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equal to twice E Mj. As the moon passes from to0 t the perturbing force gradually becomes more andmore inclined to the line E A, but continues to actoutwards with respect to the orbit M x M 2 M 3 M 4 . At0„ # however, the perturbing force is for the momenttangential to the orbit, and afterthe moon has passed 0 1;the force acts inwards. This continues until the moonhas passed to 0 2 , a point corresponding in position to 0 4hut on the left of M 2 E. At M 2 it is clear that theforce is represented by the line M 2 E, and is simplyradial. Also, in actual amount the perturbing force isless at M 2 than at any other point in the semicircleM; M 2 M s . After passing 0 2 the force is again exertedoutwards, becoming wholly outwards at M s , when it isrepresented by the line M 3 A' equal to M, A, or to thediameter of the circle M 2 M s M 4 . Passing fromM s to M 4 , and thence to M,, the moon is subjected to
* 0, is determined by the circumstance, that when O, K isdrawn square to EA (the student should pencil in the linesand letters here mentioned), and KL taken equal to twice® K, 0, L is a tangent to the circle Mi M, M„ M 4 . Since thesquare on line E 0, is equal to the rectangle under EK, EL, orto three times the square of E K, we obviously have the cosine
of the angle 0, E K equal to ^=, whence 0, M, is an arc of 54° 44';
and 0 2 M 3 , M s 0 3 , and 0 4 M, are also arcs of 54° 44'. In Herschel’sOutlines of Astronomy these arcs are given as 64° 14', and thefigure to art. 676 is correspondingly proportioned. But 54° 44' isthe correct value. Indeed it will be obvious in a moment that anarc of 60° would give a perturbing force lying witbin the tangent,since the tangent at the extremity of an arc of 60° clearly cuts theline E A at a distance from E four times as great as the distance°f the foot of a perpendicular let fall from the same extremity.
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