## 20080430

### Erasing slate: universal gravitational constant

"G = 6.67(10^11)" by Anonymous
Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Latest scribbling on the lift-and-erase slate in the hallway, outside the office door.

### Astronomy midterm question: star cluster age

Astronomy 10 Midterm 3, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.5

[15 points.] In cluster 1, the main sequence extends from spectral class O to spectral class K. In cluster 2, there are no main sequence stars cooler than spectral class F. Determine which star cluster is older, and explain the reasoning behind your choice.

• p = 15/15:
Correct. The OBAFGKM main sequence runs from massive to medium to low-mass. Massive stars become main sequence stars sooner, and spend less time as main sequence stars than medium mass stars. Since cluster 1 (OBAFGK) has more medium-mass stars on the main sequence than cluster 2 (OBAF), then cluster 1 must be older, as more time was required for its spectral class G and K stars to reach its main sequence.
• r = 12/15:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Recognizes the relationship between mass (inferred from spectral type -> temperature -> luminosity -> mass along the main sequence line), and main sequence lifetime, but rest of discussion is garbled or incomplete.
• t = 9/15:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least mentions some understanding of the relationships between mass and evolution times.
• v = 6/15:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x = 3/15:
Implementation/application of ideas, but credit given for effort rather than merit.
• y = 1.5/15:
Irrelevant discussion/effectively blank.
• z = 0/15:
Blank.
Section 4160
p: 5 students
r: 9 students
t: 10 students
v: 14 students
x: 1 student
y: 1 student
z: 0 students

A sample "p" response (from student 1223):

### Astronomy midterm question: cool star brighter than a hot star?

Astronomy 10 Midterm 3, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.5

[15 points.] How is it possible that a cool star can be more luminous than a hot star? Using an H-R diagram, explain your reasoning with Wien's law and/or the Stefan-Boltzmann law

• p = 15/15:
Correct. The Stefan-Boltzmann law states that luminosity depends on both the size (surface area) of the star, and its temperature. Thus a merely warm star can still be more luminous than a hotter star if the warm star is much bigger in size than the hotter star.
• r = 12/15:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Stefan-Boltzmann law is garbled, but at least understands how size can be independent from temperature in determining the luminosity of a star.
• t = 9/15:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. Discussion based somehow on Wien's law and/or the Stefan-Boltzmann law.
• v = 6/15:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Typically confuses luminosity with apparent magnitude by saying that the warm star could be closer to the Earth than the hotter star, when in fact luminosity (and absolute magnitude) are unaffected by distance.
• x = 3/15:
Implementation/application of ideas, but credit given for effort rather than merit.
• y = 1.5/15:
Irrelevant discussion/effectively blank.
• z = 0/15:
Blank.
Section 4160
p: 19 students
r: 10 students
t: 7 students
v: 4 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 1231):
Another "p" response (from student 4607), using what appears to be a luminosity versus wavelength graph (due to the Planck curves) rather than a luminosity versus temperature (H-R digram) graph:

## 20080429

### Astronomy midterm question: Sun as a pulsar?

Astronomy 10 Midterm 3, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal M3.5
[15 points.] Decide whether our Sun will ever become a pulsar. If so, then discuss how this process will occur. If not, then discuss why this process cannot occur. Explain using the properties and evolution of stars.

• p = 15/15:
Correct. Adequately describes either what the Sun would eventually become (white dwarf), or why the Sun cannot become a neutron star (pulsar).
• r = 12/15:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
• t = 9/15:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least demonstrates understanding that the Sun is not massive enough to become a supergiant -> type II supernova -> neutron star, and/or understands the properties of a pulsar that presumably cannot be attained by the medium-mass Sun.
• v = 6/15:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Explains that the Sun will become a pulsar, or explains that the Sun cannot become a pulsar because it is too massive.
• x = 3/15:
Implementation/application of ideas, but credit given for effort rather than merit.
• y = 1.5/15:
Irrelevant discussion/effectively blank.
• z = 0/15:
Blank.
Section 4160
p: 9 students
r: 3 students
t: 10 students
v: 9 students
x: 7 students
y: 1 student
z: 1 student

Section 5166
p: 11 students
r: 5 students
t: 14 students
v: 22 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 0223):
Another "p" response (from student 4607), perhaps looking to be disappointed when the Sun does not go out with a bang:

## 20080428

### Astronomy midterm question: H-R turnoff points

Astronomy 10 Midterm 3, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.5
[15 points.] Show what the turnoff point of a star cluster is on an H-R diagram, and explain how this can be used to estimate the age of the stars in the cluster.

