EBK LOOSE-LEAF VERSION OF UNIVERSE
11th Edition
ISBN: 9781319227975
Author: KAUFMANN
Publisher: VST
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 20, Problem 40Q
(a)
To determine
The diameter of the nebula SMC N76 (in parsec) by using the information given in the caption of the figure 20-13.
(b)
To determine
The comparison between the diameter calculated in part (a) for nebula SMC N76 and the planetary nebula depicted in figure 20-7.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
A planetary nebula expanded in radius 0.3 arc seconds in 30 years. Doppler measurements show the nebula is expanding at a rate of 35 km/s. How far away is the nebula in parsecs?
First, determine what distance the nebular expanded in parsecs during the time mentioned. Δd = vpc/sTs
So we first need to convert the rate into pc/s and the time into seconds:
vpc/s = vkm/s (1 pc / 3.09 x 1013km)
vpc/s = ?
Ts = (Tyr)(365 days/yr)(24 hrs/day)(3600 s/hr)
Ts = ? s
Δd= vpc/sTs
Therefore, Δd = ? pc
Suppose two protostars form at the same time, one with a mass of 0.5MSunSun
[Select ALL answers that are true in alphabetical order]A) The 10MSun protostar will have a smaller change in surface temperature during this phase than the 0.5MSun protostar.B) The 10MSun protostar will reach the main sequence cooler and fainter than the 0.5MSun protostar.C) The 10MSun star will end its main-sequence life before the 0.5MSun star even completes its protostar stage.D) The 10MSun protostar will have a smaller change in luminosity during the sequence shown than the 0.5MSun protostar.E) The 10MSun protostar will be much more luminous than the 0.5MSun protostar.
. The radius of the nebula is about 0.401 light-years. The gas is expanding away from the star at a rate of about 37
kilometers/second
. Considering that distance = velocity x time, calculate how long ago the gas left the star if its speed has been constant the whole time. Make sure you use consistent units for time, speed, and distance. Answer in years.
Chapter 20 Solutions
EBK LOOSE-LEAF VERSION OF UNIVERSE
Ch. 20 - Prob. 1CCCh. 20 - Prob. 2CCCh. 20 - Prob. 3CCCh. 20 - Prob. 4CCCh. 20 - Prob. 5CCCh. 20 - Prob. 6CCCh. 20 - Prob. 7CCCh. 20 - Prob. 8CCCh. 20 - Prob. 9CCCh. 20 - Prob. 10CC
Ch. 20 - Prob. 11CCCh. 20 - Prob. 12CCCh. 20 - Prob. 13CCCh. 20 - Prob. 14CCCh. 20 - Prob. 15CCCh. 20 - Prob. 16CCCh. 20 - Prob. 17CCCh. 20 - Prob. 18CCCh. 20 - Prob. 1QCh. 20 - Prob. 2QCh. 20 - Prob. 3QCh. 20 - Prob. 4QCh. 20 - Prob. 5QCh. 20 - Prob. 6QCh. 20 - Prob. 7QCh. 20 - Prob. 8QCh. 20 - Prob. 9QCh. 20 - Prob. 10QCh. 20 - Prob. 11QCh. 20 - Prob. 12QCh. 20 - Prob. 13QCh. 20 - Prob. 14QCh. 20 - Prob. 15QCh. 20 - Prob. 16QCh. 20 - Prob. 17QCh. 20 - Prob. 18QCh. 20 - Prob. 19QCh. 20 - Prob. 20QCh. 20 - Prob. 21QCh. 20 - Prob. 22QCh. 20 - Prob. 23QCh. 20 - Prob. 24QCh. 20 - Prob. 25QCh. 20 - Prob. 26QCh. 20 - Prob. 27QCh. 20 - Prob. 28QCh. 20 - Prob. 29QCh. 20 - Prob. 30QCh. 20 - Prob. 31QCh. 20 - Prob. 32QCh. 20 - Prob. 33QCh. 20 - Prob. 34QCh. 20 - Prob. 35QCh. 20 - Prob. 36QCh. 20 - Prob. 37QCh. 20 - Prob. 38QCh. 20 - Prob. 39QCh. 20 - Prob. 40QCh. 20 - Prob. 41QCh. 20 - Prob. 42QCh. 20 - Prob. 43QCh. 20 - Prob. 44QCh. 20 - Prob. 45QCh. 20 - Prob. 46QCh. 20 - Prob. 47QCh. 20 - Prob. 48QCh. 20 - Prob. 49QCh. 20 - Prob. 50QCh. 20 - Prob. 51QCh. 20 - Prob. 52QCh. 20 - Prob. 53QCh. 20 - Prob. 54QCh. 20 - Prob. 55QCh. 20 - Prob. 56QCh. 20 - Prob. 57QCh. 20 - Prob. 58QCh. 20 - Prob. 59QCh. 20 - Prob. 60QCh. 20 - Prob. 61QCh. 20 - Prob. 62QCh. 20 - Prob. 63QCh. 20 - Prob. 64QCh. 20 - Prob. 65QCh. 20 - Prob. 66QCh. 20 - Prob. 67QCh. 20 - Prob. 68QCh. 20 - Prob. 69QCh. 20 - Prob. 70QCh. 20 - Prob. 71QCh. 20 - Prob. 72QCh. 20 - Prob. 73QCh. 20 - Prob. 74QCh. 20 - Prob. 75Q
Knowledge Booster
Learn more about
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
