SizzlinShred
Astronomy · magnitude in motion

Measure the sky from the ground.

Astronomy is the art of magnitude in motion: Nicomachus gives it the part of size that moves and revolves (Introduction to Arithmetic I.3), and Boethius later named the four mathematical arts the quadrivium (De institutione arithmetica I.1). Six figures on this page let you set a latitude, a date, an angle or an eccentricity and see what follows, from the seasons to Kepler's third law. Each figure is a model, and each says what it leaves out.

Planet data come from NASA's Planetary Fact Sheet, converted with 1 AU = 149.5978707 million km and 1 year = 365.25 days. All angles are in degrees.

Part of Astronomy in the curriculum, the art of magnitude in motion.

I · The sky

Seasons come from the tilt.

The Earth's axis is tilted about 23.44 degrees from the perpendicular to its orbit, and it keeps pointing the same way as the planet goes round the Sun. For half the year the northern hemisphere leans towards the Sun and for half it leans away. The Sun's declination, the latitude where it stands overhead at noon, therefore swings between +23.44 degrees at the June solstice and −23.44 degrees at the December solstice. Outside the tropics the summer Sun stands higher at noon, so a given beam of sunlight is spread over less ground, and it stays up longer. In winter the reverse holds.

The side view shows the Earth with its tilted axis and parallel sunlight. The horizon view shows the Sun's height at noon for the latitude you choose. Two sliders set the latitude (−90 to 90 degrees) and the day of the year (1 to 365), and four buttons jump to the March equinox, the June solstice, the September equinox and the December solstice. The readouts are the Sun's declination, the noon altitude, the day length and the Earth-Sun distance that day in AU. The declination and the distance are worked out for noon, Universal Time, on that day of 2026. The noon altitude is 90° − |latitude − declination|; a result below zero means the Sun stays below the horizon at noon. The day length comes from cos H = −tan(latitude) × tan(declination), where H is the angle the Earth turns between noon and sunset. The Earth turns 15 degrees an hour, so the day lasts 2H ÷ 15 hours; the readout gives 24 hours for polar day and 0 for polar night.

Distance does not make the seasons. The Earth is nearest the Sun (perihelion) in early January, about 0.983 AU (147.1 million km), when it is winter in the north. It is farthest (aphelion) in early July, about 1.017 AU (152.1 million km), a change of about 3 per cent. Perihelion falls between 2 and 5 January (Universal Time) in every year from 2015 to 2030. Sunlight weakens with the square of the distance (the inverse-square law), so it is about 7 per cent stronger at perihelion than at aphelion. The tilt matters far more: at latitude 40 degrees north the noon Sun stands 73.4 degrees high in June and 26.6 degrees high in December. The distance does leave a mark on the calendar. By Kepler's second law (section V) the Earth moves faster near perihelion, so the half of the year that holds the northern winter is shorter. The Sun was overhead south of the equator for 178.9 days, from the September equinox of 2025 to the March equinox of 2026, and north of it for 186.4 days, from then to the September equinox of 2026. The Library lesson [The Sky as a Sphere](/library/mathematics/the-sky-as-a-sphere/) works through the geometry.

Try this

  • Set the latitude to 0 and press the four date buttons in turn. The day length stays at 12 hours. The noon altitude is about 90 degrees at the equinoxes and 66.6 degrees (90 − 23.44) at the solstices.
  • Set the latitude to 40 and compare the June solstice with the December solstice. The noon altitude is 73.4 degrees against 26.6, the day length 14.8 hours against 9.2, and the Earth-Sun distance 1.016 AU against 0.984. The Sun is nearer in the month with the short day.
  • Set the latitude to 70 and press the June solstice, then the December solstice: the day length reads 24 hours, then 0. Return to the June solstice and lower the latitude until the day drops below 24 hours. That happens at about 66.56 degrees (90 − 23.44), the latitude of the Arctic Circle.

The figure treats the Sun as a point and ignores the air, so day length is geometric. Real days around the equinox are about 7 to 11 minutes longer between the equator and latitude 52 degrees, because the air bends sunlight (refraction) and the Sun is a disc, not a point.

Sources: NASA, Earth Fact Sheet, nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html: obliquity to orbit 23.44 degrees, eccentricity 0.0167, perihelion 147.095 and aphelion 152.100 million km. Read 2026-10-04.; US Naval Observatory, seasons API (aa.usno.navy.mil/api/seasons?year=YYYY&tz=0), queried for every year from 2015 to 2030: perihelion fell between 2 and 5 January (Universal Time) each year, on 3 January 2026 at 17:15 UT; aphelion fell on 6 July 2026 at 17:30 UT. Read 2026-10-04.; US Naval Observatory, Earth's Seasons (aa.usno.navy.mil/data/Earth_Seasons; data read from aa.usno.navy.mil/api/seasons): September equinox 22 Sep 2025 18:19 UT, March equinox 20 Mar 2026 14:46 UT, September equinox 23 Sep 2026 00:05 UT; June solstice 21 Jun 2026 08:24 UT and December solstice 21 Dec 2026 20:50 UT. The four date buttons use 20 March, 21 June, 23 September and 21 December 2026. The 178.9 and 186.4 day intervals are computed here. Read 2026-10-04.; NOAA, Solar Calculator details, gml.noaa.gov/grad/solcalc/calcdetails.html: sunrise and sunset assume 0.833 degrees of atmospheric refraction (USNO, quoted in Wikipedia, Equinox, splits this into 34 arcminutes of refraction and 16 of semidiameter). The 7 to 11 minutes in the caveat are computed here with it. Read 2026-10-04.; US Naval Observatory, Computing Approximate Solar Coordinates, aa.usno.navy.mil/faq/sun_approx: low-precision formulas for the Sun's mean anomaly g = 357.529 + 0.98560028 D, mean longitude q = 280.459 + 0.98564736 D, ecliptic longitude L = q + 1.915 sin g + 0.020 sin 2g and distance R = 1.00014 − 0.01671 cos g − 0.00014 cos 2g AU (D in days from J2000.0), accurate to about 1 arcminute within two centuries of 2000. The figure evaluates them at noon UT on each day of 2026, with the obliquity from the constants (23.44 degrees, NASA) in place of the obliquity formula on the USNO page, to give the declination (sin δ = sin 23.44° × sin L) and the Earth-Sun distance. Read 2026-10-04.; Computed here with Node on 2026-10-04 from the US Naval Observatory formulas at noon UT in 2026: the declinations (−0.04 degrees on 20 March, +23.44 on 21 June, −0.19 on 23 September, −23.44 on 21 December), noon altitudes, day lengths (14.8 and 9.2 hours at latitude 40 degrees north, 23.98 hours at 66.56 degrees on 21 June) and Earth-Sun distances (0.983 AU on 3 January, 1.016 on 21 June, 0.984 on 21 December) quoted in the copy, and the 1.069 ratio of sunlight at perihelion to aphelion ((152.100 ÷ 147.095)²).

