Astronomy in-class activity: OBAFGKM poetry slam, illustrated

Astronomy 10, spring semester 2007
Cuesta College, San Luis Obispo, CA

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.

Three favorites from this semester, by virtue of including illustrations (which were projected onto an overhead screen while the students read their poems):

Oh Because A Freaking Giraffe Kicked Me Right Near Something Critical.
--J. S.

Only Beautiful Astronomers Find Gorgeous Killer Moons.
--D. S.

Oh Bummer, Another Freaky Giant Killer Monkey Raided Nearby School Classes.
--S. P.

Previous post: OBAFGKM poetry slam (Spring Session 2007).

Plus a favorite from a past semester (Fall 2005):

Oh Beautiful Astronomical Friends Go Kiss Mr. Len.
--B. B.

Astronomy in-class activity: OBAFGKM poetry slam

Astronomy 10, Spring Semester 2007
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.

Two favorites from this semester:

Oops! Britney Attacked Federline's GMC. Kevin's Mad.
--C. K.

Only Batman Always Forgets Giant Kryptonite Meteorites Resting Near Superman's Car.
--K. P.

Astronomy clicker question: protostar to main sequence evolution

Astronomy 10, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.5

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

[0.3 points.] How is it possible for the luminosity of a protostar to remain (approximately) constant as it becomes a main sequence star?
(A) Its surface temperature gets hotter as its size gets smaller.
(B) Its surface temperature gets cooler as its size gets smaller.
(C) Its surface temperature gets hotter as its size gets larger.
(D) Its surface temperature gets cooler as its size gets larger.

Correct answer: (A).

The evolutionary track of a protostar as it becomes a main sequence star can be approximated as a horizontal path from right-to-left across an H-R diagram, which means that its luminosity remains constant while its surface temperature increases. From the Stefan-Boltzmann law, constant luminosity means that size must decrease while temperature increases.

Student responses
Section 4136
(A) : 19 students
(B) : 2 students
(C) : 7 students
(D) : 1 student

Student responses
Section 5076
(A) : 7 students
(B) : 0 students
(C) : 13 students
(D) : 1 student

Astronomy quiz question: applications of the Stefan-Boltzmann law

Astronomy 10 Quiz 8, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.5

[Version 1]

[3.0 points.] Which one of the following statements best describes the relationship between a main sequence star and a supergiant that have the same luminosity?
(A) The main sequence star is cooler and smaller than the supergiant.
(B) The main sequence star is cooler and larger than the supergiant.
(C) The main sequence star is hotter and smaller than the supergiant.
(D) The main sequence star is hotter and larger than the supergiant.
(E) (None of the above choices (A)-(D), as it is not possible for a main sequence star to have the same luminosity as a supergiant.)

Correct answer: (C)
From an H-R diagram, a main sequence star must be hotter in order to have the same luminosity as a supergiant. From the Stefan-Boltzmann law, since luminosity is proportional to size and temperature^4, and with both stars having the same luminosity, the hotter star must be smaller in size.

Student responses
Section 4136
(A) : 6 students
(B) : 3 students
(C) : 22 students
(D) : 1 student
(E) : 0 students

[Version 2]
[3.0 points.] Which one of the following statements best describes the relationship between a main sequence star and a white dwarf that have the same luminosity?
(A) The main sequence star is cooler and smaller than the white dwarf.
(B) The main sequence star is cooler and larger than the white dwarf.
(C) The main sequence star is hotter and smaller than the white dwarf.
(D) The main sequence star is hotter and larger than the white dwarf.
(E) (None of the above choices (A)-(D), as it is not possible for a main sequence star to have the same luminosity as a white dwarf.)

Correct answer: (B)
From an H-R diagram, a main sequence star must be cooler in order to have the same luminosity as a supergiant. From the Stefan-Boltzmann law, since luminosity is proportional to size and temperature^4, and with both stars having the same luminosity, the cooler star must be larger in size.

Student responses
Section 5076
(A) : 3 students
(B) : 10 students
(C) : 1 student
(D) : 4 students
(E) : 2 students

Astronomy clicker question: why are white dwarfs small?

