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It is a well known fact that during a total solar eclipse the disc of the moon almost completely covers the disc of the sun. From this fact and from the information that sun's angular distance $\theta$ is measured to be $1920^{\prime \prime}$, determine the approximate diameter of the moon. Given earth-moon distance $=3.8452 \times 10^8 \mathrm{~m}$.
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During total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameters of both the sun and the moon must be equal.
$\therefore$ Angular diameter of the moon,
$\theta=$ Angular diameter of the sun
$=1920^{\prime \prime}=1920 \times 4.85 \times 10^{-6} \mathrm{rad}$
$\left[\because 1^{\prime \prime}=4.85 \times 10^{-6} \mathrm{rad}\right]$
Earth-moon distance, $\mathrm{s}=3.8452 \times 10^8 \mathrm{~m}$
Diameter of the moon, $\mathrm{D}=\theta \times \mathrm{s}$
$=1920 \times 4.85 \times 10^{-6} \times 3.8452 \times 10^8$
$=3.581 \times 10^6 \mathrm{~m}=3581 \mathrm{~km}$.
$\therefore$ Angular diameter of the moon,
$\theta=$ Angular diameter of the sun
$=1920^{\prime \prime}=1920 \times 4.85 \times 10^{-6} \mathrm{rad}$
$\left[\because 1^{\prime \prime}=4.85 \times 10^{-6} \mathrm{rad}\right]$
Earth-moon distance, $\mathrm{s}=3.8452 \times 10^8 \mathrm{~m}$
Diameter of the moon, $\mathrm{D}=\theta \times \mathrm{s}$
$=1920 \times 4.85 \times 10^{-6} \times 3.8452 \times 10^8$
$=3.581 \times 10^6 \mathrm{~m}=3581 \mathrm{~km}$.
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