much diminished, as it is increased in the quadratures. The moon's orbit is more flat in the syziges, and more curve in the quadratures; and therefore she goes farther from the earth in the quadratures. For the orbit will be more curve where the central force is greater, that is in the quadratures. 289. The motion of the moon's nodes, supposing her orbit to be nearly circular, may be thus found: In fig. 7, plate XIII, let Aq BQ be the moon's orbit, T the earth, P the moon, SAB the line of the apsides, Q, q the quadratures, m Nn the line of the nodes. PK, PH, AZ perpendiculars upon TQ and Nn. The force by which the moon is drawn out of her orbit has been found to be |