- Why do nebulae near hot stars look red? Why do dust clouds near stars usually look blue?arrow_forwardYou can estimate the age of the planetary nebula in image (c) in Figure 22.18. The diameter of the nebula is 600 times the diameter of our own solar system, or about 0.8 light-year. The gas is expanding away from the star at a rate of about 25 mi/s. Considering that distance=velocitytime , calculate how long ago the gas left the star if its speed has been constant the whole time. Make sure you use consistent units for time, speed, and distance. Figure 22.18 Gallery of Planetary Nebulae. This series of beautiful images depicting some intriguing planetary nebulae highlights the capabilities of the Hubble Space Telescope. (a) Perhaps the best known planetary nebula is the Ring Nebula (M57), located about 2000 lightyears away in the constellation of Lyra. The ring is about 1 light-year in diameter, and the central star has a temperature of about 120,000 °C. Careful study of this image has shown scientists that, instead of looking at a spherical shell around this dying star, we may be looking down the barrel of a tube or cone. The blue region shows emission from very hot helium, which is located very close to the star; the red region isolates emission from ionized nitrogen, which is radiated by the coolest gas farthest from the star; and the green region represents oxygen emission, which is produced at intermediate temperatures and is at an intermediate distance from the star. (b) This planetary nebula, M2-9, is an example of a butterfly nebula. The central star (which is part of a binary system) has ejected mass preferentially in two opposite directions. In other images, a disk, perpendicular to the two long streams of gas, can be seen around the two stars in the middle. The stellar outburst that resulted in the expulsion of matter occurred about 1200 years ago. Neutral oxygen is shown in red, once-ionized nitrogen in green, and twice-ionized oxygen in blue. The planetary nebula is about 2100 light-years away in the constellation of Ophiuchus. (c) In this image of the planetary nebula NGC 6751, the blue regions mark the hottest gas, which forms a ring around the central star. The orange and red regions show the locations of cooler gas. The origin of these cool streamers is not known, but their shapes indicate that they are affected by radiation and stellar winds from the hot star at the center. The temperature of the star is about 140,000 °C. The diameter of the nebula is about 600 times larger than the diameter of our solar system. The nebula is about 6500 light-years away in the constellation of Aquila. (d) This image of the planetary nebula NGC 7027 shows several stages of mass loss. The faint blue concentric shells surrounding the central region identify the mass that was shed slowly from the surface of the star when it became a red giant. Somewhat later, the remaining outer layers were ejected but not in a spherically symmetric way. The dense clouds formed by this late ejection produce the bright inner regions. The hot central star can be seen faintly near the center of the nebulosity. NGC 7027 is about 3000 light-years away in the direction of the constellation of Cygnus. (credit a: modification of work by NASA, ESA, and the Hubble Heritage (STScI/AURA)-ESA/Hubble Collaboration; credit b: modification of work by Bruce Balick (University of Washington), Vincent Icke (Leiden University, The Netherlands), Garrelt Mellema (Stockholm University), and NASA; credit c: modification of work by NASA, The Hubble Heritage Team (STScI/AURA); credit d: modification of work by H. Bond (STScI) and NASA)arrow_forwardPlace the following events in the formation of stars in the proper chronological sequence, with the oldest first and the youngest last. w. the gas and dust in the nebula flatten to a disk