# The Sizes and Distances of the Sun and Moon

## Aristarchus of Samos

## Hypotheses

1. That the moon receives its light from the sun.

2. That the earth is in the relation of a point and centre to the sphere in which the moon moves.

3. That, when the moon appears to us halved, the great circle which divides the dark and the bright portions of the moon is in the direction of our eye.

4. That, when the moon appears to us halved, its distance from the sun is then less than a quadrant by one-thirtieth of a quadrant.

5. That the breadth of the [earth’s] shadow is [that] of two moons.

6. That the moon subtends one-fifteenth part of a sign of the zodiac.

We are now in a position to prove the following propositions:

1. *The distance of the sun from the earth is greater than eighteen times, but less then twenty times, the distance of the moon (from the earth)*; this follows from the hypothesis about the halved moon.

2. *The diameter of the sun has the same ratio (as aforesaid) to the diameter of the moon.*

3. *The diameter of the sun has to the diameter of the earth a ratio greater than that which* 19 *has to* 3, *but less than that which* 43 *has to* 6; this follows from the ratio thus discovered between the distances, the hypothesis about the shadow, and the hypothesis that the moon subtends one-fifteenth part of a sign of the zodiac.

## Proposition 7

*The distance of the sun from the earth is greater than eighteen times, but less than twenty times, the distance of the moon from the earth.*

For let A be the centre of the sun, B that of the earth.

Let AB be joined and produced.

Let C be the centre of the moon when halved; let a plane be carried through AB and C, and let the section made by it in the sphere on which the centre of the sun moves be the great circle ADE.

Let AC, CB be joined, and let BC be produced to D.

Then, because the point C is the centre of the moon when halved, the angle ACB will be right.

Let BE be drawn from B at right angles to BA; then the circumference ED will be one-thirtieth of the circumference EDA; for, by hypothesis, when the moon appears to us halved, its distance from the sun is less than a quadrant by one-thirtieth of a quadrant [Hyphothesis 4].

Thus the angle EBC is also one-thirtieth of a right angle.

Let the parallelogram AE be completed, and let BF be joined.

Then the angle FBE will be half a right angle.

Let the angle FBE be bisected by the straight line BG; therefore the angle GBE is one-fourth part of a right angle.

But the angle DBE is also one-thirtieth part of a right angle; therefore the ratio of the angle GBE to the angle DBE is that which 15 has to 2: for, if a right angle be regarded as divided into 60 equal parts, the angle GBE contains 15 of such parts, and the angle DBE contains 2.

Now, since GE has to EH a ratio greater than that which the angle GBE has to angle DBE, therefore GE has to EH a ratio greater than that which 15 has to 2.

Next, since BE is equal to EF, and the angle E is right, therefore the square on FB is double of the square on BE.

But, as the square on FB is to the square on BE, so is the square on FG to the square on GE; therefore the square on FG is double of the square on GE.

Now 49 is less than double of 25, so that the square on FG has to the square on GE a ratio greater than that which 49 has to 25; therefore FG also has to GE a ratio greater than that which 7 has to 5.

Therefore, componendo, FE has to EG a ratio greater than that which 12 has to 5, that is, than that which 36 has to 15.

But it was also proved that GE has to EH a ratio greater than that which 15 has to 2; therefore, ex aequali, FE has to EH a ratio greater than that which 36 has to 2, that is, than that which 18 has to 1; therefore FE is greater than 18 times EH.

And FE is equal to BE; therefore BE is also greater than 18 times EH; therefore BH is much greater than 18 times HE.

But, as BH is to HE, so is AB to BC, because of the similarity of the triangles; therefore AB is also greater than 18 times BC.

And AB is the distance of the sun from the earth, while CB is the distance of the moon from the earth; therefore the distance of the sun from the earth is greater than 18 times the distance of the moon from the earth. Again, I say that it is also less than 20 times that distance.

For let DK be drawn through D parallel to EB, and about the triangle DKB let the circle DKB be described; then DB will be its diameter, because the angle at K is right.

Let BL, the side of a hexagon, be fitted into the circle.

Then, since the angle DBE is 1/_{30}th of a right angle, the angle BDK is also 1/_{30}th of a right angle; therefore the circumference BK is 1/_{60}th of the whole circle.

But BL is also one-sixth part of the whole circle.

Therefore the circumference BL is ten times the circumference BK. And the circumference BL has to the circumference BK a ratio greater than that which the straight line BL has to the straight line BK; therefore the straight line BL is less than ten times the straight line BK.

And BD is double of BL; therefore BD is less than 20 times BK.

But, as BD is to BK, so is AB to BC; therefore AB is also less than 20 times BC.

And AB is the distance of the sun from the earth, while BC is the distance of the moon from the earth; therefore the distance of the sun from the earth is less than 20 times the distance of the moon from the earth.

And it was before proved that it is greater than 18 times that distance.