Loose Leaf For Explorations: Introduction To Astronomy
9th Edition
ISBN: 9781260432145
Author: Thomas T Arny, Stephen E Schneider Professor
Publisher: McGraw-Hill Education
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Chapter 15, Problem 3TY
To determine
The reason that the spectrum of a supernova shows lots of iron and nickel, but no hydrogen.
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Betelgeuse is a nearby supergiant that will eventually explode into a supernova. Let's see
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Chapter 15 Solutions
Loose Leaf For Explorations: Introduction To Astronomy
Ch. 15 - Prob. 1QFRCh. 15 - Prob. 2QFRCh. 15 - Prob. 3QFRCh. 15 - Prob. 4QFRCh. 15 - Prob. 5QFRCh. 15 - Prob. 6QFRCh. 15 - Prob. 7QFRCh. 15 - Prob. 8QFRCh. 15 - Prob. 9QFRCh. 15 - Prob. 10QFR
Ch. 15 - Prob. 11QFRCh. 15 - Prob. 12QFRCh. 15 - Prob. 13QFRCh. 15 - What is nonthermal radiation?Ch. 15 - What happens when a gravitational wave moves? What...Ch. 15 - What is a black hole? Are they truly black? What...Ch. 15 - Prob. 17QFRCh. 15 - Prob. 18QFRCh. 15 - Prob. 19QFRCh. 15 - Prob. 20QFRCh. 15 - Prob. 1TQCh. 15 - Prob. 2TQCh. 15 - Prob. 3TQCh. 15 - Prob. 5TQCh. 15 - Prob. 6TQCh. 15 - Prob. 7TQCh. 15 - Prob. 8TQCh. 15 - Suppose you jumped into a black hole feet first....Ch. 15 - Prob. 10TQCh. 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 - Prob. 9PCh. 15 - Prob. 10PCh. 15 - Prob. 11PCh. 15 - Prob. 12PCh. 15 - Prob. 1TYCh. 15 - Prob. 2TYCh. 15 - Prob. 3TYCh. 15 - Prob. 4TYCh. 15 - Prob. 5TYCh. 15 - Prob. 6TYCh. 15 - What evidence leads astronomers to believe that...Ch. 15 - (15.3) The Schwarzschild radius of a body is (a)...Ch. 15 - Prob. 9TYCh. 15 - Prob. 10TY
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- A supernova's energy is often compared to the total energy output of the Sun over its lifetime. Using the Sun's current luminosity, calculate the total solar energy output, assuming a 1010 year main-sequence lifetime. Using Einstein's formula E = mc? calculate the equivalent amount of mass, expressed in Earth masses. [Hint: The total energy output of the Sun over its lifetime is given by its current luminosity times the number of seconds in a year times its ten billion-year lifetime; Week 5 slide 4; mass of earth = 6x1024kg; c = 3x10®m/s. Your answer should be 200-300 Earth masses.]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_forwardConsider an M-dwarf star of mass 0.1M⊙ and luminosity 10−3L⊙. When the star joins the main sequence 75% of its mass is hydrogen and 25% is helium. The star is fully convective and hence the interior is always fully mixed. Fusion reactions provide all of the luminosity of the star, and each reaction converts 4 hydrogen nuclei into 1 helium nucleus. The combined mass of 4 hydrogen nuclei is 6.690×10−27kg and the mass of one helium nucleus is 6.643×10−27kg. Estimate the main sequence life time of this star, assuming that the luminosity is constant throughout the star's life time. Express your answer in Gyr. The solar mass M⊙=2×1030kg and the solar luminosity L⊙=3.83×1026W.arrow_forward
- For a main sequence star with luminosity L, how many kilograms of hydrogen is being converted into helium per second? Use the formula that you derive to estimate the mass of hydrogen atoms that are converted into helium in the interior of the sun (LSun = 3.9 x 1026 W). (Note: the mass of a hydrogen atom is 1 mproton and the mass of a helium atom is 3.97 mproton. You need four hydrogen nuclei to form one helium nucleus.)arrow_forwardObservations show that stellar luminosity, L, and mass, M, are related by L x M3.5 for main sequence stars. Obtain an expression that relates the main sequence life time and the mass of a star. You should assume that the luminosity is constant throughout a star's main sequence life time, and that the amount of mass converted into energy by a star while it is on the main sequence is given by AM main sequence life time of a 20 Solar mass star given that the Sun is expected to spend 1010 years on the main sequence. Comment on the significance of your answer. fM, where f is a constant. Estimate thearrow_forwardIn a star of 1 solar mass (M☉), the core hydrogen burning phase, also known as the main sequence phase, lasts for approximately 10 billion years. Suppose there's a star of 15 solar masses (M☉). Stars of higher mass burn through their hydrogen at a faster rate, following an approximate relation that the lifetime of a star on the main sequence (T) is proportional to its mass (M) raised to the power of -2.5 (T ∝ M^-2.5). Calculate approximately how long this 15 solar mass star would remain in the main sequence phase, compared to the 1 solar mass star.arrow_forward
- A main sequence star of mass 25 M⊙has a luminosity of approximately 80,000 L⊙. a. At what rate DOES MASS VANISH as H is fused to He in the star’s core? Note: When we say “mass vanish '' what we really mean is “gets converted into energy and leaves the star as light”. Note: approximate answer: 3.55 E14 kg/s b. At what rate is H converted into He? To do this you need to take into account that for every kg of hydrogen burned, only 0.7% gets converted into energy while the rest turns into helium. Approximate answer = 5E16 kg/s c. Assuming that only the 10% of the star’s mass in the central regions will get hot enough for fusion, calculate the main sequence lifetime of the star. Put your answer in years, and compare it to the lifetime of the Sun. It should be much, much shorter. Approximate answer: 30 million years.arrow_forwardQUESTION 21 In a Type Ia supernova, the cause of the violent outburst is: 1) the sudden emission of a shell of stellar material from a dying low-mass star 2) the collapse of a very massive protostar to the main sequence 3) an enormous release of neutrinos during a sudden episode of hydrogen fusion 4) the transfer of so much mass from a companion star that a white dwarf goes "over the limit" and collapses, causing an enormous amount of sudden fusion 5) two neutron stars colliding with each otherarrow_forwardA 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_forward
- We will take a moment to compare how brightly a white dwarf star shines compared to a red giant star. For the sake of this problem, let's assume a white dwarf has a temperature around 10,000 K and a red giant has a temperature around 5,000 K. As for their stellar radiatin, the white dwarf has a radius about 1/100th that of the Sun, and a red giant has a radius around 100 times larger than the Sun. With this in mind, how does the luminosity of a red giant star compare to that of a white dwarf (Hint: do not try to enter all of these numbers into the luminosity equation {it won't go well}; instead, remember that you are only interested in the ratio between the two, so all common units and components can be divided out)? Please enter your answer in terms of the luminosity of the red giant divided by the luminosity of the white dwarf and round to two significant figures. Also, please avoid using commas in your answer.arrow_forward(a)What type of supernova is most luminous in light? (b) What kind of star results in such an explosion? (c) What type of supernova is most luminous in neutrino emission? (d) What kind of star results in such an explosion?arrow_forwardBetelgeuse is a nearby supergiant that will eventually explode into a supernova. At peak brightness, the supernova will have a luminosity of about 20 billion times the Sun. It is 600 lightyears away. All stellar brightnesses are compared with Vega, which has an intrinsic luminosity of about 60 times the sun, a distance of 25 lightyears away, an absolutely magnitude of 0.6 and an apparent magnitude of 0. a) At peak brightness, how many times brighter will betelgeuses 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'ss brightness will Betelgeuse be?arrow_forward
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