Check yourself · 4 cards · spaced review in the study dashboard

  1. recall

    What is the Earth's axial tilt, and how is the Sun's noon altitude found from latitude and declination?

    Show answer

    The tilt is about 23.44 degrees. The noon altitude is 90° − |latitude − declination|.

  2. explain

    Why can the changing Earth-Sun distance not be what causes the seasons?

    Show answer

    The Earth is nearest the Sun in early January, when it is winter in the north, and farthest in early July. The distance changes by only about 3 per cent (0.983 to 1.017 AU), which makes sunlight about 7 per cent stronger at perihelion. The tilt, by contrast, moves the noon Sun at latitude 40 degrees north from 73.4 degrees high in June to 26.6 degrees in December.

  3. apply

    At latitude 40 degrees north the Sun's declination is +23.44 degrees at the June solstice and −23.44 degrees at the December solstice. What is the noon altitude on each date?

    Show answer

    June: 90 − |40 − 23.44| = 73.44 degrees. December: 90 − |40 + 23.44| = 26.56 degrees.

  4. explain

    Why is the Sun overhead north of the equator for more days of the year than south of it?

    Show answer

    By Kepler's second law the Earth moves more slowly when it is farther from the Sun, around July, so the half of the year that holds the northern summer lasts longer. From the March equinox to the September equinox of 2026 took 186.4 days; from the September equinox of 2025 to the March equinox of 2026 took 178.9 days.

II · The Moon

Phases come from the angle, not the shadow.

The Sun lights half of the Moon at all times. A phase is how much of that lit half faces the Earth. At new Moon the lit half faces away and the Moon lies near the Sun in the sky. At full Moon the Moon lies opposite the Sun and the lit half faces us. The cycle takes 29.53 days, the synodic month. That is longer than the 27.32 days the Moon takes to circle the Earth once against the stars. In those 27.32 days the Earth moves about 27 degrees round the Sun, so the Sun's direction, seen from the Earth, moves too, and the Moon needs about two more days to catch up with it.

The top-down view puts the Earth at the centre, the Moon on its orbit and sunlight from one side, with each body half lit. The second view shows the Moon as the Earth sees it. One slider sets the day of the lunar month, from 0 to 29.5. Four readouts are labelled Phase, Illuminated fraction, Elongation and Highest in the sky; the last gives the approximate local time the Moon stands highest. The illuminated fraction is (1 − cos θ) ÷ 2, where θ is the angle round the orbit from the new Moon position. The elongation is the angle in the sky between the Moon and the Sun, east of the Sun while the Moon waxes and west while it wanes: 90 degrees at both quarters and 180 degrees at full Moon. The Moon is highest at about noon at new Moon, 6 pm at first quarter, midnight at full Moon and 6 am at last quarter. The Moon gains about 12.2 degrees on the Sun each day. The Earth must turn that much further, and a little more because the Moon keeps moving, to bring it back to the meridian, the north-south line across the sky where it stands highest. So the Moon is highest about 50 minutes later each day.

Phases are not caused by the Earth's shadow. The shadow falls on the Moon only in a lunar eclipse, at full Moon, and Aristotle took the always-curved edge of that shadow as evidence that the Earth is a sphere (On the Heavens II.14). There is not an eclipse every month because the Moon's orbit is tilted about 5.1 degrees to the plane of the Earth's orbit. At most new and full Moons the Moon passes above or below the line through the Sun and the Earth. An eclipse needs a new or full Moon near a node, where the Moon's orbit crosses that plane, and that happens only in eclipse seasons about 173 days apart. The figure draws the Moon's orbit flat, so the three bodies would line up at every new and full Moon; in the sky the tilt usually prevents it.

Try this

  • Set the day to 0, 7.4, 14.8 and 22.1 in turn. Illuminated fraction reads about 0, 50, 100 and 50 per cent, Elongation about 0, 90 (east), 180 and 90 (west) degrees, and the Moon is highest at about noon, 6 pm, midnight and 6 am.
  • Compare day 3.7 with day 25.8, and day 11.1 with day 18.5. Each pair lights almost the same fraction (about 15 per cent and about 85 per cent). The fraction alone does not tell you whether the Moon is waxing (growing) or waning (shrinking); the phase name does.
  • Step the slider from day 14 to day 15, then from day 15 to day 16, and read Highest in the sky. It moves about 49 minutes later at each step. The readout leaves out the Moon's own motion while the Earth turns towards it; with that included, the real delay averages about 50 minutes.

The Moon moves at a steady speed in a flat orbit here. The real orbit is tilted 5.1 degrees, and the time from one new Moon to the next runs from 29.27 to 29.83 days, so the times the Moon is highest are approximate.

