ARISTARCHO'S PROPORTIONS

Rosa M. Ros, Eder Viñuales
"EAAE Summerschools" Working Group
Technical University of Catalonia, University of Zaragoza (Spain)

Abstract

Aristarchos (310 - 230 B.C.) deduced some proportions between the distances and the radii of the Earth-Moon-Sun system. Aristarchos calculated the Sun's radius, the Moon's radius, the distance from the Earth to the Sun and the distance from the Earth to the Moon concerning the Earth's radius. Some years after Eratostenes (280-192 B.C.) determined the radius of our planet and it was possible to calculate all the distances and radii of the Earth-Moon-Sun system.

The proposal of this activity is to repeat both experiments with students. The idea is to repeat the mathematical process designed by Aristarchos and Eratostenes and, if possible to repeat the observations.

Aristarchos' experiment redone

Relationship between the Earth-Moon and Earth-Sun distances

Aristarchos determined that the angle from the Earth to the Sun and the Moon which is in the quarter phase position was 87.


Fig 1 - Relative position of the Moon's quarter

We know that he made a mistake, possibly because it was very difficult for him to determine the instant of the quarter phase. In fact this angle is 89 51', but the process used by Aristarchos is correct. In figure 1, if we use the definition of the sine, it can be deduced that sin 9' = ES/EM, where ES is the distance from the Earth to the Sun, and EM is the distance from the Earth to the Moon. Then approximately,

ES=400 EM
(although Aristarchos deduced ES = 19 EM).

Relationship between the radius of the Moon and the Sun

The relationship between the diameter of the Moon and the Sun must be similar to the previous formula, because from the Earth we observe both diameters equal to 0.5. Therefore the radius also verifies

RS=400 RM

Relationship between the Earth's distance and the radius of the Moon or the Sun

Given that the Moon's diameter observed is 0.5, with 720 times this diameter it is possible to cover the circular path of the Moon around the Earth. The longitude of this path is 2p times the distance Earth-Moon, that is to say 2 RM 720 = 2p EM, isolating,

EM = 720 /(2p) RM

and by a similar reasoning,

ES = 720 /(2p) RS

Relationship between the Earth's distances, the Moon's radius, the Sun's radius and the radius of the Earth

During a Moon's eclipse, Aristarchos observed that the time necessary in order for the Moon to cross the cone's shadow of the Earth was double the time necessary for the Moon's surface to be covered.


Fig 2 : Shadow's cone and relative position of the Earth-Moon-Sun system

Therefore he deduced that the shadow of the Earth's diameter was double the Moon's diameter, that is to say, the relation of both diameters or radii was 2:1. Really we know that this value is 2.6:1.


Fig 3 - Measuring the diameter of the shadow's cone

Then we deduce the following relationship

where x is an auxiliary variable.

Introducing in this previous expression the relationships ES = 400 EM and RS=400 RM, it is possible to eliminate x and simplifying it,

RM = RE 401/1440

is obtained, which offer the opportunity to express all the dimensions concerning the Earth's radius, so

RS = RE . 2014/18
ES = RE . 81640/p
EM= RE . 401/ 2 p

We only need to calculate the Earth's radius to obtain all the distances and radii of the Earth-Moon-Sun system.

Measurements with students

It is a good idea to repeat the measurements made by Aristarchos with students. In particular, the students first have to calculate the angle from the Earth, between the Moon's quarter and the Sun. For this measurement it is only necessary to use a theodolite and to find out the exact instant of the quarter phase. So we verify if this angle is 87 or 89 51' (really it is very difficult).

Secondly, during an eclipse of the Moon, using a chronometer, it is possible to calculate the relation of times between: "the first and the last contact of the Moon with the Earth's shadow", that is to say, measuring the diameter of the cone's shadow of the Earth (Fig. 3) and "the time necessary to cover the Moon surface", that is to say the measure the Moon's diameter.


Fig 4 - Measuring the diameter of the Moon

Finally it is possible verify if the relationship between both times is 2:1 or 2.6:1

The most important objective of this activity is not the result obtained. The most important thing is that it suggests to the students that if they use their knowledge and intelligence, they can obtain interesting results with reduced facilities. In this case the ingenuity of Aristarchos was important in obtaining some concepts of the size of the Earth-Moon-Sun system.

It is a good idea to measure the Earth's radius with students according to the process used by Eratostens. Although Eratostenes' experiment is very well-known, we present here a short version in order to complete the previous Aristarchos experiment.

Eratostenes experiment redone

We consider two sticks introduced perpendicularly in the soil, in two different cities of the Earth's surface on the same meridian. The sticks must be pointed towards the Earth's centre. Normally it is better to use a plumb where we mark a point to measure longitudes. We need to measure the longitude on the plumb from the soil to this mark, and the longitude on its shadow from the base of the plumb to the mark's shadow.

We can consider that the solar rays are parallel. Those Sun's rays produce two shadows for each plumb. We measure the longitudes of the plumb and its shadow. Using the definition of tangent we can obtain the angles a and b .


Fig 5 - Stick's situation and angles in Eratosthene's experiment.

The central angle g can be calculated because the sum of the three angles of a triangle is equal to p radians. Then :

p = a + b + g
and simplifying

g = a - b

where a and b can be obtained from measuring the plumb and its shadow.

Finally by proportionality between the angle g and the longitude of its arc d (determined by the distance on the meridian between the two cities), and 2p radians of the meridian circle and its longitude 2 p RE, that is to say g/d = 360/(2 p RE), then

RE = d/g

it can be deduced, where g is obtained from the observation and d is the distance in km between both cities. We can find d from a good map (for example the army maps are an excellent source for obtaining it).

Also the objective of this activity is not to obtain a precise result. We would only like the students to discover that by thinking and using all the resources that they can imagine, they can obtain surprising results.

References.