Determining Planetary Positions in the Sky for ±50 Years to an Accuracy Of

Home , Ecliptic, Evection, Mean longitude

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Ashok and www.physedu.in 1 ol eachieved. be would neo lnt sntcniee.Thereby of accuracy considered. not an is planets influ- of mutual ence which in positions predictplanetary to law Kepler’s use could one Hence, simple. mathematically are well, orbits reasonably planetary describe Ke- which laws, all, really After pler’s not positions. does accurate one very sky need the lo- in to planets able cate be to However, work. computer lot and a equations involving mathematical complicated task, of tedious very posi- a is planetary tions general the a calculating is It that notion task. non-trivial a be could appear Sun, around position contin- its that shifts uously Earth from seen es- sky, when the pecially in them posi- locating sky, the their in change tions continuously and stars ∼ < 1 nti ae,w mlyavr sim- very a employ we paper, this In ◦ ihacalculator a with ∼ < 2 1 ◦ npaeaypositions planetary in ePhysics, le boratory ± 50 ple method to calculate the positions of the Our task becomes simple since the orbits of planets. The technique we use enables us all planets more or less lie in the same plane, to calculate planetary positions to an accu- viz. the ecliptic plane. < racy of 1◦ for 50 years from the starting Actually for locating planets in sky, ∼ ± epoch. Moreover, this involves very simple even manual calculations could suffice, calculations and can be done using a calcula- however, to correct for the orbital elliptic- tor. All we need are the initial specifications ity, numerical values of some trigonomet- of planetary orbits for some standard epoch ric functions may be needed, otherwise one and their time periods of revolution. Al- needs tabulated values of corrections [1]. though accurate planetary positions could Also, in order to transform the planet po- be obtained easily from the internet, yet it is sition to a geocentric perspective, a plot of very instructive and much more satisfying the relative positions of Sun, Earth and the to be able to calculate these ourselves, start- planet on a graph sheet or a chart, using ing from basic principles and using a simple a scale and a protractor, may be required. procedure. However, both these tasks could be per- formed with the help of a scientific calcula- Our first step would be to calculate tor. Further, if one wants to calculate Moon’s the positions of all the planets (including position too (even though one could locate that of the Earth) in their orbits around the Moon in sky easily, without its position cal- Sun. We initially consider the planets to culation), required for knowing the New revolve around the Sun in uniform circu- Moon or the Full Moon dates, or the dates lar motions. Knowing their original posi- of possible solar or lunar eclipses in a spe- tions for the starting epoch, we calculate cific year, computations could be done on a their approximate positions for the intended scientific calculator. epoch. As a consequence of this approximation there will be some errors since the ac- In this paper, we calculate the motion tual orbits are elliptical and in an elliptical of naked-eye planets only, although the pro- orbit the angular speed of the planet is not cedure can be applied equally well for the uniform and to an extent varies. Therefore, remaining planets also. to get more accurate positions, we need to make appropriate corrections, which are de- 2 Celestial co-ordinates rived in Appendix A. These corrections account for the elliptical motion. All celestial bodies in the sky, including Knowing the positions of the planets stars, planets, Sun, Moon and other objects, around the Sun, we can then use simple co- appear to lie on the surface of a giant sphere ordinate geometry to transform their posi- called the Celestial Sphere. Due to Earth’s tion with respect to an observer on Earth. eastward rotation around its axis, the ce-

