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... which has gone through stars will have more Helium. Some of you, however, stated that all of the Hydrogen would be converted to Helium. This is not correct; as I mentioned in class, at its current luminosity, the Sun has enough Hydrogen to keep shining for something like 80 billion years, which is m ...
... which has gone through stars will have more Helium. Some of you, however, stated that all of the Hydrogen would be converted to Helium. This is not correct; as I mentioned in class, at its current luminosity, the Sun has enough Hydrogen to keep shining for something like 80 billion years, which is m ...
Document
... • The nursery of stars. This “cloud” is a combination of Hydrogen, gas, and dust (remnants of other stars and objects of the universe). • The matter used to make stars are recycled from other substances in the universe. ...
... • The nursery of stars. This “cloud” is a combination of Hydrogen, gas, and dust (remnants of other stars and objects of the universe). • The matter used to make stars are recycled from other substances in the universe. ...
Chapter 14 Our Star 14.1 A Closer Look at the Sun Why was the
... • How does nuclear fusion occur in the Sun? – The core’s extreme temperature and density are just right for nuclear fusion of hydrogen to helium through the proton-proton chain – Gravitational equilibrium acts as a thermostat to regulate the core temperature because fusion rate is very sensitive to ...
... • How does nuclear fusion occur in the Sun? – The core’s extreme temperature and density are just right for nuclear fusion of hydrogen to helium through the proton-proton chain – Gravitational equilibrium acts as a thermostat to regulate the core temperature because fusion rate is very sensitive to ...
Chapter 10: The Stars
... • 8. Why doesn’t stellar parallax work to find distances to all stars? • As the stars get further away, the parallax angle gets smaller (and the apparent shift gets smaller). At distances beyond a few hundred light years, the angle and apparent shift are so small we can’t ...
... • 8. Why doesn’t stellar parallax work to find distances to all stars? • As the stars get further away, the parallax angle gets smaller (and the apparent shift gets smaller). At distances beyond a few hundred light years, the angle and apparent shift are so small we can’t ...
AY5 Announcements
... (1) More massive stars require higher central temperatures (hydrostatic eqm.) (2) The P-P fusion rate and luminosity is proportional to T4 Therefore, more massive stars will have higher central temperature and higher Luminosity. This is what is seen along the H-R Diagram main sequence. ...
... (1) More massive stars require higher central temperatures (hydrostatic eqm.) (2) The P-P fusion rate and luminosity is proportional to T4 Therefore, more massive stars will have higher central temperature and higher Luminosity. This is what is seen along the H-R Diagram main sequence. ...
Chapter 07
... Light and Matter Spectra of stars are more complicated than pure blackbody spectra. characteristic ...
... Light and Matter Spectra of stars are more complicated than pure blackbody spectra. characteristic ...
Homework #4 Solutions ASTR100: Introduction to Astronomy
... Therefore, wavelength of maximum intensity at 12,000K is half the wavelength of maximum intensity at 6,000K. c) We know that the intensity of light emitted by a given object depends on its wavelength or frequency. Here, the wavelength at 12,000K is reduced by half which means its frequency has incre ...
... Therefore, wavelength of maximum intensity at 12,000K is half the wavelength of maximum intensity at 6,000K. c) We know that the intensity of light emitted by a given object depends on its wavelength or frequency. Here, the wavelength at 12,000K is reduced by half which means its frequency has incre ...
Astronomy - SparkNotes
... • This flux is equal to the area under the curve of intensity versus wavelength for a blackbody. 2. Atomic lines: According to quantum mechanics, electrons bound to an atom can only have particular values of energy; they are unique to that element. Absorption or emission of a photon of light by the ...
... • This flux is equal to the area under the curve of intensity versus wavelength for a blackbody. 2. Atomic lines: According to quantum mechanics, electrons bound to an atom can only have particular values of energy; they are unique to that element. Absorption or emission of a photon of light by the ...
On the Cosmic Nuclear Cycle and the Similarity of Nuclei and Stars
... stars were not made one-at-a-time in SN explosions but were more abundantly made in higher energy fragmentation events that produced our galaxy, probably in a high density region associated with active galactic nuclei (AGN), quasars, or massive neutron stars. The origin of these high-density, energe ...
... stars were not made one-at-a-time in SN explosions but were more abundantly made in higher energy fragmentation events that produced our galaxy, probably in a high density region associated with active galactic nuclei (AGN), quasars, or massive neutron stars. The origin of these high-density, energe ...
