Ancient Greeks: Distances to the Moon and Sun
1. Aristarchus of Samos (310-230? BCE) measured EM/ES, the ratio of the earth-moon
distance to the earth-sun distance. His procedure rests on three assumptions that aren’t
quite met:
a. the moon’s orbit about the earth is perfectly circular
b. the moon’s velocity is perfectly uniform
c. the sun is close enough to the earth that the tiny angle  is big enough to be useful.
M (first quarter)
90º
E
90º-
º-2


90º
S
M ’ (third quarter)
 MES =  EM ’S = 90º– because the interior angles of any triangle sum to 180º.
 MEM ’ = 2(90º–º–2
180  2 the short arclength from M ' to M

360
the whole circumference
because the arclength is
proportional to its angle.
The moon appears exactly half full (first and third quarters) when the terminator – the line
dividing the light and dark halves – is a perfectly straight line as viewed from the earth. But
the moon is spherical, and the terminator, being a line upon its surface, must be curved. Thus,
the only way it can appear straight is for us to view it exactly edge on, that is, when angles
SME and SM’E are 90º. As can be seen in the diagram, the two arcs connecting M and M’ are
not equal. If the moon’s speed in orbit is constant, the ratio of the shorter arclength to the
longer equals ratio of the time from third to first quarter, t, to the time from first to third
quarter, t’.
Aristarchus observed that t’ is one day longer than t: t’ = t + 1d.
T  1d
.
2
180  2 t (T  1d) / 2
T  1d
1d
1d
 
  2  180
 180    90  90
 3 .
Thus
360
T
T
T
T
27.5 d
Since the period of the moon’s orbit is T = t’ + t = t + 1d + t = 2t + 1d, t 
1 EM

:
19 ES
Aristarchus measured the sun’s distance as 19 times the moon’s distance.
Finally sin   sin 3 
2. Ptolemy (flourished ~ 140 CE) measured EM/R, the earth-moon distance in terms of R.
The blue circle is the earth; the white circle is the moon.
Z
B
E 


A
M
EM
MB
BE
. EM is the earth-moon


sin  sin  sin 
EM sin 
distance, center-to-center, and BE is the earth’s radius, R. So
.

R
sin 
Now, the observations: Someone at A sees the moon directly overhead, and someone at B
sees the moon not directly overhead (Z) but at angle ZBM.
 = 180º –  ZBM.
 = 180º –  – .
There are several ways to find . If A and B are both on the equator,  is the longitude
difference. If A and B are on the same longitude line,  is the latitude difference.
Alternatively, a single observer at A can note the time, t, it takes the earth’s rotation to carry
him to B. ºt/24 hours. (He would also have to correct for how far the moon itself
moved in its orbit during the rotation from A to B.)
First, the math: The law of sines says that
EM
sin(180  ZBM )
sin(180  ZBM )
.


R
sin(180  [180  ZBM ]   )
sin(ZBM   )
EM
 59 .
Ptolemy got
R
Finally,
3. Eratosthenes of Cyrene (276 – 194 BCE) measured the radius of the earth to be 4000
miles. So Ptolemy’s work gives EM = 59 R = 59×4000 miles = 200,000 miles. And
Aristarchus’s work gives ES = 19 EM = 19×59×4000 miles = 4,000,000 miles.
Reference: George Abell: Exploration of the Universe, Holt, Rinehart and Winston, 1964