2 www.physedu.in of these two circles on the celestial sphere are called the “Vernal Equinox” and the “Autumnal Equinox”. The Vernal Equinox, also known as the Spring Equinox, is the point on the celestial sphere that the Sun passes through around 21st of March every year. In astronomy, an epoch is a moment in time for which celestial co-ordinates or orbital elements are speciﬁed, while a celestial co-ordinate system is a co-ordinate system for mapping positions in the sky. There are different celestial co-ordinate systems, each using a co-ordinate grid projected on the celestial sphere. The co-ordinate systems dif- Figure 1: Celestial sphere showing the eclip- fer only in their choice of the fundamental tic co-ordinate system [2]. plane, which divides the sky into two equal hemispheres along a great circle . Each co- ordinate system is named for its choice of lestial sphere appears to rotate westward fundamental plane. around Earth in 24 hours. Inﬁnitely extend- ing the plane of Earth’s equator into space The ecliptic co-ordinate system is a ce- it appears to intersect the celestial sphere to lestial co-ordinate system that uses the eclip- form a circle, which is called the Celestial tic for its fundamental plane. The longitudi- Equator. nal angle is called the ecliptic longitude or As Earth moves around Sun - as seen celestial longitude (denoted by λ), measured from the Earth - Sun changes its position eastwards from 0◦ to 360◦ from the vernal with respect to the background stars. The equinox. The latitudinal angle is called the path that Sun takes on the celestial sphere ecliptic latitude or celestial latitude (denoted is called the “Ecliptic”. The familiar Zodiac by β), measured positive towards the north. constellations are just divisions of the eclip- This coordinate system is particularly useful tic into twelve parts. Since all other plan- for charting solar system objects. ets revolve around Sun in nearly the same The Earth’s axis of rotation precesses plane, they also appear to move on the eclip- around the ecliptic axis with a time pe- tic. riod of about 25800 years. Due to this, The celestial equator is inclined to the the equinoxes shift westwards on the eclip- ecliptic by 23.5◦. The points of intersections tic. Due to the westward shift of the Vernal

3 www.physedu.in Table 1: Mean longitude λ0 on 01/04/2020, 00:00 UT

Planet λi (◦) T (days) ω0 (◦/day) λ0 (◦) Mercury 250.2 87.969 4.09235 277.2 Venus 181.2 224.701 1.60213 150.6 Earth 100.0 365.256 0.98561 189.6 Mars 355.2 686.980 0.52403 270.9 Jupiter 34.3 4332.59 0.08309 288.8 Saturn 50.1 10759.2 0.03346 297.6

Equinox, which is the origin of the ecliptic speed is give by, ω0 = 360/T (◦/day). We co-ordinate system, the ecliptic longitude of denote the Mean Longitude of the planets in the celestial bodies increases by an amount the imaginary circular orbit for subsequent 360/258 1.4 per century or a degree in dates as λ . ∼ ◦ 0 72 years, the life span of a human. We now demonstrate how to calculate ∼ Most planets, dwarf planets, and many λ0 for Mars on April 1, 2020. small solar system bodies have orbits with The initial mean longitude, λi, of Mars small inclinations to the ecliptic plane, and on 01.01.2000 at 00:00 UT = 355.2◦. therefore their ecliptic latitude β is always Number of days between 01.01.2000 small. Due to only small deviations of the and 01.04.2020 = 7396. orbital planes of the planets from the plane Mean angle traversed duration this pe- of the ecliptic, the ecliptic longitude alone riod = 0.52403 7396 = 3, 875.7 . × ◦ may sufﬁce to locate planets in the sky. So, on 01.04.2020 at 00:00 UT, the mean longitudes is λ , = 355.2 + 3, 875.7 = 4, 230.9 = 3 Calculating planetary positions 0 360 11 + 270.9 . × ◦ 3.1 Heliocentric circular orbit We take out 11 integer number of com- plete orbits to obtain λ0 = 270.9◦. Here, we consider the planets to move In the same way, mean longitudes of all around Sun in circular orbits with a uniform planets have been calculated in Table 1 for angular speed. The initial values of mean the same epoch. longitudes (λi) of the planets given in Ta- ble 1 are for 1st of January, 2000 A.D., 00:00 3.2 Heliocentric elliptical orbit UT (adapted from [3]). In Table 1, we have also listed the period, T (days), of revolution Our next step is to correct for the elliptical of the planets [4]. Then, the mean angular shape of the orbit as for some of the plan-

4 www.physedu.in Table 2: Corrected longitude λ on 01/04/2020, 00:00 UT

Planet λ0 (◦) λp (◦) θ0 (◦) e ∆θ (◦) θ (◦) λ (◦) Mercury 277.2 77.5 199.7 0.2056 -6.0 193.7 271.5 Venus 150.6 131.6 19.0 0.0068 0.3 19.2 151.1 Earth 189.6 102.9 86.7 0.0167 1.9 88.6 191.8 Mars 270.9 336.1 294.8 0.0934 -10.2 284.6 261.0 Jupiter 288.8 14.3 274.5 0.0485 -5.6 269.0 283.6 Saturn 297.6 93.1 204.5 0.0555 -2.5 202.0 295.4

