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Astronomy 1011H:

Professor Paul Woodward:

LCSE (Laboratory for Computational Science & Engineering)125 Walter Library, Digital Technology Center(offices on 4th floor of Walter Library).

paul@lcse.umn.edu

612-625-8049

Office hours: after lecture or by appointment.

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Astronomy 1001 Web Site:

www.astro.umn.edu and follow links from there to the syllabi for the various sections of Astronomy 1001 and this course, section 003.

This syllabus contains lots of useful information,as well as a link to a web site with lecture notes(as well as the course syllabus): www.lcse.umn.edu/astronomy1001

Reading schedule is heaviest in the beginning.

Textbook is excellent, but rather heavy on physics. Older editions will do just fine, or you may relay solely on the lecture slides.

We will not do everything in the textbook, and we do some things that are not in it.

The lecture notes should be your guide, and of course this study guide, to what I think is important and therefore might be on the test.

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The second midterm exam will be given on Wednesday, April 4, from 2:30 to 3:45 PM in Willey Hall, room 125.

Be sure to bring two pencils and a picture ID.

The midterm exam will consist entirely of multiple choice questions.

No books or “cheat sheets” will be permitted at the exam.

You will not be allowed to access your cell phone at the exam.

The exam, like the course, will be graded on a curve.

A histogram of the results of the exam will be shown in class and posted on the course Web site (along with the lecture notes) in order to help you see where your own score falls in this distribution. Approximate letter grades indicating the grading curve will also be indicated on this histogram.

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Note that some material was covered in the lectures that is not in the text, such as the discussion of the difference between nuclear fission and fusion and the work going on to harness nuclear fusion to generate power on earth.

The second midterm exam will not focus on material covered on the first midterm exam. However, some points, like the principles from physics that were discussed early on, recur and are involved in the explanation of the subject matter of the second midterm.

An outline of the material to be covered on the second midterm exam follows.

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This study guide is being provided 1 week before the test.

This study guide reflects the material that I have covered through the second class this week (3/28/18).

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V. The Outer Planets (we discussed only Jupiter and Saturn, with its moons and rings in detail in class).Chapter 8 of your textbook.

A. Jupiter -- the material concerning Jupiter’s interior, atmosphere and moons that we discussed in class will be covered on this midterm exam.

1. Jupiter’s interior.a. We believe that the general composition of Jupiter is

essentially the same as the sun – namely mostly hydrogen and helium, with trace amounts of other elements.

b. Jupiter gets hotter as one goes deeper into the interior. It is still cooling off from the process of its formation. As it has cooled, its internal pressure has dropped, and then gravity has caused it to contract.

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c. This contraction releases gravitational potential energy in the form of kinetic energy of motion radially inward.

d. This is then converted into heat (thermal energy).e. In this way, Jupiter has remained hot, but we believe

that it is no longer contracting significantly as it cools, because it is has reached a regime of relatively cool temperatures where density becomes pretty much independent of temperature.

f. Still, Jupiter now emits more radiation (in the infrared of its black body spectrum) than it receives from the sun.

2. Modeling Jupiter’s interior.a. We derive our knowledge of Jupiter’s interior from

building a physical model.

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a. We assume that the material composition of Jupiter is the same as the composition that we observe near the surface – mostly hydrogen and helium.

b. We assume Jupiter is a perfectly spherical object.c. We ignore Jupiter’s rotation as having only a small

effect on its internal structure. (why can we not do this for Saturn?)

d. We assume that no nuclear reactions are underway in Jupiter’s interior, and we ignore the very small loss of heat energy at Jupiter’s surface.

3. Building the Jupiter interior model.a. We approximate Jupiter by a set of concentric spherical

shells of material.b. On the spherical surface located between any 2 such

shells, we can compute the gravitational acceleration due to the mass of all shells inside this surface by simply concentrating their masses at the center and applying Netwon’s law of gravity for this point mass

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c. The famous mathematician, Gauss, proved that all the spherical shells outside our particular surface exert NO gravitational force on us (that is, the gravitational forces from all the different regions of that outer material exactly cancel out).

d. We can compute the pressure force at our particular surface by taking the difference of the pressure in the spherical shell just inside our surface and that just outside. We expect the pressure in the inner shell to be larger, so that the net pressure force (the difference between the two pressures) will act outwards.

e. We devise a computer program that can do these calculations, finding at each spherical surface the net force and direction of the force when both gravity and pressure are considered.

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f. The computer program will very rapidly move these spherical surfaces either inward or outward, according to the direction of the net force there, until the two forces of gravity (inward) and pressure (net outward) come into perfect balance.

g. Then we have a consistent model for the interior of Jupiter.

h. An ingredient to such a model is our knowledge of how the pressure of the material making up Jupiter changes when we change its density (we have assumed the pressure is insensitive to temperature).

i. Modeling Jupiter in this fashion indicates that in its interior, hydrogen assumes a liquid metallic form.

j. We have never seen liquid metallic hydrogen in the lab, but we know from our theory of quantum mechanics that it must form inside Jupiter.

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4. Jupiter’s atmosphere.a. In upper layers of the atmosphere, different

compounds condense at different temperatures, and therefore at different heights.

b. Highest are the ammonia clouds, which are white.c. Deeper down are clouds of ammonium hydrosulfide,

which are brown.d. Below these clouds are water clouds.e. Heat from interior transported outward by

convection.

