2 Assignment (Due Septe.23)
1. Latitude Distance: Earths radius is approximately 6370 km. a. What is Earths circumference? b. What distance is represented by each degree of latitude? c. What distance is represented by each arcminute of latitude? d. Can you give similar answers for the distances represented by a degree or arcminute of longitude? Why or why not? Solution: a. We know that circumference = 2 x x radius, so we can compute the circumference of the Earth: Circumference = 2 x x (6370 km) = 40,000 km b. There are 90o of latitude between the North Pole and the equator. This distance is also one-quarter of Earths circumference. Using the circumference from part (a), the distance is equator to pole distance = circumference / 4 = 10,000 km So if 10,000 km is the same as 90o of latitude, then we can convert 1o into kilometers: 1o x 10,000 km/90o = 111 km So 1o of latitude is the same as 111 kilometers on the Earth c. There are 60 arcminutes in a degree. So we can find how many arcminutes are in a quarter-circle: 90o x 60 arcminutes/1o = 5,400 arcminutes Doing the same thing as the part (b): 1 arcminute x 10,000 km/ 5,400 arcminutes = 1.85 km Each arcminute of latitude represents 1.85 kilometers. d. We cannot provide similar answers for longitude, because lines of longitude get closer together as we near the poles, eventually meeting at the poles themselves. So there is no single distance that can represent 1o of longitude everywhere on Earth. 2. Scale of the Moon. The Moons diameter is about 3,500 km and its average distance from Earth is about 380,000 km. How big and how far from Earth is the Moon on the 1-to-10 billion scale used in Chapter 1? Compare the size of the Moons orbit to the size of the Sun on this scale. Solution: Starting with the size of the Moon, we convert to the scale model distance
by dividing by 10 billion: 3,500 km / 1010 = 3.5 x 10-7 km = 0.35 mm The Moon size on this scale is 0.35 mm. We perform the same conversion to get to the Moons scale distance: 380,000 km /1010 = 3.8 x 10-5 km = 38 mm The Moons scale distance is 38 mm. 3. Find the Suns diameter. The Sun has an angular diameter of about 5o and an average distance of about 150 million km. What is the Suns approximate physical diameter? Compare your answer the to actual value of 1,390,000 km. Solution: Physical size = 2 x (distance) x (angular size) / 360o = 1.31 x 107 km If the angular size is 0.5o, thus the physical size is 1.31 x 106 km which is comparable to the actual value of 1.4 x 106 km. 4. Find a Stars diameter. The supergiant star Betelgeuse (in the constellation Orion) has a measured angular diameter of 0.044 arcsecond. Its distance has been measured to be 427 light-years. What is the actual diameter of Betelggeuse? Compare your answer to the size of our Sun and the Earth -Sun distance. Solution: Physical size = 2 x (distance) x (angular size) / 360o = 2 x 3.14 x 427 light-years x (0.044 arcsecond) / 360o = 9.1 x 10-5 light-years = 8.6 x 108 km This is 600 times the Suns diameter of 1.39 x 106 km. It is also about 6 times the distance between the Earth and Sun (1.5 x 108 km). 5. Eclipse Conditions. The Moons precise equatorial diameter is 3,476 km, and its orbital distance from Earth varies between 356,400 km and 406,700 km. The Suns diameter is 1,390,000 km and its distance from Earth ranges between 147.5 and 152.6 million km. a. Find the Moons angular size as its minimum and maximum distance from Earth. b. Find the Suns angular size as its minimum and maximum distance from Earth. c. Based on your answers to (a) and (b), is it possible to have a total solar eclipse when the Moon and Sun are both at their maximum distance?
Explain. Solution: a. angular size = physical size x 360o / (2 x distance) Minimum: angular size = 3,476 km x 360o / (2 x 3.14 x 356,400 km) = 0.559o Maximum: Angular size = 3,476 km x 360o / (2 x 3.14 x 406,700 km) = 0.49o b. Minimum: Angular size = 1,390,000 km x 360o / (2 x 3.14 x 147.6 x106 km) = 0.54o Maximum: Angular size = 1,390,000 km x 360o / (2 x 3.14 x 152.6 x 106 km) = 0.522o c. When both objects are at their maximum distance from Earth, both objects appear with their smallest angular size. At this time, the Suns angular size is 0.522o and the Moons angular size is 0.49o. The Moons angular size is significantly smaller than the Suns. So it could not fully cover the Suns disk. Since it cannot completely cover the Sun, there can be no total eclipse under these conditions. They can be only an annular or partial eclipse under these conditions.