E2.1 Stars Stars are the things you see most of in the night sky. You already know all about the Sun, which is a pretty good example of an average star But what exactly is a star???
E2.1/2 Stars Stars are formed by interstellar dust coming together through mutual gravitational attraction. The loss of potential energy is responsible for the initial high temperature necessary for fusion. The fusion process releases so much energy that the pressure created prevents the star from collapsing due to gravitational pressure.
E2.1Nuclear fusion Very high temperatures are needed in order to begin the fusion process: usually 107 K.
E2.2A star is a big ball of gas, with fusion going on at its center, held together by gravity! Massive Sun-like Low-mass Star Star StarThere are variations between stars, but by and large they’re really pretty simple things.
E2.1What is the most importantthing about a star? MASS! The mass of a normal star almost completely determines itsLUMINOSITY and TEMPERATURE! Note: “normal” star means a star that’s fusing Hydrogen into Helium in its center (we say “hydrogen burning”).
E2.2The LUMINOSITY of a star ishow much ENERGY it gives offper second: This light bulb has a luminosity of 60 Watts The energy the Sun emits is generated by the fusion in its core…
What does luminosity have to do E2.2with mass?The mass of a star determines Pressurethe pressure in its core: Gravity pulls outer layers The core in, supports the Gas Pressure pushes weight of the them out. whole star!The more mass the star has,the higher the central pressure!
The core pressure determines E2.2 the rate of fusion… PRESSURE & RATE OFMASS TEMPERATURE FUSION …which in turn determines the star’s luminosity!
E2.3Luminosity is an intrinsic property…it doesn’t depend on distance! This light bulb has a luminosity of 60 Watts… …no matter where it is, or where we view it from, it will always be a 60 Watt light bulb.
E2.3 LuminosityThe Luminosity of a star is the energy that it releasesper second. Sun has a luminosity of 3.90x1026 W(often written as L): it emits 3.90x1026 joules persecond in all directions.The energy that arrivesat the Earth is only avery small amountwhen compared will thetotal energy released bythe Sun.
E2.4 Apparent brightness When the light from the Sun reaches the Earth it will be spread out over a sphere of radius d. The energy received per unit time per unit area is b, where: L b d 4d 2 b is called the apparent brightness of the star
E2.3 LuminosityExercise 13.1The Sun is a distance d=1.5 x 1011 m from the Earth.Estimate how much energy falls on a surface of 1m2in a year. L= 3.90x1026 W d
E2.3At a distance of d=1.5 x 1011m, the energy is “distributed”along the surface of a sphere of radius 1.5 x 1011 m The sphere’s surface area is given by: A = 4πd2 = 4 π x (1.5 x 1011)2 = d =2.83 x 1023 m2 The energy that falls on a surface area of 1m2 on Earth per second will be equal to: b = L/A = 3.90x1026 / 2.83 x 1023 = = 1378.1 W/m2 or 1378.1 J/s m2In a year there are: 365.25days x 24h/day x 60min/h x60s/min = 3.16 x 107 sSo, the energy that falls in 1 m2 in 1 year will be: 1378.1 x 3.16 x 107 = 4.35 x 1010 joules
E2.6 Black body radiation A black body is a perfect emitter. A good model for a black body is a filament light bulb: the light bulb emits in a very large region of the electromagnetic spectrum. There is a clear relationship between the temperature of an object and the wavelength for which the emission is maximum. That relationship is known as Wien’s law: maxT constant maxT 2.9x10 m K -3
E2.6 Wien Displacement lawBy analysing a star’sspectrum, we can know in whatwavelength the star emits moreenergy.The Sun emits more energy atλ=500 nm.According to Wien’s law, thetemperature at the Sun’s surfaceis inversely proportional to themaximum wavelength.So:
E2.5 Black body radiation Apart from temperature, a radiation spectrum can also give information about luminosity. The area under a black body radiation curve is equal to the total energy emitted per second per unit of area of the black body. Stefan showed that this area was proportional to the fourth power of the absolute temperature of the body. The total power emitted by a black body is its luminosity. According to the Stefan-Boltzmann law, a body of surface area A and absolute temperature T has a luminosity given by: L σAT 4 where, σ = 5.67x108 W m-2 K-4
E2.5/6 Why is this important? The spectrum of stars is similar to the spectrum emitted by a black body. We can therefore use Wien Law to find the temperature of a star from its spectrum. If we know its temperature and its luminosity then its radius can be found from Stephan-Boltzmann law.
Real spectra are more E2.7complicated than this (rememberemission and absorption lines?)Blackbody Emission andSpectrum Absorption Lines
Stars can be arranged into E2.8categories based on thefeatures in their spectra… This is called “Spectral Classification” How do we categorize stars? A few options: 1. by the “strength” (depth) of the absorption lines in their spectra 2. by their color as determined by their blackbody curve 3. by their temperature and luminosity
E2.8 First attempts to classify stars used the strength of their absorption lines… Stars were labeled “A, B, C…” in order of increasing strength of Hydrogen lines. They also used the strength of the Harvard “computers”! Williamina Fleming
Later, these categories werereordered according totemperature/color… OBAFGKM(LT)! Annie Jump Cannon
E2.8 OBAFGKM - Mnemonics O Be A Fine Girl/Guy Kiss MeOsama Bin Airlines! Flies Great, Knows Manhattan! Only Boring Astronomers Find Gratification in Knowing Mnemonics!
E2.8 Eventually, the connection was made between the observables and the theory. Observable: • Strength of Hydrogen Absorption Lines • Blackbody Curve (Color) Theoretical: • Using observables to determine things we can’t measure: Temperature and LuminosityCecilia Payne
E2.8 The Spectral SequenceClass Spectrum Color Temperature O ionized and neutral helium, bluish 31,000-49,000 K weakened hydrogen B neutral helium, stronger blue-white 10,000-31,000 K hydrogen A strong hydrogen, ionized white 7400-10,000 K metals F weaker hydrogen, ionized yellowish white 6000-7400 K metals G still weaker hydrogen, ionized yellowish 5300-6000 K and neutral metalsK weak hydrogen, neutral orange 3900-5300 K metalsM little or no hydrogen, neutral reddish 2200-3900 K metals, molecules L no hydrogen, metallic red-infrared 1200-2200 K hydrides, alkalai metals T methane bands infrared under 1200 K
E2.7 “If a picture is worth a 1000 words, a spectrum is worth 1000 pictures.” Spectra tell us about the physics of the star and how those physics affect the atoms in it
E2.11The Hertzsprung-Russell diagram This diagram shows a correlation between the luminosity of a star and its temperature. The scale on the axes is not linear as the temperature varies from 3000 to 25000 K whereas the luminosity varies from 10-4 to 106, 10 orders of magnitude.
E2.11H-R diagram The stars are not randomly distributed on the diagram. There are 3 features that emerge from the H-R diagram: Most stars fall on a strip extending diagonally across the diagram from top left to bottom right. This is called the MAIN SEQUENCE. Some large stars, reddish in colour occupy the top right – these are red giants (large, cool stars). The bottom left is a region of small stars known as white dwarfs (small and hot)