7 Definitions

Theodosius Spherics, Part I

Eratosthenes measured the Earth by supposing it a sphere. Before we can study the heavens, we must understand the sphere itself and the circles that lie upon it. The following definitions are drawn from the Spherics of Theodosius of Bithynia.

The Sphere

If a plane cuts a sphere, the section is a circle.

Spherics I.1 Take an orange and cut it with a knife. No matter how you slice — through the middle, near the top, at an angle — the flat face you expose is always a circle. This is not an accident of oranges but a property of all spheres.

Proof

Let a plane cut a sphere whose center is O and radius r. Drop a perpendicular from O to the plane, meeting it at the point N. Let P be any point where the plane meets the surface of the sphere. Then the triangle ONP has a right angle at N.

Since P lies on the sphere, $|OP| = r$. Since N is fixed, $|ON|$ is the same for every choice of P. By the Pythagorean theorem:

$|NP|^2 = |OP|^2 - |ON|^2 = r^2 - |ON|^2$

This quantity is the same for every point P in the section. Therefore every point of the section is at the same distance from N, and the section is a circle with center N. Q.E.D.

Corollary. If the cutting plane passes through O, then $|ON| = 0$ and $|NP| = r$, so the section is a great circle. Otherwise $|NP| < r$, and the section is a small circle.

Great and Small Circles

Great circles on a sphere bisect one another.

Spherics I.6 Two great circles always cross at exactly two points, and each cuts the other into two equal arcs. The equator and any meridian show this: they meet at two opposite points on the equator, and each divides the other in half.

Proof

Let two great circles lie on a sphere with center O. Each great circle lies in a plane passing through O. Two distinct planes through the same point intersect in a straight line. This line passes through O and meets the sphere at two points, call them A and B. Since O is the midpoint of the diameter AB, the chord AB is a diameter of both circles. A diameter bisects any circle. Therefore each great circle bisects the other. Q.E.D.

If a great circle passes through the pole of another circle, it cuts that circle at right angles.

Spherics I.11 Think of a meridian passing through the north pole. It crosses the equator — whose pole is the north pole — at right angles. This is true not only for the equator but for any circle on the sphere: a great circle through its pole always meets it perpendicularly.

Proof

Let the circle c have pole P, and let a great circle through P meet c at the point A. Drop a perpendicular from P to the plane of c, meeting it at N. Since P is equidistant from all points of c, and the perpendicular foot is the unique closest point in the plane to P, the point N must be the center of c (for the center is the unique point equidistant from all points of any circle).

Therefore PN is perpendicular to the plane of c and passes through its center. Moreover, by the same argument applied to the sphere: since O is equidistant from all points of c (they lie on the sphere), the foot of the perpendicular from O to the plane of c is also N, so O, N, and P are collinear. The great circle through P contains the line PN, which is perpendicular to the plane of c. Therefore the plane of the great circle contains the perpendicular to the plane of c at N. The two planes share the point A and the perpendicular direction, so they meet at right angles. Q.E.D.

Parallel Circles

Circles on a sphere whose planes are parallel have the same poles.

Spherics I.13 Slice an apple twice with parallel cuts, and the two circles you expose share the same axis — the line perpendicular to both slices through the center of the apple. The endpoints of that axis are the common poles.

Proof

Let two parallel planes cut a sphere with center O, producing circles c₁ and c₂ with centers N₁ and N₂. The perpendicular from O to one plane is also perpendicular to the other, since the planes are parallel. Therefore O, N₁, and N₂ all lie on the same line, and this line is the axis of both circles. Because the poles are defined as the endpoints of the axis on the surface of the sphere, its endpoints are the common poles. Q.E.D.

Of two parallel circles, the one nearer the center of the sphere is the larger.

Spherics I.15 Among the parallel circles of the sky — the equator, the tropics, the arctic circles — the equator is the largest, because it passes through the center. The farther a parallel circle lies from the center, the smaller it is.

Proof

From Proposition I.1, the radius of a section at distance $|ON|$ from the center is $\sqrt{r^2 - |ON|^2}$. The nearer circle has a smaller distance $|ON|$ and therefore a larger radius. Q.E.D.

Poles and Axes

The line joining the poles of a circle passes through the center of that circle and is perpendicular to its plane.

Spherics I.16 This is the converse of what we have already observed: poles determine an axis, and that axis meets the circle's plane at right angles through its center.

Proof

Let P be a pole of circle c on a sphere with center O. By definition, P is equidistant from every point of c. Let N be the foot of the perpendicular from P to the plane of c. Since P is equidistant from all points of c, and the perpendicular foot is the unique point in the plane nearest to P, the point N must be the center of c (for only the center is equidistant from all points of a circle). Therefore PN passes through the center of c and is perpendicular to its plane. Since this axis also passes through O, the opposite endpoint on the sphere is the other pole. Q.E.D.

If a point on the sphere is equidistant from three points of a circle, it is a pole of that circle.

Spherics I.21 Three points on a circle determine it. If some point on the sphere is the same distance from all three, it must in fact be the same distance from every point of that circle — it is the pole.

Proof

Let P on the sphere be equidistant from three points A, B, C of circle c. The three chords PA, PB, PC are equal. Drop a perpendicular from P to the plane of c, meeting it at N. By the Pythagorean theorem applied to each right triangle, $|NA| = |NB| = |NC|$. A point equidistant from three points of a circle is the center of that circle; therefore N is the center of c, and the line PN is perpendicular to the plane of c through its center. Hence P is a pole of c. Q.E.D.

The Celestial Sphere and Its Poles

The celestial sphere is the sphere, of indefinitely great radius, upon whose inner surface the fixed stars appear situated.

The observer stands at the center. Since the stars are so distant that no motion of the observer on Earth produces a perceptible change in their arrangement, we may treat the Earth as a point at the center of this sphere, for the purposes of observation.