shape due to gravity and a steadily increasing rate of angular rotation x. a star emerges when the mass is great enough and the temperature is high enough to trigger thermonuclear fusion in the core y. the rotation of the nebular cloud increases as gas and dust concentrates by gravity within the growing protostar in the center z. some force, perhaps from a nearby supernova, imparts a rotation to a nebular cloud y, then z, then w, then x z, then y, then w, then x w, then y, then z, then x z, then x, then w, then y x, then z, then y, then w MacBook Air on .H. O O O Oarrow_forward
- result, the -1 from a T Tauri star, estimate the mass density of the wind at a distance of 100 AU from the star. (Hint: Refer to Example 11.2.1.) Compare your answer with the density of the giant molecular cloud agpalain Example 12.2.1. 12.19 Assuming a mass loss rate of 10-7 Mo yr¹ and a stellar wind velocity of 80 km s -niem-1299 eararrow_forwardWhite Dwarf Size II. The white dwarf, Sirius B, contains 0.98 solar mass, and its density is about 2 x 106 g/cm?. Find the radius of the white dwarf in km to three significant digits. (Hint: Density = mass/volume, and the volume of a 4 sphere is Tr.) 3 km Compare your answer with the radii of the planets listed in the Table A-10. Which planet is this white dwarf is closely equal to in size? I Table A-10 I Properties of the Planets ORBITAL PROPERTIES Semimajor Axis (a) Orbital Period (P) Average Orbital Velocity (km/s) Orbital Inclination Planet (AU) (106 km) (v) (days) Eccentricity to Ecliptic Mercury 0.387 57.9 0.241 88.0 47.9 0.206 7.0° Venus 0.723 108 0.615 224.7 35.0 0.007 3.4° Earth 1.00 150 1.00 365.3 29.8 0.017 Mars 1.52 228 1.88 687.0 24.1 0.093 1.8° Jupiter 5.20 779 11.9 4332 13.1 0.049 1.30 Saturn 9.58 1433 29.5 10,759 9.7 0.056 2.5° 30,799 60,190 Uranus 19.23 2877 84.3 6.8 0.044 0.8° Neptune * By definition. 30.10 4503 164.8 5.4 0.011 1.8° PHYSICAL PROPERTIES (Earth = e)…arrow_forwardThe difference in absolute magnitude between two objects is related to their fluxes by the flux-magnitude relation: FA / FB = 2.51(MB - MA) A distant galaxy contains a supernova with an absolute magnitude of -19. If this supernova were placed next to our Sun (M = +4.8) and you observed both of them from the same distance, how much more flux would the supernova emit than the Sun? Fsupernova / FSun = ?arrow_forward
- 12.1 In a certain part of the North American Nebula, the amount of interstellar extinction in the visual wavelength band is 1.1 magnitudes. The thickness of the nebula is estimated to be 20 pc, and it is located 700 pc from Earth. Suppose that a B spectral class main-sequence star is observed in the direction of the nebula and that the absolute visual magnitude of the star is known to be My = -1.1 from spectroscopic data. Neglect any other sources of extinction between the observer and the nebula. (a) Find the apparent visual magnitude of the star if it is lying just in front of the nebula. (b) Find the apparent visual magnitude of the star if it is lying just behind the nebula. Problems 443 (c) Without taking the existence of the nebula into consideration, based on its apparent mag- nitude, how far away does the star in part (b) appear to be? What would be the percentage error in determining the distance if interstellar extinction were neglected?arrow_forwardIf the hottest star in the Carina Nebula has a surface temperature of 51,000 K, at what wavelength (in nm) does it radiate the most energy? Hint: Use Wien's law: ?max = 2.90 ✕ 106 nm · K T How does that compare with 91.2 nm, the wavelength of photons with just enough energy to ionize hydrogen? -The wavelength calculated above is shorter than 91.2 nm. Photons at this calculated wavelength will have more than enough energy to ionize hydrogen. -The wavelength calculated above is longer than 91.2 