Sources: NASA, Moon Phases, science.nasa.gov/moon/moon-phases/: the Sun always illuminates half of the Moon; the cycle repeats every 29.5 days; the first quarter rises around midday and the full Moon rises at sunset. Read 2026-10-04.; NASA, Moon Fact Sheet, nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html: synodic period 29.53 days, inclination to the ecliptic 5.145 degrees, apparent diameter 1896 arcseconds. Read 2026-10-04.; Wikipedia, Lunar month: mean synodic month 29.530589 days, sidereal month 27.321661 days, actual time between lunations from about 29.274 to about 29.829 days (citing Meeus). Read 2026-10-04.; NASA, Why Do Eclipses Happen?, science.nasa.gov/eclipses/geometry/: the Moon's orbit is tilted by about five degrees; there are two eclipse seasons a year. Read 2026-10-04.; Wikipedia, Eclipse cycle: eclipse year 346.620076 days; half of it, 173.3 days, is the spacing of eclipse seasons (computed here). Read 2026-10-04.; Aristotle, On the Heavens II.14, J. L. Stocks translation, classics.mit.edu/Aristotle/heavens.2.ii.html: 'in eclipses the outline is always curved', so the Earth's surface is spherical; offered as evidence of the senses that corroborates the argument. Read 2026-10-04.; Computed here with Node on 2026-10-04: 26.9 degrees (27.32 days × 360 ÷ 365.256), 12.19 degrees a day (360 ÷ 29.53), 48.8 minutes per day-step in the readout (12.19 ÷ 15 hours), the mean lunar day of 24.84 hours (24 × 29.5306 ÷ 28.5306, so about 50.5 minutes later each day), and the illuminated fractions and times in the try prompts (day 11.1: 85.6 per cent; day 18.5: 85.0 per cent).

Check yourself · 4 cards · spaced review in the study dashboard

  1. recall

    How long is the synodic month, and why is it longer than the Moon's orbit against the stars?

    Show answer

    It is 29.53 days. The Moon circles the Earth in 27.32 days against the stars. In that time the Earth has moved about 27 degrees round the Sun, so the Moon needs about two more days to return to the same position relative to the Sun.

  2. explain

    Why are the Moon's phases not caused by the Earth's shadow?

    Show answer

    The Sun always lights half the Moon, and a phase is how much of that half faces us. The Earth's shadow falls on the Moon only in a lunar eclipse, at full Moon.

  3. apply

    The Moon is 22.1 days past new Moon, in a month of 29.53 days. What, roughly, are its elongation from the Sun, the fraction of its disc that is lit, and the local time when it is highest in the sky?

    Show answer

    22.1 days is about three quarters of the month, so the Moon is about 270 degrees round its orbit from new Moon. The elongation is about 90 degrees, about 50 per cent is lit, and it is highest at about 6 am. This is the last quarter.

  4. explain

    Why is there not an eclipse at every new Moon and every full Moon?

    Show answer

    The Moon's orbit is tilted about 5.1 degrees to the plane of the Earth's orbit, so at most new and full Moons it passes above or below the line through the Sun and the Earth. An eclipse needs a new or full Moon near a node, which happens only in eclipse seasons about 173 days apart.

III · The Earth

Two places and one angle measure the Earth.

Cleomedes preserves the argument of Eratosthenes in The Heavens (1.7). He reports that on the summer solstice the Sun stood overhead at Syene, so the gnomon of a sundial, its upright pointer, cast no shadow there at noon. Pliny, who does not name Eratosthenes, reports a well at Syene lit to the bottom at midday on the solstice (Natural History 2.183). Cleomedes mentions no well; the figure draws one only to picture the overhead Sun. At Alexandria, at the same hour, the gnomon of a bowl-shaped sundial cast a shadow whose arc was one fiftieth of a circle, 7.2 degrees. Cleomedes lists five assumptions. The two places lie on one meridian, the same north-south line, and they are 5,000 stades apart. The Sun's rays are parallel. A line crossing two parallels makes equal alternate angles, the angles on opposite sides of it between the parallels. Arcs standing on equal angles are in the same proportion to their circles. The rule about alternate angles is Euclid, Elements I.29, the first proposition that depends on the parallel postulate (see the Academy on [Euclid's postulates](/academy/#postulates)). Because the Sun is overhead at Syene, the ray there points to the Earth's centre. The upright gnomon at Alexandria points to the centre too, so its line crosses both parallel rays, and the angle between the two places at the centre equals the shadow angle. It follows that the 5,000 stades between the places are one fiftieth of the Earth's circumference: 250,000 stades. Strabo (Geography 2.5.7) and Pliny (Natural History 2.247) report 252,000. Aristotle had cited 400,000 from earlier mathematicians (On the Heavens II.14).

The figure draws a cross-section of the Earth with parallel sunbeams, a well at Syene with the Sun overhead, a gnomon at Alexandria with its shadow angle marked, and the equal angle at the Earth's centre. Three controls set the shadow angle in degrees, the distance between the places in stades, and the stade length from 157.5 to 185 metres. The readouts give the circumference in stades (distance × 360 ÷ angle), the circumference in kilometres for the chosen stade, and its difference from the modern circumference through the poles (the meridional circumference), about 40,008 km. At the starting values, 7.2 degrees and 5,000 stades, the first readout is 250,000 stades. The Library lesson [Eratosthenes Measures the Earth](/library/mathematics/eratosthenes-measures-the-earth/) takes the argument step by step.

The accuracy of the result depends on the stade, whose length is uncertain. Estimates run from about 157.5 metres to about 185 metres, and some run wider. The lower value is 300 Egyptian royal cubits, as Hultsch took it in 1882, and the slider starts there. The upper is eight stades to the Roman mile, from Pliny, the value Rawlins and Engels adopted. At 157.5 metres, 250,000 stades is 39,375 km, 1.6 per cent below the modern figure. At 185 metres it is 46,250 km, 15.6 per cent above. Claims that Eratosthenes came within 1 or 2 per cent rest on the shorter stade, and the sources do not settle which he used. The other assumptions are rough as well. Syene (Aswan) lies at 24.09 degrees north, about 0.65 degrees (72 km) north of today's Tropic of Cancer, the farthest north the Sun stands overhead, which moves as the tilt changes. In Eratosthenes' time the tilt was about 23.72 degrees (J. Laskar's 1986 formula), so Syene then lay only about 0.36 degrees (40 km) north of the tropic. Alexandria is 7.11 degrees of latitude north of Syene and about 3 degrees of longitude west, so the two are not on one meridian. The north to south distance between their latitudes is about 788 km, and the shortest distance between them over the surface is about 842 km. At 157.5 metres, 5,000 stades is 787.5 km, close to the 788; at 185 metres it is 925 km. That agreement does not show which stade he used. The method is sound, and because every assumption is stated we can see where the error could lie.