Watch the sky for an hour. The stars move together, rigidly, as though the whole sphere were turning. They rise in the east, climb, and set in the west. But two points in the sky do not move at all. In the northern sky, one star stays very nearly fixed while every other star wheels around it — today that star is Polaris, though the pole wanders slowly among the stars over millennia due to the slow wobble of the Earth's axis. When the pyramids were built, Thuban in the constellation Draco held that position; in the era of Eratosthenes, the pole had drifted to within a few degrees of Kochab in Ursa Minor, and no single star marked it closely. Directly opposite, below the southern horizon for observers in the north, lies another such fixed point.

The celestial poles are the two points on the celestial sphere that remain stationary during the daily revolution of the stars.

By Definition 4 of the previous chapter, these are the poles of the sphere, and the line joining them is the axis about which the sphere appears to rotate.

The celestial axis is the line joining the north and south celestial poles, passing through the observer.

The angle between the celestial pole and the horizon is equal to the observer's latitude. At the north pole of the Earth, Polaris stands overhead; at the equator, it sits on the horizon. Eratosthenes, working in Alexandria at about 31 degrees north, would have seen the pole a third of the way up his northern sky.

The Horizon and Meridian

The horizon is the great circle on the celestial sphere whose plane is perpendicular to the observer's vertical and passes through the observer.

The horizon divides the celestial sphere into a visible hemisphere above and a hidden hemisphere below. Stars cross this circle when they rise and when they set.

The meridian is the great circle passing through the celestial poles and the observer's zenith.

The zenith is the point on the celestial sphere directly overhead. The nadir is the point directly below, hidden by the Earth. The meridian passes through both, and through both celestial poles. It cuts the sky into an eastern half and a western half.

Because the zenith is, by definition, the pole of the horizon, and the meridian is a great circle passing through it, Proposition I.11 guarantees that the meridian cuts the horizon at exactly right angles.

When a star crosses the meridian, it has reached its highest point in the sky. The ancients called this culmination, and it is the moment when a star's altitude is easiest to measure — the moment Eratosthenes chose for his shadow experiment.

By Proposition I.6, the horizon and meridian, being great circles, bisect each other. They cross at the north and south points of the horizon. The east and west points lie on the horizon halfway between these, where the celestial equator (which we shall define next) meets the horizon.

The Equator and the Tropics

The celestial equator is the great circle on the celestial sphere whose poles are the celestial poles.

By Proposition I.11, the celestial equator is perpendicular to every meridian. It is the projection of the Earth's equator onto the sky. A star on the celestial equator always rises due east and sets due west. Furthermore, because the horizon and the celestial equator are both great circles, they bisect each other (by Proposition I.6). Therefore, a star on the equator spends exactly half its time above the horizon, no matter where on Earth the observer stands. At an equatorial location, it will pass directly overhead.

The daily rotation of the celestial sphere carries each star along a circle parallel to the equator. By Proposition I.13, all these parallel circles share the same poles — the celestial poles. By Proposition I.15, the equator is the largest of them, and circles farther from the equator are smaller. A star near the pole traces a small circle; a star on the equator traces the greatest circle of all.

The declination of a star is its angular distance north or south of the celestial equator, measured along the great circle through the star and the celestial poles.

Declination plays the same role on the celestial sphere that latitude plays on the Earth. It is positive for stars north of the equator, negative for stars south of it, and runs from $-90°$ at the south pole to $+90°$ at the north pole. A star on the equator has declination $0°$.

The tropics are two small circles parallel to the equator, marking the farthest points north and south that the Sun reaches during the year.

The Sun does not stay on the equator. Over the course of a year, it drifts northward and then southward again. On the summer solstice — the day Eratosthenes made his measurement — the Sun reaches the tropic of Cancer, at declination $+23\frac{1}{2}°$. On the winter solstice, it reaches the tropic of Capricorn at declination $-23\frac{1}{2}°$. We will shortly see why.

The ever-visible circle is the small circle around the elevated pole that just touches the horizon. Every star within it never sets. The ever-hidden circle is its counterpart around the depressed pole.

The size of these circles depends on the observer's latitude. We will determine their boundaries precisely in the next chapter.

The Ecliptic and the Zodiac

The ecliptic is the great circle on the celestial sphere along which the Sun appears to travel during the course of a year.

Night by night, the Sun's position among the stars shifts slightly eastward. Over a full year, it traces a complete circle — the ecliptic. This circle is tilted with respect to the equator; it crosses the equator at two points and reaches its greatest distance north and south at two others.

The equinoxes are the two points where the ecliptic crosses the celestial equator. The solstices are the two points where the ecliptic is farthest from the equator.

Because the ecliptic and the equator are both great circles, Proposition I.6 dictates that they bisect each other. Therefore, the two equinoxes lie exactly opposite one another, and the Sun spends exactly half the year north of the equator and half the year south of it. When the Sun is at an equinox, it lies on the equator, and day and night are equal everywhere on Earth. When it reaches a solstice, it touches the tropic circles we defined earlier (at $23\frac{1}{2}°$ north or south), and the day is longest (summer) or shortest (winter).

The zodiac is the band on the celestial sphere, extending about $8°$ on either side of the ecliptic, within which the Moon and the planets are always found.

The Moon and the five visible planets — Mercury, Venus, Mars, Jupiter, Saturn — all move near the ecliptic, though not exactly on it. Their paths wobble slightly above and below, but always remain within the zodiac band. The ancients divided this band into twelve equal arcs of $30°$ each, named for the constellations that once occupied them.

The Obliquity of the Ecliptic

The obliquity of the ecliptic is the angle between the planes of the ecliptic and the celestial equator.

This single angle — about $23\frac{1}{2}°$ — governs the seasons. It determines how far north and south the Sun travels, how much the length of day varies through the year, and which stars are circumpolar at a given latitude.

