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

Astronomy clicker question: Charon

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

Astronomy 10 learning goal M2.5

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the end of their learning cycle, during a review session just before the midterm. Students are encouraged to collaborate openly, as their review session clicker points for responding will double if the cumulative class success rate is at or over 80% (out of 10 questions).

[0.3 points.] Which one of the following choices best explains why Charon is neither categorized as a planet, nor a dwarf planet under the new International Astronomical Union rules?
(A) Charon is one of Pluto's three satellites.
(B) Charon did not clear its orbit of other Kuiper belt objects.
(C) Charon is the remnant of a giant impact on Pluto that produced icy debris that made the Kuiper belt.
(D) Charon is not spherical in shape.
(E) Charon is mainly composed of ice.

Correct answer: (A).

Student responses
Section 4136
(A) : 30 students
(B) : 0 students
(C) : 0 students
(D) : 2 students
(E) : 0 students

Section 5076
(A) : 20 students
(B) : 0 students
(C) : 1 student
(D) : 7 students
(E) : 0 students

So far, so good. Let's see how they did during the midterm...

Astronomy clicker question: what's wrong with the IAU?

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

Astronomy 10 learning goal M2.5

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

[0.3 points.] What is problematic about the IAU (International Astronomical Union) classification scheme for planets and dwarf planets?
(A) It does not define how planets around other stars should be classified.
(B) It is not yet possible to measure the roundness of many distant, small bodies.
(C) It is too vague on how much a planet should "clear out" its orbit.
(D) The classification of a planet or dwarf planet depends on its neighbors, not on its own attributes.

Correct answer: (none, opinion only, but see comments below).

Student responses
Section 4136
(A) : 3 students
(B) : 4 students
(C) : 19 students
(D) : 6 students

Section 5076
(A) : 2 students
(B) : 0 students
(C) : 9 students
(D) : 7 students

This question prompts discussion on some of the objections to the recent International Astronomical Union (IAU) categorization scheme adopted at their August 2006 meeting in Prague.

(A) = The categorization scheme only applies to planets, dwarf planets, etc. around the Sun, and does not deign to classify the recently discovered "super-Jupiters" orbiting around other stars.

(B) = Many dwarf planet candidates are in limbo, without more information about their actual shape. The asteroids Pallas and Vesta, and as many as a hundred other Kuiper belt objects ("Santa" (2003 EL₆₁), the "Easterbunny" (2005 FY₉, et al.) may or may not be classified as dwarf planets for many years to come.

(C) = This is a real point of contention, as the IAU states that a planet must "clear out" its orbit. Students bring up the fact that the Earth definitely has not cleared out its orbit, as the dinosaurs can attest! The various Earth-crossing asteroid groups (the Apollos being the most prominent) indicate that the Earth has not "cleared out" its orbit, and thus under the letter of the law, the Earth is a dwarf planet, and not a planet! However, the spirit of the law suggests that since the Earth "dominates" its orbit, it clearly should not be demoted to dwarf planet status. Seems like the IAU needs some lawyers on retainer to help clean up their wording next time around.

(D) = This is also a real concern, as the "planetness" of an object is not an inherent property, it depends on the properties of its neighbors (whether they have been cleared out or not). Mercury is a good example, as it is in an empty part of the solar system. But placing it in the asteroid belt would mean that it is a dwarf planet, as it would not have enough gravitational influence to dominate, much less clear out its neighbors.

So stay tuned, as this topic is still very active and contentious...

Astronomy clicker question: why shouldn't Pluto be demoted?

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

Astronomy 10 learning goal M2.5

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

[0.3 points.] Why should Pluto not be demoted from a planet to dwarf planet?
(A) Planet status, once given, cannot be revoked.
(B) It is the "king of the Kuiper belt."
(C) It has satellites (moons) of its own.
(D) (Pluto should be demoted from planet to dwarf planet.)

Correct answer: (none, opinion only, but see comments below).

Student responses
Section 4136
(A) : 9 students
(B) : 2 students
(C) : 9 students
(D) : 10 students

Section 5076
(A) : 3 students
(B) : 1 student
(C) : 9 students
(D) : 7 students

Each of these choices brings up opinions that many people have about demoting Pluto from planet to dwarf planet status. After the results of this clicker question are displayed, each of these issues are discussed.

(A) = "No takebacks." However, the first few asteroids that were discovered were touted as new planets, and astrological symbols were generated for them. But as more information was known about these "planets," there was determined to be many more of them in what is known as the asteroid belt, and their small size ultimately meant that their planetary status was stripped.