Try this

  • Leave the angle at 7.2 degrees and the distance at 5,000 stades: the readout is 250,000 stades. Slide the stade length from 157.5 to 185 metres and watch the kilometres run from 39,375 to 46,250, and the difference from 40,008 km run from about −1.6 to +15.6 per cent.
  • Change only the angle to 7.0, 7.1 and 7.5 degrees: the circumference becomes 257,143, 253,521 and 240,000 stades. A tenth of a degree moves the answer by about 1.4 per cent.
  • Set the distance to 4,500 stades with the angle at 7.2 degrees. The circumference falls by 10 per cent, to 225,000 stades. The answer is proportional to the distance, so a 10 per cent error in the 5,000 stades is a 10 per cent error in the answer.

The figure assumes a sphere, one meridian, parallel rays and the Sun exactly overhead at Syene; none is exactly true, and the Earth is slightly flattened (equatorial circumference 40,075 km, meridional 40,008 km).

Sources: Cleomedes, The Heavens (On the Circular Motions of the Celestial Bodies) 1.7, translated by Alan Bowen and Robert Todd (University of California Press, 2004), pp. 81 to 84: five assumptions; Syene and Alexandria under the same meridian, 5,000 stades apart; the arc at Alexandria one fiftieth of its circle; 250,000 stades. The passage was read in full as quoted at nonagon.org/ExLibris/eratosthenes-measures-earth on 2026-10-04. The locator 1.7 (pp. 96 to 103 in Ziegler's edition) comes from the Kiwi Hellenist blog (kiwihellenist.blogspot.com/2023/06/eratosthenes-2a.html). The Greek text was not checked.; Strabo, Geography 2.5.7, H. L. Jones translation (Loeb, 1917) at LacusCurtius, penelope.uchicago.edu/Thayer/E/Roman/Texts/Strabo/2E1*.html: 'according to Eratosthenes, the equator measures two hundred and fifty two thousand stadia'; the summer tropic passes through Syene because the index of the sundial casts no shadow at noon at the summer solstice. Read 2026-10-04 (Perseus was unavailable that day).; Pliny the Elder, Natural History 2.183 (at Syene, 5,000 stades from Alexandria, no shadow at midday on the solstice and a well made for the experiment lit throughout: 'puteumque eius experimenti gratia factum totum inluminari') and 2.247 (Eratosthenes, 252,000 stades: 'CCLII milium stadiorum'). Latin text at LacusCurtius, penelope.uchicago.edu/Thayer/L/Roman/Texts/Pliny_the_Elder/2*.html; the English is paraphrased here from the Latin. Read 2026-10-04.; Euclid, Elements I.29, David Joyce's edition, mathcs.clarku.edu/~djoyce/elements/bookI/propI29.html: a straight line falling on parallel straight lines makes the alternate angles equal; 'This is the first proposition which depends on the parallel postulate.' Read 2026-10-04.; Aristotle, On the Heavens II.14, J. L. Stocks translation, classics.mit.edu/Aristotle/heavens.2.ii.html: 'those mathematicians who try to calculate the size of the earth's circumference arrive at the figure 400,000 stades'. Read 2026-10-04.; Wikipedia, Stadion (unit): historians estimate the stadion at between 150 and 210 m. Read 2026-10-04.; I. Tupikova, 'A common-sense approach to the problem of the itinerary stadion', Archive for History of Exact Sciences 76 (2022), 319 to 361, open access at link.springer.com/article/10.1007/s00407-022-00287-6, section 2 and notes 14, 15, 32, 36 and 40: from Pliny's schoenus of 40 stades, Hultsch (1882) took Eratosthenes' stade as 300 royal cubits of about 0.525 m, 157.5 m; 185 m was adopted by Dicks (1960), Rawlins (1982) and Engels (1985). N. Walkup, 'Eratosthenes and the Mystery of the Stades' (home.adelphi.edu/~bradley/HOMSIGMAA/Walkup.pdf), pp. 12 to 15, quotes Rawlins's 185 m and the conversion at eight stades to a Roman mile of about 1,479 m. Secondary reports, as used in the Library lesson; the original works were not read. Pliny, Natural History 12.53 (a schoenus of 40 stades by Eratosthenes' reckoning, 'that is, 5 miles') and 2.85 (125 paces to the stade), Latin at LacusCurtius. The Kiwi Hellenist, Part 3 (kiwihellenist.blogspot.com/2023/06/eratosthenes-2b.html), argues against the 157.5 m value and gives 177 to 192 m for the metrological stadion.; NASA, Earth Fact Sheet: equatorial radius 6378.137 km, polar radius 6356.752 km, volumetric mean radius 6371.000 km. The meridional circumference of that ellipse (40,007.86 km) and the equatorial circumference (40,075.02 km) are computed here and agree with Wikipedia, Earth's circumference (40,007.863 and 40,075.017 km).; Wikipedia, Aswan (24°05′20″N 32°53′59″E) and Alexandria (31°11′51″N 29°53′33″E). Computed here from those coordinates: latitude difference 7.11 degrees, longitude difference 3.01 degrees, north-south distance 788 km (the meridian arc) and surface distance 842 km (the geodesic, by Vincenty's formula), both on the WGS84 ellipsoid, as in the Library lesson. The 0.65 degrees uses an obliquity of 23.436 degrees for 2026. J. Laskar's 1986 polynomial for the mean obliquity, as printed in Wikipedia, Axial tilt, gives 23.72 degrees around 240 BC, so Syene then lay 0.36 degrees (40 km) north of the tropic (computed here). Wikipedia, Axial tilt: over the past 5 million years the obliquity has varied between 22°2′33″ and 24°30′16″. Read 2026-10-04.; Computed here with Node on 2026-10-04: the apply card's 800 × 360 ÷ 7.2 = 40,000 km, and the try-prompt circumferences (257,142.9, 253,521.1, 240,000 and 225,000 stades).

Check yourself · 4 cards · spaced review in the study dashboard

  1. recall

    What did Eratosthenes measure, and what circumference in stades follows from his angle and distance?

    Show answer

    The noon shadow angle at Alexandria on the day the Sun was overhead at Syene: one fiftieth of a circle, 7.2 degrees. With 5,000 stades between the places, 5,000 × 360 ÷ 7.2 = 250,000 stades (Cleomedes). Strabo and Pliny report 252,000.

  2. explain

    In the argument of Eratosthenes, why is the angle at the Earth's centre between Syene and Alexandria equal to the shadow angle at Alexandria?

