Aristarchus Measures ACTIVITY 1S the Size and Distance of the Moon In the third

Question

Aristarchus Measures ACTIVITY 1S the Size and Distance of the Moon In the third century Bc, Greek philosopher-mathematician Aristarchus, from the island of Samos in the Aegean Sea, proposed a bold rearrangement of the heavens For hundreds of years preceding Aristarchus, Greek philosophers believed that Earth occupied the hub of the universe, and that the Sun, Moon, planets, and even the star-studded celestial sphere, which they held enclosed the universe, all circled around it. Aristarchus proposed instead that the Sun holds the central position, casting its light symmetrically outward on the other celestial bodies. Ironically, Aristarchus's prime legacy to science turned out to be something other than his Sun-centered universe, which was largely forgotten until Polish mathematician Nicholas Copernicus reintroduced it some 18 centuries later. Aristarchus demonstrated for the first time how it was possible, using simple observations and elemen- tary geometry, to measure sizes and distances of celestial bodies. Figure 7-1 Geometry of a sector of a circle. The starting point for this activity is the geometry of an arc or sector of a circle-basically, a piece of pie. What is the mathematical relationship between the angle enclosed by an arc-its angular width-and the length of the arc itself? In Figure 7-1, s represents the length of the are; r is the radius of the are; and 0 (the Greek letter "theta") is the angular width of the arc, that is, how many degrees the arc spans. These quantities are related by the sector equation: s - (r0)/573, where r and s are expressed in units of length Gnches, meters, light-years, etc.) and is expressed in degrees. In astronomical applications, the sector equation can be used to deduce, say, the actual diameter of a celestial object, if its angu- lar diameter and distance are known. Or, if the equation is rewritten as-573s/, the distance of a celestial object can be computed if the object's true diameter s and angular diameter are known. It's the latter form that Aristarchus used to determine the distance to the Moon. 31

Explanation / Answer

Q1.

part a:

radius=5 inches

angle made=theta=30 degrees

then length pie crust=r*theta/57.3=2.6178 inches

part b:

angle=theta=30 degrees

sector length=4 inches

then radius=57.3*4/30=7.64 inches

Q2.

angular width of the eclipse shadow=(3 hours/720 hours)*360 degrees=1.5 degrees

Q3.

fraction of eclipse shadow’s width that the moon occupy is given by

shadow width of moon/eclipse width=0.5 degrees/1.5 degrees=1/3

Q4.

moon diameter/earth’s diameter=moon’s shadow/eclipse width=1/3

==>moon’s diameter=0.33D

Q5.

moon’s angular diameter=theta=0.5 degrees

s=0.33D

then r=57.3*s/theta

=37.818D

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