LLocating Planets in Sky Using Manual Calculations
Ashok K. Singal
Astronomy and Astrophysis Division, Physical Research LaboratoryNavrangpura, Ahmedabad 380 009, India. [email protected]
Submitted on xx-xxx-xxxx
Abstract
In this article, we describe a very simpletechnique to locate naked-eye planets in thesky, to an accuracy of ∼ ◦ . The procedure,comprising just three steps, involves verysimple manual calculations for planetary orbitsaround the Sun; all one needs are the initialspecifications of planetary positions for somestandard epoch and the time periods of theirrevolutions. After applying a small correctionfor the orbital ellipticity, appearance of a planetrelative to Sun’s position in sky, as seen by anobserver from Earth, is found using a scale and aprotractor (found inside a school geometry box). Quite often, seeing a bright star-like objectin the evening sky (or in the morning skyfor the early birds!) many of us would havewondered whether it is a planet and if so,which one. To be able to actually locate aplanet in the sky is something that couldbe thrilling to most of us and occasionally it provides us an opportunity to impressour friends and acquaintances. Althoughdaily planetary positions could be obtainedfrom the professional ephemeris [1] or sim-ply from the internet, yet it is very instruc-tive and much more satisfying to be ableto calculate these ourselves, starting from,say, one of Kepler’s famous laws, whichstates that planets go around Sun in ellipti-cal paths.The path that Earth takes in the sky iscalled the ecliptic. The familiar Zodiac con-stellations are just divisions of the eclipticinto twelve parts. Sun as well as planets, asseen from Earth, also appears to move alongthe ecliptic and pass through the Zodiacconstellations. The angle along the eclipticis called longitude (denoted by λ ), measuredeastwards, that is, anti-clockwise, from 0 ◦ to360 ◦ . Its origin, λ = ◦ , is known as the FirstPoint of Aries, the position Sun occupies onthe ecliptic around 21st of March every year.The first step in our exercise would beto calculate the longitude of a planet as wellas that of Earth around Sun. We initially1 a r X i v : . [ phy s i c s . pop - ph ] J un hysics Education Publication Dateconsider the planets to revolve around Sunin uniform circular motion. However, thiswill entail a correction since the actual orbitsare elliptical where the angular speed is notuniform. In second step we employ a pre-computed table to find the necessary correc-tions. From another table we also determinedistances of planets from Sun. In a third andfinal step, we plot the distance along the lon-gitude of chosen planet as well as that ofEarth centred around Sun on a graph sheetor a chart, using a scale and a protractor(found inside a school geometry box) andmeasure the angle between the planet andSun as seen from Earth, which allows us tolocate the planet in the sky.
First we consider the planets to movearound Sun in circular orbits with uniformangular speeds. We need longitudes of plan-ets in their orbits around Sun on some initialdate. Here we take 1st January, 2000, 00:00UT as our initial date [2] for which we havelisted the longitudes ( λ i ) of the planets (Ta-ble A1). Also listed in Table A1 is the pe-riod T (days) of revolution of each planet [3].From T we get the mean angular speed ofthe planet as ω = T ( ◦ /day). We de-note the Mean Longitude of the planets inthe imaginary circular orbit for subsequentdates as λ .As an example, we calculate the meanlongitude λ for Venus on 1st October 2018. The initial longitude of Venus (on01.01.2000), λ i = ◦ .The mean angular speed of Venus, ω = ◦ /dayNo. of days between 01.01.2000 and01.10.2018, N = × + + = ω N = × = ◦ .So, on 01.10.18 the mean longitude ofVenus, λ = + = ◦ .After taking out 30 complete orbits ininteger multiple of 360 ◦ , we get mean lon-gitude of Venus, λ = − × = ◦ .As another example we also calculatethe mean longitude of Jupiter on 01.10.18, λ = + × = ◦ . Aftertaking out one complete cycle, the mean lon-gitude of Jupiter on 01.10.18 is λ = ◦ .We also need to calculate the mean lon-gitude of Earth on 01.10.18 as λ = + × = ◦ . Or the mean lon-gitude of Earth on 01.10.18 is λ = ◦ (aftertaking out 19 complete orbits). hysics Education Publication Datebased on the formulation derived in [4]. En-tries for each planet in Table A2 also ac-count for the orientation of its elliptical or-bit within the ecliptic plane, specified by thelongitude of the perihelion, where perihe-lion is the point closest to the Sun on the el-liptical orbit of the planet.The correction from the Table A2 forVenus for λ = − ◦ . Therefore cor-rected longitude is λ = − = ◦ .The correction for Jupiter for = ◦ is − ◦ , with corrected longitude λ = − = ◦ .Similarly we calculate the corrected el-liptical longitude for Earth as λ = ◦ . The difference