(B), (C) = "Pluto is special." Eris (2003 UB₃₁₃, or "Xena") is even bigger than Pluto, and also has a satellite. So it has even more of what made Pluto so "special."

(D) = So why should Pluto be a planet, once these objections are addressed?


Periodically updated comparison chart of the sizes of the Earth, Moon, Mercury, Ceres, Vesta, Pallas, Pluto, Xena, and other TNOs ("Trans-Neptunian Objects," aka Kuiper belt objects), posted by Wm. Robert Johnson (www.johnstonsarchive.net).

Astronomy clicker question: why is Pluto not a planet?

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

Astronomy 10 learning goal M2.5

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

[0.3 points.] Why was Pluto demoted from planet to dwarf planet?
(A) It does not orbit the Sun.
(B) It did not have enough formation heat to completely melt, and enough gravity to round itself into a spherical shape.
(C) It did not clear out its orbit.
(D) (All of the above choices (A)-(C).)

Correct answer: (C).

Student responses
Section 4136
(A) : 1 student
(B) : 10 students
(C) : 10 students
(D) : 11 students

Section 5076
(A) : 2 students
(B) : 8 students
(C) : 6 students
(D) : 5 students

At the International Astronomical Union (IAU) August 2006 meeting in Prague, the solar system was categorized by three criteria, as shown at right. To be classified as a planet, Pluto must separately fulfill all three criterion--it orbits the Sun (and not a planet), it is sufficiently large enough to have retained formation heat and enough gravity to reshape itself into a sphere (and not some lumpy potato shape), and enough influence over its domain to clear its orbit (which it has not, being one of as many as 10 billion other Kuiper belt objects located 30-50 AU from the Sun, just beyond the orbit of Neptune).

The Astronomy Education Review (http://aer.noao.edu) has several articles of interest:

Fraknoi, A. 2006, "Teaching What a Planet Is: A Roundtable on the Educational Implications of the New Definition of a Planet," Astronomy Education Review, 5(2).
An overview of the IAU categorization scheme.

LoPresto, M. C. 2006, "A First Glimpse of Student Attitudes about Pluto's 'Demotion,'" Astronomy Education Review, 5(2).
Further discussion on student views of Pluto's planetary demise.

Bon mot: Burns' hog-weighing method

Physics 8A learning goal Q6.5

"Burns' Hog-Weighing Method:
(1) Get a perfectly symmetrical plank and balance it across a sawhorse.
(2) Put the hog on one end of the plank.
(3) Pile rocks on the other end until the plank is again perfectly balanced.
(4) Carefully guess the weight of the rocks."
--Robert Burns (1759-1796)

This quote is shown on the announcements section on the overhead projector just before a statics/dynamics midterm, to provide some source of levity for stressed-out students.

Physics quiz question: pivoting triangle

Physics 8A Quiz 6, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q6.1

[3.0 points.] Consider a horizontal metal right triangle plate, 0.400 m on each leg, pivoted about a vertical axis through point "O" at its 90.0° corner. Two forces are exerted as shown, resulting in a net counterclockwise torque of magnitude 0.600 N*m, and a counterclockwise acceleration of 1.90 rad/s^2. The plate and the forces are in the plane of this page.

If F_1 has a magnitude of 8.00 N, which one of the following choices best corresponds to the magnitude of F_2?
(A) 5.00 N.
(B) 8.00 N.
(C) 11.0 N.
(D) 16.0 N.

(Cf. Young and Freeman, University Physics, 11/e, Exercise 10.3.)

Correct answer: (C)
Applying Newton's second law, where net torque = +0.600 N*m = -F_1*(0.400 m) + F_2*(0.400 m), and thus F_2 = 11.0 N. Be careful in setting up all torques in the proper rotation direction (+ = ccw, - = cw), and note that the "perpendicular" lever arm for the force F_1 is the same as the lever arm for the perpendicular force F_2.

Student responses:
(A) : 2 students
(B) : 6 students
(C) : 23 students
(D) : 3 students

Physics quiz question: unwinding rolling yo-yo

Physics 8A Quiz 6, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q6.4

[3.0 points.] A yo-yo is initially at rest on a horizontal surface. A string is attached to the yo-yo and is pulled in the direction shown. There is sufficient friction for the yo-yo to roll without slipping.