    Show answer

    The Sun is overhead at Syene, so the ray there points to the Earth's centre, and the ray at Alexandria is parallel to it. The upright gnomon at Alexandria also points to the centre, so its line crosses both rays, and a line crossing two parallels makes equal alternate angles (Euclid I.29).

  3. explain

    Why can nobody say without qualification how many kilometres Eratosthenes measured, or that he was within 2 per cent?

    Show answer

    The length of the stade is uncertain. At about 157.5 metres his 250,000 stades is 39,375 km, 1.6 per cent below 40,008 km. At 185 metres it is 46,250 km, 15.6 per cent above.

  4. apply

    Two places lie on the same north-south line, 800 km apart. At noon on the same day the Sun is overhead at one, and at the other the Sun's rays make an angle of 7.2 degrees with an upright gnomon. What circumference does the method of Eratosthenes give?

    Show answer

    The angle at the Earth's centre is also 7.2 degrees, one fiftieth of a circle, so the circumference is 800 × 360 ÷ 7.2 = 40,000 km.

IV · Retrograde motion

A planet loops when the Earth overtakes it.

Watched night after night, Mars drifts eastwards among the stars, stops, moves westwards for about 72 days on average (about 60 to 81 days at the oppositions of 1997 to 2040), stops again and resumes. This backward stretch is retrograde motion, and each stop is called a station. For an outer planet, one farther from the Sun than the Earth, retrograde motion happens around opposition, when the Earth lies between the Sun and the planet. Ptolemy handled it with circles on circles (Almagest, Book XII covers stations and retrograde motion). Each planet rides on a small circle, the epicycle, whose centre is carried round the Earth on a larger circle, the deferent. The deferent's centre sits a little off the Earth. In his models for the outer planets the line from the epicycle's centre to the planet always stays parallel to the line from the Earth to the mean Sun (Linton, From Eudoxus to Einstein, p. 77). The mean Sun is an imaginary Sun that moves at a steady rate along the real Sun's yearly path (p. 71). The model builds that link in but does not explain it; that is an interpretation, not Ptolemy's own claim. Copernicus held that retrograde motion is only apparent, produced by the observer's own motion. He still used circles on circles, and the Prutenic Tables of planetary positions, computed from his models, were little different in accuracy from the older Alfonsine Tables.

In the figure both orbits are circles in one plane: the Earth at 1 AU with a period of 1 year, Mars at 1.524 AU with a period of 1.881 years. The Sun-centred view shows both planets and the line of sight from the Earth to Mars. The Earth-centred view traces the looped path Mars seems to follow. A strip plots Mars's apparent longitude, its direction among the stars as seen from the Earth, against time. The controls are play and pause and a time slider over about 800 days centred on an opposition. The readouts give the days from opposition, whether the motion is direct or retrograde, and the Earth-Mars distance in AU.

In this model the numbers can be checked. Each year the Earth gains 1 − 1/1.881, about 0.47, of a lap on Mars. It therefore laps Mars every 1 ÷ (1 − 1/1.881) = 2.135 years, about 780 days (NASA gives 779.94 for the real planet). Retrograde motion begins 36.4 days before opposition and ends 36.4 days after it, about 73 days in all, across an arc of about 16 degrees. Mars is 0.52 AU away at opposition, 0.64 AU at the two stations and 2.52 AU near conjunction, when it lies beyond the Sun. Real Mars comes closer than the model allows, 0.365 AU at its nearest, because its orbit is an ellipse. The Library lesson [Models of the Heavens](/library/mathematics/models-of-the-heavens/) follows the argument from Ptolemy to Kepler.

Try this

  • Move the slider to its left end and press play. Watch the line of sight and the strip: Mars's apparent longitude rises, levels off, falls, levels off and rises again. The readout says retrograde only from about −36 to +36 days around opposition.
  • Drag the slider to 0 and read the distance, 0.52 AU. Drag it to either end, about 400 days away: the distance is about 2.5 AU and the Sun lies roughly between the two planets. The motion is direct, and the strip shows Mars's longitude rising steeply.
  • Step the slider a day at a time near −36 and near +36. The motion flips between direct and retrograde at those two stations, and the distance there is about 0.64 AU.

The orbits here are circles in one plane, travelled at constant speed. Real orbits are ellipses (Mars has eccentricity 0.0935), and Mars's orbit is tilted 1.85 degrees to the plane of the Earth's orbit, so real loops differ in size and shape from one opposition to the next.

Sources: NASA, Mars Fact Sheet, nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html: semimajor axis 227.956 million km, sidereal period 686.980 days, synodic period 779.94 days, eccentricity 0.0935, inclination 1.848 degrees, minimum distance from the Earth 54.6 million km (0.365 AU). Read 2026-10-04.; Wikipedia, Apparent retrograde motion: table of planetary retrograde constants (Mars: synodic period 780 days, 72 days in retrogradation, the usual tabulated average); the Earth overtakes an outer planet 'like a faster car on a multi-lane highway'. Read 2026-10-04.; Retrograde durations at the oppositions of Mars from 1997 to 2040, recomputed here with Node on 2026-10-04 from Keplerian ellipses with JPL's approximate elements (ssd.jpl.nasa.gov/planets/approx_pos.html, Table 1, Mars and the Earth-Moon barycentre): 60 to 81 days, with a mean of 73.2 over those oppositions against the tabulated long-run 72, as in the Library lesson Models of the Heavens.; Wikipedia, Deferent and epicycle: neither circle was centred on the Earth; the lines from each planet through its epicycle's centre were all parallel to the line from the Sun to the Earth (Linton, below, specifies the mean Sun); Copernicus's models kept (smaller) epicycles. Read 2026-10-04.; C. M. Linton, From Eudoxus to Einstein: A History of Mathematical Astronomy (Cambridge University Press, 2004), chapter 3: in Ptolemy's models for the outer planets the line from the epicycle's centre to the planet is always parallel to the line from the Earth to the mean Sun (p. 77); the mean Sun moves uniformly around the ecliptic once each tropical year (p. 71, note 26). Read 2026-10-04.; Wikipedia, Almagest: Book XII covers stations and retrograde motion. The chapter inside Book XII and Ptolemy's own text were not checked. Read 2026-10-04.; Stanford Encyclopedia of Philosophy, Nicolaus Copernicus, plato.stanford.edu/entries/copernicus/: the Copernican model maintained epicycles; Copernicus explained that retrograde motion was only apparent, due to the observers not being at rest; Gingerich (1993, 232) quoted: 'there was relatively little to distinguish between the accuracy of the Alfonsine Tables and the Prutenic Tables'. Read 2026-10-04.; Computed here with Node on 2026-10-04 from circular orbits (Earth 1 AU and 1 year, Mars 1.524 AU and 1.881 years, 1 year = 365.25 days): synodic period 779.8 days, stations at −36.4 and +36.4 days, retrograde arc 15.9 degrees, distances 0.524, 0.636 and 2.524 AU, fastest direct motion 0.71 degrees a day at conjunction (389.9 days from opposition; 0.707 degrees a day also at ±400 days).

