Astronomy 162, Week 5 Energy and the Sun Patrick S
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Hydrostatic Shapes
7 Hydrostatic shapes It is primarily the interplay between gravity and contact forces that shapes the macroscopic world around us. The seas, the air, planets and stars all owe their shape to gravity, and even our own bodies bear witness to the strength of gravity at the surface of our massive planet. What physics principles determine the shape of the surface of the sea? The sea is obviously horizontal at short distances, but bends below the horizon at larger distances following the planet’s curvature. The Earth as a whole is spherical and so is the sea, but that is only the first approximation. The Moon’s gravity tugs at the water in the seas and raises tides, and even the massive Earth itself is flattened by the centrifugal forces of its own rotation. Disregarding surface tension, the simple answer is that in hydrostatic equi- librium with gravity, an interface between two fluids of different densities, for example the sea and the atmosphere, must coincide with a surface of constant ¡A potential, an equipotential surface. Otherwise, if an interface crosses an equipo- ¡ A ¡ A tential surface, there will arise a tangential component of gravity which can only ¡ ©¼AAA be balanced by shear contact forces that a fluid at rest is unable to supply. An ¡ AAA ¡ g AUA iceberg rising out of the sea does not obey this principle because it is solid, not ¡ ? A fluid. But if you try to build a little local “waterberg”, it quickly subsides back into the sea again, conforming to an equipotential surface. Hydrostatic balance in a gravitational field also implies that surfaces of con- A triangular “waterberg” in the sea. -
2. Astrophysical Constants and Parameters
2. Astrophysical constants 1 2. ASTROPHYSICAL CONSTANTS AND PARAMETERS Table 2.1. Revised May 2010 by E. Bergren and D.E. Groom (LBNL). The figures in parentheses after some values give the one standard deviation uncertainties in the last digit(s). Physical constants are from Ref. 1. While every effort has been made to obtain the most accurate current values of the listed quantities, the table does not represent a critical review or adjustment of the constants, and is not intended as a primary reference. The values and uncertainties for the cosmological parameters depend on the exact data sets, priors, and basis parameters used in the fit. Many of the parameters reported in this table are derived parameters or have non-Gaussian likelihoods. The quoted errors may be highly correlated with those of other parameters, so care must be taken in propagating them. Unless otherwise specified, cosmological parameters are best fits of a spatially-flat ΛCDM cosmology with a power-law initial spectrum to 5-year WMAP data alone [2]. For more information see Ref. 3 and the original papers. Quantity Symbol, equation Value Reference, footnote speed of light c 299 792 458 m s−1 exact[4] −11 3 −1 −2 Newtonian gravitational constant GN 6.674 3(7) × 10 m kg s [1] 19 2 Planck mass c/GN 1.220 89(6) × 10 GeV/c [1] −8 =2.176 44(11) × 10 kg 3 −35 Planck length GN /c 1.616 25(8) × 10 m[1] −2 2 standard gravitational acceleration gN 9.806 65 m s ≈ π exact[1] jansky (flux density) Jy 10−26 Wm−2 Hz−1 definition tropical year (equinox to equinox) (2011) yr 31 556 925.2s≈ π × 107 s[5] sidereal year (fixed star to fixed star) (2011) 31 558 149.8s≈ π × 107 s[5] mean sidereal day (2011) (time between vernal equinox transits) 23h 56m 04.s090 53 [5] astronomical unit au, A 149 597 870 700(3) m [6] parsec (1 au/1 arc sec) pc 3.085 677 6 × 1016 m = 3.262 ...ly [7] light year (deprecated unit) ly 0.306 6 .. -
Lecture 3 - Minimum Mass Model of Solar Nebula
