Note 25 — Chapter II · GREEK ASTRONOMY.

In the figure, which is taken from the De Revolutionibus of Coppernicus (chapter IV., § 85), let D, K, M represent respectively the centres of the sun, earth, and moon, at the time of an eclipse of the moon, and let S Q G, S R E denote the boundaries of the shadow-cone cast by the earth; then Q R, drawn at right angles to the axis of the cone, is the breadth of the shadow at the distance of the moon. We have then at once from similar triangles

G K - Q M : A D - G K :: M K : K D.

Hence if K D = n. M K and ∴ also A D = n. (radius of moon), n being 19 according to Aristarchus,

G K-Q M: n. (radius of moon)-G K :: 1 : n n . (radius of moon) - G K = n G K - n Q M ∴ radius of moon + radius of shadow = (1 + 1∕n) (radius of earth).

By observation the angular radius of the shadow was found to be about 40′ and that of the moon to be 15′, so that

radius of shadow = 8∕3 radius of moon; ∴ radius of moon = 3∕11 (1 + 1∕n) (radius of earth).

But the angular radius of the moon being 15′, its distance is necessarily about 220 times its radius,

and ∴ distance of the moon = 60 (1 + 1∕n) (radius of the earth),

which is roughly Hipparchus’s result, if n be any fairly large number.