#### Question

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.

#### Solution 1

The position of the Sun, Moon, and Earth during a lunar eclipse is shown in the given figure.

Distance of the Moon from the Earth = 3.84 × 10^{8} m

Distance of the Sun from the Earth = 1.496 × 10^{11} m

Diameter of the Sun = 1.39 × 10^{9} m

It can be observed that ΔTRS and ΔTPQ are similar. Hence, it can be written as:

`(PQ)/(RS) = (VT)/(UT)`

`(1.39xx10^9)/(RS) = (1.496xx 10^11)/(3.84xx10^(8))`

`RS = (1.39xx3.84)/1.496 xx 10^6 = 3.57 xx10 ^6 m`

Hence, the diameter of the Moon is 3.57× 10^{6} m.

#### Solution 2

From examples we get θ = 1920″ and S = 3.8452 x 10^{8} m.

During the total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameter of both the sun and the moon must be equal. Angular diameter of the moon, θ= Angular diameter of the sun

= 1920″ = 1920 x 4.85 x 10^{-6} rad [1″ = 4.85 x 10^{-6} rad]

The earth-moon distance, S = 3.8452 x 10^{8} m .’. The diameter of the moon, D = θ x S

= 1920 x 4.85 x 10^{-6} x 3.8452 x 10^{8} m = 35806.5024 x 10^{2} m

= 3581 x 10^{3} m 3581 km.