Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Question
Chapter 44, Problem 30P
(a)
To determine
The reason behind the increase in the radius of the visible universe by a light year in a year.
(b)
To determine
The rate of increase of volume of the visible universe.
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Suppose that the universe were full of spherical objects, each of mass m and radius r . If the objects were distributed uniformly throughout the universe, what number density (#/m3) of spherical objects would be required to make the density equal to the critical density of our Universe?
Values:
m = 10 kg
r = 0.0399 m
Answer must be in scientific notation and include zero decimal places (1 sig fig --- e.g., 1234 should be written as 1*10^3)
Recent findings in astrophysics suggest that the observable universe can be modeled as a sphere of radius R=13.7x109 light-years=13.0 x 1025m with an average total mass density of about 1x10-26 kg/m3 Only about 4% of total mass is due to “ordinary” matter (such as protons, neutrons, and electrons). Estimate how much ordinary matter (in kg) there is in the observable universe. (For the light-year, see Problem 19.)
mathematician Archimedes, responding to a claim that the number of grains of sand was infinite,
calculated that the number of grains of sand needed to fill the universe was on the order of 1063. Our
understanding of the size of the universe has changed since then, and we now know that the
observable universe alone is a sphere with a radius of 1026 m. Estimating the size of a grain of sand,
A) Approximately how many grains of sand would fill the observable universe?
B) How many times larger or smaller is this number than Archimedes' result?
Chapter 44 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 44.2 - Prob. 44.1QQCh. 44.5 - Prob. 44.3QQCh. 44.5 - Prob. 44.4QQCh. 44.8 - Prob. 44.5QQCh. 44.8 - Prob. 44.6QQCh. 44 - Prob. 1PCh. 44 - Prob. 2PCh. 44 - Prob. 3PCh. 44 - Prob. 4PCh. 44 - Prob. 5P
Ch. 44 - Prob. 6PCh. 44 - Prob. 7PCh. 44 - Prob. 8PCh. 44 - Prob. 9PCh. 44 - Prob. 10PCh. 44 - Prob. 11PCh. 44 - Prob. 12PCh. 44 - Prob. 13PCh. 44 - Prob. 14PCh. 44 - Prob. 15PCh. 44 - Prob. 16PCh. 44 - Prob. 17PCh. 44 - Prob. 18PCh. 44 - Prob. 20PCh. 44 - Prob. 21PCh. 44 - Prob. 22PCh. 44 - Prob. 23PCh. 44 - Prob. 24PCh. 44 - Prob. 25PCh. 44 - Prob. 26PCh. 44 - Prob. 27PCh. 44 - Prob. 29PCh. 44 - Prob. 30PCh. 44 - The various spectral lines observed in the light...Ch. 44 - Prob. 33PCh. 44 - Prob. 34APCh. 44 - Prob. 35APCh. 44 - Prob. 36APCh. 44 - Prob. 37APCh. 44 - Prob. 38APCh. 44 - Prob. 39APCh. 44 - Prob. 40APCh. 44 - An unstable particle, initially at rest, decays...Ch. 44 - Prob. 42APCh. 44 - Prob. 43APCh. 44 - Prob. 44APCh. 44 - Prob. 45APCh. 44 - Prob. 46CPCh. 44 - Prob. 47CPCh. 44 - Prob. 48CPCh. 44 - Prob. 49CP
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- How many galaxies like our own would it take if they were placed edge-to-edge to reach the nearest galaxy? (Hint: See Problems 11 and 12.)arrow_forwardThe nearest neutron star (a collated star made primarily of neutrons) is about 3.00 1018 m away from Earth. Given that the Milky Way galaxy (Fig. P1.81) is roughly a disk of diameter 1021 m and thickness 1019 m, estimate the number of neutron stars in the Milky Way to the nearest order of magnitude. Figure P1.81arrow_forwardI asked the following question and was given the attached solution: Suppose that the universe were full of spherical objects, each of mass m and radius r . If the objects were distributed uniformly throughout the universe, what number density (#/m3) of spherical objects would be required to make the density equal to the critical density of our Universe? Values: m = 4 kg r = 0.0407 m Answer must be in scientific notation and include zero decimal places (1 sig fig --- e.g., 1234 should be written as 1*10^3) I don't follow the work and I got the wrong answer, so please help and show your work as I do not follow along easily thanksarrow_forward
- To get an idea of how empty deep space is on the average, perform the following calculations: (a) Find the volume our Sun would occupy if it had an average density equal to the critical density of 10-26 kg / m3 thought necessary to halt the expansion of the universe. (b) Find the radius of a sphere of this volume in light years. (c) What would this radius be if the density were that of luminous matter, which is approximately 5% that of the critical density? (d) Compare the radius found in part (c) with the 4-ly average separation of stars in the arms of the Milky Way.arrow_forwardAssume the observable Universe is charge neutral, and that it contains n nuclei (hydrogen plus helium nuclei, ignoring other elements). Take the helium mass fraction as 1/4. How many electrons are there in the observable Universe? Enter your answer in scientific notation with one decimal place. Value: n = 4*1080arrow_forwardIf the current mass of the universe consists of 75% protons and 25% 4He, what is the current ratio of protons to neutrons?arrow_forward
- Consider a universe identical to ours, but where the only difference is that the mass of the proton and the one of the neutron are the same. What would have been the neutron fraction 39 seconds after the Big Bang? Consider for the neutron a mean lifetime of 886 s. Enter your answer to 3 decimal places. Answer:arrow_forwardThe Universe is approximately 13.8 Billion years old. What is the volume of the visible universe in m3?arrow_forwardApproximate values of length (in meters) 107 Diameter of Earth 1011 Distance from Earth to Sun 1016 Distance traveled by light in one year 1021 Diameter of the Milky Way Galaxy 1022 Distance from Earth to the nearest galaxy 1025 Distance from Earth to the edge of the known universearrow_forward
- Using our example from the previous unit, let's try to determine the Hubble time for this example universe. You were given that a good representative galaxy receded at a speed of 4000 km/s and was found to be 20 Mpc away. With that in mind, what would the age of that universe be in years (aka what is that universe's Hubble time)? Go ahead and take the number of kilometers per Mpc to be approximately 3.1*10^19 km/Mpc. While this problem may look scary at first, this is really just bringing you full circle to one of the unit conversion problems you encountered at the beginning of this course.arrow_forwardNow let’s determine when the universe “started its trip”. The distance is 800 Mpc, but first convert Mpcinto km because the rate, or velocity, is in km/sec.800 Mpc = _____________________kmUse equation (E) to determine how many seconds ago the universe started:_____________________secsarrow_forwardThe matter density in the Universe today is ?m=2.7×10−27kgm−3. What would be the value of the density parameter, Ω0, if the Hubble constant had the value H0 = 38 km/s/Mpc?arrow_forward
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