Lesson Plan #14     http://www.phy6.org/Stargaze/Lhipprc2.htm

(8c)   How Distant is the Moon?  

This section describes the calculation by which Aristarchus used the duration of a total lunar eclipse to deduce the distance of the Moon.
(Section (8d), covered in lesson 15, gives a later method, somewhat more difficult)

Part of a high school course on astronomy, Newtonian mechanics and spaceflight
by David P. Stern

This lesson plan supplements: "How Distant is the Moon--1", section #8c: on disk Shipprc2.htm, on the web

"From Stargazers to Starships" home page and index: on disk Sintro.htm, on the web

Goals: The student will

  • Learn about eclipses of the Moon.

  • Learn a simple application of "pre-trigonometry."

  • Discover how Aristarchus, a Greek astronomer around 230 BC, used a simple observation of the eclipse of the Moon, plus clever reasoning, to deduce the distance of the Moon.

  • [Optional Learn the need to be careful in interpreting observations. The shadow of the Earth at the Moon's distance is narrower than the Earth, because the Sun is seen not as a point but as a disk 0.5 degrees wide.
        Because this discussion complicates the presentation, it may be omitted. The matter is treated in more detail, from a different approach, in sect. 9b)

Terms: Lunar eclipse. Also: first contact, last contact

Stories and extras: About Aristarchus, an early Greek astronomer. "Stargazers" comes back to this astronomer in section (9a) in connection with his estimate of the size and distance of the Sun, which apparently led him to propose that the Earth revolved around the Sun, not vice versa, about 1800 years before Copernicus proposed it.

Start the lesson by asking--has anyone seen an eclipse of the Moon?
If so--ask if the Moon was full, or half Moon, or crescent?
If not--ask: would such an eclipse be more frequent with full Moon, half Moon or crescent?

   Of course it must be a full Moon. If Sun-Earth-Moon are lined up, the Sun illuminates the entire side of the Moon facing Earth.

    Why is it much more likely that you can see a total eclipse of the Moon from your location (in a given year, say), than a total eclipse of the Sun?

   About half the lunar eclipses are total, because the shadow of the Earth is much wider than the Moon, and the eclipsed Moon is visible from the entire night side of the Earth.

    Eclipses of the Sun, on the other hand, are usually total only in a small local area, because the Moon is barely large enough to cover the Sun. Unless you are lucky enough to be in that area, you will only see a partial eclipse, or none at all.

   Ancient astronomers knew how lunar eclipses arose--they knew the Earth was round and cast its shadow on the Moon. The Greek astronomer Aristarchus, around 270 BC, found an ingenious way to use such eclipses to estimate the Earth-Moon distance. Here is how. Then proceed with the lesson.

Guiding questions and additional tidbits

-- What is an "eclipse of the Moon?"?

    An eclipse of the Moon is the passage of the Moon through the shadow of the Earth.

--If r is the Earth's radius, how wide is the Earth's shadow, approximately, at the distance of the Moon?
    The shadow's width is approximately 2 r. [Draw on the board.]

-- If we accept the above approximation of 2r, and if t is the duration of a lunar eclipse in which the Moon passes right through the center of the shadow--what is the velocity V at which the Moon moves through the shadow?
    velocity = distance/time,     so V = 2r/t.

   (Actually, not only the Moon moves across the sky, but so does the Sun, because of the Earth's motion around it. Strictly speaking, V is the velocity of the Moon, relative to direction of the Sun.)

-- Using the same notation, if R is the distance to the Moon, express the time T of one full revolution of the Moon around the Earth.

(2 r/ t)= 2πR/T
The two expressions
for V must be equal

r (T/t) = π R

Cancel factors 2 and
    multiply both sides by T.
(T/πt) = R/ r

Divide both sides by πr.
If t = 3 hours, T= 30 days, or 720 hours. R/r ~ 76

[Optional] -- Actually, a correction must be made here: why?
    Using sketches on the board, or on projected transparencies demonstrate: Viewed from the Moon, if the Sun were a point, a point on the Moon enters the shadow of the Earth when the Earth covers the middle of the Sun, and the point comes out when the middle of the Sun is uncovered again. The width of the shadow is then indeed 2r.

        However, the Sun is not a point in the sky like a sky, but fills a circle about 0.5°wide. When a point on the Moon would observe the Earth covering the middle of the Sun, it still receives sunlight from half the Sun. To darken the Sun, all of the Sun must be covered and this only happens when the center of the Sun is already 0.25° behind the edge of the Earth.

    Similarly, the shadow ends when the center of the Sun reaches a point 0.25° short of the edge of the Earth, which is when the first bit of the Sun pokes out past the edge.
    The result is that the effective width of the Earth's shadow is less than 2r--say 2r'. The calculation gives R/r', and because, r' < r, we get something bigger than R/r. The correct ratio is about 60.

-- Who first made this calculation?
    The Greek astronomer Aristarchus of Samos first made the calculation. Aristarchus then used other observations to conclude that the Sun was much larger than Earth, and therefore proposed that the Earth circled the Sun, not the other way around. (That estimate is discussed in Section #9a). His idea was about 1800 years early, and no one took it seriously.

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Author and Curator:   Dr. David P. Stern
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Last updated: 9.13.2004