the semi-circle AGP be described, and let CL be drawn parallel to SF, and GDH perpendicular to AP. Then, the angle PCL PSF is the mean anomaly, and PSD is the true anomaly. The angle PCG is called the eccentric anomaly. Assuming AC, the mean distance of the planet, to be a unit, put SC е eccentricity PCL= PSF mean anomaly PSD true anomaly X PCG T time of describing semi-ellipse PDA t time of describing arc PD. or, By the property of the ellipse, :: area PGA : area PRA :: AC: CR area PGS : area PDS, area PGA or, But, by Kepler's second law (153), area PRA: area PDS:: T: t. Hence, area PGA area PGS :: T:t:: 180°: PCL :: area PGA : sect. PCL. Hence, area PGS sect. PCG sect. PCG (A) But, sect. GCL ≥ AC × arc GL, and triang. GCS = 1⁄2 AC × CS X sin PCG. Hence, sect. PCL sect. PCL sect. GCL arc GL CS x sin PCG e sin x. Or, for the arc GL, or angle GCL, in seconds, we have (App. 51, Cor.), angle GCL =ew sin x Also, since PCG PCL GCL, we have, + x = m + ew sin x (B) (C) For most of the planets the value of e is less than 0.1, and it is not for any of them more than about 0.25. Hence, it is evident (B), that the angle GCL is generally quite small, and that it is never large. The sector GCL differs, therefore, but very little from the triangle GCL, and we have (A), triang. GCL triang. GCS, nearly. Consequently, SL is nearly parallel to CG, and the angle PSL is nearly equal to PCG, the eccentric anomaly. Put 2d area PGS, 180° diff. of angles CSL and CLS pangle PSL = eccentric anomaly, nearly, x = p + z. By trig., we have, CLCS CL + CS:: tang or, 1 (CSL - CLS): tang e: 1+e: tang † (180° (CSL + CLS), 2d) : tang § (180° — m), or, 1—e:1+e::cot (90° tang d. Hence, But, m+d=90° CSL = 180° or, tang2 1 u = 1 + e e p 1⁄2 m + d (E) Now, substituting p+z instead of x in formula (C) and observing that, as z must be very small, we may regard cos z = 1, and w sin z = seconds, we have, z, in p + z = m +ew sin (p + %). z) = m + ew sin pez cos p, very nearly (F) + ew sin p Hence, 1. But, SH Hence, But (App. 12), 1 COS u 1 + cos u 1 + e tang d (D) 1 e 1 (CSL + CLS) +90° (CSL CLS) PSL = P; " 1 + e cos u m tang2x. e cos p X p+ very nearly, (G) e cos p And, since xp, we have, m + ew sin P +1 If x is desired with still greater accuracy, it may be obtained by taking p equal to the value of x found from formula (G), and recomputing with this value. This repetition is, however, seldom if ever necessary. 1, SC e, and PSD = u, we have, the property of Now, as AC the ellipse, SD CH CS (1 e2) cos u 1 + e cos u m): cot (90° tang m COS X 1 + e 1 and SHSD cos u = CG cos PCG , very nearly. CS d) tang im : e, or, cos u = (1 + e) cos x e + (1 − e) cos x (1 e2) cos u 1 + e cos u = COS X e. 1 + e Hence, tangu = tang x√ (H) 1 e Having found the value of p from the expressions (D and E), we find x from (G), and then the true anomaly u, from (H). 58. To determine the height of a lunar mountain. Let ABO, Fig. 68, be the enlightened hemisphere of the moon, E the situation of the earth, ES' the direction of the sun from the earth, and SM a solar ray, touching the moon in O, which will be one of the points in the curve separating the enlightened from the dark part of the moon. Also, let M be the summit of a mountain, situated near to O, and just sufficiently elevated to receive the sun's light on its top. To an observer at E, the summit M, of the mountain, will appear as a bright spot on the dark part of the moon. Let the angle MEO be measured with a micrometer, and let the angle of elongation CES', be found from the positions of the sun and moon at the time. We may, without material error, regard ES' as parallel to MS, EC as equal to EO, and the angle MES' as equal to CES'. Consequently, sin OME sin MES' sin CES'. Hence, from the triangle EMO, we have, or, Put or, Now, Ma OC. Then, since ♪ and C are both small, we tang C. Hence (A), tang C. sin OME: sin MEO :: EO: MO, MC sin CES': sin MEO :: EC: EC sin 8 tang C, tang C ang MEO sin & sin CES' & sin CES' 1 - OC sin C tang C have (97), OC • (B) OC cos C (App. 11) cos C OC tang C tang C = d tang 1⁄2 C tang 1⁄2 C = d d ( cos