Which one of the choices below best describes the direction of the static friction force of the floor on the yo-yo, and the resulting rolling direction of the yo-yo?

(A) The static friction force points to the left, and the yo-yo will begin to roll along the floor to the left.
(B) The static friction force points to the left, and the yo-yo will begin to roll along the floor to the right.
(C) The static friction force points to the right, and the yo-yo will begin to roll along the floor to the left.
(D) The static friction force points to the right, and the yo-yo will begin to roll along the floor to the right.

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

Correct answer: (D)
The forces acting on the yo-yo are the tension force (up), normal force of the floor (down), weight (down), and the static friction force (direction unknown). In order for the yo-yo to unwind the string while rolling without slipping across the floor, it must accelerate clockwise, while translationally accelerating to the right. Thus the static friction force must point to the right, in order to translationally accelerate the yo-yo to the right. Alternatively, the yo-yo would like to slip its bottom surface to the left across the floor as it unwinds, but the static friction force will prevent that from occuring by opposing that sliding, and thus point to the right.

Student responses:
(A) : 3 students
(B) : 18 students
(C) : 1 student
(D) : 11 students

Physics quiz question: vertical cable on diagonal beam

Physics 8A Quiz 6, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q6.5

[3.0 points.] A uniform, 150 kg beam is supported using a cable connected to the ceiling, as shown at right. The lower end of the beam rests on the floor, which is not frictionless.

Which one of the choices below best corresponds to the static friction force (magnitude and direction) of the floor on the beam?

(A) 735 N, to the left.
(B) 735 N, to the right.
(C) (There is zero static friction force exerted by the floor on the beam.)
(D) (There is not enough information given to determine the static friction force of the floor on the beam.)

(Cf. Young and Freeman, University Physics, 11/e, Problem 11.59.)

Correct answer: (C)
The forces acting on the beam are the tension force (up), normal force of the floor (up), weight (down), and the static friction force (direction unknown). The beam is in equilibrium, so by Newton's first law these forces must sum to zero (along the x- and y-directions), and their torques must sum to zero (about any arbitrary origin). Note that the static friction force, if it exists, is the only x-direction force, and would not be able to be canceled out. Therefore it cannot exist.

Student responses:
(A) : 10 students
(B) : 10 students
(C) : 9 students
(D) : 5 students

Physics quiz question: circular saw blade

Physics 8A Quiz 5, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q5.2

[3.0 points.] Consider a circular saw blade 0.120 m in radius that requires 10.0 s to rotate counterclockwise through 200 radians. The initial angular velocity w_0, at the beginning of this time interval, is +21.0 rad/s. (The blade is always rotating counterclockwise.)

Which one of the following statements best describes the magnitude of the final angular velocity w of the blade at the end of the 10.0 s interval, compared to the magnitude of the initial angular velocity w_0?
(A) w < w_0.
(B) w = w_0.
(C) w > w_0.
(D) (Not enough information is given to determine how the magnitude of the final angular velocity compares to the magnitude of the initial angular velocity.)

(Cf. Young and Freeman, University Physics, 11/e, Exercise 9.15.)

Correct answer: (A)
Assuming that the saw blade turns at a constant rate, it will rotate through (21.0 rad/s)*(10.0 s) = 210 radians. Since it actually rotates through 200 radians during this time, it must be slowing down, and thus its final angular velocity w is slower than its initial angular velocity w_0.

Student responses:
(A) : 23 students
(B) : 2 students
(C) : 9 students
(D) : 2 students

Dynamic-axis ("rolling") system

grasswheel
Originally uploaded by jessehudson.

Physics 8A learning goal Q6.4

Example of a dynamic-axis system where there is rolling without slipping. In this case, allowing the user to enjoy a barefoot stroll through the grass in the middle of the city.

Thinking about blinking

"Under the Tuscan Sun," Touchstone Pictures movie, 2003

Astronomy 10 learning goal Q8.1

Students can demonstrate the method of stellar parallax for themselves that holding up a finger, and looking at it with one eye, and then the other, will make the finger appear to shift with respect to a distant background (such as the opposite classroom wall). Holding their finger closer will result in a larger shift; holding their finger farther away will result in a smaller shift. Thus the shift angle will be inversely correlated with the distance.

Here Diane Lane is using the method of stellar parallax to determine the distance to the wine bottle. Several Cuesta College students recognize her visiting one of the many Starbucks coffeeshops around San Luis Obispo, CA. One student reports that her cousin (a restaurant server) claims that Diane Lane has different colored eyes...