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e. Rapid rotation of Jupiter creates convection cells that form horizontal bands.The law of conservation of angular momentum confines the gas to these bands, preventing the gas near the surface from moving very far in latitude.(Another way to understand this confinement is through the coriolis effect, which makes the gas veer to the right or left as it moves in latitude. As the gas moves toward the pole, the conservation of its angular momentum makes it rotate more rapidly about the spin axis and therefore be deflected from a direct, poleward path. See illustration in lecture notes.)There are many horizontal bands of circulating gas on Jupiter because it spins so rapidly.Earth has just 3 such bands in each hemisphere, called meridional circulation cells, because it spins less rapidly.Venus, which spins very slowly, has only one such circulation band in each hemisphere.

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For the earth, the meridianal circulation cells, 3 per hemisphere (see diagram in lecture slides), are driven by the unequal deposition of heat from sunlight.The sun heats the equatorial region much more than it heats the polar regions.The meridional circulation carries this heat from the equator to the poles.The heated air rises near the equator, because it is buoyant, and it is replaced by cooler air that slides in under it from higher latitudes (see diagram in lecture).The Coriolis force (conservation of angular momentum) forces the air moving northward aloft to veer to the east as it travels, so that before it has gone too far, it is traveling eastward. By now it is cooler, it descends, and then flows back toward the equator, making a meridional circulation cell.

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The earth’s meridional circulation cells are responsible for the “trade winds” that blow consistently in predictable directions that were exploited by sailing ships in the early era of global trade.Because Jupiter rotates so rapidly, its Coriolis forces are much greater, and gases traveling northward from the equator cannot get nearly as far in latitude before they find themselves deflected.For this reason, Jupiter has many horizontal bands, which are its meridional circulation cells.In addition to these bands, Jupiter, like the earth, also has storms in its atmosphere in which the atmosphere spins around a point in latitude and longitude, like hurricanes do on earth.The largest such storm on Jupiter is the Great Red Spot.Here the atmosphere is rising at the center of the spot.

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B. Jupiter’s moon Io.1. About the size of the earth’s Moon, closest to Jupiter.2. Densest of the Galilean moons of Jupiter, rocky interior.3. Active volcanoes.4. Geysers of liquid sulfur and sulfur dioxide.5. Heat energy in interior that drives the volcanism comes

from tidal flexing.6. Io rotates just once per revolution about Jupiter, so that

it always presents the same face to Jupiter. See diagrams in lecture notes illustrating how this comes to be as a result of tidal forces, with the diagrams using the Earth-Moon system to provide this explanation.

7. Tidal flexing is result of Io’s elliptical orbit, so Io is always moving closer and further from Jupiter, and tides raised by Jupiter increase and decrease.

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8. Elliptical orbit of Io comes from its resonant motion relative to Europa and Ganymede (see diagram in lecture notes or in textbook).

9. For each single orbit of Ganymede, Europa does exactly 2 orbits, and Io does exactly 4 orbits.

10. Every Ganymede orbit, the 3 moons line up, so Io gets a gravitational tug, in the same direction each time.

11. Cause for tidal torque:a. Tidal force of Jupiter distorts shape of spinning moon, raising 2

tidal bulges, one on the near and one on the far side of the moon.

b. If the moon rotates more than once per revolution about Jupiter, then that rotation will tend to have the tidal bulges point along a line that does not go directly from the center of the moon to the center of Jupiter but is twisted slightly in the direction of rotation.

c. Gravitational force on near bulge is greater than that on far bulge, and the difference between these forces creates a net twisting force on the moon that is a torque and slows its spinning.

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12. Cause for tidal flexing.a. After moon’s rate of spin is slowed to that of its rate of

revolution, so that it rotates once per revolution and always presents the same face to Jupiter, then the tidal torque on the moon ends.

b. Now what remains is periodic motion closer and further away from Jupiter. Strength of tidal distortion of moon now varies periodically, causing moon to flex repeatedly and heat up.

C. Jupiter’s moon Europa.1. Second Galilean moon out from Jupiter.2. Dense, with rocky interior.3. Tidal flexing by Jupiter should be heating interior.4. Molten ice mantle, and frozen ice crust.5. Could have liquid water ocean about 1 km below

surface. Possible place where there could be life, in the liquid water layer.

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D. Jupiter’s moon Ganymede.1. Third Galilean moon out from Jupiter.2. Largest moon in the solar system.3. Less dense, with some rock but more ice in interior.4. Some tidal flexing by Jupiter, but less than for Europa.5. Water or ice mantle, and frozen ice crust.6. Weird, grooved terrain.

E. Jupiter’s moon Callisto.1. Furthest Galilean moon from Jupiter.2. Less dense, icy moon.3. No orbital resonances with other moons.4. Surface covered by dark powder of unknown origin.

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F. Jupiter’s Galilean moons form a sort of miniature solar system (without Jupiter, of course, being a star).

1. Central object, Jupiter, contains the bulk of the mass of the system.

2. All four Galilean moons revolve in the same direction about Jupiter as Jupiter rotates about its axis.

3. All 4 moons orbit in roughly the same plane, the equatorial plane of Jupiter.

4. All 4 moons rotate in the same direction as their orbital motion.

5. The densities of these moons decrease steadily with distance from Jupiter, with the inner moons rocky and the outer ones icy.

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G. Saturn and its rings.1. Composition is roughly same as sun, mainly hydrogen

and helium.2. Helium is depleted near surface, because separates out

as helium rain, falls, and grav. potential en. gives heat.3. Rotation so rapid that Saturn is visibly oblate from the

centrifugal force (day is about 10 hours).4. More oblate than Jupiter, because its radius is about the

same as Jupiter’s, but its mass is only about a third as great. Therefore gravity at its surface is much less than on Jupiter. Nevertheless, its similar radius and rotational period mean that the centrifugal force at the equator of both planets is roughly the same. Consequently the centrifugal force reduces the inward pull of gravity in the equatorial region of Saturn by a much larger fraction (about 3 times larger) than for Jupiter.

