Question

2. Aristarchus of Samos (310-230 BC) was a Greek mathematician and astronomer who first proposed a heliocentric model for our solar system, thousands of years before Copernicus. His ideas were based on his calculations of the sires of the Earth, Moon, and Sun. The approach below is a simplified version of his approach that includes knowledge of the work of contemporaries, such as the calculation of the Earth's radius by Eratosthenes of Cyrene (276-194 BC). (Note: We showed in class how Eratosthenes was able to calculate the Earth's radius. Use as an approximate value 6,400 km.) The symbols Ry, Ry, Rs denote the radii of the Earth, Moon, and Sun, respectively. The symbols dgs, dem denote the distances between Earth-Sun and Earth-Moon, respectively. a) Aristarchus was aware that during a total solar eclipse (that is, when the Moon is between the Sun and the observer on Earth) the Moon completely covers the Sun. He reasoned that this is so because the Sun is farther from the Earth than the Moon is, and that the two objects fit in the same cone from the point of view of the observer () Show that given two circles (here corresponding to the cross sections of the Moon and Sun), the two common external tangents meet at the same point as the line connecting the centers of the two objects. RM dEM Show that Rs des (1) E - 11 1:1 A

Answer 1

3) Considering the bigger circle.

Construction: Draw a line joining the center and the tangent

Name the point

So, for $\triangle EBS$ and $\triangle EMS$

LEBS = LEMS = 90° (by property)

$ES=ES \,\,(\text{common})$

$BS=MS \,\,(\text{radius of circle})$

So, $\triangle EBS\cong \triangle EMS\,\, (\text{By SAS})$

So, $\angle BES=\angle MES\,\, (\text{cpct})$

Thus the point is common. And the centre lie on the same line.

SImilarly working out for the smaller circle , we can show that is common. And the centre lie on the same line.

In general adding the two proofs we can prove as required.

b) Now considering $\triangle EBS$ and $\triangle EAM$

$\angle EBS=\angle EAM=90^\circ\,\, (\text{by property})$

$\angle BES=\angle AEM\,\, (\text{common})$

$\angle BSE=\angle AME\,\, (\text{BS}\mid\mid\text{AM})$

So, $\triangle EBS\sim \triangle EAM\,\, (\text{by AAA theorem})$

So, $\frac{BS}{AM}=\frac{ES}{EM}=\frac{EB}{EA}$

Thues, $\frac{R_{S}}{R_M}=\frac{d_{ES}}{d_{EM}}$

$\Rightarrow \frac{R_M}{R_S}=\frac{d_{EM}}{d_{ES}}\,\, (\text{proved})$