∆ ets, depending upon the eccentricity (e), the The correction θ to be added to θ0 (Ap- corrections could be substantial. In a circu- pendix A, Equation 4) is lar motion, the angular speed of the planet 5 is constant, however, in an elliptical orbit, ∆θ = 2 e sin θ + e2 sin2θ , 0 4 0 the angular speed of the planet varies with time. Due to this varying angular speed, ac- where e is the eccentricity of the ellipse. tual longitude λ of the planet will be some- In Table 2, we have listed values of the what different from the mean longitude λ0. longitude of perihelion (λp) and eccentricity (e) for all planets, taken from [3]. Before we could make corrections for Let’s consider Mercury’s position on the elliptical shape of the orbit we need to 01/04/20 at 00:00 UT. know the orientation of the ellipse within Mean longitude, λ = 277.2 the ecliptic and that can be deﬁned by the 0 ◦ Perihelion Longitude, λp = 77.5◦ longitude (λp) of the perihelion. The dis- Mean anomaly, θ = λ λ = 199.7 tance of the planet from Sun varies in an 0 0 − p ◦ elliptical orbit and perihelion is the point The 1st order correction then is ∆θ = 2e sin θ = 2 0.2056 on the elliptical orbit that lies closest to 1 0 × × sin(199.7 )= 0.13861 rad = 7.9 , Sun (the orbital point farthest from Sun is ◦ − − ◦ called aphelion). Longitudinal distance of while the 2nd order correction is ∆θ = 5 e2 sin2θ = 1.25 (0.2056)2 the planet from the perihelion along the el- 2 4 0 × × liptical orbit is known as its “anomaly” (de- sin(39.4◦)= 0.03354 rad = 1.9◦. noted by θ), while angular distance of mean The total correction then is ∆ ∆ ∆ position of planet with respect to the perihe- θ = θ + θ2 = 7.9 + 1.9 = 6.0 1 − − ◦ lion is called the “mean anomaly” (denoted The anomaly, θ = θ + ∆θ = 199.7 0 − by θ0). As has been discussed in Appendix 6.0 = 193.7◦ A, there is a one–to–one correspondence be- Precession of vernal equinox in 20.25 tween θ and θ . years = 20.25 360/25800 0.3 . 0 × ≈ ◦ 5 www.physedu.in Table 3: Geocentric longitude λg on 01/04/2020, 00:00 UT

Planet a (A.U.) e r (A.U.) λ (◦) rg (A.U.) λg (◦) Mercury 0.387 0.2056 0.463 271.5 1.024 345.3 Venus 0.723 0.0068 0.718 151.1 0.653 57.6 Earth/Sun 1.00 0.0167 0.999 191.8 0.999 11.8 Mars 1.52 0.0934 1.472 261.0 1.457 300.9 Jupiter 5.20 0.0485 5.192 283.6 5.318 294.4 Saturn 9.55 0.0555 10.04 295.4 10.32 300.8

∆ The longitude, λ = λ0 + θ + preces- the Vernal Equinox, and Y to be perpendic- sion of vernal equinox = 277.2 6.0 + 0.3 = ular to X in the ecliptic plane in such a way − 271.5◦. that the longitude is a positive angle. We can obtain corrections for the elliptical orbits of the remaining planets in the 3.4 An example same way. The calculated anomaly (θ) and As an example, this procedure is demon- longitude (λ) are listed in Table 2. The resid- < strated for Mercury’s position on 01/04/20 ual errors in λ values are 1◦. ∼ at 00:00 UT.

3.3 Geocentric perspective 3.4.1 Heliocentric co-ordinates

Until now, we have calculated the longi- Distance, r, of Mercury from Sun can be ob- tudes, λ, of the planets on the celestial tained from the anomaly θ as, sphere centred on the Sun. We can also cal- a(1 e2) culate radii r of their orbits around the sun, r = − = 0.463A.U., for that epoch, giving their positions in po- 1 + e cos θ lar form (r, λ). To get the positions of planets where a = 0.387 A.U. is the length of semi– on the celestial sphere centred on the Earth, major axis of its elliptical orbit. Distances to we ﬁrst convert the polar co-ordinates into planets in the Solar system are convention- rectangular form with origin on Sun, and af- ally measured in astronomical units (A.U.), ter shifting the origin from Sun to Earth, we the mean distance of Earth from Sun, which change them back into polar form. is nearly 1.5 108 km. × For converting into a rectangular form, The heliocentric polar co-ordinates of we have to decide upon the direction of the Mercury thus are (0.463 A.U., 271.5◦). Then X and Y axes. We assume X to be in the pos- we can get heliocentric rectangular co- itive direction along the line joining Sun to ordinates of Mercury as,