Check yourself · 4 cards · spaced review in the study dashboard

  1. recall

    When does Mars appear to move backwards among the stars, and for roughly how long?

    Show answer

    Around opposition, when the Earth lies between the Sun and Mars and overtakes it. In reality it averages about 72 days and varies from about 60 to 81 days, because Mars's orbit is an ellipse. In the circular model it lasts about 73 days, from 36.4 days before opposition to 36.4 days after.

  2. explain

    How does the Sun-centred view explain a retrograde loop without Mars ever reversing?

    Show answer

    The Earth orbits faster (1 year against 1.881 years), so it overtakes Mars. As it passes, the line of sight from the Earth to Mars swings backwards against the stars, although Mars keeps moving the same way round the Sun.

  3. apply

    The Earth's period is 1 year and Mars's is 1.881 years. How long is it between successive oppositions of Mars?

    Show answer

    Each year the Earth gains 1 − 1/1.881 of a lap on Mars, so it laps Mars every 1 ÷ (1 − 1/1.881) = 2.135 years, about 780 days. NASA gives 779.94 days.

  4. explain

    Did Copernicus's Sun-centred system do away with epicycles or predict the planets much better than Ptolemy's?

    Show answer

    No. Copernicus still used circles on circles, and the tables computed from his models were little different in accuracy from the older ones. What he changed was the explanation of retrograde motion: it is apparent, produced by the observer's own motion.

V · Equal areas

A planet sweeps equal areas in equal times.

Kepler worked out the orbit of Mars from Tycho Brahe's observations. His circular model, the 'vicarious hypothesis', placed Mars at opposition within 2 minutes of arc of Tycho's positions (a minute of arc is a sixtieth of a degree), but its distances were wrong. When he changed it to fit the distances, it missed Tycho's positions by 8 minutes of arc, about a quarter of the Moon's apparent width and more than Tycho's observations allowed. Kepler kept the observations and gave up the circle: 'these eight minutes alone will lead us along a path to the reform of the whole of Astronomy' (Astronomia Nova, chapter 19). The result was Astronomia Nova (1609). In it the orbit of Mars is an ellipse with the Sun at one focus, which is the first law, stated there for Mars alone. The book also gives the second law in two forms: that the line from the Sun to the planet sweeps equal areas in equal times, and that the speed varies inversely with the distance. By 1621 Kepler knew that the two differ and that the area rule is the right one (J. L. Russell, 1964).

The figure draws an ellipse with the Sun at one of its two foci, the points inside it whose distances to any point on the curve always add up to the same total. The planet moves by Kepler's equation, M = E − e sin E. Here M is an angle that grows in step with time. E is an angle that fixes the planet's place on the ellipse (the eccentric anomaly). The eccentricity e is 0 for a circle and closer to 1 the more stretched the ellipse. Kepler posed the problem of finding the position from the time and judged that no exact geometrical solution exists. The equation is transcendental and has no solution in closed form, so the figure finds E by iteration, improving a guess until it stops changing. The shaded sectors are equal-time sectors, so they have equal areas, though their shapes differ: short and wide near perihelion, where the planet is close to the Sun, and long and narrow near aphelion. The controls are an eccentricity slider from 0 to 0.9 and play and pause. The readouts give the current distance as a fraction of the semi-major axis a, which is half the longest diameter of the ellipse. They also give the speed ratio at perihelion and aphelion, (1 + e) ÷ (1 − e), and real eccentricities for comparison: Earth 0.0167, Mars 0.0935, Mercury 0.2056 and Halley's Comet 0.968.

The speed ratio follows from the area rule. At perihelion and aphelion the line from the Sun to the planet is perpendicular to the motion, so equal areas in equal times means that distance × speed is the same at both. The distances there are a(1 − e) and a(1 + e), so speed at perihelion ÷ speed at aphelion = (1 + e) ÷ (1 − e). For Mars that is 1.206, which matches NASA's maximum and minimum orbital speeds (26.50 and 21.97 km/s). For the Earth it is 1.034. Real planetary orbits are close to circles: the Earth's orbit has the Sun 0.0167 AU, about 2.5 million km, off centre, and its short axis is 99.986 per cent of its long axis. Drawn to scale, a planet's orbit looks like a circle, so a diagram with a clear oval exaggerates it. Eccentricities near the top of the slider, and beyond it, are typical of comets, not planets: Halley's Comet has 0.968, and its speed ratio is about 61.

Try this

  • Set the eccentricity to 0: the orbit is a circle, the speed ratio is 1.00 and the shaded sectors are identical. Raise it to 0.1, close to Mars's 0.0935: the speed ratio is 1.22.
  • Set the eccentricity to 0.5 and press play. The speed ratio is 3.00 and the distance runs from 0.5 to 1.5 of a. The shaded sectors near perihelion are short and wide, those near aphelion long and narrow, and all have the same area.
  • Set the eccentricity to 0.9 and press play. The speed ratio is 19 and the distance runs from 0.1 to 1.9 of a. The planet moves fast near perihelion and slowly near aphelion. In this model it sweeps about 150 degrees round the Sun in the first twelfth of its period after perihelion, and only about 4 degrees in the twelfth centred on aphelion.

The figure has two bodies only, a planet of negligible mass and a fixed Sun. Real orbits are disturbed slightly by the other planets.

