I estimated the time for the moon to completely transit the disk of the sun during a solar eclipse using no tools beyond a water bottle with calibrated markings and a ball-point pen and no facts beyond everyday common knowledge. A remarkable agreement with the observation of 0.4% indicates that this strategy is successful at estimating the relative motion of the sun and moon as witnessed from the earth.

The sun's diameter is about 400 times larger than that of the moon and is also about 400 times farther away—this makes for an unusual coincidence that the moon and sun subtend the same angle in the sky when seen from the earth.

The new moon can occasionally come between the sun and the earth, completely covering the disk of the sun. This stunning visual effect, known as a total solar eclipse, can be observed every 2–3 years at some location on the earth. There are some excellent resources available for learning more about solar eclipses.1

This anecdote is from a physicist sitting in the path of totality in the wilderness of the Cascade Mountain Range in Oregon on the morning of the August 21, 2017, total solar eclipse. Waiting for the eclipse to happen, curiosity struck. What could we estimate based on observation alone, perhaps along with a few guiding principles from the last few hundred years of astronomic inquiry? Going further, if we had a few simple numbers or facts about our universe, what could we calculate? Momentarily setting aside historical questions such as how we “know” that the earth is spherical, I tried to estimate the time it should take for the moon to transit the entire disk of the sun, as shown in Fig. 1.

Fig. 1.

Perceived position of the moon relative to the sun (shaded) at five separate moments in time. Times 1 and 5 correspond to the beginning and end, respectively, of the partial solar eclipse. The transit “distance” $2 d moon$ shows the full distance traveled by the moon over this duration.

Fig. 1.

Perceived position of the moon relative to the sun (shaded) at five separate moments in time. Times 1 and 5 correspond to the beginning and end, respectively, of the partial solar eclipse. The transit “distance” $2 d moon$ shows the full distance traveled by the moon over this duration.

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These were the facts I started with:

• The moon has approximately the same spatial extent in the sky as the sun (same solid angle or apparent size).

• My pen, held at arm's length, had a diameter nearly exactly blocking the sun (or the moon).

• It takes ∼27 days for the moon to orbit the earth.

• We orbit the sun every 365 days.

It seemed that the primary effect responsible for the relative motion of the position in the sky of the moon with respect to the sun was the orbital speed of the moon. Even though the positions of both the sun and the moon move across the sky every day, they both move at a rate determined essentially by earth's rotational speed. Thus, we can be concerned only with the relative motion and ignore the common motion of the positions of the sun and moon across the sky.

The time that it takes the moon to cross the sun, $T$ (from moments 1–5 in Fig. 1), can be estimated by considering the arc of angle the moon crosses in moving two of its own diameters ($2 d moon$)

$θ = ω moon T = 2 d moon r eye − moon,$
(1)

where $ω moon$ is the angular velocity of the moon moving around the earth (see Fig. 2 for the geometrical parts). Solving this relationship for $T$, we find

$T = 2 ω moon ⋅ ( d moon r eye − moon ).$
(2)

The angular frequency of the moon can be neatly calculated by knowing that the period of the moon's orbit is $T moon = 27$ days, which means an angular frequency of $ω moon = 2 π / T moon = 2.7 μ rad / s$. Next, we needed to determine the ratio of some distances.

Fig. 2.

Using similar triangles to estimate the ratio of distance to diameter for three objects: my pen, the moon, and the sun.

Fig. 2.

Using similar triangles to estimate the ratio of distance to diameter for three objects: my pen, the moon, and the sun.

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I noticed that one could use the geometric properties of similar triangles to estimate the ratio above in parentheses, as related to the other distances as shown in Fig. 2

$d moon r eye − moon = d pen r eye − pen = d sun r eye − sun.$
(3)

This estimation technique works nicely as long as one knows some of the other distances, such as the distance from my eye to the pen or the diameter of the pen. But, being out in the wilderness, I had forgotten to bring a ruler ☺. However, I did have a cylindrical 1-liter water bottle, with accurate markings of the water level in units of milliliters. Trusting these markings and recalling that a liter of water is 1000 cm3 (thank you, metric system!), I proceeded to “make” a ruler by marking the circumference of the bottle, $C bottle$, on a strip of paper wrapped around the water bottle and relating this length to the diameter of the bottle using $C bottle = π d bottle$ (or less accurately, by holding the paper to the base of the bottle to measure the diameter directly). By laying the diameter length of paper on the volume markings, I found that the diameter length corresponds to a volume of 0.58 L of water. I then used the cylindrical geometry of the water bottle to find the bottle diameter, $d bottle$, in centimeters