To measure the obliquity, observe the Sun's noon altitude at the summer solstice and again at the winter solstice. At the summer solstice, the Sun culminates at its highest point of the year; at the winter solstice, its lowest. Half the difference between these two altitudes is the obliquity.

Worked Example

1. At Alexandria (latitude $\approx 31°$N), the noon altitude of the Sun at the summer solstice is about $82\frac{1}{2}°$.
2. At the winter solstice, the noon altitude is about $35\frac{1}{2}°$.
3. The difference is $82\frac{1}{2}° - 35\frac{1}{2}° = 47°$.
4. Half the difference: $47° \div 2 = 23\frac{1}{2}°$.

Eratosthenes himself is credited with a measurement of $\frac{11}{83}$ of a full circle for the obliquity, which works out to about $23°51'$ — remarkably close to the modern value of $23°26'$.

Now that the great circles of the sky are defined, we can ask: which stars rise and set, which never set, and which never rise? The answers depend on the observer's latitude and on the geometry of parallel circles established in the previous chapter. This is the subject of Chapter 3.

Stars That Rise and Set

If a star's parallel circle is cut by the horizon, the star rises and sets each day.

Spherics II.1, II.2  |  Phaenomena Prop. 1, 2

Most stars behave this way. The daily rotation carries the star along its parallel circle. When the circle dips below the horizon, the star sets; when it climbs back above, the star rises. The horizon divides the parallel circle into two arcs: a visible arc above and a hidden arc below.

Proof

Let the parallel circle of a star be cut by the plane of the horizon into two arcs. As the celestial sphere rotates, the star traverses its parallel circle uniformly. When the star is on the arc above the horizon, it is visible. When it passes onto the arc below, it disappears. The two crossing points are the star's rising point and setting point. Because the sphere rotates at a constant, uniform speed, the physical length of these arcs directly dictates the amount of time the star spends visible in the sky versus hidden below the Earth. Q.E.D.

The lengths of the two arcs are generally unequal. When the visible arc is longer, the star is above the horizon for more than half the day. When it is shorter, less than half. Only when the horizon bisects the parallel circle are the arcs equal, and the star spends exactly twelve hours above the horizon. We shall see when this occurs.

Stars That Never Set

If a star's parallel circle lies entirely above the horizon, the star never sets. Such a star is called circumpolar.

Spherics II.5  |  Phaenomena Prop. 3

Look at the northern sky on any clear night. The stars nearest the pole — Ursa Minor, Ursa Major, Cassiopeia — wheel around the pole without ever touching the horizon. They are always visible. The ancients watched them trace complete circles, never dipping out of sight.

Proof

The horizon is a great circle. The celestial pole is at an angular altitude above the horizon equal to the observer's latitude $\phi$. A star at angular distance $\delta$ from the equator (its declination) traces a parallel circle at angular distance $90° - \delta$ from the pole.

For this parallel circle to lie entirely above the horizon, every point of it must be above the horizon. The lowest point of the circle is reached when the star crosses the meridian below the pole. At that moment, the star's altitude is:

$\phi - (90° - \delta) = \phi + \delta - 90°$

This altitude is positive — meaning the star remains above the horizon — whenever:

$\delta > 90° - \phi$

That is, whenever the star's declination exceeds the observer's co-latitude (the angular distance from the zenith to the pole, or $90° - \phi$). The boundary is the ever-visible circle — the small circle at angular distance $90° - \phi$ from the pole. Every star within this circle is circumpolar. Q.E.D.

Worked Example

1. At Alexandria, $\phi \approx 31°$N.
2. The co-latitude is $90° - 31° = 59°$.
3. Stars with declination greater than $59°$N never set.
4. Polaris has declination $\approx 89°$N — well within the circumpolar zone.
5. The bright star Capella has declination $\approx 46°$N — it rises and sets at Alexandria, since $46° < 59°$.

Stars That Never Rise

If a star's parallel circle lies entirely below the horizon, the star never rises. It is permanently invisible.

Spherics II.8  |  Phaenomena Prop. 4

Just as there are stars that never set near the visible pole, there are stars that never rise near the hidden pole. For Eratosthenes in Alexandria, the stars of the far south — the Southern Cross, the bright star Canopus at its lowest — grazed or never cleared the southern horizon.

Proof

By the same reasoning as the previous proposition, a star's highest point above the horizon occurs when it crosses the meridian on the south side. The south pole lies at an angle $\phi$ below the horizon. Let the star be at an absolute angular distance $|\delta|$ south of the equator. It therefore traces a parallel circle at angular distance $90° - |\delta|$ from the south pole.

Its altitude at upper culmination (the highest point of its parallel circle) is the distance it reaches above the south pole, minus the angle the south pole is depressed:

$(90° - |\delta|) - \phi = 90° - \phi - |\delta|$

This altitude is negative — meaning the star's highest point remains below the horizon — whenever:

$|\delta| > 90° - \phi$

That is, whenever the star's southern declination exceeds the co-latitude. The boundary is the ever-hidden circle around the depressed pole. Every star within this circle is permanently invisible. Q.E.D.

The symmetry is exact. The zone of permanently visible stars around the visible pole and the zone of permanently hidden stars around the hidden pole are the same angular size. Between them lies the zone of stars that rise and set.

Unequal Day and Night

For a star not on the celestial equator, the visible and hidden arcs of its parallel circle are unequal. The inequality increases with the star's distance from the equator.

Spherics II.13  |  Phaenomena Prop. 7, 8

This is why summer days are long and winter days are short. The Sun, moving along the ecliptic, spends part of the year north of the equator and part south. When it is north, its parallel circle is tipped so that more of it lies above the horizon (for northern observers), and the day is longer. When it is south, less lies above, and the day is shorter.

Proof

The horizon is a great circle. The equator is a great circle. By Proposition I.6, they bisect each other at the true East and West points of the horizon. Therefore the horizon divides the equator into two equal arcs, and any star on the equator spends exactly half the day visible.