6 www.physedu.in Xh = r cos(λ)= 0.012 A.U. the perihelion, semi–major axis and eccen- Y = r sin(λ)= 0.463 A.U. tricity etc. change so slowly with time that h − Similarly we get heliocentric rectangu- for the accuracy we are interested in, these lar co-ordinates of Earth as, can be considered constant for 50 years. ± X = 0.978 A.U. 0 − Y0 = 0.204 A.U. − 4 Locating planets in sky 3.4.2 Geocentric co-ordinates After having computed the geocentric lon- Geocentric rectangular co-ordinates of Mer- gitudes of the planets, we are now in a posi- cury then are tion to locate them in the sky. Any one famil- X = X X = 0.990 A.U. g h − 0 iar with the Zodiac constellations could lo- Y = Y Y = 0.259 A.U. g h − 0 − cate a planet from its position in the constel- Converting these into polar form, we lation in which it lies. The ecliptic is divided get the geocentric distance and longitude as, into 12 Zodiac signs – Aries, Taurus, Gemini, 2 2 rg = (Xg + Yg )= 1.024 A.U. Cancer, Leo, Virgo, Libra, Scorpio, Sagittar- q 1 λg = tan− (Yg/Xg)= 345.3◦. ius, Capricorn, Aquarius, Pisces. The Ver- In Table 3 we have given the calculated nal equinox, at zero ecliptic longitude, is geocentric longitudes of various planets on the start of the first Zodiac sign and is also 01.04.2020 at 00:00 UT. In the Earth/Sun row known as the First Point of Aries. But there in Table 3, rg, λg are the geocentric values is a caveat attached. Because of the pre- for the Sun’s position. The position of Sun cession the vernal equinox has shifted west- on the celestial sphere, as seen from Earth, ward by almost the full width of a constel- is in a direction exactly opposite to that of lation in the last 2000 years since when ∼ Earth as seen from the Sun. Therefore the the Zodiac signs and constellation were per- geocentric longitude of Sun is the heliocen- haps first identified. As a consequence, the tric longitude of Earth plus 180◦. First Point of Aries now lies in the constel- We have ignored any perturbations on lation Pisces. For example, on 01/04/2020, the motion of a planet due to the effect of geocentric longitude 294.4◦ of Jupiter im- other planets which may distort its ellipti- plies it is in the 10th Zodiac sign Capri- cal path. We are able to get the accuracy of corn, but actually lies in the Sagittarius con- < 1◦ for long periods ( 50 years) because stellation, taking into account the shift by ∼ ± most of the terms ignored in the heliocentric one constellation due to precession. There longitude calculations are periodic in nature are further complications. The twelve con- and do not grow indefinitely with time (see stellations are not all of equal length of arc e.g. [5]). The other parameters characteriz- along the ecliptic longitude. Moreover there ing the elliptical orbit, like the longitude of is another constellation, viz. Ophiuchus,

7 www.physedu.in through which the ecliptic passes. However Table 4: Elongations of the planets on these complications are somewhat set aside 01/04/2020 by the fact that there are only about half a dozen stars in the Zodiac with an appar- λg(◦) ψ(◦) λg(◦) ψ(◦) Planet ent brilliance comparable to the naked–eye 05:30IST (17:30IST) planets, therefore with some familiarity of Sun 11.8 - 12.3 - the night-sky, one could locate the planets Mercury 345.3 -26.4 346.0 -26.3 easily from their geocentric longitude val- Venus 57.6 45.9 58.1 45.8 ues. It further helps to remember that un- Mars 300.9 -70.9 301.3 -71.0 like stars, the planets, because of their large Jupiter 294.4 -77.4 294.4 -77.8 angular sizes, do not twinkle. Saturn 300.8 -71.0 300.8 -71.4