Sources: J. L. Russell, Kepler's Laws of Planetary Motion: 1609 to 1666, British Journal for the History of Science 2(1), 1964, pp. 1 to 24, PDF at astro.rug.nl/~vdkruit/jea3/homepage/Russell.pdf, pp. 2 to 3: 'In 1609 it was explicitly formulated only for Mars'; the second law 'originally formulated, in 1609, in two different forms'; 'By 1621 ... he had come to realize that the two laws were not identical and that the area law was correct'; Kepler 'rightly surmised that no exact geometrical solution is possible'. Read 2026-10-04.; MacTutor, Johannes Kepler: quotations, mathshistory.st-andrews.ac.uk/Biographies/Kepler/quotations/: 'Now, because they could not be disregarded, these eight minutes alone will lead us along a path to the reform of the whole of Astronomy, and they are the matter for a great part of this work', located there as Astronomia nova (Heidelberg, 1609), Chapter 19, 113 to 14; KGW 3, 177 to 78. Read 2026-10-04. Grade: Attested, in translation; the Latin was not checked.; Wikipedia, Astronomia nova, account of chapters 18 to 21: the vicarious hypothesis fits the oppositions within Tycho's two minutes of arc; with the eccentricity bisected, as the distances required, the error at the oppositions rises to 8 minutes of arc. The same article places the distance law in chapter 32 and the area law in chapter 59. Kepler's text was not checked for these chapters. Read 2026-10-04.; Wikipedia, Kepler's laws of planetary motion: Kepler's equation M = E − ε sin E, a transcendental equation. Read 2026-10-04.; NASA, Mars Fact Sheet: maximum orbital velocity 26.50 km/s, minimum 21.97 km/s, eccentricity 0.0935. NASA, Earth Fact Sheet: eccentricity 0.0167. NASA, Mercury Fact Sheet: eccentricity 0.2056. NASA, Moon Fact Sheet: apparent diameter 1896 arcseconds (31.6 arcminutes), so 8 arcminutes is 0.25 of it (computed here). Read 2026-10-04.; NASA/JPL Small-Body Database API, ssd-api.jpl.nasa.gov/sbdb.api?sstr=1P&full-prec=1: Halley's Comet eccentricity 0.96794 (osculating, epoch 1968-01-20). Read 2026-10-04.; Computed here with Node on 2026-10-04: speed ratios (1 + e) ÷ (1 − e) (Earth 1.034, Mars 1.206, Halley 61.4), the ellipse's b ÷ a (0.99986), the Sun's offset (0.0167 × 149.598 = 2.50 million km), the distance range 1 − e to 1 + e (0.1 to 1.9 of a at e = 0.9), and the true anomaly after the first twelfth of the period for e = 0.9 (149.9 degrees) and across the twelfth centred on aphelion (3.6 degrees), from Kepler's equation.

Check yourself · 4 cards · spaced review in the study dashboard

  1. recall

    State Kepler's first two laws and the book in which they appeared.

    Show answer

    First, a planet's orbit is an ellipse with the Sun at one focus. Second, the line from the Sun to the planet sweeps equal areas in equal times. Both appear in Astronomia Nova (1609), which states the first law for Mars alone and also gives the second in a less accurate form, that speed varies inversely with distance.

  2. explain

    Why must a planet move faster at perihelion than at aphelion?

    Show answer

    Equal areas must be swept in equal times. Near the Sun the line from the Sun to the planet is short, so the planet must cover more of its path in the same time to sweep the same area.

  3. apply

    A body has an orbital eccentricity of 0.5. What is the ratio of its speed at perihelion to its speed at aphelion?

    Show answer

    (1 + e) ÷ (1 − e) = 1.5 ÷ 0.5 = 3.

  4. explain

    The Earth's orbital eccentricity is 0.0167. Why does its orbit look circular?

    Show answer

    The Sun lies only 0.0167 AU, about 2.5 million km, from the centre, and the short axis is 99.986 per cent of the long axis. The speed at perihelion is only 1.034 times the speed at aphelion.

VI · The third law

Period squared is distance cubed.

Kepler's third law says that the squares of the periods of the planets are as the cubes of their mean distances from the Sun: T² ÷ a³ is the same for every planet. Kepler recorded that the relation occurred to him on 8 March 1618, that he rejected it after a faulty calculation, and that he returned to it on 15 May 1618 (MacTutor quotes his account). He published it in Book V of Harmonices Mundi (1619). Newton showed in the Principia (1687, Book I, Proposition 15) that it follows from an attraction that weakens with the square of the distance; the Academy's [Prop. XII](/academy/#prop-kepler) sets it out, including where the simple form stops being exact.

The plot is logarithmic on both axes, so each equal step along an axis multiplies the value by the same factor. It shows the period T in years against the semi-major axis a in AU for the eight planets. A power law T = K × a^m is a straight line on such a plot, and its slope is m. Kepler's law says m = 3/2 and, in years and AU, K = 1. The table lists a, T and T² ÷ a³ for each planet, with T² ÷ a³ to 4 decimal places. Checkboxes choose which planets enter the fit, which is the straight line that best fits their points (by least squares). An input predicts T from any a. The readouts give the fitted exponent and constant.

With all eight planets (NASA values) the fit gives m = 1.499 and K = 0.9997. Mercury, Venus, Earth and Mars have T² ÷ a³ within 0.03 per cent of 1. Jupiter gives 0.9985 and Uranus 1.0027. Saturn and Neptune give 0.9885 and 0.9878. Most of the gap comes from the semi-major axes in the table. The fact sheet gives each distance as the midpoint of perihelion and aphelion and states no date for it. The same pages also list mean orbital elements for the year 2000 (J2000), with Saturn at 9.537 AU and Neptune at 30.069 AU. With those and the same periods the ratio is 0.9997 for Saturn and 0.9988 for Neptune. Without Saturn and Neptune the fit gives m = 1.500 and K = 0.9999. The Library lesson [Models of the Heavens](/library/mathematics/models-of-the-heavens/) puts the three laws in order.