Now consider a parallel circle north of the equator, observed from a northern latitude. The celestial pole is elevated above the horizon. Because the pole is elevated, the entire northern half of the sphere is tilted upward. Therefore, any small circle parallel to the equator and north of it must intersect the horizon at points that are north of the true East and West points.

Because the circle crosses the horizon north of the East-West line, the segment of the circle lying above the horizon must be larger than the segment lying below it. The visible arc exceeds the hidden arc.

The farther the parallel circle is from the equator (greater declination), the farther north it intersects the horizon, and the greater the asymmetry, until at the ever-visible circle the hidden arc vanishes entirely and the star never sets. Q.E.D.

Worked Example

At the summer solstice, the Sun has declination $\approx 23\frac{1}{2}°$N. At Alexandria ($\phi \approx 31°$N):

1. The Sun's noon altitude is $90° - 31° + 23\frac{1}{2}° = 82\frac{1}{2}°$ — nearly overhead.
2. The visible arc of its parallel circle spans roughly $14$ hours.
3. At the winter solstice ($\delta \approx 23\frac{1}{2}°$S), the noon altitude is $90° - 31° - 23\frac{1}{2}° = 35\frac{1}{2}°$.
4. The visible arc spans roughly $10$ hours.
5. At the equinox ($\delta = 0°$), the day is exactly $12$ hours.

These are the same solstice altitudes we used in Section 2.5 to measure the obliquity. The same geometry that explains the tilt of the ecliptic also explains the inequality of the seasons.

The Equinox

When the Sun is at an equinox, day and night are equal for every observer on Earth.

Phaenomena Prop. 9

Twice a year the Sun crosses the celestial equator — once heading north (the vernal equinox), once heading south (the autumnal equinox). On these days, the Sun behaves as a star on the equator.

Proof

When the Sun lies on the celestial equator, its parallel circle for that day is the equator itself. The horizon, being a great circle, bisects the equator (by Proposition I.6). Therefore half the equator lies above the horizon and half below. The Sun traverses each half in equal time. Day equals night. Q.E.D.

The word equinox means precisely this: aequus nox, equal night.

Moreover, at the equinox the Sun rises due east and sets due west. This is because the equator, being perpendicular to the celestial axis (by Definition 5 and Proposition I.11), crosses the horizon at the points that are $90°$ from the celestial poles — the east and west points.

With the geometry of risings and settings complete, we have exhausted what can be learned from the fixed stars and the daily rotation alone. To proceed further, we must study the bodies that move against the starry background: the Sun and the Moon. This is the subject of Part II.

The Moon as an Illuminated Sphere

The Moon is a sphere that shines by reflecting the light of the Sun.

The Moon produces no light of its own. This was understood early: Anaxagoras taught it, and Parmenides before him likely knew it. The evidence is direct. The bright portion of the Moon always faces the Sun. When the Moon is in the east at sunset, the bright side faces west, toward the Sun on the horizon. When the Moon is high at midnight, the bright side faces downward, toward the Sun below the Earth. The bright face always points Sunward.

Furthermore, the boundary between the bright and dark portions of the Moon is always a curve — sometimes a circular arc, sometimes nearly a straight line, but never irregular. This is precisely what we would expect from a sphere lit from one side. A sphere always presents a circular outline, and the shadow boundary on a sphere (called the terminator) is always an ellipse as seen from a distance — a circle seen obliquely.

Because the Sun illuminates exactly half of the sphere, this boundary is a great circle (as defined in Chapter 1). When we view a great circle from an angle, perspective compresses it into an ellipse.

Crescent, Quarter, Gibbous, Full

The phases of the Moon arise from the changing angle between the Sun, the Moon, and the observer.

Half the Moon is always illuminated — the half facing the Sun. But how much of that illuminated half we can see depends on where we are looking from. As the Moon orbits the Earth, we see the illuminated half from different angles.

When the Moon is roughly between the Earth and the Sun, the illuminated half faces away from us. We see little or no light: the new Moon.

A few days later, the Moon has moved eastward. We now see a thin sliver of the lit side — a crescent, with the horns pointing away from the Sun.

When the Moon has traveled a quarter of its orbit, we see exactly half of the illuminated side. The terminator appears as a straight line across the disk (because we are viewing its great circle exactly edge-on): the first quarter.

Past the quarter, we see more than half the lit side — a gibbous Moon, swelling toward full.

When the Earth lies roughly between the Moon and the Sun, we look straight at the illuminated half: the full Moon. It rises at sunset and is visible all night.

The cycle then reverses — gibbous, third quarter, crescent — until the Moon returns to conjunction with the Sun and the phases begin again. The full cycle, from new Moon to new Moon, takes about $29\frac{1}{2}$ days. This is the synodic month.

The Half-Moon and the Right Angle

At the moment of half Moon, the angle at the Moon in the triangle Sun–Moon–Earth is a right angle.

Aristarchus Prop. 1–2

This observation is the key to the entire method of Aristarchus. It seems simple — almost too simple — but from this one geometric fact, the relative distances and sizes of the Sun and Moon can be deduced.

Proof

At the moment we call half Moon, the terminator — the boundary between the bright and dark hemispheres — passes through the center of the visible disk. The terminator is the circle on the Moon's surface where the Sun's rays are tangent. At any point on this circle, the line from that point toward the Sun is perpendicular to the Moon's surface.

When we see the terminator as a straight line bisecting the disk, we are looking at it edge-on. This means our line of sight to the Moon lies in the plane of the terminator. But the line from the Moon to the Sun is perpendicular to this plane (since the terminator is the set of points where sunlight strikes tangentially).