• p = 15/15:
Correct. "Turnoff point" is where stars from a cluster deviate from the main-sequence line. A turnoff point near the top left indicates a young cluster, as the massive stars have not yet reached their supergiant phases, turnoff point near the bottom right indicates an older cluster, as only the low-mass stars are still on the main sequence. Clearly indicates/draw/describes what a typical turnoff point looks like, and how its position is correlated with age. May instead describe how the different evolution tracks/rates of different mass stars that are observed on the main sequence line can be used to determine the age of a star cluster, without explicit mention of the "turnoff point."
• r = 12/15:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
• t = 9/15:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least understands mass determines evolution tracks/rates.
• v = 6/15:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x = 3/15:
Implementation/application of ideas, but credit given for effort rather than merit.
• y = 1.5/15:
Irrelevant discussion/effectively blank.
• z = 0/15:
Blank.
Section 5166
p: 9 students
r: 4 students
t: 15 students
v: 10 students
x: 7 students
y: 7 students
z: 3 students

A sample "p" response (from student 9518):
A sample "y" response (froms student ), who like many others felt compelled to write something down (for minimal partial credit) rather than leave the answer blank:

## 20080427

### Astronomy midterm question: cooler, brighter star versus dimmer, hotter star

Astronomy 10 Midterm 3, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.5
[15 points.] How is it possible that a more luminous star can be cooler than a less luminous star? Using an H-R diagram, explain your reasoning with Wien's law and/or the Stefan-Boltzmann law.

• p = 15/15:
Correct. The Stefan-Boltzmann law states that luminosity depends on both the size (surface area) of the star, and its temperature. Thus a cooler star can still be more luminous than a hotter star if the cooler star is much bigger in size than the hotter star.
• r = 12/15:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Stefan-Boltzmann law is garbled, but at least understands how size can be independent from temperature in determining the luminosity of a star. May confound mass with size in discussion.
• t = 9/15:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. Discussion based somehow on Wien's law and/or the Stefan-Boltzmann law.
• v = 6/15:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Typically confuses luminosity with apparent magnitude by saying that the cooler star could be closer to the Earth than the hotter star, when in fact luminosity (and absolute magnitude) are unaffected by distance.
• x = 3/15:
Implementation/application of ideas, but credit given for effort rather than merit.
• y = 1.5/15:
Irrelevant discussion/effectively blank.
• z = 0/15:
Blank.
Section 5166
p: 21 students
r: 10 students
t: 9 students
v: 11 students
x: 2 students
y: 1 student
z: 1 student

A sample "p" response (from student 1315):
This same student, like many others, made a thumbnail note of Wien's law and the Stefan-Boltzmann law on their exam paper at the start of the midterm. Note the explicit use of the word "proportional" instead of the symbol, which some students this semester called "The Pliers."
Another student 1652, after writing a "p" response, decried the abstractness and irrelevance of astronomy at this point in the semester.

## 20080426

### Education research: "guessing/don't know" clicker response

Comment was made on the previous post: Astronomy clicker question: type II supernova energy source.
Thank you for providing such an extraordinary resource through your blog of questions, etc!

Question: Do you or other professors ever use (or have you considered using) a response of "? - I don't want to guess."

I used this response frequently for formative assessment questions when I taught math at the college level and integrated the use of clickers in my teaching. Students were quite willing to choose this response because they weren't embarrassed after choosing it.

By providing this response for questions used in formative assessment I was able to get better data to inform my instruction. Sometimes (much to my chagrin) 30-50% of my students would choose this response following instruction on a topic so that I then knew I should reteach the topic and then reassess.

Thanks.

Tim Fahlberg

Though not an option on this semester's (Spring 2008) Astronomy 10 clicker questions, there is an "I'm lost, and I don't know how to answer this" response on Physics 5A clicker questions (cf. previous post, Physics clicker question: average versus instantaneous velocity for an example), which is a much more common response by students earlier in the semester, than later in the semester, most likely due to an unfamiliarity with both clickers and the material in general. (Whenever the "I'm lost" category does become the majority of responses in sporadic cases, then yes, the instructor is obliged to diagnose the cause of the disconnect, and recover using more exposition/examples.)

Also for some clicker questions in Physics 5A, students enter a numerical value for their answer following a calculation. If students are lost and don't know how to answer the question, they are instructed to enter a "nonsense" numerical answer (e.g., "-999") rather than an approximate guess. These answers will then fall outside the correct answer (which can be extended to within a +/- uncertainty), and also amuses the class as to the creativity of "wild" answers.

