Understanding Our Universe
3rd Edition
ISBN: 9780393614428
Author: PALEN, Stacy, Kay, Laura, Blumenthal, George (george Ray)
Publisher: W.w. Norton & Company,
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Chapter 12.2, Problem 12.2CYU
To determine
The reason why Sun will become larger and more luminous when there is no hydrogen in its core.
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Betelgeuse is a nearby supergiant that will eventually explode into a supernova. Let's see
how awesome it would look. At peak brightness, the supernova will have a luminosity of
about 10 billion times the Sun. It is 600 light-years away. All stellar brightnesses are
compared with Vega, which has an intrinsic luminosity of about 60 times the Sun, a distance
of 25 light-years, an absolute magnitude of 0.6 and an apparent magnitude of 0 (by
definition).
a) At peak brightness, how many times brighter will Betelgeuse be than Vega?
b) Approximately what apparent magnitude does this correspond to?
c) The Sun is about -26.5 apparent magnitude. What fraction of the Sun's brightness will
Betelgeuse be?
Using solar units, we find that a star has 4 times the luminosity of the Sun, a mass 1.25 times the mass of the Sun, and a surface temperature of 4090 K (take the Sun's surface temperature to be 5784 K for the sake of this problem). This means the star has a radius of.................... solar radii and is a .................... star (use the classification).
As a star forms out of gravitational collapse of a cloud:
it gains gravitational energy and loses kinetic energy.
the star is much hotter than the initial cloud. The kinetic energy gain equals the loss in gravitational energy.
the star is much hotter than the initial cloud, however, the kinetic energy gain is only half the loss in
gravitational energy.
the interior of the star is much hotter than the surface layers because the central density is higher.
Chapter 12 Solutions
Understanding Our Universe
Ch. 12.1 - Prob. 12.1CYUCh. 12.2 - Prob. 12.2CYUCh. 12.3 - Prob. 12.3CYUCh. 12.4 - Prob. 12.4CYUCh. 12.5 - Prob. 12.5CYUCh. 12.6 - Prob. 12.6CYUCh. 12 - Prob. 1QAPCh. 12 - Prob. 2QAPCh. 12 - Prob. 3QAPCh. 12 - Prob. 4QAP
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- Let's calculate how much mass will be lost by the Sun during the course of its main-sequence lifetime. While it is on the main sequence, a star converts about 10% of the hydrogen initially present into helium (remember that it is only the core of the star that is hot enough for fusion). During nuclear fusion, the Sun converts about 0.7% of the core hydrogen mass into energy. The total mass of the Sun is 2 × 1030 kg. How many kilograms of mass will be converted to energy during the main sequence stage of the Sun's life? What is the ratio of this lost mass to the Earth's mass (6 × 1024 kg)? In other words, how many Earths of mass will be turned into energy?arrow_forwardAfter the hydrogen in the core of a star is used up the star moves off the main-sequence, and as the core becomes denser and hotter other fusions could happen. When the core mass density is on the order of 106 kg m−3, at what temperature would electron degeneracy pressure become important in the core? For the electron degeneracy pressure to be more important, should the temperature increase or decrease?arrow_forwardOne way to calculate the radius of a star is to use its luminosity and temperature and assume that the star radiates approximately like a blackbody. Astronomers have measured the characteristics of central stars of planetary nebulae and have found that a typical central star is 16 times as luminous and 20 times as hot (about 110,000 K) as the Sun. Find the radius in terms of the Sun’s. How does this radius compare with that of a typical white dwarf?arrow_forward
- A Crude Analysis: In about 5 billion years, the Sun is going to look a lot different. Our sun is going to turn into a red-giant, a bigger star whose core temperature is much higher than the Sun's current core temperature (you will learn about the red giants in the coming weeks). Assume the core temperature of the red-giant phase of the Sun does not go beyond 100 million degrees. Do you think the temperature is high enough for helium fusion to occur? Note that this question is about helium fusion not hydrogen fusion. How are you going about proving your claim? Question: What temperature in degrees Kelvin must the red-giant sun be at to allow for the helium-helium interactions to take place not considering the Quantum Mechanical effects (i.e. what temperature would allow helium atoms to breach the helium-helium potential wall without help from Quantum Mechanics)? Use wolfram alpha to find the values for the constants. Round your answer to two decimal places. Your answer i [ Select ] 1.47…arrow_forwardA star such as our Sun will eventually evolve to a “red giant” star and then to a “white dwarf” star. A typical white dwarf is approximately the size of Earth, and its surface temperature is about 2.4 × 104 K. A typical red giant has a surface temperature of 3.2 × 103 K and a radius ~90000 times larger than that of a white dwarf. Take the radius of the red giant to be 6 × 1010 m. What is the average radiated power per unit area of the red giant?