It is a well known fact that during a 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 a is measured to be 1920", determine the approximate diameter of the moon. Given earth-moon distance = 3.8452 x 10⁸ m

The correct option is 3581 km.

Step-by-Step Explanation:

During a total solar eclipse, the moon completely covers the disc of the sun. This implies that the angular diameter (or angular size) of both the sun and the moon as seen from the earth is approximately equal.

Let the angular diameter be denoted by $\theta$. We are given:
$\theta =1920"$
where $"$ represents arcseconds.

First, we convert the angle $\theta$ from arcseconds to radians. We know that:
$1"=\frac{1}{3600}\text{degrees}=\frac{\pi }{180\times 3600}$

Thus, the angular diameter in radians is:
$\theta =1920\times 4.85\times {10}^{-6}\text{rad}\approx 9.312\times {10}^{-3}\text{rad}$

The relationship between the actual diameter of the moon $d$, the earth-moon distance $D$, and the angular size $\theta$ is given by the formula:
$d=D\times \theta$

Given the earth-moon distance:
$D=3.8452\times {10}^8\text{m}$

Substitute the values of $D$ and $\theta$ into the formula:
$d=(3.8452\times {10}^8\text{m})\times (9.312\times {10}^{-3}\text{rad})$
$d\approx 3.581\times {10}^6\text{m}$

Converting the diameter to kilometers:
$d\approx 3581\text{km}$

Therefore, the approximate diameter of the moon is 3581 km.