## 20080425

### Astronomy clicker question: type II supernova energy source

Astronomy 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal M3.4

Students were asked the following clicker question (Classroom Performance System, einstruction.com) in the middle of their learning cycle:

[0.3 points.] A supergiant will eventually explode as a type II supernova. What provides the energy for this explosion?
(A) Radioactive decays of unstable heavy elements.
(B) Fusion of light elements into heavy elements.
(C) Sudden gravitational contraction.
(D) Dark energy.

Student responses
Section 4160
(A) : 5 students
(B) : 9 students
(C) : 13 students
(D) : 3 students

Section 5166
(A) : 7 students
(B) : 16 students
(C) : 25 students
(D) : 6 students

Response (D) is merely a ruse (as it had not yet been covered yet in this course sequence). Response (A) is fission, which is not a source of energy for a star. Response (D) is fusion, which can no longer provide energy for a star at the end of its supergiant phase, as it core is iron at this point, past which requires more energy to be put in than can be released by fusion. Thus with no energy source to balance gravity, the core undergoes runaway contraction (becoming a neutron star in the process), and the resulting energy from this gravitational collapse of the core is transferred to the outer layers of the supergiant in an "implosion-explosion" sequence--a type II supernova.

Previous post: Type II supernova simulators.

### Erasing slate: Elmo, presumably

"Untitled" by Anonymous
Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Latest scribbling on the lift-and-erase slate in the hallway, outside the office door.

## 20080424

### Physics quiz question: constant cross-section pipe flow

Physics 5A Quiz 5, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 9.51

[Version 1]

[3.0 points.] Water flows through a pipe with a speed of 0.80 m/s through a pipe of 2.0 cm inside radius, at point [1]. The pipe has the same radius of 2.0 cm, at point [2], at a height lower than point [1]. How does the speed of the water change it flows from [1] to [2]?
(A) Speed increases.
(B) Speed remains the same.
(C) Speed decreases.
(D) (Not enough information is given to determine this.)

From the continuity equation, if the pipe maintains a constant cross-sectional area, then the speed does not change as it flows from point [1] to point [2]. (However, in this case since the height drops, such that according to Bernoulli's equation there is a pressure increase for the water as it flows from point [1] to point [2].) Students apparently were inclined to say that water automatically increases speed as it flows "downhill" (but in this case, water is confined to within a pipe, and must obey continuity), or that perhaps viscosity (not covered in this course) is a factor in slowing down the fluid.

Student responses
Sections 4987, 4988
(A) : 9 students
(B) : 8 students
(C) : 2 students
(D) : 0 students

[Version 2]

[3.0 points.] Water flows through a pipe with a speed of 0.80 m/s through a pipe of 2.0 cm inside radius, at point [1]. The pipe has the same radius of 2.0 cm, at point [2], at a height higher than point [1]. How does the speed of the water change it flows from [1] to [2]?
(A) Speed increases.
(B) Speed remains the same.
(C) Speed decreases.
(D) (Not enough information is given to determine this.)

From the continuity equation, if the pipe maintains a constant cross-sectional area, then the speed does not change as it flows from point [1] to point [2]. (However, in this case since the height increases, such that according to Bernoulli's equation there is a pressure decrease for the water as it flows from point [1] to point [2].) Students apparently were inclined to say that water automatically decreases speed as it flows "uphill" (but in this case, water is confined to within a pipe, and must obey continuity), or that perhaps viscosity (not covered in this course) is a factor in slowing down the fluid.

Student responses
Sections 4987, 4988
(A) : 1 student
(B) : 8 students
(C) : 7 students
(D) : 0 students

Previous post: Physics quiz question: ideal laminar flow through horizontal pipe.

## 20080423

### Physics quiz question: floating block of wood

Physics 5A Quiz 5, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 9.32

[3.0 points.] A block of oak wood (880 kg/m^3) floats in oil with 70.0% of its volume submerged. What is the density of the oil?
(A) 264 kg/m^3.
(B) 616 kg/m^3.
(C) 795 kg/m^3.
(D) 1.26 x 10^3 kg/m^3

The bouyant force on a partially submerged object is equal to the weight of the fluid displaced, F_B = rho_fluid*g*Vol_displaced, where Vol_displaced = 0.700*Vol_object. Since the object is in static equilibrium:

w = F_B
rho_object*g*Vol_object = rho_fluid*g*0.700*Vol_object,
rho_object/0.700 = rho_fluid,

which is response (D). Response (B) = 0.700*rho_object; response (A) = (1 - 0.700)*rho_object, and response (C) is 700,000/rho_object. Note that for the wood to float in oil, the density of oil must be greater than that of the wood--this condition can only be satisfied by response (D).