_________W/m2 What is the average radiated power per unit area of the white-dwarf?________W/m2 What is the total power radiated by the red giant? _________W What is the total power radiated by the white dwarf? ________W Please show full work! Thank you!arrow_forwardWhen the Sun becomes a red giant, it's luminosity will be 2000 times its current value. The solar flux at Earth will also increase by a factor of 2000. Neglecting the greenhouse effect, the surface temperature of the earth is determined by thermal equilibrium: the flux of radiation absorbed equals the flux of radiation emitted. This means the Earth's surface flux must also increase by a factor of 2000. If the current average surface temperature is 58 degrees F, what will the average surface temperature be when the Sun is a red giant. Express your answer in units of degrees Fahrenheit. [Hint: Recall that the Stefan-Boltzmann law says that the flux F emitted by a blackbody is related to its surface temperature T (measured in Kelvins) is F=σT4 . Use this law in the form of a ratio, expressing T in Kelvins. Then convert back to Fahrenheit.]arrow_forward
- A red giant star might have radius = 104 times the solar radius, and luminosity = 1730 times solar luminosity. Use the data given below to calculate the temperature at the surface of the red giant star. Data: solar radius R = 7 x 108 meters solar luminosity L = 4 x 1026 watts Stefan-Boltzmann constant a = 5.67 x 10-8 W m² K-4 (in K) A: 1226 OB: 1434 OC: 1678 OD: 1963 OE: 2297 OF: 2688 OG: 3145 OH: 3679arrow_forwardThe apparent magnitude of a star is observed to vary between m = +0.4 and m = +0.1 because the star pulsates and hence continuously changes its radius and temperature. When at its peak brightness, the star’s radius has increased by a factor of two compared to its value at the mini- mum brightness. Determine the value of T+/T−, where T+ is the temperature when the star is at its peak brightness and T− is the temperature when the star is at it minimum brightness. Note: we expect T+/T− < 1 because the star’s temperature decreases as its radius increases.arrow_forwardA star population is composed of stars with masses in the range between 1M and 150M. The initial mass function is = (M/M)-2.3, where = (Mo). The luminosity of a star scales with its mass as L/L = (M/M) 3.3. Calculate the percentage of the total luminosity of the stars in the population which is produced by stars with mass between 120M and 150M.arrow_forward
- The mass-luminosity relation describes the mathematical relationship between luminosity and mass for main sequence stars. It describes how a star with a mass of 4 M⊙ would have a luminosity of ______ L⊙. If a star has a radius 1/2 that of the Sun and a temperature 4 that of the Sun, how many times higher is the star's luminosity than that of the Sun? (If it is smaller by a factor of 8, you would write 0.125 because 1/8=0.125) If a star has a radius 2 times larger than the Sun's and a luminosity 1/4th that of the Sun, how many times higher is the star's temperature than that of the Sun? (If it is smaller by a factor of 8, you would write 0.125 because 1/8=0.125) If a star has a surface temperature 2 times lower than the Sun's and a luminosity the same as the Sun, how many times larger is the star than the Sun? (If it is smaller by a factor of 8, you would write 0.125 because 1/8=0.125)arrow_forwardA star begins its life with a mass of 5 MSunbut ends its life as a white dwarf with a mass of 0.8 MSun. List the stages in the star’s life during which it most likely lost some of the mass it started with. How did mass loss occur in each stage?arrow_forwardThe text says a star does not change its mass very much during the course of its main-sequence lifetime. While it is on the main sequence, a star converts about 10% of the hydrogen initially present into helium (remember it’s only the core of the star that is hot enough for fusion). Look in earlier chapters to find out what percentage of the hydrogen mass involved in fusion is lost because it is converted to energy. By how much does the mass of the whole star change as a result of fusion? Were we correct to say that the mass of a star does not change significantly while it is on the main sequence?arrow_forward
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