C Observing that Ma is the height of the mountain, we have, by substituting the value of tang C (B), Height of the mountain When the moon's apparent diameter was 32′ 5′′.2 and her elongation CES' 125° 8', the angle MEO for one of the mountains was found by Dr. Herschel to be 40′′.625. Hence, taking the moon's diameter 2160 miles, we easily obtain, from the preceding formula, the height of the mountain = 1.45 miles. ang MEO \2 (A) & sin CES' INTERPOLATION. EC sin 8 59. Interpolation is a method by which, from several given consecutive values of a quantity, corresponding to given values of another quantity, on which it depends or with which it is connected, an intermediate value of the first may be found, corresponding to a given intermediate value of the second. Thus, if the values of an astronomical quantity x, whose variation is not very irregular, be given for the times T, T + 1hr., T + 2hrs., &c., its value for an intermediate time, as for instance, for the time T+ 13hr., may be found by interpolation. 60. Let the values of the quantity x, at the times T, T + 1, T + 2, &c., be a, a', a", &c. If the first of these values be subtracted from the second, the second from the third, and so on, the remainders are called first differences. If the first of these be subtracted from the second, the second from the third, and so on, the remainders are called second differences. In like manner we obtain third differences, fourth differences, &c. Let the values of x, at the times T, T + 1, &c., and the successive orders of differences be arranged as in the following table; in which the first differences are denoted, by ▲a, sa', &c., the second by ▲3ɑ, ▲2ɑ', &c., and the others in a similar manner. vals. of X 1st. diffs. A3a′′=A3a'+A*a's3a+2s*a+s3a sa = Δα sa' + s2a' = Δα!!! sa"! + A2a" Δα!!!! sa""' + A3a""' = a + sa 2nd. diffs. Δα Δα Δα A3a' Δα" A2a" Δα!!! A2a'!!! 3rd. diffs. Δα A3a' A3a" s2a's2a+s3α Δ 43a""sa"+3a" =s3a+3s3a+3s*a+s3a Δ s3a′′=s3a'+s3a'=s3a+203a+s1a 3 Hence, since a = a + ▲a, a' : a' + sa', &c., we have, α a" a"" a!!!! a!!!!! &c. 4 = a + 2sa + s2a = a + 3sa + 3▲3a + s3a = a + 4▲a + 643a + 4▲3α + s1a = a + 5sa + 10▲a + 10▲3a + 5A1a + ▲3a'. &c. The co-efficients of the terms in the values of a, a', a", &c., which are the values of x at the times T, T1, T2, &c., are, therefore, the same as the co-efficients of the terms in the first, second, third, &c., powers of a binomial. Hence, if we take t for an interval of time, expressed in hours, or hours and parts of an hour, we shall have, for the value of x at the time Tt, t(t-1) x = a + t.sa + t(t-1) (t-2) (t-3) 2.3.4. x = a + tsa + (A) 61. The process of interpolation can only be employed with advantage when the differences of some order, not very high, become equal, or very nearly equal. If we suppose the fourth differences to be equal, then the the fifth and higher differences must 0. Hence, we would have, in this case, for the value of x at the time Tt, the rigorous formula, t(t-1) (t-2) Δία + t (t-1) (t-2) (t3) 2.3.4 T 2 T 1 T T + 1 T + 2 T 3 T + 4 3) sta Δια ▲1a + &c. t (t − 1) vals. of X (B) If the differences of the fifth and higher orders, though not absolutely 0, are very small, the formula will still be very nearly accurate for values of t not exceeding, or not much exceeding, 4hrs. For, in this case, the co-efficients of the terms in which these higher differences enter, being each less than a unit, the terms omitted will be very small. s2a + 62. Another Formula. We may obtain a formula, that is frequently more convenient than those above, by taking the times and values of x, as in the following table. 1st. diffs. Δα t (t − 1) (t − 2) s9a + Δα Δα Δα Aa" Δα!!! 2nd diffs. A9a Δα A2a Δα" Put sasa, 2b, sa, c, s3α, + s3α, 2d, and ▲1a, e. Then, since half the difference of two quantities, added to half the sum, gives the greater quantity, we have, sa = b+c, and ▲3a, = d + e. 1⁄2 Hence, assuming the fourth differences to be equal, Atae, A3a = A3a, +e= d+ 3e, and s2α Δα͵ + Δα͵ c+d+e. Substituting these values of ▲α, A2ɑ, A3a, and ▲1a, in the formula (B), it becomes, |
