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# It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Q 2.32: It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

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From examples 2.3 and 2.4, we have,

Diameter of the Earth = $1.276 \times 10^7 m$
Distance between the Moon and the Earth,$D_{moon}$ = $3.84\times10^8 m$

Distance between the Sun and the Earth, $D_{sun}$ = $1.496\times10^{11} m$

Diameter of the Sun, $d_{sun}$ = $1.39\times10^9 m$

Let, Diameter of the Moon be  $d_{moon}$

Now, During Solar eclipse, the angle subtended by Sun's diameter on Earth = angle subtended by moon's diameter

$\frac{1.39\times10^9}{1.496\times10^{11}} = \frac{d_{moon}}{3.83\times10^8}$            $(\because \Theta = d/D)$

$\Rightarrow d_{moon} = \frac{1.39\times10^9}{1.496\times10^{11}}\times 3.83\times10^8 = 3.56\times10^6 m$

Therefore, the diameter of the moon = $3.56\times10^3 km$

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