Starlight and Atoms - School District of Clayton
... passes through a cool, low-density gas, the result will be an absorption spectrum. Light excites electrons in atoms to higher energy states ...
... passes through a cool, low-density gas, the result will be an absorption spectrum. Light excites electrons in atoms to higher energy states ...
Our Star, the Sun
... universe! • By comparison there are about 0.0000005 protons per cm3 in the universe. ...
... universe! • By comparison there are about 0.0000005 protons per cm3 in the universe. ...
Powerpoint Presentation (large file)
... universe! • By comparison there are about 0.0000005 protons per cm3 in the universe. ...
... universe! • By comparison there are about 0.0000005 protons per cm3 in the universe. ...
Astrophysics
... mass interior to radius r and ρ is the mass density. b. (1 pt) Rewrite the equation with m as the independent variable, i.e, dP/dm = ... c. (1 pt) Use the dP/dm equation to obtain an approximate expression for the pressure at the center of the sun, in terms of G, M, and R, where M is the total mass ...
... mass interior to radius r and ρ is the mass density. b. (1 pt) Rewrite the equation with m as the independent variable, i.e, dP/dm = ... c. (1 pt) Use the dP/dm equation to obtain an approximate expression for the pressure at the center of the sun, in terms of G, M, and R, where M is the total mass ...
Chapter 21: Energy and Matter in the Universe
... Energy going into the gravitational force would have taken energy away from the photons, presumably distributing this energy among gravitons, the carrier particles for the gravitational force. The less energetic photons would have had a lower average temperature, and the cooling process of the Unive ...
... Energy going into the gravitational force would have taken energy away from the photons, presumably distributing this energy among gravitons, the carrier particles for the gravitational force. The less energetic photons would have had a lower average temperature, and the cooling process of the Unive ...
Type II supernova
A Type II supernova (plural: supernovae or supernovas) results from the rapid collapse and violent explosion of a massive star. A star must have at least 8 times, and no more than 40–50 times, the mass of the Sun (M☉) for this type of explosion. It is distinguished from other types of supernovae by the presence of hydrogen in its spectrum. Type II supernovae are mainly observed in the spiral arms of galaxies and in H II regions, but not in elliptical galaxies.Stars generate energy by the nuclear fusion of elements. Unlike the Sun, massive stars possess the mass needed to fuse elements that have an atomic mass greater than hydrogen and helium, albeit at increasingly higher temperatures and pressures, causing increasingly shorter stellar life spans. The degeneracy pressure of electrons and the energy generated by these fusion reactions are sufficient to counter the force of gravity and prevent the star from collapsing, maintaining stellar equilibrium. The star fuses increasingly higher mass elements, starting with hydrogen and then helium, progressing up through the periodic table until a core of iron and nickel is produced. Fusion of iron or nickel produces no net energy output, so no further fusion can take place, leaving the nickel-iron core inert. Due to the lack of energy output allowing outward pressure, equilibrium is broken.When the mass of the inert core exceeds the Chandrasekhar limit of about 1.4 M☉, electron degeneracy alone is no longer sufficient to counter gravity and maintain stellar equilibrium. A cataclysmic implosion takes place within seconds, in which the outer core reaches an inward velocity of up to 23% of the speed of light and the inner core reaches temperatures of up to 100 billion kelvin. Neutrons and neutrinos are formed via reversed beta-decay, releasing about 1046 joules (100 foes) in a ten-second burst. The collapse is halted by neutron degeneracy, causing the implosion to rebound and bounce outward. The energy of this expanding shock wave is sufficient to accelerate the surrounding stellar material to escape velocity, forming a supernova explosion, while the shock wave and extremely high temperature and pressure briefly allow for theproduction of elements heavier than iron. Depending on initial size of the star, the remnants of the core form a neutron star or a black hole. Because of the underlying mechanism, the resulting nova is also described as a core-collapse supernova.There exist several categories of Type II supernova explosions, which are categorized based on the resulting light curve—a graph of luminosity versus time—following the explosion. Type II-L supernovae show a steady (linear) decline of the light curve following the explosion, whereas Type II-P display a period of slower decline (a plateau) in their light curve followed by a normal decay. Type Ib and Ic supernovae are a type of core-collapse supernova for a massive star that has shed its outer envelope of hydrogen and (for Type Ic) helium. As a result, they appear to be lacking in these elements.