Lecture 3 - Minimum mass model of solar nebula o Topics to be covered: o Composition and condensation o Surface density profile o Minimum mass of solar nebula PY4A01 Solar System Science Minimum Mass Solar Nebula (MMSN) o MMSN is not a nebula, but a protoplanetary disc. Protoplanetary disk Nebula o Gives minimum mass of solid material to build the 8 planets. PY4A01 Solar System Science Minimum mass of the solar nebula o Can make approximation of minimum amount of solar nebula material that must have been present to form planets. Know: 1. Current masses, composition, location and radii of the planets. 2. Cosmic elemental abundances. 3. Condensation temperatures of material. o Given % of material that condenses, can calculate minimum mass of original nebula from which the planets formed. • Figure from Page 115 of “Physics & Chemistry of the Solar System” by Lewis o Steps 1-8: metals & rock, steps 9-13: ices PY4A01 Solar System Science Nebula composition o Assume solar/cosmic abundances: Representative Main nebular Fraction of elements Low-T material nebular mass H, He Gas 98.4 % H2, He C, N, O Volatiles (ices) 1.2 % H2O, CH4, NH3 Si, Mg, Fe Refractories 0.3 % (metals, silicates) PY4A01 Solar System Science Minimum mass for terrestrial planets o Mercury:~5.43 g cm-3 => complete condensation of Fe (~0.285% Mnebula). 0.285% Mnebula = 100 % Mmercury => Mnebula = (100/ 0.285) Mmercury = 350 Mmercury o Venus: ~5.24 g cm-3 => condensation from Fe and silicates (~0.37% Mnebula). =>(100% / 0.37% ) Mvenus = 270 Mvenus o Earth/Mars: 0.43% of material condensed at cooler temperatures. -
Astronomy 201 Review 2 Answers What Is Hydrostatic Equilibrium? How Does Hydrostatic Equilibrium Maintain the Su
Astronomy 201 Review 2 Answers What is hydrostatic equilibrium? How does hydrostatic equilibrium maintain the Sun©s stable size? Hydrostatic equilibrium, also known as gravitational equilibrium, describes a balance between gravity and pressure. Gravity works to contract while pressure works to expand. Hydrostatic equilibrium is the state where the force of gravity pulling inward is balanced by pressure pushing outward. In the core of the Sun, hydrogen is being fused into helium via nuclear fusion. This creates a large amount of energy flowing from the core which effectively creates an outward-pushing pressure. Gravity, on the other hand, is working to contract the Sun towards its center. The outward pressure of hot gas is balanced by the inward force of gravity, and not just in the core, but at every point within the Sun. What is the Sun composed of? Explain how the Sun formed from a cloud of gas. Why wasn©t the contracting cloud of gas in hydrostatic equilibrium until fusion began? The Sun is primarily composed of hydrogen (70%) and helium (28%) with the remaining mass in the form of heavier elements (2%). The Sun was formed from a collapsing cloud of interstellar gas. Gravity contracted the cloud of gas and in doing so the interior temperature of the cloud increased because the contraction converted gravitational potential energy into thermal energy (contraction leads to heating). The cloud of gas was not in hydrostatic equilibrium because although the contraction produced heat, it did not produce enough heat (pressure) to counter the gravitational collapse and the cloud continued to collapse. -
Hertzsprung-Russell Diagram
Hertzsprung-Russell Diagram Astronomers have made surveys of the temperatures and luminosities of stars and plot the result on H-R (or Temperature-Luminosity) diagrams. Many stars fall on a diagonal line running from the upper left (hot and luminous) to the lower right (cool and faint). The Sun is one of these stars. But some fall in the upper right (cool and luminous) and some fall toward the bottom of the diagram (faint). What can we say about the stars in the upper right? What can we say about the stars toward the bottom? If all stars had the same size, what pattern would they make on the diagram? Masses of Stars The gravitational force of the Sun keeps the planets in orbit around it. The force of the Sun’s gravity is proportional to the mass of the Sun, and so the speeds of the planets as they orbit the Sun depend on the mass of the Sun. Newton’s generalization of Kepler’s 3rd law says: P2 = a3 / M where P is the time to orbit, measured in years, a is the size of the orbit, measured in AU, and M is the sum of the two masses, measured in solar masses. Masses of stars It is difficult to see planets orbiting other stars, but we can see stars orbiting other stars. By measuring the periods and sizes of the orbits we can calculate the masses of the stars. If P2 = a3 / M, M = a3 / P2 This mass in the formula is actually the sum of the masses of the two stars. -
