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 1, Problem 8P
How long does it take light to cross the diameter of our Milky Way Galaxy?
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Chapter 1 Solutions
Horizons: Exploring the Universe (MindTap Course List)
Ch. 1 - What is the largest dimension of which you have...Ch. 1 - What is the difference between our Solar System,...Ch. 1 - Why are light-years more convenient than miles,...Ch. 1 - Why is it difficult to detect planets orbiting...Ch. 1 - Prob. 5RQCh. 1 - What is the difference between the Milky Way and...Ch. 1 - What are the largest known structures in the...Ch. 1 - Prob. 8RQCh. 1 - How Do We Know? How does the scientific method...Ch. 1 - You and three of your friends have won an...
Ch. 1 - Think back to the last time you got a new phone...Ch. 1 - The diameter of Earth across the equator is 7928...Ch. 1 - The diameter of the Moon across its equator is...Ch. 1 - One astronomical unit is about 1.50108 km. Explain...Ch. 1 - Venus orbits 0.72 AU from the Sun. What is that...Ch. 1 - Light from the Sun takes 8 minutes to reach Earth....Ch. 1 - The Sun is almost 400 times farther from Earth...Ch. 1 - If the speed of light is 3.00105 km/s. how many...Ch. 1 - How long does it take light to cross the diameter...Ch. 1 - The nearest large galaxy to our n is about 2.5...Ch. 1 - How many galaxies like our own would it take Laid...Ch. 1 - In Figure 1-4, the division between daylight and...Ch. 1 - Prob. 2LTLCh. 1 - Prob. 3LTLCh. 1 - Prob. 4LTL
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- The distance from the Sun to the nearest star is about 4 1016 m. The Milky Way galaxy (Fig. P1.31) is roughly a disk of diameter 1021 in and thickness 1019 m. Find the order of magnitude of the number of stars in the Milky Way. Assume the distance between the Sun and our nearest neighbor is typical. Figure P1.31 The Milky Way galaxy.arrow_forwardEarth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m^2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. a) At the orbit of Venus (67 million km from the Sun). b) At the orbit of Jupiter (780 million km from the Sun). c) At the mean distance of Pluto (40 Astronomical Units).arrow_forwardOn a 1-to-10^19 scale our Milky Way galaxy would just about fit on a soccer field. On this scale, how far is the distance from the sun to alpha centauri (one of the closest stars to the sun, at a distance of 4.4 light years).arrow_forward
- Earth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. a) At the orbit of Jupiter (780 million km from the Sun).arrow_forwardThe closest star to Earth, after the Sun, is Proxima Centauri, which is just over 4 light years away from us. A light year is the distance traveled in one year by light, which travels at about 300,000 kilometers per second. What is the order of magnitude of the number of kilometers that separate us from Proxima Centauriarrow_forwardEarth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. a) At the orbit of Venus (67 million km from the Sun)arrow_forward
- Recall that Hubble’s Law is given by V=HR; this means that H has units of inverse seconds (1/sec). A convenient laboratory set of units is to give H in km per sec per megaparsec. A parsec is 3.26 light years and the speed of light is 3 X 105 km/sec. Use 3.156 X 107 sec/yr. The first data off the then new Hubble Space telescope suggested a value of H equal to 108 km per sec per megaparsec. What is H in inverse seconds? Hint divide by the number of km in a megaparsec.arrow_forwardThe Sun and the planets of the solar system are located in a spiral galaxy , called the Milky Way . Which numbered area on the model best represents the location of the Sun in this galaxy ?arrow_forwardEarth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m^2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. b) At the orbit of Jupiter (780 million km from the Sun).arrow_forward
- Suppose you need to find the height of a tall building. Standing 15 meters away from the base of the building, you aim a laser pointer at the closest part of the top of the building. You measure that the laser pointer is 7 degreestilted from pointing straight up. The laser pointer is held 2 meters above the ground. How tall is the building?arrow_forwardOur galaxy is approximately 100,000 light years in diameter and 2,000 light years thick through the plane of the galaxy. If we were to compare the ratio of the diameter galaxy and its thickness to the ratio of the diameter of a CD and its thickness (CD has a diameter of 12 cm and thickness of 0.6 mm), what would be the factor differentiating those ratios? Put differently, if the galaxy were scaled down to the diameter of a CD, how many times thicker or thinner would the galaxy be than the CD? (For example if it would be twice as thick, you would answer 2 and if it were twice as thin you would answer 0.5 (aka 1/2))arrow_forwardThe distance between the Earth and the Sun is ca. 150,000,000 km. If the Sun suddenly stopped shining, how long will it take before we find out?arrow_forward
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