Student responses
Sections 4987, 4988
(A) : 0 students
(B) : 10 students
(C) : 3 students
(D) : 20 students

### Erasing slate: smiley eyebrow girl

"Untitled" by Anonymous
Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Latest scribbling on the lift-and-erase slate in the hallway, outside the office door.

## 20080422

### Physics quiz question: different densities in a U-tube

Physics 5A Quiz 5, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Question 9.9, Problem 9.24

[3.0 points.] A manometer contains two different fluids of different densities. Both sides are open to the atmosphere. Which location has the greater absolute pressure?
(A) Location [1].
(B) Location [2].
(C) (Both locations have the same absolute pressure.)
(D) (Not enough information is given to determine this.)

Consider the fluid pressure at location [A], the interface between the darker fluid and the lighter fluid in the left side of the U-tube. This is the same pressure as location [B], in the lighter fluid in the right side of the U-tube, thus P_A = P_B. Then with d as the depth from locations [1] and [2] down to locations [A] and [B]:

P_A = P_B,
P_1 + rho_dark*g*d = P_2 + rho_light*g*d.

However, since the darker fluid has a lighter density than that of the darker fluid, rho_dark < rho_light, such that P_1 > P_2.

Student responses
Sections 4987, 4988
(A) : 12 students
(B) : 15 students
(C) : 7 students
(D) : 0 students

## 20080421

### Astronomy quiz question: stellar evolution times

Astronomy 10 Quiz 9, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.5

[Version 1]

[3.0 points.] Shown at right is an H-R diagram of the evolutionary tracks of stars of different masses. Which one of the following choices best describes the evolutionary track that takes the least amount of time?
(A) The track taken by a massive protostar as it becomes a main sequence star, and then later becomes a supergiant.
(B) The track taken by a medium mass protostar as it becomes a main sequence star.
(C) The track taken by a low mass protostar as it becomes a main sequence star.
(D) (All of the above tracks (A)-(C) take the same amount of time.)
(E) (Not enough information is given to determine which of these evolutionary tracks takes the least amount of time.)

Due to their greater gravitational forces, massive protostars take a relatively short amount of time to contract and heat up to initiate fusion; and also their rapid fusion rates will cause them to have a relatively short main sequence lifetime (cf. previous post on main sequence lifetimes). In contrast, medium-mass stars will take longer than that just to get from the protostar to the main sequence stages, while low-mass stars take a even longer amount of time.

Student responses
Section 4160
(A) : 25 students
(B) : 3 students
(C) : 7 students
(D) : 0 students
(E) : 0 students

[Version 2]

[3.0 points.] Shown at right is an H-R diagram of the evolutionary tracks of stars of different masses. Which one of the following choices best describes the evolutionary track that takes the most amount of time?
(A) The track taken by a low mass protostar as it becomes a main sequence star.
(B) The track taken by a medium mass protostar as it becomes a main sequence star.
(C) The track taken by a massive protostar as it becomes a main sequence star, and then later becomes a supergiant.
(D) (All of the above tracks (A)-(C) take the same amount of time.)
(E) (Not enough information is given to determine which of these evolutionary tracks takes the most amount of time.)

Section 5166
(A) : 34 students
(B) : 2 students
(C) : 11 students
(D) : 5 students
(E) : 3 students

## 20080420

### Astronomy quiz question: heavy nuclei fusion

Astronomy 10 Quiz 9, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.4

[3.0 points.] Which one of the following choices best explains why nuclei of elements heavier than hydrogen require higher temperatures to undergo fusion?
(A) Stronger repulsive forces between nuclei.
(B) Stronger gravitational forces between nuclei.
(C) Less energy is contained in heavier nuclei.
(D) More energy is contained in heavier nuclei.
(E) Stronger degeneracy pressures between nuclei.

All atomic nuclei repel each other, due to their positively charged protons. The relative amount of repulsion between two same-element nuclei depends on the square of the protons contained in a nucleus; it is this repulsion that must be overcome in order for fusion to be initiated.

Student responses
Section 4160
(A) : 17 students
(B) : 5 students
(C) : 2 students
(D) : 4 students
(E) : 8 students

Section 5166
(A) : 30 students
(B) : 5 students
(C) : 2 students
(D) : 16 students
(E) : 4 students

## 20080419

### Astronomy quiz question: rapid fusion rates

Astronomy 10 Quiz 9, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.3

[3.0 points.] Which one of the following choices best explains why the fusion rate in the core of a massive main sequence star is more rapid than the core of a less massive main sequence star?
(A) The core of a massive main sequence star contains more hydrogen.
(B) The core of a massive main sequence star is younger.
(C) The core of a massive main sequence star has a higher temperature and pressure.
(D) The core of a massive main sequence star does not have as much convection currents to stir up hydrogen.
(E) The core of a massive main sequence star does not have as much degeneracy pressure.