STARS in HYDROSTATIC EQUILIBRIUM Gravitational Energy
STARS IN HYDROSTATIC EQUILIBRIUM Gravitational energy and hydrostatic equilibrium We shall consider stars in a hydrostatic equilibrium, but not necessarily in a thermal equilibrium. Let us define some terms: U = kinetic, or in general internal energy density [ erg cm −3], (eql.1a) U u ≡ erg g −1 , (eql.1b) ρ R M 2 Eth ≡ U4πr dr = u dMr = thermal energy of a star, [erg], (eql.1c) Z Z 0 0 M GM dM Ω= − r r = gravitational energy of a star, [erg], (eql.1d) Z r 0 Etot = Eth +Ω = total energy of a star , [erg] . (eql.1e) We shall use the equation of hydrostatic equilibrium dP GM = − r ρ, (eql.2) dr r and the relation between the mass and radius dM r =4πr2ρ, (eql.3) dr to find a relations between thermal and gravitational energy of a star. As we shall be changing variables many times we shall adopt a convention of using ”c” as a symbol of a stellar center and the lower limit of an integral, and ”s” as a symbol of a stellar surface and the upper limit of an integral. We shall be transforming an integral formula (eql.1d) so, as to relate it to (eql.1c) : s s s GM dM GM GM ρ Ω= − r r = − r 4πr2ρdr = − r 4πr3dr = (eql.4) Z r Z r Z r2 c c c s s s dP s 4πr3dr = 4πr3dP =4πr3P − 12πr2P dr = Z dr Z c Z c c c s −3 P 4πr2dr =Ω. Z c Our final result: gravitational energy of a star in a hydrostatic equilibrium is equal to three times the integral of pressure within the star over its entire volume. -
1. A) the Sun Is in Hydrostatic Equilibrium. What Does
AST 301 Test #3 Friday Nov. 12 Name:___________________________________ 1. a) The Sun is in hydrostatic equilibrium. What does this mean? What is the definition of hydrostatic equilibrium as we apply it to the Sun? Pressure inside the star pushing it apart balances gravity pulling it together. So it doesn’t change its size. 1. a) The Sun is in thermal equilibrium. What does this mean? What is the definition of thermal equilibrium as we apply it to the Sun? Energy generation by nuclear fusion inside the star balances energy radiated from the surface of the star. So it doesn’t change its temperature. b) Give an example of an equilibrium (not necessarily an astronomical example) different from hydrostatic and thermal equilibrium. Water flowing into a bucket balances water flowing out through a hole. When supply and demand are equal the price is stable. The floor pushes me up to balance gravity pulling me down. 2. If I want to use the Doppler technique to search for a planet orbiting around a star, what measurements or observations do I have to make? I would measure the Doppler shift in the spectrum of the star caused by the pull of the planet’s gravity on the star. 2. If I want to use the transit (or eclipse) technique to search for a planet orbiting around a star, what measurements or observations do I have to make? I would observe the dimming of the star’s light when the planet passes in front of the star. If I find evidence of a planet this way, describe some information my measurements give me about the planet. -
Astronomy 241: Review Questions #1 Distributed: September 20, 2012 Review the Questions Below, and Be Prepared to Discuss Them I
Astronomy 241: Review Questions #1 Distributed: September 20, 2012 Review the questions below, and be prepared to discuss them in class on Sep. 25. Modified versions of some of these questions will be used in the midterm exam on Sep. 27. 1. State Kepler’s three laws, and give an example of how each is used. 2. You are in a rocket in a circular low Earth orbit (orbital radius aleo =6600km).Whatve- 1 locity change ∆v do you need to escape Earth’s gravity with a final velocity v =2.8kms− ? ∞ Assume your rocket delivers a brief burst of thrust, and ignore the gravity of the Moon or other solar system objects. 