Horizons: Exploring the Universe (MindTap Course List)
14th Edition
ISBN: 9781305960961
Author: Michael A. Seeds, Dana Backman
Publisher: Cengage Learning
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Textbook Question
Chapter 2, Problem 4P
By what factor is sunlight brighter than moonlight? (Hint: See Figure 2.6 and Table 2-1.)
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2
Part 3
1. The diameter of the Sun is 1,391,400 km. The diameter of the Moon is 3,474.8 km. Find
the ratio, r= Dsa/Dsvan between the sizes.
2. From the point of view of an obs erver on Eanth (consider the Earth as a point-like object),
during the eclipse, the Moon covers the Sun exactly. Sketch a picture to illustrate this
fact. Use a nuler to get a straight line. Your drawing does not need to be in scale.
3. The Sun is 1 Astronomical Unit (AU) away from the Earth. Find the distance between the
Earth and the Moon in AU's using the ratio of similar triangles. Show your work.
DEM=
AU.
Convert this to kilometers. Use 1 AU = 149,600,000 km.
DEM =
km.
Chapter 2 Solutions
Horizons: Exploring the Universe (MindTap Course List)
Ch. 2 - Why have astronomers added modern constellations...Ch. 2 - What is the difference between asterism d a...Ch. 2 - What characteristic do starts in a constellation...Ch. 2 - Do people from other cultures on Earth see the...Ch. 2 - Prob. 5RQCh. 2 - Prob. 6RQCh. 2 - What does the word apparent mean in apparent...Ch. 2 - In what ways is the celestial sphere a scientific...Ch. 2 - Why do astronomers use the word on to describe...Ch. 2 - Earth did not rotate, could you define the...
Ch. 2 - Where would you go on Earth if you wanted to be...Ch. 2 - Prob. 12RQCh. 2 - Explain h to make a simple astronomical...Ch. 2 - 14. Why does the number of circumpolar...Ch. 2 - How could you detect Earths precession t examining...Ch. 2 - How Do We know? How can a scientific model be...Ch. 2 - Using stars from one or more of the “official”...Ch. 2 - Prob. 2DQCh. 2 - Prob. 1PCh. 2 - If two stars differ by 8.6 magnitudes, what is...Ch. 2 - Prob. 3PCh. 2 - By what factor is sunlight brighter than...Ch. 2 - If you are at a latitude of 35° north of Earths...Ch. 2 - Prob. 1LTLCh. 2 - Prob. 2LTL
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- Next you will (1) convert your measurement of the semi-major axis from arcseconds to AU, (2) convert your measurement of the period from days to years, and (3) calculate the mass of the planet using Newton's form of Kepler's Third Law. Use Stellarium to find the distance to the planet when Skynet took any of your images, in AU. Answer: 4.322 AU Use this equation to determine a conversion factor from 1 arcsecond to AU at the planet's distance. You will need to convert ? = 1 arcsecond to degrees first. Answer: 2.096e-5 AU (2 x 3.14 x 4.322 x (.000278/360) = 2.096e-5) Next, use this number to convert your measurement of the moon's orbital semi-major axis from arcseconds to AU. A) Calculate a in AU. B) Convert your measurement of the moon's orbital period from days to years. C) By Newton's form of Kepler's third law, calculate the mass of the planet. D) Finally, convert the planet's mass to Earth masses: 1 solar mass = 333,000 Earth masses.arrow_forwardAs we discuss in class, the radius of the Earth is approximately 6370 km. Theradius of the Sun, on the other hand, is approximately 700,000 km. The Sun is located,on average, one astronomical unit (1 au) from the Earth. Imagine that you stand near Mansueto Library, at the corner of 57th and Ellis.Mansueto’s dome is 35 feet (10.7 meters) high. Let’s imagine we put a model of theSun inside the dome, such that it just fits — that is, the model Sun’s diameter is 35 feet The nearest star to the Solar System outside of the Sun is Proxima Centauri,which is approximately 4.2 light years away. Given the scale model outlined above,how far would a model Proxima Centauri be placed from you? Give your answer inmiles and kmarrow_forwardCalculate the total amount of radiative energy per second intercepted by Mars from the Sun using the flux of radiation from the Sun at Mars' orbital radius. Flux of radiation from the Sun at Mars' orbital radius is 597 W m-2. The luminosity of the Sun Ls = 3.8×1026 W. Mars orbits at a distance of 2.25×1011 m (1.5 AU) from the Sun. Note: Consider carefully the cross-sectional area Mars presents to the outwards flow of radiative energy when answering this question.arrow_forward
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