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4. Horizontal bands in atmosphere are convection cells, as on Jupiter.

5. Helium appears to be settling toward center of Saturn, and is depleted in the surface layers. This process liberates gravitational potential energy and provides heat. This explains present temperature of Saturn.

6. Rings are inside the Roche zone.a. Roche zone is region near Saturn where tidal forces

are so great that small objects cannot accumulate into a single large object from their gravitational attraction.

b. No large moons inside Roche zone, but can have very small objects, held together by the strength of the material from which they are made (and not by gravity).

7. Rings made from chunks of ice.

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8. Inelastic collisions (can chip and not bounce) dissipate all but ordered velocities of circular motion around Saturn.

9. Rings very thin, due to dissipation of velocities out of the ring plane by inelastic collisions.

10. Theories of ring origin

a. Primordial. Have always been there.

1) Hard to see why rings have not spiraled into planet by now, given dissipation of orbital motions by inelastic collisions.

b. Ring material supplied by progressive disintegration of small moons that wander into Roche zone.

1) May be a transient phenomenon, lasting only, say, 100 million years, and we are just luckyto see it.

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Cassini Division in Saturn’s ringsa. Fairly wide gap in rings where appears from

distance of earth to be few if any ring particles.1) Close-up views from space probes show that

Cassini Division does contain ring particles, but they have been ground into a fine dust or powder

b. Particles in Cassini Division orbit Saturn precisely twice for each orbit of the moon Mimas.

1) This orbital resonance of particles in the Cassini Division gives them periodic tugs from Mimas always at the same point in their orbits.

2) Result is that these ring particles develop elliptical orbits, and these then necessarily cross each other, causing collisions that break the particles up into very small pieces.

3) Is same mechanism that makes Io’s orbitaround Jupiter elliptical, causing its tidal flexing.

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11. Enke Gap in Saturn’s rings.a. Less prominent than Cassini Division.b. Caused by gravitational perturbations of the tiny

moon Pan, which orbits Saturn within the Enke Gap.c. Ring particles inside orbit of Pan orbit faster, are

tugged to slower orbital speeds by Pan’s gravity as they pass by, have their centrifugal force reduced by this, and fall into orbits closer to Saturn.

d. Ring particles outside orbit of Pan orbit slower, are tugged to faster orbital speeds by Pan’s gravity, have their centrifugal force increased by this, and move into orbits further from Saturn.

e. The critical feature of the gravity-centrifugal-force relationship:

1) Explains why, after orbit change from gravitational tug from passing moon Pan,ring particle finds new, stable orbit.

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2) A ring particle orbiting Saturn just inside the orbit of Pan overtakes Pan repeatedly, because Kepler’s third law says the period of the particle’s orbit must be smaller than that of Pan.

3) This particle is therefore tugged by Pan to orbit more like Pan does, namely slower.

4) This is a torque. The angular momentum of the particle is reduced, but conserved after Pan passes by

5) The centrifugal acceleration (v2/r) is reduced on the particle, but gravitational acceleration is unchanged.

6) Particle now falls in toward Saturn, because gravity is now stronger than centrifugal acceleration.

7) Falls from radius r0 to smaller radius r1 , but, in falling, its angular momentum, mvr, is conserved.

8) Must therefore have: v0r0 = v1r1 , so that the new centrifugal acceleration at r1 is (v1

2/r1), and this must = (v1

2r12/r1

3) = (v02r0

2/r13) = (v0

2/r0)×(r03/r1

3)

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8) Must therefore have: v0r0 = v1r1 , so that the new centrifugal acceleration at r1 is (v1

2/r1), and this must = (v1

2r12/r1

3) = (v02r0

2/r13) = (v0

2/r0)×(r0 /r1 )3

9) This means, since r1 is smaller than r0 , that the centrifugal acceleration has increased as the particle fell in toward Saturn by the cube of the ratio of the original radius to the smaller final radius.

10) This tends to fix the problem that the centrifugal acceleration on the ring particle was too small after the passage of Pan to balance gravity from Saturn.

11) But during the ring particle’s fall, the gravitational acceleration toward Saturn increased as the centrifugal acceleration also increased.

12) The key point is that the centrifugal acceleration on the ring particle increased faster than the gravitational acceleration increased.

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12) The key point is that the centrifugal acceleration on the ring particle increased faster than the gravitational acceleration increased.

13) The gravitational acceleration is proportional to the inverse square of the distance to the planet.

14) Therefore it began as (something/r02) and it ended as

(something/r12).

15) At the end, it is larger by the factor (r02/r1

2), because the somethings are the same and cancel in the ratio.

16) However, the centrifugal acceleration got larger by the factor (r0

3/r13).

17) This is why the ring particle settles into a stable orbit closer to Saturn. As it falls inward toward the planet, the centrifugal acceleration it feels grows faster than the gravitational acceleration it feels. Consequently, it must eventually stop falling inward.Similar argument applies to particles orbiting outside Pan.

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13. Action of shepherd moons in keeping a thin ring in place. Example of thin F ring.

a. Difference between solid-body rotation and Keplerian rotation and its role in allowing shepherd moons to keep a thin ring in place (see above notes on Pan and Enke gap).

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VI. Solar system formation. Chapter 6 of your textbook.

A. Four challenges to a theory of solar system formation.

1. Patterns of motion.

2. Progression of planet average densities and composition.

3. Asteroids and comets.

4. Various exceptions to the generally obeyed rules or trends.

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B. Nebular collapse theory.

1. Gas cloud collapsed under its self gravity to produce the solar system.

2. Gas cloud was probably located in region in which many stars were forming from sections of large amount of interstellar gas.

3. Collapse may have been stimulated by pressure from the wind or ionizing radiation from a nearby, young, bright, very massive star.

