Skygazing/Solar eclipse lab on a sunny day

Purpose

Today we did a lab that creates a scale model of a solar eclipse. In his left hand is an iPhone that can take the shadow's photograph.

Is this shadow of the 1-inch ball the umbra or the penumbra? The vertical shadow is the support that held the Styrofoam ball in place.

A 1-inch diameter "Moon" must be held 110 inches above the ground for a proper scale model

The digital image of the penumbra was imported into Inkscape and converted into grey-scale objects. A map of USA was found on commons and placed next to it at the proper scale. This evidence suggests that the penumbra associated with a solar eclipse is larger than the United States.

To measure the size of the Moon's penumbra during a solar eclipse using a scale model.

Why this is a scale model

A scale model is a drawing that preserves all the angles. Here we represent the Moon as a small Styrofoam ball. The Earth is most easily represented as a map of a region where a solar eclipse is expected. Here we use the United States.

For the Sun, we use the actual object, which of course is not to the same scale. But for our purposes, any object that produces light with the proper angular distribution can be used. In other words, the actual Sun provides the required angular spread of light rays (i.e., light spread over a range of about half a degree). The actual size of the Sun is irrelevant (otherwise one could use a solar eclipse to calculate the size and distance to the Sun).

Equipment

• 10 ft rod to hold up the model "Moon" (we clamped a two-meter stick to a one-meter stick)
• One inch diameter ball to server as the Moon (rod and ball dimensions may vary proportionally)
• Tape measure or other way to measure distance on the ground (we used 2 metersticks)
• Camera (we used an Apple iPhone)
• White screen. We used about a dozen sheets of paper. It would also be worthwhile to print onto this paper a properly scaled image of a region on Earth, or perhaps the Earth itself.
• At least two people (three works better)
• The Sun on a sunny day
• The instructor and students should be mentally prepared to do calculations and proportional reasoning on three different scales:

1. The solar system (especially Earth/Moon system)
2. A scale where the Moon is reduced to a small ball
3. A scale on which photos of the shadow are analyzed

Procedure

1. The Moon was a one inch diameter styrofoam ball attached to a stick that was 110 inches long (chosen to match the Earth-Moon distance).
2. We forgot to measure the altitude of the Sun. Fortunately the images on commons contain a time stamp and we know where we were (Celina, Ohio). Somewhere a website will give us that angle.
3. Photograph the shadow alongside a ruler for scale.
4. Record the location of the shadow so that you may later calculate the distance from the "Moon" at the end of the stick and the "Earth", represented by the white paper on the ground.

How far from the model "Earth" should we hold the model "Moon"?

• The Moon is about 60 earth-radii from Earth.
• The Moon has a diameter about 1/4 that of the Earth.
• Our Styrofoam "Moon" had a diameter of one inch.

Keeping track of scaled objects can be confusing. One trick is to use capital letters for the real thing and lower case for the scale:

$d$ equals scaled distance from Earth to Moon

$D$ is the actual distance.

$r_{E}$ equals scaled radius of Earth

$R_{E}$ is the actual radius.

$r_{M}$ equals scaled radius of the Moon

$R_{M}$ is the actual radius.

Focusing on only astronomical data, we have:

$D\approx 60R_{E}{\text{ and }}R_{E}\approx 4R_{M}\rightarrow D\approx (100)(2R_{M})$

Since $2r_{M}=1{\text{ inch}}$ we conclude that the Styrofoam ball should be held at a distance of

$d=120{\text{ inches}}$

away from the Earth. Our "model" Earth was just a piece of paper, but we can add a scale to it using the fact that a one inch diameter ball represents the 2159 mile diameter Moon.

Calculating the scale

The scale is:

$1\ {\widehat {\text{in}}}=2159{\text{ miles }}$

where we have used the "hat" notation to define the "scaled" inch. This "scaled" inch equals 2159 miles, but is also one (actual) inch on the scale model.

$1\ {\widehat {\text{in}}}=2159{\text{ miles }}\left({\frac {5280\ {\text{ft }}}{1{\text{ mile }}}}\right)\left({\frac {12\ {\text{in }}}{1{\text{ft}}}}\right)=1.37\times 10^{8}{\text{ inches}}$

In other words, our scale is 1.37 x 10⁸ to 1.

