CWRU ASTRO 101 EXAM 1. QUESTIONS WITH 100% CORRECT ANSWERS.

The Sun has an angular diameter of about 0.5◦ and an average distance of about 150 million km. What is the Sun's approximate physical diameter? Compare your answer to the actual value of 1 390 000 km. Given that the Sun has an angular diameter of about 0.5◦ and an average distance of about 150 million kilometers, we can use the angular diameter equation from Mathematical Insight 2.1: θ = d × 360◦ 2πD ,where θ is the angular diameter, d is the physical diameter, and D is the physical distance. Therefore, θ = d × 360◦ 2πD d = θ × 2πD 360◦ = 0.5 ◦ × 2π( km) 360◦ = 1 310 000 km. This is the Sun's approximate diameter. This is smaller than the Sun's actual diameter of 1 390 000 km by about 5.7%. Eclipse Conditions The Moon's precise equatorial diameter is 3476 km, and its orbital distance from Earth varies between 356 400 and 406 700 km. The Sun's diameter 5 is 1 390 000 km, and its distance from Earth ranges between 147.5 and 152.6 million km. (a) Find the Moon's angular size at its minimum and maximum distances from Earth. (b) Find the Sun's angular size at its minimum and maximum distances from Earth. (c) Based on your answers to parts (a) and (b), is it possible to have a total solar eclipse when the Moon and Sun are both at their maximum distance? Explain. (a) The Moon's angular size at its minimum distance is θ = d × 360◦ 2πD = 3476 km × 360◦ 2π(356 400 km) = 0.559◦ . The Moon's angular size at its maximum distance is θ = d × 360◦ 2πD = 3476 km × 360◦ 2π(406 700 km) = 0.49◦ (b) The Sun's angular size at its minimum distance is θ = d × 360◦ 2πD = 1 390 000 km × 360◦ 2π( km) = 0.54◦ . The Sun's angular size at its maximum distance is θ = d × 360◦ 2πD = 1 390 000 km × 360◦ 2π( km) = 0.522◦ (c) When the Moon is at its maximum distance, its angular size is 0.49◦ . The Sun's angular size at its maximum distance is 0.522◦ . Since these two are slightly different, there will not be a total solar eclipse. Instead, it will be an annular or partial solar eclipse.