4. Collapse may have been stimulated by pressure from the blast wave from the supernova explosion of a nearby, very massive star, that would have formed only a bit earlier but lived out its life very fast.

a. This scenario supported by nuclear isotope evidence from a meteorite, the oldest material yet found on Earth or on the Moon.

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5. Conservation of angular momentum means that as the gas cloud collapses under gravity, it spins faster.

6. Collisions of different parts of the gas cloud dissipate motions, except for the spinning motion that is demanded by the conservation of angular momentum.

7. Flat, rotating disk is formed, with protostar in center.

a. Conservation of angular momentum demands that as material moves inward in such a rotating disk, it must spin faster.

b. Faster rotation as fall inward in disk means that centrifugal force outward grows stronger.

c. Gravity also grows stronger, but does so less rapidly.

d. Result is that always will eventually stop inward motion, and this is why the disk persists.

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e. To form the protosun, angular momentum must therefore be somehow removed from the disk material, allowing it to fall further inward.

f. Magnetic fields act to form a kind of coupling between material in different parts of the spinning disk that acts like a viscosity and can transport angular momentum outward so disk can fall inward.

8. In the protoplanetary disk around the protosun, all is rotating in same plane and in same direction, which explains the “patterns of motion” in our solar system.

9. Planetesimals condense out of gas and accumulate, through inelastic collisions, to form larger objects.

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10. In central part of disk, where temperatures are greater, only rocky materials or metals can condense, so that planets formed here are dense and rocky.

a. As go out from protosun in the disk, temperatures decline and a progression of materials in the gas become able to condense into solid particles.

11. Far out in disk, where is much colder, ices can alsocondense, so that there is much more material available to accumulate into planets.

a. Larger amount of planetesimal material and weaker wind from forming star at center (the protosun) allows these planets to sweep up nebular gas.

b. Growing planets are gaseous, large, and have gas disks orbiting around them, in same direction as their orbital motion about the protosun.

c. From these disks, systems of moons form.

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14. Common misconceptions:a. The inner planets are made of rock and metals,

because these are heavier and settled to the center of the protoplanetary nebula as a result of gravity, leaving the more volatile, gaseous material behind.This is not our present belief, because the conservation of angular momentum prevented this material from moving substantially inward through the gas of the nebula. The inner planets are made of rock and metals because only those materials could condense into solid particles in the hotter part of the nebula close to the sun.

b. Be sure that you are aware of the present theory for how the earth acquired its oceans and atmosphere.

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VII. Sun (discussed in Chapter 10 of your textbook).

A. Build interior model, as did for Jupiter.1. Assume composition same as on surface or as found in

icy meteorites.2. Balance gas pressure forces (outward) and gravity

(inward).3. In core, temperature so high that hydrogen fuses to form

helium, producing heat, light radiation, and neutrinos.4. Hydrogen fusion can power the sun for billions of years,

while chemical burning processes (like burning coal), or gradual liberation of gravitational potential energy by gradual contraction cannot.

5. In central 2/3 to ¾ of the sun’s radius, energy moves outward carried by light radiation bouncing back and forth in the highly opaque gas.

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6. Light diffusing outward from the center of the sun takes about a million years to reach the surface.

7. In outer 1/3 or ¼ of the sun’s radius, energy is carried outward by material convection: heated, buoyant gas rises, gives off some of its heat, sinks again, only to be reheated.

8. Heat energy thus carried outward from the sun’s interior nuclear furnace is emitted at the solar surface as the sunlight that warms the earth.

9. The very high temperatures and densities near the center of the sun are required to get hydrogen to fuse into helium, because the colliding protons must overcome their mutual electrical repulsion in order to get close enough for the stronger but short-range nuclear force to pull them together.

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10. Creating similar conditions under controlled circumstances in fusion reactors on the earth is one of the major technical challenges facing us today.

a. Two major approaches are being pursued.

b. First is to hold the hydrogen gas together using magnetic forces. This would produce fusion energy continuously.

c. Second is to push the hydrogen to high density and temperature with laser beams trained on it from all directions. This generates a microexplosion.

11. Controlled nuclear fusion produces radioactive by-products, like controlled nuclear fission. However, the fusion by-products decay away in weeks, while some fission by-products remain radioactive for up to a million years.

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12. Fission is the splitting up of a large atomic nucleus to form smaller nuclei, transforming mass energy into other, more useful forms of energy in the process.

13. The nuclear power plant down the Mississippi River from Minneapolis uses fission of Uranium nuclei to produce heat, which it converts into electricity.

14. Fusion is the combining of two small nuclei into a single, larger nucleus, transforming mass energy into other, more useful forms of energy in the process.

15. Getting fusion to happen in a controlled fashion on earth is very difficult. The reason is that to fuse, two protons must collide so strongly and so directly head-on that their mutual electrical repulsion can be overcome and the much stronger, but very short range nuclear force can bind them together. High density gives us many collisions, some of which can be head-on. Very high temperature gives these collisions enough strength.

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16. For this reason, controlled fusion is much more difficult and challenging than controlled fission. Essentially, we must create conditions similar to those at the center of the sun, and maintain these conditions long enough for the hydrogen to burn to give helium.

17. The main effort in this area in the U.S. is at a department of energy lab in California, where the world’s largest laser system is used to compress tiny capsules of hydrogen fuel for a short time to extreme conditions.

18. Here the difficulty is figuring out how to get the bulk of the laser energy deposited in the capsule confining the hydrogen fuel in such a way that the fuel is compressed radially to high density and temperature without it squirting out and without parts of the capsule protruding into the fuel and contaminating it.

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19. The main effort in controlled fusion in Europe is to build a huge magnetic system to confine the hydrogen fuel under less extreme conditions but for a longer time.