Astronomy current events question: start of daylight savings time

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 movitvates students to show up promptly to lab, as the time cut-off for the quiz is strictly enforced!)

[0.2 points.] How should San Luis Obispo, CA clocks be set for daylight savings time at 2:00 AM on Sunday morning, March 11? Circle your answer below.
(A) Clocks should be set forward one hour, such that an hour is "lost."
(B) Clocks should be set back one hour, such that an hour is "gained."
(C) (California will not begin daylight savings time until 2:00 AM on Monday morning, March 12.)
(D) (California will not begin daylight savings time until the first Sunday in April.)
(E) (California will not observe daylight savings time for 2007.)

Correct answer: (A).

Student responses
Section 4137
(A) : 20 students
(B) : 3 students
(C) : 0 students
(D) : 0 students
(E) : 0 students

Section 4138
(A) : 13 students
(B) : 5 students
(C) : 1 student
(D) : 0 students
(E) : 0 students

Section 4139
(A) : 14 students
(B) : 2 students
(C) : 0 students
(D) : 0 students
(E) : 1 student

Most students memorized, "spring forward, fall back."

Pop quiz on Monday, March 12--will students show up to their first morning class on time? Let's see if cell phone service providers do this properly, as many students rely on their cell phones to tell time. Note to instructors: do not schedule midterms on the first Monday following a daylight savings time switch!

Moon and Mercury confusion

"The Earth's Moon" movie (frame from 06'01")
Starry Night Pro 5.0 DVD, starrynight.com


Santa Maria Rupes (detail)
NASA, JPL, Mariner 10, © Calvin J. Hamilton (LANL)
Astronomy Picture of the Day, "Mercury's Faults," January 21, 1996

Astronomy 10 learning goals Q5.1, Q5.2

Apparently the editors at Starry Night Software goofed, as they have a rupes (a lobate scarp, or compression fault line) from Mercury's surface prominently shown in their Moon video. Have students to watch out for the Mercury feature in this movie!

Astronomy clicker question: hot or not (the cornish hen/turkey effect)

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

Astronomy 10 learning goal Q5.5

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

[0.3 points.] Which terrestrial planet has the hottest core?
(A) Mercury, which is closest to the Sun.
(B) Venus, which has the hottest atmosphere.
(C) Earth, which is the most massive.
(D) Mars, which has the tallest volcanos.

Correct answer: (C).

Student responses
Section 4136
(A) : 6 students
(B) : 3 students
(C) : 14 students
(D) : 10 students

Section 5076
(A) : 9 students
(B) : 7 students
(C) : 2 students
(D) : 4 students

The terrestrial planets are subject to the "cornish hen/turkey effect." Suppose a cornish hen and a turkey are both taken out of the oven, where they were at the same temperature. Allowed to cool off on the counter for an hour, the cornish hen (being much smaller) has cooled off considerably, while the interior of the turkey still retains a significant amount of heat. Thus small things cool off faster than larger things. The Earth has a hotter core than any of the other terrestrial planets because it is the most massive, and this results in its current of volcanic and tectonic plate motion activity, compared to the dormant or dead geological activity on the smaller terrestrial planets.

Terrestrial world icosahedra
Originally uploaded by Waifer X.

Physics quiz question: pilot in looping airplane

Physics 8A Quiz 4, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q4.4

[3.0 points.] An airplane flies in a loop (a circular path in a vertical plane) of radius 200 m. The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop, and fastest at the bottom. Which one of the free-body diagrams (A)-(F) best corresponds to forces acting on the pilot when she is at the bottom of the loop? The forces on these diagrams are unlabeled, but are drawn approximately to scale. Clearly circle your answer.



(Cf. Young and Freeman, University Physics, 11/e, Exercise 5.51.)

Correct answer: (C)
Choices (B) and (D) are zero net force cases, while choices (E) and (F) have net forces that point downwards. By process of elimination, the only remaining choices are (A) and (C), which have net forces that point upwards, which is radially inwards when the pilot is at the bottom of the loop. There are two forces that act on the pilot, the weight force of the Earth on the pilot (downwards), and the normal force of the seat on the pilot (upwards), with the normal force having a (much) larger magnitude than the weight force.

Student responses:
(A) : 3 students
(B) : 10 students
(C) : 14 students
(D) : 2 students
(E) : 1 student
(F) : 4 students

Overheard: what is the fish?