(Cf. previous post on main sequence lifetimes.)

Student responses
Section 5166
(A) : 12 students
(B) : 1 student
(C) : 39 students
(D) : 2 students
(E) : 1 student

## 20080418

### Astronomy quiz question: massive and luminous stars

Astronomy 10 Quiz 9, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.3

[3.0 points.] Which one of the following statements best explains why the most massive main sequence stars are also the most luminous main sequence stars?
(A) They fuse hydrogen more rapidly.
(B) They are the youngest stars.
(C) They contain more unstable radioactive isotopes.
(D) They release more energy from gravitational contraction.
(E) (None of the above choices (A)-(D), as the most massive main sequence stars are not the most luminous main sequence stars.)

(Cf. previous post on main sequence lifetimes.)

Student responses
Section 4160
(A) : 25 students
(B) : 3 students
(C) : 2 students
(D) : 5 students
(E) : 1 student

## 20080417

### Astronomy clicker question: main sequence to giant/supergiant evolution

Astronomy 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal M3.1

Students were asked the following clicker question (Classroom Performance System, einstruction.com) in the middle of their learning cycle:

[0.3 points.] According to the Stefan-Boltzmann law, how does the luminosity of a medium mass or massive main sequence star change as its outer layers expand and cool off, as it becomes a giant or supergiant?
(A) It becomes dimmer.
(B) It remains the same.
(C) It becomes brighter.
(D) (Any of the above (A)-(C) choices, depending on how old the star is.)

Student responses
Section 4160
(A) : 10 students
(B) : 14 students
(C) : 6 students
(D) : 2 students

Section 5166
(A) : 26 students
(B) : 5 students
(C) : 20 students
(D) : 3 students

According to the Stefan-Boltzmann law, the luminosity of a star is proportional to its size (its surface area) and the fourth power of its temperature. As a medium-mass or massive main sequence star becomes a giant or supergiant, its outer layers expand and cool. Thus the size increases while the temperature decreases, resulting in approximately the same luminosity as it makes a horizontal track to the right across a Hertzsprung-Russell diagram.

## 20080416

### Vacuum in an air-filled room

Dinosaur Comics, by Ryan North
www.qwantz.com
April 16, 2008 (excerpt)

Physics 8A learning goal Q12.1

Ryan North comments on popular unlikely disasters, among them the statistical improbability of experiencing a vacuum in a corner of an air-filled room, and swallowing a black hole.

Previous post: comment on Ryan North's Dinosaur Comics discussion of the the dark energy-accelerated heat death of the universe.

## 20080415

### Astronomy in-class activity: energy flow within stars

Astronomy 10 In-class activity 20 v.07.04.03, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.2

Students find their assigned groups of three to four students, and work cooperatively on an in-class activity worksheet describing the energy flow within a main sequence star such as our Sun. Students are instructed to look for connections and similarities, then to concentrate on specific differences.

The answers for the second page are shown below.

## 20080414

### Astronomy in-class activity: OBAFGKM poetry slam

Astronomy 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.5

Students were instructed to use at least OBAFGKM, and/or part or all of the additional OBAFGKMRNSC or OBAFGKMLT extensions to individually write an original, coherent and an appropriate (nothing worse than "PG-13" rated!) mnemonic, and to give a rousing reading of their OBAFGKM mnemonic poem for the class.

One Bottle of Absolut Frequently Gives Killer Mornings
--A. B.

Only Boys Act Foolish Girls Keep Mellow
--T. D.

Only Bubba's Admiration Fueled Gump's Killer Modern Recipe: New Shrimp Chocolate
--R. E.

Oh Bright Awareness Forgive Good Kind Minds
--R. F.

Originally: Beatles Are Four Guys Keeping Music Love-Themed
--C. G.

Once Boys Are Fat Girls Kiss More
--K. G.

Occasionally Bob Argues For Guavas, Kiwis & Mangos, Regularly Never Succeeds Consistently
--S. L.

Our Best Autism Friends Got Knowledgeable Minds
--A. L. M.

Obama Beat A Fat Girl Known As M[rs.] Clinton
--A. M.

Offer Brotherly Advice For Good Karma, Man
--C. M.

Only Big Angry Feminists Go Kicking Men
--V. P.

0s (Zeros) Belong Away From Grades...Kindly Make Less Tests
--E. R.

Out Back A Farmer Grows Killer Mushrooms
--C. S.

Only Bandits And Filchers Gather Kindling & Matches Right Near Stephen Colbert's Lime Trees
--A. B.