20. Both efforts have not yet succeeded, but the researchers are very hopeful.

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Chapter 11 of your textbook:VIII. Observable properties of stars and the H-R diagram.

A. Luminosity, the amount of energy radiated per unit time.1. Distinction between apparent brightness and luminosity.2. If know apparent brightness and distance, then can get

luminosity, an intrinsic property of the star which is independent of its location.

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B. Distance1. Most direct measurement is by parallax.2. If star in a cluster, can get distance by main sequence

fitting (see below, and remember your lab on this).C. Surface Temperature

1. Related to color through the properties of the black body spectrum.

2. Spectral types are even more directly related to surface temperature.

3. Spectral types classified according to which lines are in the spectrum and how strong they are.

4. Lines from highly ionized elements indicate high surface temperatures, while lines from molecules indicate low surface temperatures.

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5. Annie Jump Cannon created the spectral classifications we use today, and classified 400,000 stars.

6. Cecelia Payne-Gaposhkin, applying the then (1925) new atomic theory, explained the relationship of these spectral types to ionization states, and hence to surface temperature.

D. Mass1. Determining the mass unambiguously is very difficult.2. Most direct determination is from orbital period and

average separation of a (relatively) low mass satellite.3. Observation of planets too difficult, so use eclipsing

binary stars.4. We observe distant binary stars by noting the light

variations from their eclipsing of each other and/or the periodic Doppler shifts from their orbital motions.

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5. From Doppler shifts, get orbital velocities, and from these, along with the orbital periods, we can deduce orbital separation, independent of the distance to the double star.

6. From fact that a binary is eclipsing, we know that the orbits are seen edge-on, so that the line-of-sight velocities from Doppler shifts are indeed the full orbital velocities, not just projected components of these velocities.

7. Orbital periods plus orbital separation, from the Doppler shifts, give us, through Newton’s law of gravitation, the sum of the two stars’ masses.

8. The size of the individual stars’ Doppler shifts, together with the necessary balance of momentum in the system, from Newton’s 3 laws of motion, give us the ratio of the masses of the two stars.

9. The sum of the masses, the ratio of the masses, and a little algebra give us the two mass values individually.

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10. Note that we can do all this without knowing the distance to the binary star. The key is the information about the velocities that comes from the Doppler shifts, and the knowledge that we see the system along a line of sight in the orbital plane because the binary is eclipsing.

11. By measuring the time intervals of the eclipses, and knowing the orbital velocities and separations, we can even deduce the radii of the two stars.

12. Very few binary stars out there allow all this wonderful data to be determined. Eclipsing binaries are rare.

E. Apparent brightness and color (hence surface temperature) are the quantities most easily determined from observations.

F. Plot of apparent brightness against surface temperature is called the H-R (Hertzsprung-Russell) diagram.(Which quantity on which axis, which way increasing?)If we know distances, we can convert brightness to luminosity, the amount of energy radiated per unit time by a star.

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1. Approximate locations on H-R diagram of stars burning hydrogen in their cores (main sequence), red giants, white dwarfs.

2. Orientation of lines of equal stellar radius in H-R diagram and how this orientation comes from the properties of the black body spectrum.

3. Trend of stellar mass along main sequence in diagram.

4. Trend of main sequence lifetime along main sequence.

G. To reduce scatter from errors in luminosity, stemming from distance measurements, plot H-R diagrams for star clusters, all of which presumably formed from a single large gas cloud and are therefore nearly the same distance from us.

1. Direct relationship between apparent brightness and luminosity for stars in a star cluster. Same conversion factor for all the cluster stars.

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2. Cluster stars all approximately same age.

3. Cluster stars all presumably have approximately the same composition (relative abundances of elements).

4. Fit main sequence of star cluster H-R diagram to that of H-R diagram of stars near enough to us to have accurately known distances.This gives conversion factor between apparent brightness and intrinsic luminosity, and from this we can compute the distance to the cluster.This is called main sequence fitting.

5. Use theory of stellar evolution to compute main sequence lifetime of the brightest star in the cluster that is still on the main sequence, and this lifetime will nearly equal the age of all the stars in the cluster.

6. Note that for very young cluster, the lowest mass stars may not yet have reached the main sequence.

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7. H-R diagrams of star clusters of different ages give us our best observational information on stellar evolution.

a) Stars spend the longest part of their lives on the main sequence. This is why most of the stars that we see in a cluster are on the main sequence.

b) After their main sequence stage of evolution, stars move to the right in the H-R diagram.

1) This rightward motion in the H-R diagram tells us that their surface temperatures decrease without their luminosities also decreasing.

2) The properties of the black body spectrum then demand that these stars must be increasing in radius. They are becoming giant stars.

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IX. Life History of a Low-Mass Star like the Sun.This is discussed in Chapter 12 of your textbook.

From gas cloud to main sequence.1. Collapsing gas cloud first identifiable as protostar when

has a visible surface.2. This happens when cloud becomes dense enough to be

opaque, so we see its surface and do not simply see right through it.

3. Gravity of cloud pulls it together, so that it shrinks.4. Decreasing size liberates gravitational potential energy.5. Liberated energy appears as heat.6. Escaping light from surface carries away heat energy.

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7. This tends to reduce gas pressure, reinforcing collapse.

8. Continued collapse causes more energy to be liberated, and central region gets hotter.

9. Eventually, core hot enough (a few million degrees K) to fuse hydrogen into helium via proton-proton chain of reactions (see diagram in your text or the class notes).

10. Hydrogen fusion generates heat, which enhances gas pressure and halts the gravitational collapse.

11. The star is now on the main sequence.

12. Many more low mass stars seem to form from a given large cloud than high mass stars.

13. There could be a large number of very low mass “stars” (or perhaps, more properly, “Jupiters”) out there, and they would be so faint that we would not see them.

