Lesson Plan #15     http://www.phy6.org/Stargaze/Lhipparc.htm

(8d)  How Distant is the Moon?--2  

  This lesson is optional. It presents one of two methods by which ancient Greek astronomers estimated the distance to the Moon. The other method, earlier and simpler, is covered in section (8c) and in the preceding lesson plan. The section covered here was written in connection with the total solar eclipse of 11 August 1999. Strictly speaking it requires using a cosine to determine the effect of the Sun not being directly overhead, though the unit also allows a rough estimate to be substituted.

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?--2," section #8d: on disk Shipparc.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 total eclipses of the Sun and their limited regions of totality.

  • Understand that this limited view occurs because the Moon is close enough to us for different points on Earth to view it differently.

  • Learn about the way Hipparchus used a solar eclipse to estimate the distance to the Moon--and get some appreciation for the ingenuity of early astronomers. The method uses "pre-trigonometry" as described in the section on the parallax.

  • Use simple trigonometry--or a crude estimate--to correct for the fact that the eclipsed Sun was not directly overhead, but at some angle to the zenith.

  • As a working example, apply the method of Hipparchus to the solar eclipse of August 11, 1999. Note: The teacher may here use the shortcut to the solution described at the end of this lesson plan.

Terms: Solar eclipse (total and partial) (terms umbra and penumbra may be used by the teacher for total and partial shadow), baseline, parallax.

;Start by reviewing eclipses.
   A lunar eclipse enabled the ancient Greeks to estimate how far the Moon was--over 2000 years ago. Such an eclipse tends to be long, and is often total, because the Earth is big and its shadow is much larger than the Moon, easily able to cover the Moon entirely, for hours.

   A solar eclipse occurs when the Moon's shadow falls on the Earth, and is completely different. Because the Moon is small, its shadow will only cover part of the Earth.

   You might think the shadow is as big as the Moon, but it is actually much smaller. If the Sun were a point-like object, the Moon's shadow would indeed have the size of the Moon. But actually, the Sun covers in the sky a disk of 0.5°, about as big as the Moon. As a result, only in a small region on Earth, maybe 100-150 km across, is it completely covered by the Moon. The Moon is close enough to Earth that if we move only a few hundred kilometers, our line of sight changes sufficiently to uncover some of the Sun, so that we only see a partial eclipse.

   The Earth-Sun distance and Earth-Moon distance vary due to the elliptical orbits of the Earth and the Moon. Sometimes the Moon is too far to cover the Sun, and we do not get any totality. At such eclipses, even in the center of the shadow zone, a "ring of fire" is seen around the dark Moon, which is not quite big enough to cover the entire Sun.

       [Another way of looking at this--but perhaps too confusing to the class, it to note that the shadow forms a cone. Near the Moon, the Moon covers half the sky, and its shadow is as wide as the Moon. A bit further away, the Moon covers only part of the sky, and the shadow-cone is narrower. By the time we reach Earth, the Moon is barely big enough to cover the Sun--we are near the tip of the cone and get a very small shadow. Still further, past the tip of the cone, the Moon appears smaller than the Sun and we don't see a total shadow anywhere.
        This point is discussed in more detail in section 9c.]

   The fact that moving a distance as small as 100-150 km makes the difference between a total and a partial eclipse suggests that the Moon cannot be too distant.

   Such a move essentially shifts the angle at which we observe the Moon. It only takes 100-150 km to move the Moon by a noticeable fraction of the width of the Sun's disk, so that part of that disk, covered during totality, is now exposed. Hipparchus, about 140 years after Aristarchus, realized that this too could tell us how far the Moon was, and would therefore give a way of checking what Aristarchus had claimed.

   Let us try it here:

Go over section #8b of "Stargazers", without the mention of the 1999 eclipse, which should be discussed separately later, perhaps with the shortcut mentioned earlier. The questions below are for both lesson and review:

Guiding questions and additional tidbits

-- The Greek astronomer Hipparchus used an eclipse of the Sun to estimate the distance of the Moon. When did that eclipse happen?

    The eclipse probably occurred during the year 129 BC.

-- What were the observations Hipparchus used?
    He used the fact that the eclipse was total at the Hellespont, the sea passage leading from the Mediterranean to the Black Sea. However, only 80% of the Sun was covered in Alexandria, the main seaport and at the time, the capital of Egypt. [The calculation below is presented on the board as its points are discussed.]

-- Suppose for a moment that the eclipse happened when the Sun was right overhead. The event gives us a tall narrow triangle whose side is the approximate distance to the Moon. What would be the "baseline"--the base of that triangle?

    The baseline would be the distance from Alexandria to the Hellespont.

-- If the Hellespont is exactly north of Alexandria, and its latitude is 9° more--in terms of the radius r of the Earth, what is the baseline?

    360 degrees around the Earth are equal to 2πr= 6.28 r

    So the distance covered by 9° is (9/360) 6.28 r = (6.28/40) r = 0.157r

-- What is then the angle at the top of the triangle?

       From the Hellespont, at the peak of the eclipse, the direction to the edge of the Moon overlaps the edge of the Sun.
       From Alexandria, the same edge falls short of the edge of the Sun by 20% of the disk of the Sun. The disk covers 0.5°, so the edge of the Moon is shifted by 0.1°.

-- Suppose the Moon's distance is R. The baseline covers 0.1° in a circle of radius R, centered on the Moon. What is the length of the baseline, in terms of R?

    360° around the Moon are equal to 2πR = 6.28 R

    So 0.1° equals (0.1/360) 6.2 R - (6.28/3600) R = 0.0017444 R

--So what is the ratio R/r?
    We have     0.0017444 R = 0.157 r    so     R/r = 90

-- If the Sun is not straight overhead but (say) overhead at the equator, how does this change the calculation?
    The baseline is no longer the distance between the Hellespont and Alexandria, but the projection of that distance on the direction perpendicular to line to the Moon, i.e. to a line paralleling the equator--say, 0.75 or 2/3 of the baseline we used earlier. That reduces all dimensions of the triangle by the same factor, making R equal to 67.5 or 60 Earth radii. The latter figure is close to the observed average distance.

Present the 1999 eclipse--perhaps with other details, on the eclipse path. Can the calculation of Hipparchus be repeated here?

   Shortcut to the solution: in 129 BC, Alexandria had 80% of the Sun covered while the Hellespont had totality, so the distance between the two corresponded to the angle subtended by 20% of the Sun.
   In 1999, from the lines on the map given by the web site, Alexandria had 71%, the Hellespont 90%. The distance therefore corresponded to 19% of the Sun--almost exactly the same as 2000 years ago.

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Author and Curator:   Dr. David P. Stern
     Mail to Dr.Stern:   stargaze("at" symbol)phy6.org .

Last updated: 12.17.2001