Therefore, at the moment of half Moon, the line Moon–Earth and the line Moon–Sun are perpendicular:

$\angle \text{Sun–Moon–Earth} = 90°$

The triangle formed by the Sun, Moon, and Earth is a right triangle, with the right angle at the Moon. Q.E.D.

This is the foundation of everything that follows in this chapter. With a right triangle whose vertices are the three most prominent bodies in the sky, we can use the geometry learned in Part I of our geometry course to compare their distances — if we can measure one more angle.

The Lunar Eclipse and the Earth's Shadow

A lunar eclipse occurs when the Moon passes through the shadow of the Earth.

The Earth, like any opaque sphere illuminated by the Sun, casts a shadow into space on the side opposite the Sun. Because the Sun is larger than the Earth, this shadow (the umbra) is shaped like a long, narrowing cone. Most of the time the Moon's orbit carries it above or below this shadow, and the full Moon shines unobstructed. But occasionally the alignment is exact, and the Moon enters the shadow.

During a lunar eclipse, the Moon does not vanish entirely. It turns a deep red, because the Earth's atmosphere bends some sunlight into the shadow — but the ancients could observe clearly that the bright disk was being covered by something. What they saw was the circular edge of the Earth's shadow creeping across the Moon.

Aristotle noted that the shadow cast by the Earth on the Moon is always circular, regardless of which part of the Earth faces the Moon at the time. The only solid that casts a circular shadow from every direction is a sphere. This was one of the ancient proofs that the Earth is round.

A lunar eclipse can only occur at full Moon, when the Moon is directly opposite the Sun. It is visible from everywhere on the night side of the Earth — everyone who can see the Moon sees the same eclipse at the same time.

The Solar Eclipse

A solar eclipse occurs when the Moon passes between the Earth and the Sun, blocking the Sun's light.

A solar eclipse can only occur at new Moon, when the Moon is roughly between the Earth and the Sun. If the alignment is exact, the Moon's disk covers the Sun's disk, and the Sun's light is blocked.

Unlike a lunar eclipse, a solar eclipse is not seen from the entire day side of the Earth. The Moon's shadow is narrow — a cone that tapers to a point — and only observers within this shadow see a total eclipse. Others nearby see a partial eclipse, and most of the Earth sees nothing at all. The shadow sweeps rapidly across the Earth's surface as the Moon moves in its orbit.

That the Moon can cover the Sun at all tells us something important: the Moon and the Sun appear to be nearly the same angular size as seen from Earth — about half a degree. Since the Sun is far more distant (as we shall prove), it must be far larger than the Moon. The near-equality of apparent size is a coincidence of distance and proportion.

The Shape of the Shadow

The shadow of the Earth tapers to a cone. At the distance of the Moon, the cross-section of this shadow is larger than the Moon.

Aristarchus Hyp. 5

When a sphere is illuminated by a larger sphere, the shadow cast is a cone that converges to a point beyond the smaller sphere. The Sun is larger than the Earth (a fact Aristarchus will prove), so the Earth's shadow is a tapering cone.

Aristarchus observed that during a lunar eclipse, the Moon takes appreciable time to cross the shadow. From the duration and the Moon's known angular speed, the ancients estimated that the shadow at the Moon's distance is about two to three times the Moon's own diameter. Aristarchus took the shadow to be twice the Moon's diameter — his Hypothesis 5.

This ratio constrains the geometry. The shadow cone's width at the Moon's distance depends on three things: the size of the Sun, the size of the Earth, and the distance from the Earth to the Moon. With one more measurement — the angular size of the Sun and Moon — the system of proportions can be solved. Aristarchus had all the pieces.

The Method of Aristarchus

At the moment of half Moon, the Sun's distance from the Earth is greater than 18 but less than 20 times the Moon's distance from the Earth.

Aristarchus Prop. 3–4

In Section 4.3 we proved that at half Moon, the triangle Sun–Moon–Earth has a right angle at the Moon. Call the angle at the Earth — the angle between the lines of sight to the Moon and to the Sun — $\alpha$.

Aristarchus attempted to measure this angle and reported it as $87°$ — that is, $3°$ less than a right angle. (The true value is about $89°50'$, or $10'$ less than a right angle — an extraordinarily difficult measurement to make with the naked eye.)

Proof

In the right triangle Sun–Moon–Earth, with the right angle at the Moon:

$\cos \alpha = \dfrac{\text{distance to Moon}}{\text{distance to Sun}}$

Therefore:

$\dfrac{\text{distance to Sun}}{\text{distance to Moon}} = \dfrac{1}{\cos \alpha}$

Aristarchus did not have the modern cosine function. Instead, he bounded the ratio using the geometry of the triangle directly. He showed:

$18 < \dfrac{1}{\cos 87°} < 20$

by constructing comparison triangles and using inequalities about the ratios of angles to sides. His geometric argument relies on a lemma: if two angles are unequal, the ratio of the larger angle to the smaller is less than the ratio of the tangent of the larger to the tangent of the smaller.

The conclusion: the Sun is between 18 and 20 times farther than the Moon. Q.E.D.

With the true angle of $89°50'$, the same method gives a ratio of about 390 — the Sun is roughly 390 times farther than the Moon. The method is sound; only the measurement was imprecise. A fraction of a degree separates Aristarchus's estimate from the truth, but even his conservative result proved the Sun was much larger than anyone had previously imagined.

The Ratio of Distances

The distance of the Sun from the Earth is greater than 18 but less than 20 times the distance of the Moon from the Earth.

Aristarchus Prop. 7

Proposition 7 refines and confirms the result of Propositions 3–4 by a second approach, using the observed angular sizes of the Sun and Moon together with the eclipse geometry.

The Sun and Moon subtend nearly the same angle in the sky — about $\frac{1}{2}°$. This is not a coincidence we can explain, but it is a fact we can use. If two objects appear the same angular size, their actual sizes are proportional to their distances: the larger object must be farther away by the same factor that it is larger.