nm. Photons at this calculated wavelength will have more than enough energy to ionize hydrogen. -The wavelength calculated above is shorter than 91.2 nm. Photons at this calculated wavelength will not have enough energy to ionize hydrogen. -The wavelength calculated above is longer than 91.2 nm. Photons at this calculated wavelength will not have enough energy to ionize hydrogen.arrow_forwardUse the following formula (fitted to data) M = -4x10-13n gR Mo/year for the mass loss of asymptotic giant branch stars to: a) explain why L, g (gravity on surface), and R enter the equation the way they do (nominator or denominator). b) show that the expression for M is equivalent to LR M = -4x10-13n Mo/year M c) estimate the mass loss rate of a star with M = 1 Mo, L = 7000 Lo, T = 3000 K. Assume n = 1 and use the Stefan-Boltzmann equation to calculate R (in Ro).arrow_forward
- Assuming that at the end of the He burning phase of the stellar core (r < R_core) has no H or He or other metals and is composed completely of Carbon, X=Y=0, X_c = 1 ; The envelope above the core has a normal stellar composition ( r > R_core). Calculate the length of time in years that a 1M_sol and 10M_sol star will live on the horizontal branch or the time between the start and end of the He burning phase. Assume that the normal relationship between mass and luminosity holds for horizontal branch stars. Please be as detailed as possiblearrow_forwardLook at the four stages shown in Figure 21.8. In which stage(s) can we see the star in visible light? In infrared radiation? Figure 21.8 Formation of a Star. (a) Dense cores form within a molecular cloud. (b) A protostar with a surrounding disk of material forms at the center of a dense core, accumulating additional material from the molecular cloud through gravitational attraction. (c) A stellar wind breaks out but is confined by the disk to flow out along the two poles of the star. (d) Eventually, this wind sweeps away the cloud material and halts the accumulation of additional material, and a newly formed star, surrounded by a disk, becomes observable. These sketches are not drawn to the same scale. The diameter of a typical envelope that is supplying gas to the newly forming star is about 5000 AU. The typical diameter of the disk is about 100 AU or slightly larger than the diameter of the orbit of Pluto.arrow_forwardStars that have masses approximately 0.8 times the mass of the Sun take about 18 billion years to turn into red giants. How does this compare to the current age of the universe? Would you expect to find a globular cluster with a main-sequence turnoff for stars of 0.8 solar mass or less? Why or why not?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Foundations of Astronomy (MindTap Course List)PhysicsISBN:9781337399920Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
Stars and Galaxies (MindTap Course List)PhysicsISBN:9781337399944Author:Michael A. SeedsPublisher:Cengage Learning
AstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStax
Horizons: Exploring the Universe (MindTap Course ...PhysicsISBN:9781305960961Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
Stars and GalaxiesPhysicsISBN:9781305120785Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
Foundations of Astronomy (MindTap Course List)
Physics
ISBN:9781337399920
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Stars and Galaxies (MindTap Course List)
Physics
ISBN:9781337399944
Author:Michael A. Seeds
Publisher:Cengage Learning
Astronomy
Physics
ISBN:9781938168284
Author:Andrew Fraknoi; David Morrison; Sidney C. Wolff
Publisher:OpenStax
Horizons: Exploring the Universe (MindTap Course ...
Physics
ISBN:9781305960961
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Stars and Galaxies
Physics
ISBN:9781305120785
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning