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, Problem 30QAP
To determine
The reason why does a degenerate core heat up as the star continues burning the shell around it.
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Consider 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.
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?
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: 3679
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
Ch. 12 - Prob. 5QAPCh. 12 - Prob. 6QAPCh. 12 - Prob. 7QAPCh. 12 - Prob. 8QAPCh. 12 - Prob. 9QAPCh. 12 - Prob. 10QAPCh. 12 - Prob. 11QAPCh. 12 - Prob. 12QAPCh. 12 - Prob. 13QAPCh. 12 - Prob. 14QAPCh. 12 - Prob. 15QAPCh. 12 - Prob. 16QAPCh. 12 - Prob. 17QAPCh. 12 - Prob. 18QAPCh. 12 - Prob. 19QAPCh. 12 - Prob. 20QAPCh. 12 - Prob. 21QAPCh. 12 - Prob. 22QAPCh. 12 - Prob. 23QAPCh. 12 - Prob. 24QAPCh. 12 - Prob. 25QAPCh. 12 - Prob. 26QAPCh. 12 - Prob. 27QAPCh. 12 - Prob. 28QAPCh. 12 - Prob. 29QAPCh. 12 - Prob. 30QAPCh. 12 - Prob. 32QAPCh. 12 - Prob. 33QAPCh. 12 - Prob. 34QAPCh. 12 - Prob. 35QAPCh. 12 - Prob. 36QAPCh. 12 - Prob. 37QAPCh. 12 - Prob. 38QAPCh. 12 - Prob. 39QAPCh. 12 - Prob. 40QAPCh. 12 - Prob. 42QAPCh. 12 - Prob. 43QAPCh. 12 - Prob. 44QAPCh. 12 - Prob. 45QAP
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- In 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_forwardIn the H-R diagram we see that stellar masses__________ downward along the main sequence. At the upper end of the main sequence, the hot, luminous O stars can have masses as high as _________ or more times that of the Sun. On the lower end, cool, dim M stars may have as little as __________ times the mass of the Sun. Many more stars fall on the lower end of the main sequence than on the upper end, which tells us that _________stars are much more common than __________ stars.arrow_forwardA 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_forward
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