$\dfrac{\text{diameter of Sun}}{\text{diameter of Moon}} = \dfrac{\text{distance to Sun}}{\text{distance to Moon}}$

This means that whatever the ratio of distances turns out to be, it is also the ratio of diameters. The Sun is not merely farther than the Moon — it is larger by the same factor. If the Sun is 19 times farther than the Moon, it must also be 19 times larger in diameter.

The Relative Sizes

The diameter of the Sun is to the diameter of the Moon as the Sun's distance is to the Moon's distance — between 18 and 20 to one.

Aristarchus Prop. 9, 10

Aristarchus now brings in the eclipse observations. From Section 5.3, the Earth's shadow at the Moon's distance is about twice the Moon's diameter. This constrains the size of the Earth relative to the Moon.

Argument

The shadow cone tapers from the Earth's diameter to a width of two Moon-diameters at the Moon's distance. The rate of taper is dictated by the angular size of the Sun. Using the distance ratio (between 18 and 20) and the shadow ratio (about 2), Aristarchus establishes by a chain of proportions:

$\dfrac{\text{diameter of Earth}}{\text{diameter of Moon}}$ is between $\dfrac{108}{43}$ and $\dfrac{60}{19}$

— roughly between $2\frac{1}{2}$ and $3\frac{1}{6}$, or about 3 to 1. The Earth is about three times the Moon's diameter.

The modern value is 3.67 — Aristarchus was close, despite his distance ratio being far too small, because the shadow measurement partially compensates for the distance error.

The Sun is Much Larger Than the Earth

The diameter of the Sun is to the diameter of the Earth as a quantity greater than 19 to 3 but less than 43 to 6.

Aristarchus Prop. 15

This is the culminating result of the treatise. Combining the distance ratio (Prop. 7) with the Earth-to-Moon size ratio (Prop. 9–10), Aristarchus can now compare the Sun to the Earth directly.

Argument

The Sun is between 18 and 20 times the Moon's diameter (from the equal apparent size and the distance ratio). The Earth is between $\frac{108}{43}$ and $\frac{60}{19}$ times the Moon's diameter. Dividing:

$\dfrac{\text{diameter of Sun}}{\text{diameter of Earth}}$ is between $\dfrac{19}{3}$ and $\dfrac{43}{6}$

— roughly between $6\frac{1}{3}$ and $7\frac{1}{6}$. The Sun's diameter is more than six times the Earth's.

In volume, a sphere scales as the cube of the diameter ($V \propto d^3$). A body $6\frac{1}{3}$ times wider contains more than $250$ times the volume. The Sun is enormously larger than the Earth.

This result, more than any other, may have led Aristarchus to his heliocentric hypothesis. If the Sun is vastly larger than the Earth, it is natural to suppose that the smaller body revolves around the larger, not the reverse. We take up this hypothesis in Part III.

Definitions

1. The Cosmos

The cosmos is the whole containing the Sun, the Earth, the Moon, the planets, and the fixed stars.

2. The Celestial Sphere

The celestial sphere is the sphere of immense radius upon which the fixed stars appear situated, as defined in Chapter 2.

3. The Ecliptic

The ecliptic is the great circle on the celestial sphere traced by the apparent annual motion of the Sun among the stars, as defined in Section 2.4.

4. The Zodiac

The zodiac is the band surrounding the ecliptic, within which the Moon and planets are always found, as defined in Section 2.4.

5. Retrograde Motion

Retrograde motion is the apparent reversal of a planet's usual eastward drift among the fixed stars. For a time the planet appears to move westward, before resuming its eastward course.

All five visible planets exhibit retrograde motion at regular intervals. In any geocentric system, this reversal must be explained by adding secondary circles (epicycles) to each planet's motion. In the Aristarchan system, as we shall show, it arises naturally from the geometry of combined orbits.

Hypotheses

Archimedes Sand Reckoner I.4–7

The following hypotheses are attributed to Aristarchus. They replace the ordinary assumption that the Earth is stationary at the center of the cosmos.

Hypothesis 1

The Sun remains at rest near the middle of the cosmos.

Hypothesis 2

The Earth moves in a circle about the Sun, completing one revolution each year.

Hypothesis 3

The Earth rotates about its own axis, completing one rotation each day.

Hypothesis 4

The sphere of the fixed stars is so distant that the Earth's orbit bears no perceptible ratio to it.

Hypothesis 5

The planets also move in circles about the Sun, each with its own period.

Hypotheses 1 and 2 are motivated by the result of Part II: the Sun is vastly larger than the Earth. Hypothesis 3 replaces the daily rotation of the entire cosmos with the daily rotation of the Earth alone — a far simpler motion for a smaller body. Hypothesis 4 explains why the stars show no observable shift (parallax) as the Earth moves. Hypothesis 5 extends the system to include all the wandering stars.

We now show that these hypotheses, together with the geometry of Books I and II, account for every major appearance in the sky.

The Rotation of the Earth

The daily rising and setting of the stars is caused by the rotation of the Earth, not by the revolution of the celestial sphere.

Proposition I

Proof

In Part I we established that the stars appear to move uniformly from east to west, completing a full revolution in one day. Two explanations are consistent with this appearance:

(a) The celestial sphere rotates from east to west about the observer, carrying all the stars with it.

(b) The Earth rotates from west to east, while the stars remain fixed.

In both cases, the observer's horizon sweeps across the celestial sphere at the same rate, producing the same risings, settings, and culminations. Every result proved in Part I — the circumpolar stars, the inequality of day and night, the equinox — holds identically under either hypothesis.

But consider the alternative. Under (a), the entire cosmos — including the immensely distant stars — must complete a full revolution every day. A star near the celestial equator, at the distance Aristarchus attributes to the fixed stars, would have to travel a circle of inconceivable circumference in twenty-four hours. Under (b), only the Earth must rotate, and its circumference is known — Eratosthenes measured it.