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B. Main sequence life: Hydrogen burning in core.

1. Proton-proton chain (mainly) builds helium from hydrogen in core .

2. Boring. Long. Good for life forms on planets.

C. Red Giant Phenomenon.

1. Hydrogen fuel exhausted in core.

2. Helium in core does not burn to liberate energy and provide pressure support.

3. As in initial contraction to main sequence, heat energy is radiated from star’s surface, reducing pressure support, and leading to contraction under gravity.

4. Contraction liberates gravitational potential energy, which causes core and surrounding layers to actually heat up.

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5. Unburned hydrogen just outside of helium core gets so hot that it starts to fuse to form more helium.

6. This phase is called hydrogen shell burning.

7. The hydrogen burning shell is so hot and dense, that it generates much more (up to 10,000 times more) heat than did the hydrogen burning core when the star was on the main sequence.

8. The escaping heat flux puffs up the outer layers of the star by about a factor of 100 in radius to form a “red giant.”

9. As the hydrogen burning shell produces more inert helium, added to the core, the core’s increasing gravity makes the central region, including the hydrogen burning shell, shrink, which of course makes it heat up, which of course makes the hydrogen burn more rapidly.

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10. Eventually, core reaches 100 million degrees K needed to fuse helium into carbon by the “triple alpha process” (two helium nuclei collide to form a beryllium nucleus, and before it can come apart another helium nucleus fuses with it to form a carbon nucleus, which is stable).

D. Core helium burning stage.

1. Helium fusion in core provides heat and pressure, so core expands, and hydrogen shell lifted up, so now less dense & less hot, so total luminosity is diminished.

2. Puffed out outer layers of red giant phase now contract to form much smaller, but still very large, helium burning star.

3. At this high luminosity, helium is exhausted in core in only about 100 million years.

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E. Second red giant phase.

1. Core shrinks.

2. Helium burning shell around core.

3. Hydrogen burning shell around helium burning region.

4. Luminosity shoots up again.

5. Outer layers puff way up again, because of increased luminosity, which produces radiation pressure.

6. This stage lasts only about one million years.

7. Carbon cannot burn in core unless temperature rises to 600 million degrees K.

8. For one solar mass stars, electron degeneracy pressure (pressure resulting from the impossibility of two electrons being in the same place at the same time) halts the core contraction before it gets this hot.

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9. The outer layers surrounding the inert carbon core are ejected to form what is called a planetary nebula(when first seen, using small telescopes, these objects resembled planets).

10. The carbon core, still very hot, then very gradually cools, growing steadily dimmer.

11. This exposed core is called a white dwarf, because its high surface temperature makes it appear white.

12. Its very small size, due to its enormous density, gives it a small surface radiating area, so that it is dim.

Evolution of solar mass star past main sequence (a review).Many were confused, so worth a recap.

Core and burning shells right near it drive the dynamics.

Outer layers are passive, except insofar as their weight bears down upon the core region, helping to establish the pressure there.

Outer layers do whatever they have to do to carry luminosity of the inner region out to surface, where is radiated into space as a black body spectrum. Key is that balance, so that have same energy passing outward through each spherical surface at each radius, is established very quickly compared to lifetime of star in any of its phases (except explosions).

Evolution of solar mass star past main sequence.Core region does whatever comes naturally under the conditions in which it finds itself:

1. Main sequence: hydrogen burns in core.

a) If burns too fast, heat and pressure cause core to expand, and this causes density and temperature to drop, and this reduces reaction rate (and heat generation rate) toward equilibrium.[For a massive star, reactions must go faster in core to provide enough energy to lift all the material above it against gravity in order to produce the lessening of central density and pressure needed to make the mechanism above work by which the reaction rates adjust to an equilibrium. This equilibrium is therefore reached at a higher density and temperature than for a low mass star, and this enables the CNO catalytic reaction cycle to produce helium faster than the p-p reaction chain that dominates for low mass stars like the sun.]

b) If burns too slowly, does not generate enough heat and pressure to balance gravity, so contracts, liberates gravitational potential energy, so heats up and gets denser, so reactions go faster, leads to equilibrium.

Evolution of solar mass star past main sequence.Core region does whatever comes naturally under the conditions in which it finds itself:

1. Main sequence: hydrogen burns in core.

c) Quickly adjusts so that a steady amount of energy is generated in core.

d) This energy is carried outward through the star by either radiation making its way outward or by convection or both.

e) If the energy escaping from the surface is not great enough to balance that being generated in the core, then the surface layers expand, producing a larger surface area and enabling more energy to escape, until equilibrium is reached.

f) If the energy escaping from the surface is greater than that being generated in the core, then the surface layers are cooled, they collapse inward, gravitational potential energy is liberated, and an equilibrium is reached with a higher surface temperature but a smaller radiating surface, so that we have balance of energy generation and loss.

Evolution of solar mass star past main sequence.Core region does whatever comes naturally under the conditions in which it finds itself:

1. Expanding subgiant: hydrogen burns in shell around core.

a) Burning rate is self-adjusting, just as it is on main sequence.

b) Cannot burn in core, because H depleted there.

c) As energy lost through surface and not replenished from core burning, core region contracts as gravity now overwhelms pressure there.

d) As core region contracts, grav. potential energy liberated, and core region heats up.

e) Keeps contracting and heating until hydrogen ignites in shell around core.

f) As more and more inert helium ash accumulates in core, providing less pressure (4 H replaced by 1 He, and pressure proportional to temperature times number of particles, which is now smaller) and not providing heat from burning, core shrinks and heats up due to release of gravitational potential energy.