Scaling the USA

http://www.distance.to/New-York/San-Francisco says that the distance between NYC and SF is 2566 miles = 1430 km. Convert this to centimeters and divide by the scaling factor to get:

${\frac {1.43\times 10^{8}{\text{ cm}}}{1.37\times 10^{8}}}=1.04\approx 1{\widehat {\text{cm}}}={\text{ scaled length of the US}}$

This says that the penumbra is larger than the USA? This seems to big, but see below. This experiment should be repeated and the results reported here.

Other photos from the experience

showing how we did the lab

photos of shadow
300 cm 250 cm 200 cm 150 cm 100 cm 50 cm 0 cm

Understanding the size of the penumbra

This diagram explains why the penumbra of a solar eclipse is approximately twice the size of the Moon. Since the Sun and Moon have nearly the same angular diameter, the umbra is reduced to nearly a single point in length. Therefore the three angles marked in red are nearly equal, all with a base equal to the Moon's diameter.

Conclusion

If you go here http://eclipse.gsfc.nasa.gov/SEgoogle/SEgoogle2001/SE2017Aug21Tgoogle.html, you will see that the penumbra is very large and extends far outside the USA. Perhaps our results are correct.

We measured the length of the shadow on the photograph marked 0cm. It was projected on a classroom screen at a scale where 4.9 cm* equals 1 cm. In other words, the classroom projected image of the photo expanded the meter stick so that 1 cm on the stick measured 4.9cm* on the screen.. The shadow's length on the screen was 21 plus/minus 1 cm*. We concluded that the umbra was

1.69 ± 0.08 inches

This much smaller than the 2 inch umbra we expected from a 1 inch moon. Perhaps the brightest part of the penumbra was indistinguishable from the illuminated portion of the paper.

Test the hypothesis:How big is the umbra in this nasa gif?

2017 August 21, a total eclipse of the Sun occurs in United States of America.

File:Total solar eclipse NASA TSE2017-1.gif has been annotated to show the outline of the USA, as well as the 50% and 0% partial eclipse contours.

It looks to me that the height of the umbra is approximately equal to the radius of the Earth. According to https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html, the ratio of diameters for the Earth to the Moon is:

${\frac {R_{M}}{R_{E}}}=0.2725$

Let $d_{U}$ be the diameter of the umbra, which we estimate to be equal to the radius of Earth and also twice the diameter of the Moon, but since the diameter of the Moon is twice its radius, we have:

$d_{U}\approx R_{E}\approx 4R_{M}=(4)(0.2725)\cdot R_{E}=1.09R_{E}$

I think the greyscale stuff was wrong (see above)

Not bad!

W:On the Sizes and Distances (Aristarchus)

The article on Wikipedia is excellent:

copied from w:Special:Permalink/773165157

Half Moon Aristarchus began with the premise that, during a half moon, the moon forms a right triangle with the Sun and Earth. By observing the angle between the Sun and Moon, φ, the ratio of the distances to the Sun and Moon could be deduced using a form of trigonometry. From the diagram and trigonometry, we can calculate that ${\frac {S}{L}}={\frac {1}{\cos \varphi }}=\sec \varphi .$ The diagram is greatly exaggerated, because in reality, S = 390 L, and φ is extremely close to 90°. Aristarchus determined φ to be a thirtieth of a quadrant (in modern terms, 3°) less than a right angle: in current terminology, 87°. Trigonometric functions had not yet been invented, but using geometrical analysis in the style of Euclid, Aristarchus determined that $18<{\frac {S}{L}}<20.$ In other words, the distance to the Sun was somewhere between 18 and 20 times greater than the distance to the Moon. This value (or values close to it) was accepted by astronomers for the next two thousand years, until the invention of the telescope permitted a more precise estimate of solar parallax. Aristarchus also reasoned that as the angular size of the Sun and the Moon were the same, but the distance to the Sun was between 18 and 20 times further than the Moon, the Sun must therefore be 18-20 times larger. Lunar eclipse Aristarchus then used another construction based on a lunar eclipse: By similarity of the triangles, ${\frac {D}{L}}={\frac {t}{t-d}}\quad$ and $\quad {\frac {D}{S}}={\frac {t}{s-t}}.$ Dividing these two equations and using the observation that the apparent sizes of the Sun and Moon are the same, ${\frac {L}{S}}={\frac {\ell }{s}}$, yields ${\frac {\ell }{s}}={\frac {t-d}{s-t}}\ \ \Rightarrow \ \ {\frac {s-t}{s}}={\frac {t-d}{\ell }}\ \ \Rightarrow \ \ 1-{\frac {t}{s}}={\frac {t}{\ell }}-{\frac {d}{\ell }}\ \ \Rightarrow \ \ {\frac {t}{\ell }}+{\frac {t}{s}}=1+{\frac {d}{\ell }}.$ The rightmost equation can either be solved for ℓ/t ${\frac {t}{\ell }}(1+{\frac {\ell }{s}})=1+{\frac {d}{\ell }}\ \ \Rightarrow \ \ {\frac {\ell }{t}}={\frac {1+{\frac {\ell }{s}}}{1+{\frac {d}{\ell }}}}.$ or s/t ${\frac {t}{s}}(1+{\frac {s}{\ell }})=1+{\frac {d}{\ell }}\ \ \Rightarrow \ \ {\frac {s}{t}}={\frac {1+{\frac {s}{\ell }}}{1+{\frac {d}{\ell }}}}.$ The appearance of these equations can be simplified using n = d/ℓ and x = s/ℓ. ${\frac {\ell }{t}}={\frac {1+x}{x(1+n)}}$ ${\frac {s}{t}}={\frac {1+x}{1+n}}$ The above equations give the radii of the Moon and Sun entirely in terms of observable quantities. The following formulae give the distances to the Sun and Moon in terrestrial units: ${\frac {L}{t}}=\left({\frac {\ell }{t}}\right)\left({\frac {180}{\pi \theta }}\right)$ ${\frac {S}{t}}=\left({\frac {s}{t}}\right)\left({\frac {180}{\pi \theta }}\right)$ where θ is the apparent radius of the Moon and Sun measured in degrees. It is unlikely that Aristarchus used these exact formulae, yet these formulae are likely a good approximation to those of Aristarchus.