For a more precise location of a planet we can calculate its relative angular distance 57.6 11.8 = 45.9◦. Thus Venus has an east- from the Sun along the ecliptic. The differ- − ern elongation 45.9◦ on 01.04.2020, 05:30 IST, ence between the geocentric positions of a and would be visible only in the evening, af- planet and Sun (Table 4) is called the elon- ter the sunset, in the western sky. gation (ψ) of the planet and it tells us about As Earth completes a rotation in 24 planet’s position in the sky with respect to hours, the westward motion of the sky is at a that of the Sun. The longitude increases rate 360/24 = 15◦/ hour. This rate is strictly eastwards, therefore, if the longitude of the true for the celestial equator, but we can use planet is greater than that of the Sun, then this as an approximate rotation rate even for the planet lies to the east of the Sun. That the ecliptic, which is inclined at 23.5◦ to the means, in the morning the Sun will be rising equator. Therefore Mercury, on 01.04.2020, before the planet rises but in the evening the will rise about 26.4/15 1.5 hours before planet will be setting after the sunset. So the ∼ the sunrise, and could be seen in the eastern planet will be visible in the evening sky in sky in the morning. the west. On the other hand, if the geocen- It is, however, possible to determine the tric longitude of the planet is smaller than planetary positions for any other time of that of the Sun, then it lies to the west of the the day. For instance, in the case of Venus, Sun and will rise before the sunrise and will which is visible in the evening hours on our be visible in the morning in the eastern sky. chosen date of 1/4/2020, it might be prefer- In Table 4, positions of planets with re- able to calculate the position on that date for spect to Sun on 01.04.2020 are given for 00:00 12:00 UT, corresponding to 17:30 hr IST, lo- UT, which corresponds to 05:30 IST (Indian cally an evening time. For this, we use the Standard Time). For example, the geocen- number of days in Step 1 as 7396.5. It should tric longitude of Venus with respect to Sun is be noted that not only the longitude of each

8 www.physedu.in cult job, which can be accomplished only by complex scientific computations, using fast computers. Our motive here has been to dispel such a notion and bring out the fact that such complex and accurate computations are not always really necessary. One can calculate the position of planets using the method derived here and get the thrill of finding the planet at the predicted position in the night sky. We have been able to obtain the position Figure 2: A schematic representation of the < of planets within an accuracy of 1◦., us- elongations of planets at the times of sunrise ∼ ing a calculator. This method can be used to and sunset on April 1, 2020. reckon planetary positions up to 50 years ± of the starting epoch. planet around Sun might change by a cer- tain amount, even the longitude of Earth ad- vances by 1 in a day, thus affecting the ∼ ◦ Appendix A: Correction for the elongation of even Jupiter and Saturn (Ta- orbital ellipticity ble 4), whose angular speeds are relatively small (Table 1). In Table 4, we have listed We compute the correction for motion of a the elongations of all five naked-eye planets planet in an actual elliptical orbit from that on 01-04-2020 at 00:00 hr UT (5:30 IST) and at in an imaginary circular orbit. Period of rev- 12:00 hr UT (17:30 IST). Venus with an east- olution in circular orbit is taken to be exactly ern elongation 46 on that evening, will ∼ ◦ the same as that in the elliptical orbit. The be setting approximately three hours after origin of the mean longitude in circular or- the sunset. This way, one can easily locate bit is chosen such that λ0 coincides with the the planets in the sky from their elongations. longitude λ of the planet when it is at the Figure 2 is a schematic representation perihelion in its elliptical orbit. For mathe- of the relative positions of various planets matical convenience, we take t = 0 at that with respect to the Sun at the horizon, on the instant. Then λ0(0) = λ(0). Let λp be the morning and evening of April 1, 2020. longitude of the perihelion of the planet’s el-

liptical orbit. We subtract λp from λ0 and 5 Conclusions λ to obtain what is called respectively, the mean anomaly θ0 and the anomaly θ of the It is generally thought that calculation of po- planet. Thus θ0 = λ0 - λp, and θ = λ - λp. sition of planets in the night sky is a difﬁ- Then θ0(0)= θ(0)

9 www.physedu.in Figure 3: A schematic diagram of the planet in circular and elliptical motion.

