Foundations of Astronomy (MindTap Course List)
14th Edition
ISBN: 9781337399920
Author: Michael A. Seeds, Dana Backman
Publisher: Cengage Learning
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Chapter 15, Problem 2SOP
To determine
The length that represents the time for Sun to orbit the Milky Way Galaxy.
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given:
a (distance to center of galaxy in AU) = 1,717,914,439 AU
P (suns orbital period in years) = 203,782,828.3 years
M (mass of milky way galaxy in solar masses) = 1.22 x 10^11 Msun
Question:
Assume the Milky Way Galaxy is made up entirely of stars like the Sun, i.e. on average each star has the mass of 1 MSun. Under this assumption, approximately how many stars are there in our galaxy? Express this answer in billions of stars (1 billion = 109).
How many times longer than the length of recorded history is the age of the universe?
I was also given that the length of recorded history is 10E11 s, and that the age of the universe is 10E18 s. But, when putting 10E18/10E11 = 10E7, I get the wrong answer.
Suppose we find an Earth-like planet around one of our nearest stellar neighbors, Alpha Centauri (located only 4.4 light-years away). If we launched a "generation ship" at a constant speed of 1500.00 km/s from Earth with a group of people whose descendants will explore and colonize this planet, how many years before the generation ship reached Alpha Centauri? (Note there are 9.46 ××1012 km in a light-year and 31.6 million seconds in a year.
Chapter 15 Solutions
Foundations of Astronomy (MindTap Course List)
Ch. 15 - What evidence can you give that we live in a...Ch. 15 - Prob. 2RQCh. 15 - Why didnt astronomers before Shapley realize how...Ch. 15 - Prob. 4RQCh. 15 - Prob. 5RQCh. 15 - Prob. 6RQCh. 15 - Which parts of a spiral galaxy comprise the...Ch. 15 - Prob. 8RQCh. 15 - Prob. 9RQCh. 15 - Prob. 10RQ
Ch. 15 - Prob. 11RQCh. 15 - Prob. 12RQCh. 15 - Prob. 13RQCh. 15 - Prob. 14RQCh. 15 - Prob. 15RQCh. 15 - Prob. 16RQCh. 15 - Prob. 17RQCh. 15 - Prob. 18RQCh. 15 - Prob. 19RQCh. 15 - Prob. 20RQCh. 15 - Prob. 21RQCh. 15 - Prob. 22RQCh. 15 - Prob. 23RQCh. 15 - Prob. 24RQCh. 15 - Prob. 25RQCh. 15 - Prob. 26RQCh. 15 - Rank these objects from oldest to youngest the...Ch. 15 - What evidence contradicts the top-down hypothesis...Ch. 15 - Prob. 29RQCh. 15 - The story of a process makes the facts easier to...Ch. 15 - Prob. 1PCh. 15 - Prob. 2PCh. 15 - Prob. 3PCh. 15 - Prob. 4PCh. 15 - Prob. 5PCh. 15 - Prob. 6PCh. 15 - Prob. 7PCh. 15 - Prob. 8PCh. 15 - If the Sun is 4.6 billion years old, how many...Ch. 15 - Prob. 10PCh. 15 - Prob. 11PCh. 15 - Prob. 12PCh. 15 - Prob. 13PCh. 15 - Prob. 14PCh. 15 - Prob. 15PCh. 15 - Prob. 1SOPCh. 15 - Prob. 2SOPCh. 15 - Prob. 2LTLCh. 15 - Prob. 3LTLCh. 15 - Prob. 4LTLCh. 15 - Prob. 5LTL
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- As 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_forwardUse the circumference and speed to figure out how long the Sun takes to go once around the Galaxy (the Sun's period, sometimes called the "galactic year.") Give your answer in Earth years. Be careful with your units.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_forward
- Try Now ... .... Listed below are some distances from Earth to other objects in the Milky Way galaxy. Convert each distance to light-years. (Each of these distances is less than one light-year. For an added challenge, convert each distance to light minutes or light seconds.) 1. The distance from Earth to the Moon is about 384,400 km. How many light-years is this? 2. The distance from Earth to Mars is about 784,000,000 km. How many light-years is this? 3. The distance from Earth to Pluto is about 5,750,000,000 km. How many light-years is this?arrow_forwardScience homework. Having troublearrow_forwardI attempted to answer this question and I'm not sure what I am doing wrong. My formula says A.S. = 206265 (separation/distance from observer) I know to convert to the same units, so I ended up with 80 Million Km being 8 x 10 ^ -6 LY Could you please explain each step especially for the part that I got wrong for both A and B?arrow_forward
- How far (in km) is 1 lightyear (ly) – the distance travelled by light in one Earth year? Use 299,732 km/s for the speed of light (c) and 1 year = 365 days. Show your solution and write your answer in both regular notation and scientific notation.arrow_forwardScientists agree that the age of the universe is about 4 billion years old, does the age of the rock referred in this problem support that? Why? Support your answer in at most 3 sentences.arrow_forwardCalculate the number of miles in a light-year, using 1.86 105 mi/s as the speed of light. (Hint: The number of seconds in a year, 365 days, will be useful.) Incorrect: Your answer is incorrect. mi/yarrow_forward
- - How far (in km) is 3.5 lightyears(ly) – the distance traveled by light in one Earth year? - How much is this same value in parsecs and (C) in astronomical units (AU)? Use 299,732 km/s for the speed of light (c) and 1 year = 365 days. Show your solution and write your answer in both regular notation and scientific notation.arrow_forwardIn the reading, you were told that there were roughly 10,000 galaxies in the image of the Hubble Ultra Deep Field alone. The image is roughly 10 square arcminutes and there are roughly 1.5*10^8 square arcminutes composing the entire sky. With that in mind and assuming that the Hubble Ultra Deep Field represents an average part of the sky, roughly how many galaxies may exist in the observable universe? (Please include commas for every factor of 1,000; for example 2,343,567,890)arrow_forwardRecall 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_forward
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