Only Blue Army Fans Get Killaminjaros Mate
--C. B.

Outerspace Beings Are Freaking Going to Kill Me
--R. E.

Out Back Against Farthest Galaxies Killer Monkey Reign Negligently Supreme
--D. R.

Octopus Babies Are Full-Grown Killing Machines
--J. R.

Previous post: Astronomy in-class activity: OBAFGKM poetry slam instructions.

## 20080413

### Astronomy in-class activity: OBAFGKM poetry slam instructions

Astronomy 10 In-class activity 19 v.07.04.05, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.5

Students are instructed for homework to use at least OBAFGKM, and/or part or all of the additional OBAFGKMRNSC or OBAFGKMLT extensions to individually write an original, coherent and an appropriate (nothing worse than "PG-13" rated!) mnemonic, and to give a rousing reading of their OBAFGKM mnemonic poem for the class.

Previous post: OBAFGKM poetry slam (Fall Semester 2007).

## 20080412

### Bon mots: space, time, and spacetime

"Space by itself and time by itself must sink into the shadows, while only a union of the two preserves independence."
--Hermann Minkowski

"There was a blithe certainty that came from first comprehending the full Einstein field equations, arabesques of Greek letters clinging tenuously to the page, a gossamer web. They seemed insubstantial when you first saw them, a string of squiggles. Yet to follow the delicate tensors as they contracted, as the superscripts paired with subscripts, collapsing mathematically into concrete classical entities--potential; mass; forces vectoring in a curved geometry--that was a sublime experience. The iron fist of the real, inside the velvet glove of airy mathematics."
--Gregory Benford, Timescape (1992)

"Haunted by ill angels only,
Where an Eidolon, named NIGHT,
On a black throne reigns upright,
I have reached these lands but newly
From an ultimate dim Thule-
From a wild clime that lieth, sublime,
Out of SPACE- out of Time..."
--Edgar Allen Poe, "Dreamland"

"I heard that in relativity theory space and time are the same thing. Einstein discovered this when he kept showing up three miles late for his meetings."
--Steven Wright

"You can measure distance by time. 'How far away is it?' 'Oh about 20 minutes."'But it doesn't work the other way. 'When do you get off work?' 'Around 3 miles.'"
--Jerry Seinfeld

"My grandmother started walking five miles a day when she was sixty. She's ninety-seven now, and we don't know where the hell she is."
--Ellen DeGeneris

## 20080411

### Astronomy clicker question: main sequence lifetimes

Astronomy 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal M1.4

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the end of their learning cycle:

[0.3 points.] The main sequence lifetime of a star is how long it will be able to release energy from hydrogen fusion in its core. Which type of main sequence star will have the longest lifetime?
(A) A massive main sequence star.
(B) A medium-mass main sequence star.
(C) A low-mass main sequence star.
(D) The lifetime of a main sequence star does not depend on its mass.

Student responses
Section 4160
(A) : 11 students
(B) : 2 students
(C) : 17 students
(D) : 1 student

Section 5166
(A) : 19 students
(B) : 4 students
(C) : 18 students
(D) : 1 student

Explanations for typical student responses are elicited in a whole-class discussion after results have been posted.

Some students offer that they chose response (A) because the massive stars have more hydrogen to fuse, and thus will last the longest. Other students remark that they chose response (B) because low-mass stars have much slower fusion rates, and thus will be the most frugal and have the longest main sequence lifetime. (If both effects are important, perhaps the medium-mass star would live the longest?)

"Wasteful!" or "Live Fast, Die Young"
Due to their rapid fusion rates, massive stars have the shortest main sequence lifetimes, after the hydrogen in their cores has been depleted in hundreds of thousands to a few million years. They are the "wasteful" stars in the sense that there is still plenty of hydrogen in their outer layers, but this is unavailable to the core of the massive star to maintain its main sequence lifetime.

"Churn and Burn" or "Be Mellow, Live Forever"
However, low mass stars ("red dwarfs") have such slow fusion rates that they are cool and opaque enough for convection to stir up the entire star, such that fresh hydrogen is continuously cycled between the core and the rest of the star. Due to these two effects (slow fusion and a disproportionally larger amount of available hydrogen to fuse), low mass stars have extremely long main sequence lifetimes, measured in tens of billions of years. As the currently accepted age of the universe is 13-14 billion years, no low-mass stars are expected to have ever reached the end of their main sequence lifetimes yet!

## 20080410

### Education research: ultimate goal

"The most valuable role of an expert is not to simply tell students what they know; rather, it is to use their unique expertise to build rich scenarios for students to analyze using novel ideas."
--Timothy F. Slater
From the "Advice to New Astronomy Professors" guest opinion on the Teaching Astronomy blog, run by Paul E. Robinson.