In circular motion, the angular speed, riod of T. Hence, θ0(t) and θ(t) have a one– ∆ ω0, of the planet is constant. However, in to–one relation. Hence, θ also repeats after the elliptical motion, the angular speed of time T. The uniform circular motion thus is the planet varies with time. a useful approximation because the error ∆θ Let atime t has passed after t = 0. Then, is periodic with time and does not keep ac- the change in θ0 of the planet is ω0t whereas cumulating with time to grow to very large the change in θ of the planet won’t be the values. same because of the variation in the angular To ﬁnd the correction, ﬁrst consider an speed along elliptical trajectory. elliptical orbit of a planet around the Sun as

Let ∆θ(t) = θ(t) - θ0(t). shown in Figure 3. We use the equation of the ellipse in polar co-ordinates (r, θ) where We know that θ0(t) and θ(t) are periodic by the same time interval, T, as T is the θ is the anomaly. The equation of the ellipse time period of revolution in both the cases then is (elliptical and circular motion). Hence, all l a(1 e2) r = = − , (1) value of θ0(t) and θ(t) repeat after a time pe- 1 + e cos θ 1 + e cos θ

10 www.physedu.in where l = a(1 e2) is the semi–latus rec- After integration we get − tum with a as the semi–major axis and e the 3 θ = θ 2 e sin θ + e2 sin2θ + , (3) eccentricity of the ellipse. The semi–minor 0 − 4 ··· axis of the ellipse is b = a√1 e2. − which can be written as Now, total area of the ellipse A = πab 3 ∆θ = θ θ = 2 e sin θ e2 sin2θ is swept in T, the time period of revolution. − 0 − 4 From Kepler’s second law we know that the + ··· rate of area swept out by the position vector However we want the right hand side of of planet (with respect to Sun) is a constant. Equation (3) to be expressed in terms of θ0. Therefore the rate of area swept is ∆ For that we can substitute θ = θ0 + θ on dA r2 dθ πab the right hand side to get, = = . dt 2 dt T 3 ∆θ = 2 e sin(θ + ∆θ) e2 sin[2(θ + ∆θ)] Substituting from Equation (1), we get 0 − 4 0 + 2 3 2π (1 e ) 2 dθ ··· = − . T (1 + e cos θ)2 dt Expanding in powers of ∆θ and neglecting ∆ 2 2 ∆ We notice that 2π/T is nothing but the mean terms of order e ( θ) , e θ and higher we get angular speed ω0. Therefore

