Approximately 2300 years ago, Aristarchus proposed a method for determining the relative distances of the sun and the moon in relation to the earth. Specifically, he asserted that when the moon is in its half-phase, a right triangle is formed between the Sun, the Moon, and the Earth. (This is quite true.)

From there, he determined that $\angle ESM = 3^{\circ}$, $\angle SEM = 87^{\circ}$, and $\angle SME = 90^{\circ}$ and proceeded to calculate that the Sun is approximately 19 times farther away from the Earth than the Moon is. (Actually, a more precise value for $\angle SEM$ is $89^{\circ}50'$ which would put the Sun approximately 400 times farther away from the Earth than the Moon is.)

But my question is, what forms of trigonometry were available to Aristarchus to solve this problem, given that it was about this time that trigonometric methods were being developed for geometrical and astronomical applications? Also, does anyone know how Aristarchus came to discover that a right triangle exists between the Sun, Moon, and Earth when the moon is in half-phase?

Truly, today, a problem like this might appear in a pre-Calculus book as a homework exercise and is a straightforward computation with a calculator. But I would like to know how an ancient astronomer would have arrived at the answer. Thank you.

  • 1
    $\begingroup$ See the reconstruction of On the Sizes and Distances (Aristarchus). $\endgroup$ – Mauro ALLEGRANZA Jul 31 '19 at 6:48
  • 1
    $\begingroup$ I would phrase the latter question as something about "how did A. come to realize that the illumination of the moon depends on its angle relative to the sun?" , since that's what leads to the premise that half-moon means 90 degrees. $\endgroup$ – Carl Witthoft Jul 31 '19 at 13:21
  • 1
    $\begingroup$ Before I edit -- you did mean that the angle MES is $89^{\circ} 50'$ , right? $\endgroup$ – Carl Witthoft Jul 31 '19 at 13:21
  • 1
    $\begingroup$ He would not have needed to do any trigonometry. A string and three pieces of wood in the ground would have been enough. $3^\circ$ may be his way of being $100\%$ sure it is error-free, i.e., an upper bound. Alternatively, Hipparchos, who lived at the same time period, had a list of chords with which any triangle can be 'solved'. In this case one just needs the ratio of the chords associated with $87^\circ$ and $3^\circ$. $\endgroup$ – Chrystomath Jul 31 '19 at 16:42
  • $\begingroup$ I have revised the naming of the angles so that they are consistent now. Thank you @Carl Witthoft for pointing this out. $\endgroup$ – mlchristians Jul 31 '19 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.