EARTH'S ATTRACTION ON MOON. But the moon in one second describes an angle during which the approach to the earth = 60 × 4000 × 5280 (vers. w) feet 60 × 4000 × 5280.2π2 (271)3. (24)*. (60)* =00448 feet. feet Therefore, the space through which the moon is deflected in one second from her straight path, is just the quantity through which she would fall towards the earth, supposing her to be subject to the earth's attraction, and we may, therefore, conclude that she is retained in her orbit by the force of gravity. When first Newton, in 1666, attempted to verify this result, he found a difference between the two values equal to one-sixth of the less: the reason of his failure was the incorrect measures of the earth, which he made use of in his computation; and it was not till about 16 years later that he was led to the true result, by using the more correct value of the earth's radius obtained by Picart. Principia, lib. III., prop. 4. 92. The moon's orbit is everywhere concave to the sun. Let S, E, and M (fig. 11) be the centres of the sun, earth, and moon. We must bring the sun to rest by applying to the three bodies forces equal and opposite to those which act on the sun; but these are so small that we may neglect them and consider the moon as moving round the sun fixed, and disturbed by the earth alone. The forces on M are, therefore, m' SM2 E and ᎬᎷ ... in MS, ...in ME. This last must be resolved into two, one in MS, the other per pendicular to it. Therefore, the whole central force on the moon in and the proposition will be proved if we shew that this force is but the least value of the central force corresponds to cos M=-1, and is then E 2. m' It is, therefore, always positive, or the path always concave to the sun. At new moon the force with which the moon tends to the sun is, therefore, greater than that with which she tends to the earth: the earth being itself in motion in the same direction, and, at that instant, with greater velocity, will easily explain how, notwithstanding this, the moon still revolves about it. Central and Tangential Disturbing Forces. 93. We have hitherto considered the effects of the central and tangential disturbing forces in combination; but it will be interesting to determine to which of them the several inequalities principally owe their existence. (1) To determine the effect of the central disturbing force. (1— § k2 — † m2 + § m2e cos(c0— a) — §m2 cos{(2—2m)0—2ß} ( 1 − 3k2 — 1m2 + e cos(c0 − a) + {m2 cos{ (2 — 2m) 0 — 2ß} + me cos{(2 - 2m − c) 0 − 2B + a} — 4k2 cos2(g0—y)— §m3e'cos(m0+ß−¿). If we compare this with the value of u found Art. (48), we see that the elliptic inequality, the reduction, and the annual equation are due to the central or radial force, as also one half of the variation and about a third of the evection. It would perhaps be proper to separate the absolute central force from the central disturbing force; the terms due to the latter are those which contain m; therefore, the elliptic inequality and the reduction are the effects of the former, except that in the elliptic inequality the introduction of c, or the motion of the apse, is due to the disturbing force. (2) To determine the effect of the tangential disturbing force. Let the central disturbing force be zero; = = 2m2ae cos((2-2m — c) 0 - 28+ a}, 4m2 cos{(2 – 2m) 0 – 2ẞ} - 3m2e cos((2-2m - c) 0 - 2ẞ+ a}, +ua to the first order. (1 3m2 cos{(2 – 2m) 0 — 2ß} + 21 m2e cos{(2 – 2m − c) 0 − 2ß + a}; ́1 + e cos(c✪ − a) + †m2 cos{(2 − 2m) 0 — 2ß} { We have here the remaining half of the variation and rather more than two-thirds of the evection as the effects of the tangential disturbance. Also c=1, or, to the second order, the tangential force has no effect on the motion of the apse. The inequalities in the longitude could be easily obtained from the relation but they would lead to the very same conclusions as the discussion of the values of u. To calculate the value of c to the third order. 94. We must here make use of the results which the approximations to the second order have furnished; but as the value of c is determined by that term of the differential equation whose argument is co-a, we need only consider those terms which by their combinations will lead to that one without rising to a higher order than the fourth. We shall simplify the arguments by omitting e, a, B, which can easily be supplied by remarking that ce a and m✪ – ß always enter as one symbol, c and m will therefore be sufficient to distinguish them. This only applies to the arguments. We have, Arts. (48), (23), u = a{1 + e cos(c) + m2 cos(2 − 2m) + 15 me cos(2 − 2m — c)}, P h2u2 =α- m2a {1 + 3 cos(2 – 2m)} {1 — 3e cos(c) T h2u3 T du h&us de = a + &m2ae cos(c) + 35m3ae cos(c), = = m* sin(2-2m) {1 - 4e cos(c) - 4m*cos(2-2m) - §m2 sin(2 − 2m) + 3m2e sin(2 − 2m − c) + 45m3e sin(c), + {3m2e sin(2 – 2m — c)} {— 2m3a sin(2 – 2m)} 15m3ae cos(c), the other term is of the fifth order, 2m2 cos(2—2m) — 3m3e cos(2 — 2m — c) — 45 m3e cos(c), +u)=2a{1+...- 3m3 cos(2 – 2m)+ 15 m2e cos(2-2m−c)}; થ) d2u ( a{1 + (3m2e + 13,5 m3e — 15m3e + 45m3e) cos(c) + ...} · a{1 + (3m2 e +225 m3e) cos(c) + ...}. a{1 + e cos(c) + .....}; therefore ae(1-c2) = (3m2e + 225 m3e) a; |