# 19.1: Aristarchus's Method

• • Contributed by CK12
• CK12

Around 270 BC, Aristarchus derived the Moon's distance from Earth from the duration of a lunar eclipse (Hipparchus later found an independent method).

It was commonly accepted in those days that the Earth was a sphere (although its size was calculated only a few years later, by Eratosthenes--See the chapter “The Round Earth and Columbus"). Astronomers also believed that the Earth was the center of the universe, and that Sun, Moon, planets, and stars all orbited around it. It was only natural, then, that Aristarchus assumed that the Moon moved in a large circle around Earth.

Let $$R$$ be the radius of that circle and $$T$$ the time it takes the Moon to go around once, about one month. In that time the Moon covers a distance of $$2\pi R$$, where $$\pi\approx 3.1415926\ldots$$ (pronounced “pi") is a mathematical constant, the ratio of the circumference of any circle to its diameter.

An eclipse of the Moon occurs when the Moon passes through the shadow of the Earth, on the opposite side from the Sun (therefore, it must be a full Moon). If $$r$$ is the radius of the Earth, the shadow's width is close to the Earth's diameter, or $$2r$$. Let $$t$$ be the time it takes the mid-point of the Moon to cross the center of the shadow, about 3 hours (in eclipses of the longest duration, when the Moon crosses the center of the shadow).

If the moon moves around the Earth at a constant speed --- and it takes time $$T$$ (again, about a month) to cover $$2\pi R\approx 6.28R$$, its speed can be expressed as the ratio of distance traveled to the time it takes as $V_m=\frac{2\pi R}{T}$ Since it takes the moon about 3 hours to travel a distance of $$2r$$, we can also express its speed as: $V_m=\frac{2r}{t}$ Setting these equal to each other, we find: $\frac{6.28R}{2r}=\frac{T}{t}$ From this Aristarchus obtained $\frac{R}{r}\approx 60$ This result fits the average distance of the Moon accepted today, 60 Earth radii.