Since both hypotheses produce the same appearances, and (b) requires only the small body to move rather than the entire cosmos, Hypothesis 3 accounts for the daily motion. Q.E.D.

All the propositions of Chapters 1 through 3 remain valid. We proved them in terms of the apparent motion of the celestial sphere. Whether the sphere moves or the Earth rotates beneath it, the geometric relationships between the circles — equator, horizon, meridian, ecliptic — are unchanged. The geometry is the same; only the cause of the motion is reassigned.

The Annual Revolution

The apparent annual motion of the Sun through the zodiac is caused by the revolution of the Earth around the Sun.

Proposition II

Proof

In Section 2.4 we observed that the Sun appears to move eastward along the ecliptic, completing one full circuit of the zodiac in a year. Under the geocentric view, the Sun orbits the Earth. Under Hypothesis 2, the Earth orbits the Sun.

Let the Sun be at rest and the Earth move in a circle around it. As the Earth moves, the direction from the Earth to the Sun changes continuously. Projected onto the distant celestial sphere, the Sun appears to shift its position among the stars.

Because the Earth and Sun are always on opposite sides of the orbit, the Sun always appears in the constellation of the zodiac that is exactly opposite the Earth’s current position. After the Earth has completed half its orbit, the Sun appears on the opposite side of the sky from where it appeared six months earlier. After a full orbit, the Sun has appeared to travel through every part of the zodiac and returned to its starting position.

The ecliptic, then, is not the Sun's orbit around the Earth. It is the projection of the Earth's orbit onto the celestial sphere. The tilt of the ecliptic with respect to the equator (the obliquity, measured in Section 2.5) arises because the Earth's axis of rotation is tilted with respect to the plane of its orbit around the Sun.

This tilt also explains the seasons. When the north pole tilts toward the Sun, the Sun appears north of the equator, days are long, and it is summer. When the north pole tilts away, the Sun appears south, days are short, and it is winter. The obliquity does not change from year to year — the axis points in a nearly fixed direction among the stars — so the cycle of seasons repeats. Q.E.D.

If the Earth Moves, Why No Parallax?

If the Earth moves in a circle about the Sun, the direction from the Earth to a fixed star must change during the year. This change is called stellar parallax.

Proposition III

This is the most serious objection to the heliocentric hypothesis, and the ancients knew it. If the Earth is here in January and there in July, on opposite sides of an orbit some 300 million stadia across, then a nearby star should appear to shift its position against the background of more distant stars — just as a nearby tree appears to shift against distant mountains when you walk along a road.

Proof

Let S be the Sun, and let the Earth occupy positions E₁ and E₂ at opposite ends of a diameter of its orbit. Let F be a fixed star. The directions E₁F and E₂F differ by the angle $p$ subtended at the star by the diameter of the Earth's orbit:

$p = 2 \arctan\!\left(\dfrac{r}{d}\right)$

where $r$ is the radius of the Earth's orbit and $d$ is the distance to the star. This angle $p$ is the annual parallax. Because the distance $d$ is so large, the angle $p$ is extremely small, often described by the approximation $p \approx \frac{2r}{d}$. If it is large enough to observe, the star will appear to trace a small circle (or ellipse) on the celestial sphere over the course of a year. Q.E.D.

No such shift was observed in antiquity. This does not refute the hypothesis — it constrains it. The parallax is inversely proportional to the star's distance: the farther the star, the smaller the shift. If no shift can be detected, the stars must be extremely far away. Aristarchus’s Hypothesis 4 is a geometric necessity: the cosmos must be vastly larger than the orbit of the Earth.

The Immensity of the Stellar Sphere

The sphere of the fixed stars is so distant that the Earth's orbit is as a point in comparison.

Archimedes Sand Reckoner I.6

Archimedes records that Aristarchus's hypothesis requires the stellar sphere to be incomparably larger than the Earth's orbit. Archimedes interprets this carefully:

It is clear that he means this: the ratio of the Earth's orbit to the distance of the fixed stars is the same as the ratio of the center of a sphere to its surface — that is, the ratio of a point to a finite quantity, which is no ratio at all.

This is a mathematical way of saying: the stars are so far away that the Earth's orbit, vast as it is, is negligible in comparison. The parallax angle $p$ is not zero, but it is too small to measure.

Aristarchus was right. Stellar parallax was not detected until 1838, when Friedrich Bessel measured the parallax of the star 61 Cygni at about $\frac{1}{3}$ of an arcsecond — a shift roughly 200 times smaller than what the naked eye can resolve.

The eye's angular resolution is about 1 arcminute (or 60 arcseconds). The inability to detect parallax was, for eighteen centuries, the strongest argument against the heliocentric system. It was not a flaw in the argument but a limitation of the instrument — the unaided human eye. The geometry of Aristarchus demanded a cosmos far larger than anyone was prepared to accept.

Inferior Planets Remain Near the Sun

Mercury and Venus are never seen far from the Sun because their orbits lie within the orbit of the Earth.

Proposition IV

Proof

Let the Sun be at the center and the Earth move on a circle of radius $R$. Let Mercury or Venus move on a smaller circle of radius $r$, with $r < R$, also centered on the Sun.

The greatest angular separation between the planet and the Sun, as seen from the Earth (called maximum elongation), occurs when the line from the Earth to the planet is tangent to the planet's orbit. At that moment, the triangle Sun–planet–Earth has a right angle at the planet (since a tangent is perpendicular to the radius), and the angular separation is:

$\alpha_{\max} = \arcsin\!\left(\dfrac{r}{R}\right)$

Since $r < R$, this angle is always less than $90°$. The planet can never appear on the opposite side of the sky from the Sun. It oscillates east and west of the Sun, appearing as an "evening star" after sunset or a "morning star" before sunrise, but never straying far. Q.E.D.

