ACTIVITY 8 Aristarchus Measures the Size and Distance of the Sun Following his determination of the Moon's distance and size, Aristarchus applied the forces of geometry, logic, and observation to the Sun. He envisioned how Earth, the Moon and the Sun must be arranged in space at the instant when the Moon appears precisely half-illuminated-in its first-quarter or last-quarter phase. A photograph of the quarter-phase Moon is shown in Figure 8-1. At that time, Earth, the Moon and the Sun must form a right triangle, as shown in Figure 8-2. It's no different than if you viewed an illuminated ball from the side, perpendicular to the light rays illuminating it. In a right triangle, the cosine- abbreviated cos―of an angle is defined as the length of the side adjacent to the angle divided by the length of the hypotenuse. Content removed due to copyright restrictions Figure 8-1 Quarter-phase Moon. Moon Earth d Sun Figure 8-2 Earth-Moon-Sun triangle when the Moon is in the first quarter phase. For the Earth-Moon-Sun triangle of Figure 8-2: cos(E), where E is the angle between the Moon and Sun as viewed from Earth, r is the Moon's distance, and d is the Sun's distance. To envision angle E, imagine standing outside and extend- ing one arm toward the Sun and your other arm toward the Moon; the angle formed by your out- stretched arms is E. A bit of algebra makes the cosine equation look like this: d=r/cos (E). Having already deter- mined the Moon's distance, all Aristarchus needed to do to find the Sun's distance was measure the angle E during the Moon's quarter phase. Easier said than done! If Figure 8-2 were drawn to actual scale, with the Sun very, very far away—that is, d much greater than r-angle E would be very close to, although not quite, a right angle. (In fact, trigonometry had not yet been invented in Aristarchus's time; he used analogous geometrical methods to accomplish the procedure described here.) 1. Aristarchus estimated (guessed?) that during the quarter-Moon phase the angle E was 87 degrees. Use the cosine equation in the previ- ous paragraph to determine how many times farther away the Sun is than the Moon. Your an- swer should take the form d = n xr, where n is a number. For now, leave r unspecified; it's merely the symbol that represents the Moon's distance. For example, if angle E were 30 degrees, then cos (E) 0.866, and d = 1.15 × r. 2. The solar distance d derived from Aristarchus's method is exquisitely sensitive to the adopted value for angle E. In other words, even a slight change in angle E produces a whopping change in the solar distance d. To illustrate, recompute the solar distance if Aristarchus had assumed an angle merely one degree larger, that is, 88 degrees instead of 87 degrees. Again express your answer in the form d = n × r. 3. Modern measurement reveals that Aristarchus was way off in his estimation of angle E (so far off, in fact, that we suspect he just plucked a value "out of the hat"). The true value of E is 89.85 degrees. Recompute the solar distance now, again in the form d = n× r. 4. Now it should be easy to find the true diameter of the Sun in units of Moon-diameters, in other words, how many Moons would fit across the face of the Sun? The Moon almost precisely covers the Sun during a solar eclipse—the Moon and the Sun appear the same angular diameter in the sky. But the Sun is much farther away than the Moon; to appear to be the same diam- eter as the Moon, it must, in fact, be a much larger body than the Moon. For example, if the Sun were 3 times farther than the Moon, yet they appear to be the same diameter, the Sun must really be 3 Moon-diameters wide. Or in general, if the Sun is n times farther than the Moon, yet they appear to be the same diameter, the Sun must be n Moon-diameters wide. a. Using your value for n from Part 1, write down an expression for the Sun's diameter in units of Moon-diameters, according to Aristarchus. b. Aristarchus went on to find the Sun's diameter in units of Earth-diameters. To do this, he used his estimate from lunar eclipse observations that the Moon is one third as wide as Earth. Considering this information and your answer to Part 4(a), write down an expression for the Sun's diameter in units of Earth-diameters, according to Aristarchus. 5. To the layperson, Aristarchus's report on his findings is about as exciting as a pamphlet on mixing cement. Not a syllable is wasted on commentary or personal reflection. Yet we can hardly imagine that Aristarchus was unmoved by the extraordinary result of his calculations. Based on your answers to Part 4, can you ex- plain why Aristarchus might have concluded that the Sun, and not Earth, lay at the center of the cosmos? What other unique and important feature of the Sun might have supported this conclusion? Curiously, Aristarchus did not carry out the next logical step: finding the Sun's and Moon's distances in units of Earth-diameters. With such informa- tion, he could have formed a true scale model of the Sun-Earth-Moon system, in the same way that a globe shows a scaled-down version of con- tinents and oceans. The essential question is this: given that the Sun is actually. ters across and the Moon is, Earth-diame- Earth-diameters across, how far must each object be from Earth to appear as they do, a half-degree across, in the sky? To answer this question, we apply the sector equation from the previous activity: r = 57.3 s/ 0, where r is the Sun's or Moon's distance, s is the Sun's or Moon's true diameter (expressed in Earth- diameters), and 0 is the Sun's or Moon's angular diameter in the sky. For example, if an object is actually 10 Earth-diameters wide and it spans an angle of 4 degrees in the sky, its distance is equiv- alent to about 143 Earth-diameters. 6. Using the sector equation plus Aristarchus's diameters for the Moon and the Sun, com- pute the distance of (a) the Moon and (b) the Sun in units of Earth-diameters, according to Aristarchus. 7. Now we're ready to construct the scale model of the Sun-Earth-Moon system, as it might have been envisioned by Aristarchus more than 2000 years ago. Suppose Earth is represented by a Ping-Pong ball, which is approximately 1 inch across. Using your answers to Parts 4 and 6, de- termine to scale the following measurements: a. the diameter (in inches) of a ball represent- ing the Moon b. the distance (in feet) of that Moon ball from the center of the Ping-Pong ball "Earth" c. the diameter (in inches) of a ball represent- ing the Sun d. the distance (in feet) of that Sun ball from the center of the Ping-Pong ball "Earth" Modern measurements reveal that Aristarchus was not too far off in his determination of the Moon's diameter and distance from Earth. However, he erred significantly in his estimates of the Sun's diameter and distance: the Sun is ac- tually more than 100 Earth-diameters across and 11,000 Earth-diameters distant. There's nothing wrong with Aristarchus's methods; they were just impossible to carry out reliably in his day. Nevertheless, Aristarchus's erroneous solar dis- tance became the de facto standard in the Middle Ages. For the first time, someone had calculated the size and distance of a celestial body using observational data gathered from Earth. In bold strokes, Aristarchus had combined geometry, logic, and observation to effectively free himself of his earthly bonds and to show astronomers that it was possible to "lay down a ruler" in the heavens. He had subjected a small portion of the universe to the most basic kind of scientific scru- tiny: measurement. Here lay the genesis of mod- ern astronomy.