Astronomy 10, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.5

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

[0.3 points.] Why is a white dwarf star smaller than a main-sequence star that has the same white-hot color?
(A) It is less luminous than the main-sequence star.
(B) It is more luminous than the main-sequence star.
(C) It is cooler than the main-sequence star.
(D) It is hotter than the main-sequence star.

Correct answer: (A).

The fact that both stars have the same white-hot color tells you that they must have the same temperature (Wien's law). From the Stefan-Boltzmann law, luminosity is proportional to size and temperature^4, thus with both stars having the same temperature, the less luminous star is the smaller star.

Student responses
Section 4136
(A) : 9 students
(B) : 4 students
(C) : 7 students
(D) : 11 students

Section 5076
(A) : 2 students
(B) : 1 student
(C) : 10 students
(D) : 3 students

Physics midterm problem: up-tugged yo-yo

Physics 8A Midterm 2, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q6.4

[20 points.] A thin ring of mass 2.40 kg and radius 0.100 m has a string wrapped around it that is pulled vertically upwards with an applied force 7.20 N in magnitude, and as a result it begins to roll (clockwise) to the right without slipping. Find the minimum coefficient of static friction that will allow this to happen.

(Cf. Young and Freeman, University Physics, 11/e, Problem 10.70.)

Solution and grading rubric:

• p = 20/20:
  Correct. Draws an extended-body diagram, correctly identifying the directions and locations of F_applied, w, n, and f_s (horizontally to the right, at the point of contact). Then applies N1 in the y-direction, and finds that n = 16.3 N.
  
  Applies N2 in the x-direction, and finds that
  
  a_x = f_s/m = u_s*n/m.
  
  Applies N2 rotational, and finds that
  
  -m*(r^2)*alpha = -F_applied*r + f_s*r.
  
  Constrains translational and rotational motion using a_x = r*alpha, and finds that
  
  u_s = f/(2*(m*g) - F_applied) = 0.221.
  
  
• r = 16/20:
  Nearly correct, but includes minor math errors.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least methodically applies Newton's laws for x, y, and rotations, and constrains translational and rotational motion, but typically demands n = mg, or has similar misapplications between using N1 or N2.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. At least some attempt at using Newton's laws, but with serious omissions or complications.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit. Typically sets up a diagram, and/or finds I = m*(r^2).
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
p: 1 student
r: 2 students
t: 12 students
v: 16 students
x: 5 students
y: 1 student
z: 1 student

Physics midterm problem: upward-sliding sponge

Physics 8A (currently Physics 208A) Midterm 2, spring semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Young and Freeman, University Physics, 11/e, Problem 3.63

A window washer pushes a sponge up a vertical window at constant speed by applying a force as shown at right. The sponge has a mass of 0.800 kg, and the coefficient of kinetic friction between the sponge and window is μ_k = 0.253. Determine the magnitude of the applied force, and the magnitude of the normal force exerted by the window on the sponge. Show your work and explain your reasoning using a free-body diagram, and the properties of forces, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Breaks up F_applied into x- and y-components. Applies Newton's first law in the x-direction, where:

  F_applied·cos(20.0°) = N.

  Then applies Newton's first law in the y-direction ("constant speed"), such that:

  F_applied·sin(20.0°) = m·g + μ_k·N.

  With two equations for two unknowns, solves for:

  F_applied = m·g/(sin(20.0°) - μ_k·cos(20°)) = 75.2 N,
  
  N = m·g·cos(20.0°)/(sin(20.0° - μ_k·cos(20.0°) = 70.6 N.

• r:
  Nearly correct, but includes minor math errors. And/or has correct numerical value for only one of the forces.
  
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Correctly identifies all forces acting on the sponge. At least methodically applies Newton's first law in the x- and y-directions, with F_applied broken up into x- and y-components.
  
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Typically omits f_k acting downwards, and/or demands N = m·g while applying Newton's laws.
  
• x:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y:
  Irrelevant discussion/effectively blank.
  
• z:
  Blank.
  