3. Consider a airless, rapidly-spinning planet whose axis of rotation is perpendicular to the plane of its orbit around the Sun. How will the local equilibrium surface temperature temperature T depend on latitude ! (measured, as always, in degrees from the planet’s equator)? You may neglect heat conduction and internal heat sources. 3 4. Given that the density of seawater is 1025 kg m− ,howdeepdoyouneedtodivebefore the total pressure (ocean plus atmosphere) is 2P0,wherethetypicalpressureattheEarth’s 5 1 2 surface is P =1.013 10 kg m− s− ? 0 × 5. An asteroid of mass m M approaches the Earth on a hyperbolic trajectory as shown " E in Fig. 1. This figure also defines the initial velocity v0 and “impact parameter” b.Ignore the gravity of the Moon or other solar system objects. (a) What are the energy E and angular momentum L of this orbit? (Hint: one is easily evaluated when the asteroid is far away, the other when it makes its closest approach.) (b) Find the orbital eccentricity e in terms of ME, E, L,andm (note that e>1). -
ASTRONOMY 220C ADVANCED STAGES of STELLAR EVOLUTION and NUCLEOSYNTHESIS Spring, 2015
ASTRONOMY 220C ADVANCED STAGES OF STELLAR EVOLUTION AND NUCLEOSYNTHESIS Spring, 2015 http://www.ucolick.org/~woosley This is a one quarter course dealing chiefly with: a) Nuclear astrophysics and the relevant nuclear physics b) The evolution of massive stars - especially their advanced stages c) Nucleosynthesis – the origin of each isotope in nature d) Supernovae of all types e) First stars, ultraluminous supernovae, subluminous supernovae f) Stellar mass high energy transients - gamma-ray bursts, novae, and x-ray bursts. Our study of supernovae will be extensive and will cover not only the mechanisms currently thought responsible for their explosion, but also their nucleosynthesis, mixing, spectra, compact remnants and light curves, the latter having implications for cosmology. The student is expected to be familiar with the material presented in Ay 220A, a required course in the UCSC graduate program, and thus to already know the essentials of stellar evolution, as well as basic quantum mechanics and statistical mechanics. The course material is extracted from a variety of sources, much of it the results of local research. It is not contained, in total, in any one or several books. The powerpoint slides are on the web, but you will need to come to class. A useful textbook, especially for material early in the course, is Clayton’s, Principles of Stellar Evolution and Nucleosynthesis. Also of some use are Arnett’s Supernovae and Nucleosynthesis (Princeton) and Kippenhahn and Weigert’s Stellar Evolution and Nucleosynthesis (Springer Verlag). Course performance will be based upon four graded homework sets and an in-class final examination. The anticipated class material is given, in outline form, in the following few slides, but you can expect some alterations as we go along. -
News & Views Research
RESEARCH NEWS & VIEWS measure the dwarf planet’s size, shape and density more accurately than ever before. Compared with other bodies in the Solar System, Haumea rotates quickly (making one rotation in about four hours), and is strangely shaped, like an elongated egg. Ortiz et al. calculate that the object’s longest axis is at least 2,300 km, which is larger than earlier estimates6–8. In turn, given known values for Haumea’s mass and brightness, the dwarf planet’s density and reflectivity are both lower CSI DE ANDALUCIA DE ASTROFISICA INSTITUTO than the unusually high values previously considered. The authors’ results suggest that Haumea might not be in hydrostatic equilibrium, and this touches on the still-sensitive topic of how planets and dwarf planets should be defined. Remarkably, the authors show that blinks in the starlight, detected at multiple observa- tion sites both before and after a distant star was blocked by Haumea, are consistent with a 70-km-wide ring of material encircling the body, approximately 1,000 km away from Haumea’s surface (Fig. 1). Saturn’s are the most studied of all rings, Figure 1 | Artist’s impression of Haumea and its ring. Ortiz et al.4 have measured the size, shape and and yet they remain enigmatic. Data from the density of the dwarf planet Haumea with unprecedented accuracy and found that it has a planetary ring. Cassini spacecraft revealed that gravitational This is the first time that a ring has been discovered around a distant body in the Solar System. interactions between the planet’s rings and moons shepherd the ring material, and that Ring systems represent microcosms of the transition from the outer Solar System, one of the rings is produced entirely from larger-scale rotating structures, such as including close encounters with the giant matter that spews from the moon Encela- galaxies and proto-planetary disks — the disks planets14. -