## 20080409

### Astronomy quiz question: photon emission/absorption

Astronomy 10 Quiz 8, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.4

[Version 1]

Which one of the following choices best describes what will happen to an electron in an atom that emits a photon?
(A) Transform into a proton.
(C) Undergo a redshift or a blueshift.
(D) Move from an inner to an outer orbital.
(E) Move from an outer to an inner orbital.

Student responses
Section 4160
(A) : 0 students
(B) : 3 students
(C) : 1 student
(D) : 5 students
(E) : 6 students

Section 5166
(A) : 0 students
(B) : 4 students
(C) : 3 students
(D) : 7 students
(E) : 9 students

[Version 2]

Which one of the following choices best describes what will happen to an electron in an atom that absorbs a photon?
(A) Transform into a proton.
(C) Undergo a redshift or a blueshift.
(D) Move from an inner to an outer orbital.
(E) Move from an outer to an inner orbital.

Student responses
Section 4160
(A) : 0 students
(B) : 1 student
(C) : 2 student
(D) : 14 students
(E) : 1 students

Section 5166
(A) : 4 students
(B) : 3 students
(C) : 3 students
(D) : 14 students
(E) : 3 students

## 20080408

### Astronomy quiz question: stellar distances

Astronomy 10 Quiz 8, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.4

[Version 1]

Which one of the following choices best corresponds to an absolute magnitude +3 star that is the farthest away from the Earth?
(A) A star with an apparent magnitude of –6.
(B) A star with an apparent magnitude of –3.
(C) A star with an apparent magnitude of 0.
(D) A star with an apparent magnitude of +3.
(E) A star with an apparent magnitude of +6.

Out of the above (A)-(E) choices, the star with a +6 apparent magnitude appears to be the dimmest, as seen from the Earth. Thus a star with a given absolute magnitude value (here, +3) would be farthest away if it appeared to be as dim as possible.

Student responses
Section 4160
(A) : 4 students
(B) : 3 students
(C) : 0 students
(D) : 1 students
(E) : 8 students

Section 5166
(A) : 4 students
(B) : 2 students
(C) : 1 student
(D) : 3 students
(E) : 13 students

[Version 2]

Which one of the following choices best corresponds to an absolute magnitude +3 star that is the closest to the Earth?
(A) A star with an apparent magnitude of –6.
(B) A star with an apparent magnitude of –3.
(C) A star with an apparent magnitude of 0.
(D) A star with an apparent magnitude of +3.
(E) A star with an apparent magnitude of +6.

Out of the above (A)-(E) choices, the star with a -6 apparent magnitude appears to be the brightest, as seen from the Earth. Thus a star with a given absolute magnitude value (here, +3) would be nearest if it appeared to be as bright as possible.

Student responses
Section 4160
(A) : 6 students
(B) : 5 students
(C) : 3 students
(D) : 2 students
(E) : 3 students

Section 5166
(A) : 16 students
(B) : 2 students
(C) : 3 students
(D) : 3 students
(E) : 3 students

## 20080407

### Physics quiz question: bounced bullet

Physics 5A Quiz 4, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 7.42

[3.0 points.] A 0.030 kg bullet traveling at 220 m/s in the +x direction hits a motionless 1.4 kg block and bounces off it, retracing its original path with a velocity of 170 m/s in the –x direction. What is the final speed of the block? Assume the block rests on a perfectly frictionless surface.
(A) 1.1 m/s.
(B) 3.6 m/s.
(C) 4.7 m/s.
(D) 8.4 m/s.

Since there are no external horizontal impulses, momentum is conserved. The total initial momentum is due only to the bullet, p_1,i = m_1*v_1,i = 6.6 kg*m/s. The total final momentum is due to the bullet and the block, where p_1,f + p_2,f = m_1*v_1,f + m_2*v_2,f = -5.1 kg*m/s + m_2*v_2,f; such that equating p_1,i = p_1,f + p_2,f gives:

v_2,f = (m_1*v_1,i – m_1*v_1,f)/m_2,

which is response (D). Response (A) is (m_1*v_1,i + m_1*v_1,f)/m_2; response (B) is m_1*v_1,f/m_2; while response (C) is m_1*v_1,i/m_2.

Student responses
Sections 4987, 4988
(A) : 10 students
(B) : 2 students
(C) : 12 students
(D) : 12 students

## 20080406

### Physics quiz question: catching a sled

Physics 5A Quiz 4, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 7.52

[3.0 points.] A Physics 5A student of mass 69.0 kg is at rest on ice skates. A 4.30 kg sled slides across the ice towards her. The sled is moving horizontally at 15 m/s just before the student catches and holds onto it. Neglect friction and drag. What is conserved for this collision?
(A) Momentum only.
(B) Kinetic energy only.
(C) Both momentum and kinetic energy.
(D) (Neither momentum nor kinetic energy are conserved for this collision.)

