<p>What you need, as Aristarchus, is a frame of reference in which the Earth's shadow is stationary. Otherwise you are trying to deal with a moving thing (the Moon) relative to another moving thing (the shadow cast by the light of the Sun as it orbits the Earth).</p>
<p>In a stationary-shadow frame of reference, the period you need is the time it takes the Moon to orbit from the middle of the Earth's shadow to the middle of the Earth's shadow. That is to say, from new moon to new moon: which is to say, the synodic month.</p>

CC BY-SA 4.0
2020-11-06T09:20:14.467Z
2020-11-06T09:20:14.467Z
Martin Kochanski
<p>Aristarchus famously used a lunar eclipse to determine the diameter and the distance of the Moon, given that a lunar eclipse can last up to 3 hours and the angular diameter of the Moon is 0.5°.</p>
<p>Using this method, the diameter of the Moon is</p>
<p><span class="math-container">$
\frac{(\textrm{earth diameter}) \times (\textrm{moon angular diameter})}{\frac{\textrm{duration eclipse}}{\textrm{orbital period of the Moon}} \times 360}.
$</span></p>
<p>Is it correct to use the orbital period here, rather than the synodic period? Should the orbital period or the synodic period of the Moon or some other period be used in this expression, and why?</p>

CC BY-SA 4.0
2020-11-05T22:08:26.183Z
See answers to [Is there a timekeeping word for the orbit of a moon?](https://astronomy.stackexchange.com/q/21238/7982) It's possible that the best choice is not one of the two you are offering. Also potentially related (but not duplicate): [How many Moon's Topocentric Conjunction occurs per lunar month?](https://astronomy.stackexchange.com/q/33425/7982)
uhoh
<observational-astronomy><the-moon><angular-diameter>
Should the synodic period or the orbital period be used to determine the diameter of the Moon?
usernumber
# Should the synodic period or the orbital period be used to determine the diameter of the Moon?

(Asked by: username_0 on 2020-11-05)

Aristarchus famously used a lunar eclipse to determine the diameter and the distance of the Moon, given that a lunar eclipse can last up to 3 hours and the angular diameter of the Moon is 0.5°.

Using this method, the diameter of the Moon is

$
\frac{(\textrm{earth diameter}) \times (\textrm{moon angular diameter})}{\frac{\textrm{duration eclipse}}{\textrm{orbital period of the Moon}} \times 360}.
$

Is it correct to use the orbital period here, rather than the synodic period? Should the orbital period or the synodic period of the Moon or some other period be used in this expression, and why?

**Tags:** observational-astronomy, the-moon, angular-diameter

> **Comments:**
> **username_1**: See answers to [Is there a timekeeping word for the orbit of a moon?](https://astronomy.stackexchange.com/q/21238/7982) It's possible that the best choice is not one of the two you are offering. Also potentially related (but not duplicate): [How many Moon's Topocentric Conjunction occurs per lunar month?](https://astronomy.stackexchange.com/q/33425/7982)  

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## 1 answers

### Answer by username_2 on 2020-11-06. Score: 6 | Accepted answer

What you need, as Aristarchus, is a frame of reference in which the Earth's shadow is stationary. Otherwise you are trying to deal with a moving thing (the Moon) relative to another moving thing (the shadow cast by the light of the Sun as it orbits the Earth).

In a stationary-shadow frame of reference, the period you need is the time it takes the Moon to orbit from the middle of the Earth's shadow to the middle of the Earth's shadow. That is to say, from new moon to new moon: which is to say, the synodic month.


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