Evolution of solar mass star past main sequence.Core region does whatever comes naturally under the conditions in which it finds itself:

1. Expanding subgiant: hydrogen burns in shell around core.

f) As more and more inert helium ash accumulates in core, providing less pressure (4 H replaced by 1 He, and pressure proportional to temperature times number of particles, which is now smaller) and not providing heat from burning, core shrinks and heats up due to release of gravitational potential energy.

g) Increasing temperature and density in H-burning shell increases nuclear reaction rate, so that luminosity rises.

h) Outer layers must expand from increasing radiation pressure from energy in form of light that is forcing its way outward to surface.

i) Size of star is set by the radius needed to emit enough energy in black body radiation in order to balance nuclear energy generation in H-burning shell deep inside star. Outer layers are now passively reacting to what is going on in deep interior.

Evolution of solar mass star past main sequence.Core region does whatever comes naturally under the conditions in which it finds itself:

1. Expanding red giant: hydrogen burns in shell around core.

a) Temperature of surface is set approximately at the value where hydrogen gas tends to become largely neutral (and not ionized), which allows light radiation to easily escape.

b) Size of star’s surface is set by the luminosity in the deep interior, which is constantly increasing as more and more inert helium accumulates in the core, pulling the H-burning shell further and further inward and causing it to get hotter and denser, hence more luminous.

c) This continues until temperature of the core reaches 100 million degrees, the temperature at which helium will ignite in the core.

d) Helium ignition is in a flash or not depending upon mass of star, which determines whether or not the helium atoms are effectively “touching” when they ignite or (if massive enough) whether they are instead supported by normal gas pressure (zooming, not touching).

Evolution of solar mass star past main sequence.Core region does whatever comes naturally under the conditions in which it finds itself:

1. Helium burning star.

a) Helium has ignited in the core, and created enough heat there to expand the core region by its pressure against gravity.

b) Expansion of core region causes H-burning shell region also to expand and be lifted upward.

c) Greater volume of expanded H-burning shell has caused its temperature to drop (expanding a gas cools it – this is how an air conditioner works) as well as its density, so the nuclear reaction rate drops quite a lot.

d) Net effect is lower total energy generation rate through nuclear reactions overall.

e) Less energy escaping through outer layers reduces radiation pressure there, these layers shrink until balance of escaping energy is established at much smaller size.

Evolution of solar mass star past main sequence.Core region does whatever comes naturally under the conditions in which it finds itself:

1. Asymptotic giant branch (AGB) star.

a) Helium has been depleted in core, where carbon (and a little oxygen) collect, again producing less pressure at same temperature, because there are now fewer particles.

b) Core region contracts, gravitational potential energy is released, and core region heats up.

c) Increasing temperature and density in helium shell around core causes helium to ignite there and to burn to create still more carbon.

d) Nuclear reaction rate in helium burning shell increases as more inert carbon collects in core, causing it to keep contracting and heating up.

e) Outer layers puffed out by radiation pressure as huge energy generation around core makes more and more radiation trying to escape. Surface grows to the size required to allow this much energy to escape as black body radiation.

Evolution of solar mass star past main sequence.Core region does whatever comes naturally under the conditions in which it finds itself:

1. Asymptotic giant branch (AGB) star.f) Helium burning shell and hydrogen burning shell consume material

above themselves at different rates.g) Result is that have series of “thermal pulses” in which helium shell

ignites, burns up all the available helium below the hydrogen burning shell, and then helium burning ceases for a few thousand years until have the next thermal pulse.

h) After each pulse, some of the processed material below the hydrogen burning shell, which consists of a mixture mainly of helium and carbon, is “dredged up” by convection and carried outward toward the star’s surface layers, where the temperature is very much cooler.

i) Dredged up carbon can condense to form tiny grains of graphite dust near the star’s surface. This dust is dark, absorbs the copious outward streaming light radiation, and carries the gas outward with itself.

j) The huge luminosity at this stage drives the dust and gas of the star’s vast outer envelope outward to form a planetary nebula.

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X. Life History of a High-Mass Star.This is discussed in Chapter 12 of your textbook.

A. Same phases as for low mass star up to formation of inert carbon core, but all these phases happen much faster.

B. Hydrogen fusion on main sequence involves not only the proton-proton chain of reactions, as in the sun, but also the CNO cycle.

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C. CNO cycle is a catalytic reaction chain, involving the relatively rare nuclei Carbon, Nitrogen, and Oxygen, and leaving the concentrations of these heavy nuclei unchanged.

D. CNO cycle produces energy more rapidly in high mass stars than proton-proton chain.

E. CNO cycle requires the greater temperatures that are found in cores of high mass main sequence stars in order for the protons to overcome the higher electromagnetic repulsion of these heavier nuclei and to fuse with them.

F. High mass stars produce so much energy this way that the radiation carrying the heat to the surface exerts a pressure comparable to or greater than the normal gas pressure.

G. Light is massless, but it has momentum, and hence can exert pressure, which we call radiation pressure.

H. Stars above about 100 solar masses would exert so much radiation pressure that we think they would tear themselves apart, and this explains why no such stars are observed.

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I. Pick up evolutionary story at the generation of an inert carbon core, with helium and hydrogen burning in shells around it.

1. Greater gravity of high mass star can cause carbon core to reach the 600 million degrees K required for it to burn.

2. Complex fusion reactions.

3. Among fusion reactions are helium capture reactions, which add helium nuclei to elements like carbon, oxygen, and neon (see textbook diagram or diagram in class notes) to ultimately produce magnesium.

4. Heavy nuclei can also combine with each other.

5. Final product is iron, since no energy is liberated by adding any nucleons to an iron nucleus.

6. For stars less than 8 solar masses, carbon-oxygen core is supported by electron degeneracy pressure, outer layers are ejected, and the exposed core is a white dwarf.