Try this

  • Leave all eight planets ticked: the exponent is 1.499 and the constant 0.9997. Untick Saturn and Neptune: the exponent becomes 1.500 and the constant 0.9999.
  • Tick only Mercury and Venus: the line passes through both points and the exponent is 1.500. Tick only Uranus and Neptune: it is 1.484. Neptune's T² ÷ a³ is 1.5 per cent below Uranus's, and with only two points close together on the plot that small difference tilts the line. A fit to many points spread widely is steadier.
  • In the predictor enter a = 2.766 AU, the semi-major axis of Ceres: the period is about 4.59 years with all eight planets and 4.60 without Saturn and Neptune; JPL gives 4.599. Enter 17.93 AU, Halley's Comet: about 75.7 years with all eight, against JPL's 75.9. Halley's period changes from one return to the next, so no single value is exact.

The constant is 1 only in years and AU, and only for bodies orbiting the Sun. Each central body has its own constant, and Kepler applied the same rule to the four known moons of Jupiter (Epitome, Book IV). Even for the planets the simple rule is not exact, because a planet's own mass and the pull of the others shift it slightly.

Sources: NASA, Planetary Fact Sheet, individual planet pages (nssdc.gsfc.nasa.gov/planetary/factsheet/mercuryfact.html and the seven others): semimajor axes and sidereal orbit periods used for the table and fit, and each page's 'Mean Orbital Elements (J2000)' (Saturn 9.53707032 AU, Neptune 30.06896348 AU); see dataSource. The notes page (planetfact_notes.html) defines the distance as the semi-major axis, 'midway between' perihelion and aphelion. Read 2026-10-04.; NASA JPL, Keplerian Elements for Approximate Positions of the Major Planets, Table 2a (valid 3000 BC to 3000 AD), ssd.jpl.nasa.gov/planets/approx_pos.html: Saturn a = 9.54149883 AU, Neptune a = 30.06952752 AU; periods from the mean longitude rates (36000 ÷ 1222.115 and 36000 ÷ 218.465 years). T² ÷ a³ computed here with JPL's own periods: Saturn 0.9989, Neptune 0.9988, Jupiter 0.9993, Uranus 0.9991. A corroborating check, not used in the copy. Read 2026-10-04.; NASA/JPL Small-Body Database API, ssd-api.jpl.nasa.gov/sbdb.api (full-prec=1): Ceres a = 2.76555 AU, period 1679.853 days (4.599 years), osculating at epoch 2026-06-09; Halley's Comet a = 17.9286 AU, period 27728.05 days (75.92 years), osculating at epoch 1968-01-20. Read 2026-10-04.; MacTutor, Johannes Kepler, mathshistory.st-andrews.ac.uk/Biographies/Kepler/: Kepler's account, 'conceived mentally on 8th March in this year one thousand six hundred and eighteen, but submitted to calculation in an unlucky way, and therefore rejected as false, and finally returning on the 15th of May and adopting a new line of attack, stormed the darkness of my mind'; Harmonices mundi, Linz, 1619. Read 2026-10-04.; J. L. Russell, Kepler's Laws of Planetary Motion: 1609 to 1666, British Journal for the History of Science 2(1), 1964, p. 3: 'The third law was enunciated for the planets in Harmonices Mundi (1619) and was repeated in the following year in Book IV of the much more widely-read Epitome, where Kepler extended it also to the four known satellites of Jupiter.' Read 2026-10-04.; Newton, The Mathematical Principles of Natural Philosophy, trans. A. Motte (American edition, 1846), Wikisource, Book I, Proposition 15: the periodic times in ellipses are in the sesquiplicate ratio of the greater axes, the third law. Checked for the Library lesson Models of the Heavens.; SizzlinShred Academy, Prop. XII (/academy/#prop-kepler): Newton derived the relation from an inverse-square attraction in the Principia (1687), Book I, Proposition 15. Read in the repository 2026-10-04.; Computed here with Node on 2026-10-04: least-squares fits of log T against log a (all eight planets: m = 1.4990, K = 0.99973; without Saturn and Neptune: m = 1.5002, K = 0.99994; Mercury and Venus 1.5000; Uranus and Neptune 1.4835), T² ÷ a³ for each planet (Uranus 1.00267, Neptune 0.98778, a difference of 1.5 per cent), the J2000 ratios (Saturn 0.99966, Neptune 0.99885), the predictions, and the Mars check for the connect card (1.881² ÷ 1.524³ = 3.538 ÷ 3.540 = 0.9996).

Check yourself · 5 cards · spaced review in the study dashboard

  1. recall

    State Kepler's third law, and the units in which the constant is 1.

    Show answer

    The squares of the periods of the planets are as the cubes of their mean distances from the Sun: T² = a³ when T is in years and a, the semi-major axis, is in AU.

  2. explain

    What does a straight line of slope 3/2 on a log-log plot of period against semi-major axis tell you?

    Show answer

    log T = (3/2) log a + log K, so T = K × a^(3/2) and T² ÷ a³ = K², the same for every point on the line. That is the third law.

  3. apply

    A body orbits the Sun with a semi-major axis of 9 AU. What is its period?

    Show answer

    T = 9^(3/2) = 27 years.

  4. recall

    When did Kepler first think of his third law, and where and when did he publish it?

    Show answer

    He first thought of it on 8 March 1618, rejected it after a faulty calculation and returned to it on 15 May 1618. He published it in Book V of Harmonices Mundi (1619).

  5. connect

    The retrograde-motion figure on this page puts Mars at 1.524 AU from the Sun with a period of 1.881 years. Do those two numbers agree with Kepler's third law?

    Show answer

    Yes. 1.524³ = 3.540 and 1.881² = 3.538, so T² ÷ a³ = 0.9996, within 0.04 per cent of 1. Put the other way, 1.524^(3/2) = 1.881.

From the Library

3 pages build on the Astronomy lab

  1. Eratosthenes Measures the Earth: Shadows, Angles and Stated AssumptionsMathematics, Systems & LanguagesHow Eratosthenes, as Cleomedes reports him, turned a shadow angle and a distance into the size of the Earth, with every assumption listed and the uncertain length of the stade stated plainly.
  2. Models of the Heavens: Ptolemy, Copernicus, Tycho and KeplerMathematics, Systems & LanguagesHow four models of the heavens were built to save the same sky, what each explained, and four ways to judge one model better than another (an interpretation).
  3. The Sky as a Sphere: Days, Seasons and the Phases of the MoonMathematics, Systems & LanguagesThe celestial sphere as a model of the sky, and how it explains the day, the seasons, the phases of the Moon and why eclipses do not happen every month.