$π ( d bottle 2 ) 2 d bottle = 0.58 l = 580 cm 3 .$
(4)

Thus, I could solve for $d bottle$ as 90 mm. Using my “ruler” to measure both the diameter $d pen =$ 6.9 mm and $r eye − pen =$ 600 mm, I proceeded to solve Eq. (1), substituting in the distances for my pen

$T = 2 ω moon ⋅ ( d pen r eye − pen ) = 2.37 h.$
(5)

This value seemed realistic, but I wondered if my model was incomplete.

The position of the sun in the sky is moving as well, independent of the earth's rotation. Including this effect is subtler—we are only concerned with the revolution of the earth around the sun, ignoring the rotation of the earth during the eclipse since this affects both the sun's and moon's apparent motion in the sky in the same way. The earth's rotational direction and revolution direction are both counter-clockwise, looking from “above” the north pole and the orbital plane. The effect of the earth's rotation is that the sun rises later each day than it would if the earth was not revolving around the sun. Thus, the position of the sun in the sky “moves” (for a non-rotating earth) in the same direction as the moon (the moon rises about 50 min later each solar day). Since the apparent relative motions of the sun and moon are in the same direction, we can consider an effective relative angular velocity, $ω eff$, which is the difference of the two angular velocities

$ω eff = ω moon − ω sun.$
(6)

This “same direction” motion means that the eclipse should last a slightly longer time. The sun's “relative angular velocity,” $ω sun$, can be calculated simply through the revolution time for the earth to orbit the sun or 365 days. Since the moon's orbital period is ∼7% of this time, we expect a 7% correction to our calculation. Including this correction, we calculate

$T = 2 ω eff ⋅ ( d pen r eye − pen ) = 2.56 h.$
(7)

The actual time that I observed for the moon to transit the distance $2 d moon$ was 2.55 h (9:05 am–11:38 am, also confirmed using NASA's Eyes interactive).2 The remarkable agreement with observation to within 0.4% helped confirm this mental picture of the relative motions of the positions of the sun and moon in the sky.

This story celebrates from an unusual perspective a celestial event that brought millions of people “together”—in a line across the country—for the largest migration of humans in a few days' time that North America has seen. While all of this is “known” physics, I found it incredibly fun to be able to predict the transit time with such accuracy. This exercise inspired me to wonder how ancient civilizations performed similar calculations.

The transit time estimation could be used in a classroom environment by asking students to come up with some “facts” such as the orbit periods of the moon and earth, and demonstrating the concept of similar triangles as a way of estimating spatial extent. Please use appropriate eye safety when viewing the sun, as detailed in Ref. 1. A number of other “back-of-the-envelope” calculations could be made in a classroom environment, where students are given limited “facts”. An example would be estimating the mass of the earth or the radius of the moon from knowing the acceleration due to gravity on the earth and that gravitational acceleration is one sixth of that of the earth on the surface of the moon. Some other excellent articles explore ways to determine the earth-moon distance from lunar eclipse photos3 and revisit the strategies of the ancient Greeks to determine the sizes of the sun and moon.4 There are some additional resources available in a text on ancient astronomy.5 Similar analysis to this article could be performed to estimate, for example, the transit time of Mercury across the sun, which will next happen in November 2019. In order to make this estimate, it would be required to know that the orbit time of Mercury is 88 days.

1.
J. M.
Pasachoff
and
A.
Fraknoi
, “
Resource letter OSE-1: Observing solar eclipses
,”
Am. J. Phys.
85
,
485
494
(
2017
).
2.
K.
Hussey
, “
NASA' eye
,” <https://eyes.nasa.gov/eyes-on-eclipse-detail.html> (
2017
), Accessed 8 January 2019.
3.
D. H.
Bruning
, “
Determining the Earth–Moon distance
,”
Am. J. Phys.
59
,
850
(
1991
).
4.
F.
Momeni
et al “
Determination of the Sun's and the Moon's sizes and distances: Revisiting Aristarchus' method
,”
Am. J. Phys.
85
,
207
215
(
2017
).
5.
D. H.
Kelley
and
E. F.
Milone
,
Exploring Ancient Skies: A Survey of Ancient and Cultural Astronomy
, 2nd ed. (
Springer
,
New York
,
2011
).