2 3 ∆ 3 2 t t (1 e ) 2 θ (1 2 e cos θ0) = (2 e sin θ0 e sin2θ0), − − 4 θ0(t)= ω0 dt = − 2 dθ. (2) Z0 Z0 (1 + e cos θ) or We want to get the equation in the form, θ = ( e θ 3 e2 θ ) ∆ 2 sin 0 4 sin2 0 θ0 + ∆θ, so that by adding the longitude of θ = − . (1 2 e cos θ0) the perihelion on both sides of the equation, − we could get the relation between the correct Again Expanding in powers of e and ignor- ing terms of order e3 or higher, we get longitude λ and the mean longitude λ0. A direct integration of Equation (2) may ∆ 5 2 θ = 2 e sin θ0 + e sin2θ0 (4) not be possible, but we can expand the inte- 4 grand as a series and integrate only a few which is the required expression for the cor- first most significant terms. A binomial se- rection term. ries expansion is possible because the eccentricity of an ellipse, e < 1. During the ex- Appendix B: Position of the Moon pansion we drop terms having higher than e2 factor. Here we determine Moon’s position for any t 3 2 given epoch, say, April 1, 2020, 00 UT, start- θ0(t)= (1 e + )(1 2 e cos θ Z0 − 2 ··· − ing from the initial epoch 1st January 2000, +3 e2 cos2 θ + ) dθ 00 UT. ··· 11 www.physedu.in Moon moves in a geocentric orbit of θ ). Substituting for ψ = 104.4 9.6 = 0 0 − mean eccentricity e = 0.0549 and a tropical 94.8 , we get, ∆θ = 1.27 sin(2 94.8 ◦ ev × − revolution period of T = 27.32158 days, cor- 277.1)= 1.3 . Thus a more accurate value − ◦ responding to a mean angular speed ω0 = of Moon’s geocentric longitude on April 1, 13.17640 /day with respect to the vernal 2020, 00:00 UT is λ = 98.1 1.3 = 96.8 , ◦ g − ◦ equinox. From the initial value λgi = 211.7◦ and the elongation ψ calculated thence is on 1st of January, 2000 A.D., 00:00 UT, 85.0◦. Moon’s mean geocentric longitude is calcu- There is another feature of the Moon’s lated as, λ = 211.7 + 13.17640 7396 = g0 × motion which has a direct bearing on the 104.4 . ◦ time of occurrences of solar and lunar But before we correct for the ellipticity eclipses. Moon’s orbit is inclined to the of the Moon’s orbit, we need to consider ecliptic with a mean inclination of 5.16◦, in- that unlike in a planetary orbit where the tersecting it at points known as the ascend- position of the perihelion change so slowly ing and descending nodes. The ascending with time that it can be considered constant node is the one where Moon crosses the for 50 years, in Moon’s orbit the perigee ecliptic in a northward direction. Gravita- ± rotates forward with respect to the vernal tional pull of the Sun causes a precession equinox with a period of 3231.4 days ( 8.85 of the axis of Moon’s orbital plane around ∼ years), corresponding to an angular speed that of the ecliptic with a tropical period of 0.11141 /day. With an initial value of 83.3 , 6798.4 days ( 18.6 years). Consequently ◦ ◦ ∼ the longitude of the perigee, on April 1, the nodes of the Moon’s orbit have a ret- 2020, 00:00 UT is then given by, λ = 83.3 + rograde motion of 0.05295 /day around p ∼ ◦ 0.11141 7396 = 187.3 . From this we get the ecliptic. With an initial value of 125.1 × ◦ ◦ Moon’s mean anomaly for a circular motion on 1st January 2000, 00 UT, the longitude of as, θ = λ λ = 277.1 . the ascending node for the epoch April 1, 0 g0 − p ◦ 2020, 00:00 UT is 125.1 0.05295 7396 = We can now use Equation (4) in the − × 266.5 , implying ψΩ = 81.7 , while the de- same way as for the planets to get the cor- − ◦ ◦ scending node is 180 away at ψ℧ = 98.3 . rection for the ellipticity ∆θ = 6.3 , which ◦ − ◦ − ◦ gives Moon’s corrected geocentric longitude A solar eclipse can take place near the ∆ λg = λg0 + θ = 98.1◦. However, the value New Moon time, when Moon’s elongation obtained thus is accurate only up to 2 . is 0 and it could block sunlight to reach ∼ ◦ ∼ ◦ The reason being that there is an impor- some parts of the Earth’s surface, while a lu- tant perturbation term, known as Evection, nar eclipse can take place around the Full which depends upon Moon’s mean elon- Moon time, when Moon’s elongation is ∼ gation ψ0 as well as its mean anomaly θ0, 180◦ and Earth may be blocking sunlight and has a value, ∆θ = 1.27 sin(2ψ from reaching the Moon. However, these ev ◦ 0 − 12 www.physedu.in events could occur only when Sun’s longi- of the corresponding year. The horizontal tude is close to that of one of the nodes be- axis displays the elongation in degrees, and cause it is then only Earth and Moon both is centred around the Sun, which by defi- get nearly aligned in a straight line with Sun, nition has a zero elongation. The vertical so that one of these celestial bodies could axis marks the day of the year. Thus to lo- block sunlight to cause a shadow on the cate a planet on any given date of the year, other celestial body. we select that date on the vertical axis and then move in a horizontal direction till we find the planet. From the elongation of the Appendix C: The astronomical planet we can easily locate it in the sky. Dif- calendar ferent paths of various planets as well as those of Sun, Moon and both nodes, can be In all the examples presented above, the cal- identified from their conventional symbols, culations were made using a scientific cal- which are explained at the bottom of the fig- culator. This procedure is appropriate for ure. We can use the calendar to find what quickly getting planetary positions, to lo- all planets are above the horizon at any time cate one or more planets in sky, occasion- of the day. Suppose for a given date of the ally, on some specific day. However if one year we want to locate all planets visible in wants to do many computations, say cal- the sky at, say, dawn. The Sun, at 0◦ elon- culate positions for planets for many more gation, will at that time be just rising near days of the year, the process could become the eastern horizon and the 180 elonga- − ◦ tedious and the chances of a numerical mis- tion point in the calendar will be near the take occurring in manual calculations could western horizon. The intermediate elonga- become high. Since the process of comput- tion points will be at in-between positions ing planetary positions is a repetitive one, on the celestial hemisphere, e.g., the 90 − ◦ it could then be much more convenient to elongation point will be close to the culmi- write a simple computer programme, us- nation point (the point nearest to the zenith). ing the algorithm described above, to carry This way going along the horizontal direc- out calculations. We have written such a tion from 0 to 180 at the chosen date, ◦ − ◦ programme to compute positions of all the we will find the celestial position of all plan- planets as well as of the Moon for each day ets visible in the morning sky on that date. of a specified year and present the results in At dusk, with sun setting near the western the form of an astronomical calendar which horizon, the visible sky will stretch eastward gives elongations of different planets for all from 0◦ to 180◦ elongation on the calendar. days of the year. Similarly one can locate planets on the ce- The calendar allows us to locate the lestial sphere at other hours of the day. At naked-eye planets in the sky for any time