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7. For stars above 8 solar masses, fusion goes all the way to iron, because of greater heat in core. Pressure on iron core becomes so great that electrons are forced into the iron nuclei, where they fuse with the protons to form neutrons, emitting neutrinos in this process.

8. Electron degeneracy pressure that originally supported the iron core arises because electrons are, essentially, touching each other and can be brought no closer together.

9. When electrons combine with protons, they become neutrons, which are very, very much smaller than electrons (that is, they take up much less space).

10. Iron core collapses, in a fraction of a second, until the neutrons are, essentially, touching.

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11. Neutron degeneracy pressure then supports core, which is now essentially just one humongous nucleus held together by gravity rather than by the nuclear force.

12. When a solar mass or so of iron collapses in less than a second to a ball of neutrons about the size of New York City, an astronomical amount of energy is released, mostly in the form of energetic neutrinos.

13. The neutrinos stream outward, heating and accelerating the outer layers to produce a supernova explosion.

14. In the final moments, the material of the star is “bathed in neutrons,” so that a host of very heavy elements can be produced by neutron capture, absorbing some of the copious energy and making the building blocks for future planets like the earth.

15. The supernova shines with the brilliance of 10 billion suns, but only for a few days.

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XI. Modifications of Stellar Evolution in Binary Star Systems(refer to notes from Lect14-stars2-3-26-18.pdf)

A. Roche diagram.1. Drawn in frame of reference where both stars stationary.2. Always have such a frame because tidal friction causes

orbits to decay to circles, with stars always presenting same faces to each other (as the Moon does to Earth).

3. To move along equipotential lines in diagram requires no energy and liberates no energy.(is like moving along surface of a frictionless earth with no hills or valleys).

4. Moving perpendicular to equipotential lines in diagram either requires energy or liberates it (it is like jumping up from or falling down toward earth’s surface).

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5. One special equipotential surface encompasses both stars, in the shape of a figure 8.

6. A star’s “Roche lobe” is the volume contained in it’s side of this figure-8-shaped equipotential surface.

7. There is no net gravitational and centrifugal force at the intersection point of the figure-8-shaped equipotential.(This special point is called the L1 Lagrangian point.)

8. If an expanding star “fills its Roche lobe,” so that gas on its surface finds itself at this special point, only a slight further expansion of the star will push this gas over into the Roche lobe of the companion star, and it will fall down onto that star (eventually).

B. Binary star evolution.

1. Evolution begins with 2 close but separate stars, each well within its own Roche lobe.

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2. The more massive star exhausts hydrogen fuel in its core and begins to expand into its red giant phase.

3. This expansion can make the star fill its Roche lobe.

4. Further expansion has the effect of transferring material from the star’s outer layers onto the companion star via a thin stream emanating from the L1 Lagrangian point (the cusp in the Roche lobe that faces the other star).

5. Increasing mass of companion star accelerates burning of hydrogen in its core, bringing it to its red giant phase.

6. Multiple possible scenarios from this point.

a) Proceed to common envelope phase (contact binary)[We did not discuss the common envelope phase][It will not be on the test]

1) Second star swells up as a red giant and fills its Roche lobe.

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2) Both stars fill Roche lobes and continue to swell, creating a common envelope star.

3) Two stellar cores generate heat energy at different rates, but they are both surrounded by a common envelope that equalizes the common surface temperature.

b) Proceed to white dwarf with accretion disk (nova)[this scenario could be on the test, we did get to it]

1) The first star ejects its outer envelope, leaving only a white dwarf.

2) The second star exhausts hydrogen in its core and enters its red giant phase.

3) The second star swells up, fills its Roche lobe, and begins to transfer mass back to the first star, which is now a white dwarf.

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4) The gas falling down toward the white dwarf does not strike its surface directly, because it has too much angular momentum.

5) Infalling gas forms an accretion disk around the white dwarf.

6) Gas in the disk slowly spirals inward, as its angular momentum is carried outward by mechanisms not yet understood by astronomers.(Mechanisms could be friction or spiral waves.)

7) This gas, rich in hydrogen, accumulates on the surface of the white dwarf until there is enough to burn (to form helium).

8) Burning of the gas is explosive, giving a nova outburst and ejecting a shell of gas from the system.

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9) The mass of the white dwarf gradually increases.10) Ultimately, the mass of the white dwarf becomes

too large to be supported against gravity by electron degeneracy pressure, it collapses, and a supernova explosion results.

11) We believe that all these white dwarf supernova explosions are very similar, since they are all initiated in the same way, by a gradual approach to the largest mass white dwarf possible, and then beyond that, producing the explosion.

12) For this reason, we believe that with only small variations, the luminosities of these white dwarf supernova outbursts are all the same, so that we can use them as “standard candles” to obtain distances to far away galaxies.(We compare the apparent brightness of the supernova to its “standard” luminosity.)

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c) Proceed to neutron star with accretion disk (X-ray binary) [Did not discuss. Not on test..]

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d) Black hole scenario[This will not be on the test either.]

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C. Black hole properties.[This will not be on the test. We will encounter this in next segment.]

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The second midterm exam will be given on Wednesday, April 4, from 2:30 to 3:45 PM in Willey Hall, room 125.

Be sure to bring two pencils and a picture ID.

The midterm exam will consist entirely of multiple choice questions.

No books or “cheat sheets” will be permitted at the exam.

You will not be allowed to access your cell phone at the exam.

The exam, like the course, will be graded on a curve.

A histogram of the results of the exam will be shown in class and posted on the course Web site (along with the lecture notes) in order to help you see where your own score falls in this distribution. Approximate letter grades indicating the grading curve will also be indicated on this histogram.