Grading distribution:
p: 6 students
r: 4 students
t: 7 students
v: 11 students
x: 10 students
y: 0 students
z: 0 students

Astronomy current events question: my so-called "equinox"

Astronomy 10L, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Students are assigned to read online articles on current astronomy events (skytonight.com, from Sky & Telescope magazine), and take a short current events quiz during the first 10 minutes of lab. (This motivates students to show up promptly to lab, as the time cut-off for the quiz is strictly enforced!)

[0.2 points.] The 2007 vernal equinox occured on Tuesday, March 20 for San Luis Obispo, CA, signaling the start of spring. Why weren't there exactly 12 hours between sunrise and sunset on that date?
(A) The Earth's orbit around the Sun is not a perfect circle.
(B) Daylight savings time shifts astronomical events one calendar day ahead.
(C) The date of the vernal equinox is misaligned because of accumulated leap year days.
(D) Sunrise and sunset times are defined when the top edge of the Sun is on the horizon, and not when the center of the Sun is on the horizon.
(E) The start of spring in the northern hemisphere marks the start of fall in the southern hemisphere, on the other side of the International Date Line.

Correct answer: (D).

Student responses
Section 4137
(A) : 3 students
(B) : 3 students
(C) : 2 students
(D) : 13 students
(E) : 0 students

Section 4138
(A) : 2 students
(B) : 2 students
(C) : 0 students
(D) : 13 students
(E) : 4 students

Section 4139
(A) : 2 students
(B) : 1 student
(C) : 1 student
(D) : 6 students
(E) : 4 students

When prompted, students know that the "equinox" refers to equal hours of day and night, thus exactly 12 hours between sunrise and sunset. However, on March 20, 2007 for San Luis Obispo, CA, the U. S. Naval Observatory times for sunrise and sunset are:

Sunrise 07:07
Sunset 19:15

This is due to how sunrise and sunset are defined, which is not when the center of the Sun is on the horizon, but when the very top edge of the Sun is on the horizon.

Pluto demotion humor

Pluto-nium by harlanm
Eyewitless News: Save Pluto Contest
(worth1000.com)

Astronomy 10 learning goal M2.5

Well, Pluto is not classifed as an asteroid, but did get assigned a minor planet number in 2006, to coincide with its new designation as a dwarf planet.

Pluto Is Now Just a Number: 134340 (Space.com)

Astronomy midterm question: Eris and Ceres

Astronomy 10 Midterm 2, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal M2.5

[Version 1]

[3.0 points.] Which one of the following choices best explains why Eris (formerly called "Xena") is categorized as a dwarf planet instead of a Kuiper belt object under the new International Astronomical Union rules?
(A) Eris has a spherical shape.
(B) Eris cleared its orbit of other Kuiper belt objects.
(C) Eris has a satellite orbiting around itself.
(D) Eris has a tilted orbit around the Sun.
(E) Eris has a different composition than the other Kuiper belt objects.

Correct answer: (A)
Eris is a member of the Kuiper belt, but due to its spherical shape, it is considered a dwarf planet rather than solar system debris (of which Kuiper belt objects are included).

Student responses
Section 4136
(A) : 10 students
(B) : 10 students
(C) : 9 students
(D) : 3 students
(E) : 3 students

[Version 2]
[3.0 points.] Which one of the following choices best explains why Ceres is categorized as a dwarf planet instead of an asteroid under the new International Astronomical Union rules?
(A) Ceres has a spherical shape.
(B) Ceres did not clear its orbit of other asteroids.
(C) Ceres has satellites orbiting around itself.
(D) Ceres has a tilted orbit around the Sun.
(E) Ceres has a different composition than the other asteroids.

Correct answer: (A)
Ceres is a member of the asteroid belt, but due to its spherical shape, it is considered a dwarf planet rather than solar system debris (of which asteroids are included). Note that (B) would be the correct answer as to why Ceres is considered a dwarf planet instead of a planet.

Student responses
Section 5076
(A) : 6 students
(B) : 9 students
(C) : 6 students
(D) : 1 student
(E) : 3 students