Physics 8A, Fall Semester 2006
Cuesta College, San Luis Obispo, CA

Overheard during an in-class activity (rotational kinematics with a constant angular acceleration):

Student 1: "What is the 'fish?'"
Student 2: "What?"
Student 1: "This 'fish'...what is it?"
Student 2: "Oh...that's alpha."
Student 1: "Okay... What is 'alpha?'"

The conversation is somewhat embellished, of course.

Physics quiz question: box under box, with pulley

Physics 8A Quiz 4, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q4.5

[3.0 points.] A 5.00 kg box and a 1.75 kg box are attached by an ideally stretchless, massless cable that passes around a frictionless, massless pulley. The coefficient of kinetic friction between all surfaces is 0.250. A force is applied to the 1.75 kg box, as shown, and as a result, the 1.75 kg box moves to the left at a constant speed.

(Cf. Young and Freeman, University Physics, 11/e, Problem 5.81.)

[Version 1]
Which one of the following choices best corresponds to the force that has the smallest magnitude?
(A) F_applied.
(B) NetF_{on 1.75 kg box}.
(C) T_{5.00 kg box on 1.75 kg box}.
(D) f_{k, 5.00 kg box on 1.75 kg box}.

Correct answer: (B)
The net force on the 1.75 kg box must be zero, according to Newton's first law, because it is moving at constant speed.

Student responses:
(A) : 2 students
(B) : 10 students
(C) : 3 students
(D) : 5 students

[Version 2]
Which one of the following choices best corresponds to the force that has the largest magnitude?
(A) F_applied.
(B) NetF_{on 1.75 kg box}.
(C) T_{5.00 kg box on 1.75 kg box}.
(D) f_{k, 5.00 kg box on 1.75 kg box}.

Correct answer: (A)
The applied force (exerted to the left), according to Newton's first law, must be equal to both the kinetic frictional force of the 5.00 kg box on the 1.75 kg box and the tension force of the 5.00 kg box on the 1.75 kg box added together (which are both exerted to the right).

Student responses:
(A) : 7 students
(B) : 4 students
(C) : 0 students
(D) : 3 students

Astronomy current events question: no west coast U.S. lunar eclipse

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.] On March 3, 2007 there will be a lunar eclipse for observers in Europe, and observers on the east coast of North America. Why will observers on the west coast of North America not be able to see this happening? Circle your answer below.
(A) Observers on the west coast of North America are in the northern hemisphere.
(B) There will not be enough horizon illusion.
(C) The lunar eclipse will end before the Moon rises.
(D) The Moon will be too dark.
(E) The Moon will be too far away.

Correct answer: (C).

Student responses
Section 4137
(A) : 3 students
(B) : 3 students
(C) : 17 students
(D) : 1 student
(E) : 1 student

Section 4138
(A) : 2 students
(B) : 1 students
(C) : 15 students
(D) : 3 students
(E) : 1 student

Section 4139
(A) : 0 students
(B) : 2 students
(C) : 16 students
(D) : 1 student
(E) : 0 students

However, the August 28, 2007 lunar eclipse will be visible for nearly its entire duration for observers on the west coast of North America. Students (here at Cuesta College in San Luis Obispo, CA) will be looking forward to that event...hopefully.

Astronomy quiz question: bubble, bubble, toil and trouble

Astronomy 10 Quiz 4, fall semester 2006
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q4.2

Consider a sample with differing amounts of unstable isotopes, embedded gaseous decay products, and inert material (which is not involved in the radioactive decay process), schematically shown at right above. This sample is then melted completely, and then cooled and solidified. Which one of the following choices (A)-(E) schematically shown below best corresponds to the sample after it has resolidified? Clearly circle your answer.


Correct answer (highlight to unhide): (B)

The solidification age of a sample (how long ago has it been since it was last molten) is determined by the ratio of decay products to its unstable isotopes. A larger ratio of decay products to unstable isotopes corresponds to a sample with a very old solifidification age. Because these decay products are typically gaseous, when a sample is heated up to a molten state, the decay products bubble out, and thus the ratio of decay products to unstable isotopes is "zeroed out," corresponding to a zero-age solification age as it cools back down and resolidifies.

The plurality of (A) responses seems to indicate that students are thinking that heating up and melting a sample "resets" the decay products back into the original unstable isotopes.

Student responses:
(A): 11 students
(B): 7 students
(C): 5 students
(D): 4 students
(E): 6 students

Success level: 21%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.86