13 www.physedu.in Figure 4: Astronomical calendar for locating planets in sky for the year 2020. midnight, with culmination point being at month represent Moon’s path. Their dates 180 (which is the same as 180 ), the sky of intersection with the Sun’s path at 0 elon- ◦ − ◦ ◦ towards west will stretch from 180◦ to 90◦ gation mean New Moon days, while inter- and that towards east will be from 180 sections with the midnight lines (at 180 − ◦ ± ◦ to 90 elongation, while at 9 p.m., with elongation) imply Full Moon days, with the − ◦ the culmination point at 135◦, the celestial intermediate phases at the in-between dates. hemisphere will stretch from west to east The faint lines in the calendar represent the between 45 and 225 (equivalently 135 ) relative positions of the ascending node and ◦ ◦ − ◦ elongations. descending node of Moon’s orbital plane. Intersection of the sun’s path (at 0◦ elonga- The slant dotted lines running across tion) or of the midnight lines (at 180◦) with the calendar from west to east almost every ±

14 www.physedu.in one of the lines of nodes, indicate the pos- nomical vision in the year 2020. Jupiter and sibility of occurrence of an eclipse. In the Saturn will be moving closer to each other neighbourhood of these intersection points, in sky with the two being closest (within a at the time of a New moon there might pos- degree – the accuracy limits of our calcula- sibly occur a solar eclipse while at the Full tions), an event known as a great conjunc- moon time there is a possibility of lunar tion, around 23rd of December. For about 10 eclipse. days, near the end of March, when Jupiter For many planetary phenomena, the as- and Saturn will be within 7 of each other, ∼ ◦ tronomical calendar can act as a quick indi- Mars will be lying in between the two. Not cator. However, because of the limited res- only will there be three bright objects seen in olution of the display in the calendar, for close proximity of each other in sky, Mars, a better accuracy one might go back to the seen within a degree of Jupiter on 22nd of actual tabulated data from which the calen- March will move to within a degree of Sat- dar has been generated. From our computed urn by 1st of April. But this spectacular sight data for the longitudes of the Sun, Moon will be for the eyes of the early birds only, as and the lines of nodes for the year 2020, this would be visible before sunrise, in the we ﬁnd that the possible dates for the solar eastern sky. eclipses in the year 2020 are 21st June and 14th December, which are the New Moon References date close to the intersection point of the lines of nodes with Midday line (Sun’s path [1] Singal, A. K., Phys. Ed. 34, 03, 02 (2018) at 0◦). Similarly, possible lunar eclipses on Full Moon dates near the intersection points [2] http://www.hps.cam.ac.uk/starry/mathtechniqueslrg.jpg of the lines of nodes with Midnight lines [3] Fr¨anz M. and Harper D., Planetary and (at 180◦) would in 2020 fall on January ± Space Science 50, 217 (2002) 10, June 5, July 5 and November 30. How- ever, as seen in Figure 4, these might only be [4] Nicholson I., “Unfolding Our Uni- Penumbral Lunar eclipses as the Full Moon verse”, Cambridge University Press positions do not fall very close to any of the (1999) intersection points. From the astronomical calendar (Figure [5] Simon J. L. et al, Astronomy Astro- 4), we can expect some interesting astrophysics 282, 663 (1994)

15 www.physedu.in