How Stars Work: • Basic Principles of Stellar Structure • Energy Production • the H-R Diagram
Ay 122 - Fall 2004 - Lecture 7 How Stars Work: • Basic Principles of Stellar Structure • Energy Production • The H-R Diagram (Many slides today c/o P. Armitage) The Basic Principles: • Hydrostatic equilibrium: thermal pressure vs. gravity – Basics of stellar structure • Energy conservation: dEprod / dt = L – Possible energy sources and characteristic timescales – Thermonuclear reactions • Energy transfer: from core to surface – Radiative or convective – The role of opacity The H-R Diagram: a basic framework for stellar physics and evolution – The Main Sequence and other branches – Scaling laws for stars Hydrostatic Equilibrium: Stars as Self-Regulating Systems • Energy is generated in the star's hot core, then carried outward to the cooler surface. • Inside a star, the inward force of gravity is balanced by the outward force of pressure. • The star is stabilized (i.e., nuclear reactions are kept under control) by a pressure-temperature thermostat. Self-Regulation in Stars Suppose the fusion rate increases slightly. Then, • Temperature increases. (2) Pressure increases. (3) Core expands. (4) Density and temperature decrease. (5) Fusion rate decreases. So there's a feedback mechanism which prevents the fusion rate from skyrocketing upward. We can reverse this argument as well … Now suppose that there was no source of energy in stars (e.g., no nuclear reactions) Core Collapse in a Self-Gravitating System • Suppose that there was no energy generation in the core. The pressure would still be high, so the core would be hotter than the envelope. • Energy would escape (via radiation, convection…) and so the core would shrink a bit under the gravity • That would make it even hotter, and then even more energy would escape; and so on, in a feedback loop Ë Core collapse! Unless an energy source is present to compensate for the escaping energy. -
Stellar Structure
Stellar structure For an isolated, static, spherically symmetric star, four basic laws / equations needed to describe structure: • Conservation of mass • Conservation of energy (at each radius, the change in the energy flux equals the local rate of energy release) • Equation of hydrostatic equilibrium (at each radius, forces due to pressure differences balance gravity) • Equation of energy transport (relation between the energy flux and the local gradient of temperature) Basic equations are supplemented by: • Equation of state (pressure of a gas as a function of its density and temperature) • Opacity (how transparent it is to radiation) • Nuclear energy generation rate as f(r,T) ASTR 3730: Fall 2003 Conservation of mass Let r be the distance from the center Density as function of radius is r(r) dr r m dm Let m be the mass interior to r, then r conservation of mass implies that: R 2 M dm = 4pr rdr Write this as a differential equation: dm st † = 4pr 2r 1 stellar structure dr equation † ASTR 3730: Fall 2003 Equation of hydrostatic equilibrium P(r+dr) Consider small cylindrical r+dr element between radius r and radius r + dr in the star. Surface area = dS r Mass = Dm P(r) Mass of gas in the star at smaller gravity radii = m = m(r) Radial forces acting on the element: GmDm Gravity (inward): F = - g r2 gravitational constant G = 6.67 x 10-8 dyne cm2 g-2 † ASTR 3730: Fall 2003 Pressure (net force due to difference in pressure between upper and lower faces): Fp = P(r)dS - P(r + dr)dS È dP ˘ = P(r)dS - P(r) + ¥ dr dS ÎÍ dr ˚˙ dP = - drdS dr Mass