Aristarchus simplified

I have a theory that great minds not only do much, they omit much. Suppose Aristarchus was more interested in knowing approximate facts and less interested in exact theories of geometry. Let's replace what he either knew, or should have known by the following facts:

1. The angular size of the Moon and Sun are half a degree.
2. The Earth's shadow (umbra) at the Moon's distance is twice the diameter of the Moon.
3. The angle between the Sun and the Moon when visible Moon is 50% illuminated is effectively 90 degrees.

The third of these statements would have permitted Aristarchus to set the Sun at "infinity" (i.e. very far away), and used parallel lines to bypass a lot of complicated geometry (by "complicated" I mean too sophisticated for the average mind of a 21st century educated person.) Here is how Aristarchus might have concluded that the Earth is three times larger than the Moon (in diameter).

How Aristarchus might have calculated the size of the Moon if he knew the Sun is extremely far away. To the left we see a full moon, and to the right we see two adjacent Moons that represent a statement attributed to Aristarchus to the effect that the Earth's umbra is twice as large as the Moon at one lunar distance from the Sun. At two lunar distances the base of the triangle is one lunar diameter, and the apex of the Earth's umbra is situated three lunar distances from Earth (seen at extreme right of figure). All triangles are similar if the Sun and Moon have the same angular size. Hence, Aristarchus might have calculated that Earth is 3 times as large as the Moon (the actual ratio is closer to 3.67).The bottom figure was created by compressing the top one in the vertical direction by a factor of 3. These figures are not too scale in that they assume that the angular diameter of the Sun and Moon are 6.7 degrees (instead of 0.5 degrees.) Hence the Moon is represented as being over 10 times closer to Earth than it really is.

The distance to the Moon in Earth diameters

Since one radian is 57.3 degrees, the distance to the Moon in Earth diameters can be written as:

$\theta ={\frac {2\cdot {\text{Moon's radius}}}{\text{Earth/Moon distance}}}={\frac {2R_{\text{Moon}}}{\text{Earth/Moon distance}}}={\frac {1}{2}}{\text{deg}}={\frac {1{\text{ deg}}}{2}}{\frac {\pi {\text{ rad}}}{180{\text{ deg}}}}={\frac {\pi }{360}}$

Since the Moon is three times smaller than Earth, we have:

$R_{Moon}={\frac {1}{3}}R_{Earth}$

Solving, we obtain the distance to the Moon in Earth radii:

${\text{Earth/Moon distance}}={\frac {2}{3}}{\frac {360}{pi}}R_{Earth}=76R_{Earth}$

The accepted distance is closer to 60 Earth radii. Taking the ratio to extimate the percent error that Aristarchus would have made:

${\frac {76}{60}}\approx 1.3$

which corresponds to only a 30% error that Aristarchus might have made if only he understood the vast distance to the Sun.