between the geocentric posi-tion of a planet and Sun is called the elon-gation ( ψ ) of the planet and it tells us aboutplanet’s position in the sky relative to thatof the Sun. As we want to find the sky po-sition of a planet, as seen by an observer lo-cated on Earth, we need to find the positionof Earth too in the ecliptic. Since Earth lon-gitude changes by ∼ ◦ per day, we needEarth position for the same date and time asthat of the planet we are interested in.We have already calculated the eclipticlongitudes of Venus, Jupiter and Earth. Foreach of these planets we also need to findradii r , the distance from Sun, which is listedagainst λ in Table A3 in AU (AstronomicalUnit - the mean distance between Earth and Sun). For our chosen date of 01/10/2018, r is 0.73 A.U. for Venus, 5.37 for Jupiter and1.00 A.U. for Earth.Although one could employ a calcula-tor to compute geocentric longitude and theelongation of the planet [4], however, a man-ual geometric construction could be muchmore illuminating. All one needs is a scaleand a protractor, usually found in a schoolgeometry box.Now we plot on a graph sheet or a chartpaper the position of Venus at its respec-tive distance 0.73 A.U. (on a suitable scalechosen for 1 A.U.) along its corrected longi-tude λ = ◦ (increasing anti-clockwise)around Sun. Similarly we also need to plotthe position of Earth on this diagram (seeFig. 1).To locate a planet in the sky we deter-mine its elongation ψ which is the angu-lar distance measured eastward (that is anti-clockwise) from Sun‘s position on the eclip-tic, as seen from Earth (Fig. 1).That is, from the chart we determine theangle ψ between the line joining Earth to theplanet and that from Earth to Sun. If ψ > ψ <
0, then the planet willrise before the Sun and will be visible in themorning sky above the eastern horizon.From Fig. 1 we find that on 1/10/2018Venus is ∼ ◦ hysics Education Publication DateFigure 1: Elongation of Venus on 1-10-2018.visible at sunset time about 33 ◦ away fromSun’s position in the western sky.From Fig. 2 we see that on 1/10/2018Jupiter is ∼ ◦ east of Sun, and it willbe therefore visible in the evening about44 ◦ away from Sun’s position. Thus inthe evening of 1/10/2018, one will see twobright objects (Venus and Jupiter) separatedabout 11 ◦ in the western sky.As Earth completes a full rotation in 24hours, the westward motion of the sky is ata rate 15 ◦ / hour. This rate is strictly truefor the celestial equator, but we can use thisas an approximate rotation rate even for theecliptic, which is inclined at a 23.5 ◦ to theequator. Therefore Venus with an easternelongation ∼ ◦ , will be setting a little morethan two hours after sunset, while Jupiter at ∼ ◦ hysics Education hysics Education Publication DateTable 1: positions of planets on 01.10.2018 at00:00 hr and 12:00 hr UTPlanet 00:00 hr UT 12:00 hr UT λ ( ◦ ) ψ ( ◦ ) λ ( ◦ ) ψ ( ◦ )Earth 7.8 8.3Mercury 213.4 7.7 214.9 8.0Venus 352.3 32.8 353.1 32.4Mars 345.6 117.9 345.9 117.6Jupiter 239.5 44.3 239.6 43.9Saturn 278.9 85.4 278.9 84.92018 at 00:00 hr UT (5:30 IST) and at 12:00hr UT (17:30 IST). It should be noted thatnot only the longitude of each planet aroundSun changes by a certain amount, even thelongitude of Earth advances by ∼ ◦ in aday, thus affecting the elongation of evenJupiter and Saturn (Table 1), whose angularspeeds are relatively small (Table A1). We tried to dispel a general notion that tobe able to determine positions of planets inthe night sky one requires complex scientificcomputations, using fast computers. Themotive of this article has been to impressupon the reader that such accurate calcu-lations are not really necessary for locatingnaked-eye planets in sky. It was demon-strated that in just three steps, one can findthe positions of planets manually with sim-ple arithmetic calculations. All one needsare the initial specifications of planetary po-sitions for some standard epoch and thetime periods of their revolutions. Then after applying a small correction for the orbital el-lipticity, the location of a planet in sky, fromthe point of view of an Earth-based observer,could be found and thus one could get thethrill of locating a planet at the predicted po-sition in the night sky.
References [1] “The Indian Astronomical Ephemerisfor the year 2018”, The Indian Meteo-rological Department, Kolkata (2018)[2] Fr¨anz M. and Harper D., Planetary andSpace Science 50, 217 (2002)[3] Nicholson I., “Unfolding Our Uni-verse”, Cambridge University Press(1999)[4] Singal T. and Singal A. K., Prayas, 3, 176(2008); arXiv:0910.2778v1
Appendix
Table A1: Parameters of the planetary orbitson 01.01.2000, 00:00 UTPlanet λ i ( ◦ ) T (days) ω ( ◦ hysics Education Publication DateTable A2: Correction to the longitude for elliptical orbits λ hysics Education Publication DateTable A3: Planetary distances (in AU) corresponding to λ values λ0