Since there are no external horizontal impulses, momentum is conserved. However, since there is deformation with no rebound and separation in this perfectly inelastic collision, kinetic energy is not conserved.

Student responses
Sections 4987, 4988
(A) : 27 students
(B) : 2 students
(C) : 5 students
(D) : 2 students

Related post: Physics quiz question: bullet into block

## 20080405

### Physics quiz question: same momentum, different mass objects

Physics 5A Quiz 4, spring semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 7.10

[Version 1]
Two objects with different masses have the same magnitude momentum. The object with __________ mass has more kinetic energy.
(A) less.
(B) more.
(C) (There is a tie.)
(D) (Not enough information is given to determine this.)

Starting with the condition that two objects with different masses M > m, and different speeds, v < V, respectively, have the same momentum magnitudes:

M·v = m·V,

then:

(1/2)·M·v = (1/2)·m·V.

Since v < V, then:

(1/2)·M·v = (1/2)·m·V,

(1/2)·M·v·v < (1/2)·m·V·V,

such that:

(1/2)·M·v2 < (1/2)·m·V2,

KM < Km,

and so the object with the smaller mass m will have more (translational) kinetic energy than the object with the larger mass M.

Student responses
Sections 4987, 4988
(A) : 4 students
(B) : 9 students
(C) : 5 students
(D) : 0 students

[Version 2]
Two objects with different masses have the same magnitude momentum. The object with __________ mass has less kinetic energy.
(A) less.
(B) more.
(C) (There is a tie.)
(D) (Not enough information is given to determine this.)

Student responses
Sections 4987, 4988
(A) : 12 students
(B) : 3 students
(C) : 3 students
(D) : 0 students

## 20080404

### Mastering the Sky, Time, and Space with Starry Night(TM)

Presentation at the Central Coast Astronomical Society monthly meeting, March 27, 2008, 7:30-8:15 PM, Science Forum 2402, Cuesta College, San Luis Obispo, CA.

Starry Night(TM) from Imaginova Software (currently version 6.0) is a program that is bundled with many textbooks (usually as a previous version).

On the surface, Starry Night(TM) is a basic planetarium program that simulates viewing of the night sky, but is capable of much more than that--you can also bend time and space with it!

This is a simulated view of tonight's sky from San Luis Obispo, CA, using one of the pre-packaged local horizon panoramas, which can be viewed in any direction, and zoomed in or out. Starry Night(TM) can export a star chart, or even an interactive Quicktime VR file that preserves the three-dimensional aspect of viewing the celestial sphere from within.

You can also toggle star and constellation labels...

...as well as azimuth/altitude...

... and right ascension/declination grids.

Time can be run forwards or backwards at various rates, and a specific date and time can be selected.

Polaris is the current pole star, but what was the (nearest) pole star in 10,000 B.C.?

Alas, no true pole star in 10,000 B.C., but Vega is the closest brightest match.

After mastering the manipulation of the sky and time as with most other planetarium software, Starry Night(TM) also allows you to change your viewing position. This is made possible by generating an entire "sandbox" universe for you to move around in, and you can select (in the preferences) to render the trip from you old to your new location, which is often done at superluminal speeds.

Selecting Buenos Aires, Argentina will allow you to see southern hemisphere stars and constellations.

After flying there over the surface of the Earth at supersonic speeds, we find that in the Buenos Aires sky there is no obvious "South Star" analog to Polaris.

Next we can travel to Mare Serenatatis (Sea of Serenity) on the Moon. In this case since the Moon is above the local horizon, we do not have to travel through the interior of the Earth to get there.

Since we are on the near side of the Moon, due to its synchronous rotation, we can observe the Earth, which undergoes phases as it remains nearly stationary in its position high up in the sky, while the stars rise and set over a sidereal month.

Next let's travel to Nereid, an outer satellite of Neptune.

Nereid's rotation is not synchronous with Neptune, but is apparently in resonance. We can observe the stars rising in the east and setting in the west, as Neptune (orbited by Triton) rises in the west, stalls near the meridian, and the eventually sets in the west!

You are not just limited to traveling around the Solar System. The up/down "rocketship" controls allow you to fly up and away, to outside the Solar System...

...to outside the Milky Way...

...to beyond the mapped cubic volume of galaxy superclusters. It's like having your own Powers of Ten machine!