Worked Example

1. Venus's greatest elongation from the Sun is observed to be about $46°$.
2. Therefore $\sin 46° = \frac{r}{R} \approx 0.72$.
3. Venus's orbit has a radius about $72\%$ that of the Earth's orbit.
4. Mercury's greatest elongation is about $23°$, giving $\sin 23° \approx 0.39$.
5. Mercury's orbit has a radius about $39\%$ that of the Earth's orbit.

From a single angular observation, the heliocentric model yields the relative size of each planet's orbit. No geocentric model can do this; in that system, the planets' distances are arbitrary assumptions rather than geometric necessities.

Retrograde Motion of an Outer Planet

An outer planet appears to reverse its motion among the stars when the Earth, moving faster, passes between the planet and the Sun.

Proposition V

Proof

Let Mars move on a circle about the Sun larger than the Earth's orbit. The Earth, being nearer the Sun, completes its orbit in less time and therefore moves faster along its path.

When the Earth and Mars are on the same side of the Sun but far apart, both appear to move eastward against the stars. The Earth moves faster, but the changing geometry has not yet produced a reversal.

As the Earth approaches Mars, the line from Earth to Mars begins to swing. Consider the direction from Earth to Mars, projected onto the celestial sphere. When the Earth is well behind Mars, Mars appears ahead — to the east. As the Earth draws alongside, Mars appears to slow. When the Earth passes Mars — at the moment called opposition — the line from Earth to Mars sweeps backward. Mars appears to move westward: retrograde.

After the Earth has moved well past, the geometry reverses. The line from Earth to Mars resumes its eastward drift, and Mars returns to its normal motion.

The sequence is:

1. Direct (eastward) motion — the planet drifts slowly through the zodiac.
2. First stationary point — the planet appears to stop.
3. Retrograde (westward) motion — lasting weeks or months.
4. Second stationary point — the planet appears to stop again.
5. Direct motion resumes.

This entire sequence arises from the Earth overtaking the planet. No reversal of the planet's actual motion occurs. The planet moves steadily forward in its orbit at all times. Q.E.D.

The duration and extent of retrograde motion depend on the planet's distance. Mars, nearest of the outer planets, shows the largest retrograde arc (about $16°$). Jupiter's arc is smaller (about $10°$) and Saturn's smaller still (about $7°$), because the more distant the planet, the less the Earth's motion affects the line of sight.

Retrograde Motion of an Inner Planet

An inner planet appears to reverse its motion among the stars when it passes between the Earth and the Sun.

Proposition VI

Proof

Let Venus move on a circle about the Sun inside the Earth's orbit. Venus completes its orbit faster than the Earth. Most of the time, Venus appears to move eastward through the zodiac.

When Venus approaches the near side of its orbit — the region between the Earth and the Sun — its orbital velocity is directed in such a way that it overtakes the Earth's line of sight. This causes the direction from Earth to Venus to sweep rapidly westward. Against the background of distant stars, Venus appears to reverse its motion.

The retrograde occurs near inferior conjunction, when Venus is closest to Earth. At this moment Venus is passing between the Earth and the Sun, so it is often lost in the Sun's twilight and the retrograde is partly obscured. It emerges on the other side of the Sun — switching from "evening star" to "morning star" — having traced its retrograde loop.

As with the outer planets, Venus never actually reverses its orbital direction. The apparent reversal is a consequence of the combined motions of the Earth and Venus. Q.E.D.

The Order of the Planets

The order of the planets from the Sun can be determined from their observed motions.

Proposition VII

Argument

The heliocentric hypothesis, combined with observation, determines the order of the planets by two criteria:

Inferior planets are identified by their confinement near the Sun. Mercury, with a greatest elongation of $23°$, has the smaller orbit. Venus, with $46°$, has the larger. Both lie inside the Earth's orbit.

Superior planets are identified by their synodic periods — the time between successive oppositions. A planet farther from the Sun moves more slowly and is overtaken by the Earth more frequently relative to its own period. From the observed synodic periods (the time between alignments), the sidereal periods (true orbital periods relative to the stars) can be computed. A longer sidereal period means a larger orbit.

The observed sidereal periods are approximately:

The order, from nearest the Sun to farthest, is therefore:

1. Mercury
2. Venus
3. Earth (with the Moon)
4. Mars
5. Jupiter
6. Saturn

This order was known to the ancients from observation. What the heliocentric hypothesis adds is an explanation: proximity to the Sun determines orbital speed, and orbital speed determines the observed phenomena — elongations, synodic periods, retrograde arcs — from which the order is deduced. The system is self-consistent.

Concluding Theorem

The appearances of the heavens may be explained if the Earth rotates daily and revolves annually about a stationary Sun, while the planets move in circles around the Sun and the stars lie at immense distances.

We have shown:

1. The daily motion of the stars arises from the rotation of the Earth (Proposition I).
2. The annual motion of the Sun arises from the revolution of the Earth (Proposition II).
3. The absence of stellar parallax is consistent with the immense distance of the stars (Proposition III).
4. The confinement of Mercury and Venus near the Sun arises from their smaller orbits (Proposition IV).
5. The retrograde motion of the outer planets arises from the Earth overtaking them (Proposition V).
6. The retrograde motion of the inner planets arises from their overtaking the Earth (Proposition VI).
7. The order and periods of the planets follow from their distances from the Sun (Proposition VII).

Moreover, the heliocentric model does what no geocentric model can: it determines the relative distances of the planets from a single type of observation (greatest elongation for inferior planets, synodic period for superior planets), without arbitrary assumptions.

The system rests on five hypotheses, the geometry of the sphere (Part I), and the measured sizes and distances of the Sun and Moon (Part II). From Eratosthenes measuring the Earth with a shadow, through Aristarchus measuring the cosmos with a right triangle, to the reorganization of the heavens around the Sun — the progression is one of geometry, applied with increasing ambition to the world above us.