Quiz Answers

[asterroc]

In case anyone's interested, I thought I'd post some answers to the quiz questions I posted the other day asking people for help in timing online quizzes.

Regarding the timing issue, everyone reported that it took them a very short time to type their answers - admittedly it's a self-selected group (I doubt slow typers would enjoy blogging all that much, and people who hate science probably wouldn't be answering these questions for me), but still it gives me a starting point. My plan with my class is to (a) ask students during the first week about their typing skills, (b) unless any of their answers worry me, I'll give the same amount of time for the first quiz as I would give in a face-to-face class, (c) I'll look at the statistics of how long it took them to complete the quiz and adjust subsequent quizzes accordingly.

So for the answers, below are what I was looking for, key words/concepts bolded, along with some items that would likely lose the person points and why.

1) Explain the cause of the seasons.

As the Earth orbits the Sun, its axis is tilted by 23.5º. This results in sunlight hitting the surface of the Earth more or less directly (resulting in more or less intense sunlight) depending upon which part of the Earth is pointed towards the Sun at that particular time of year. The tilt also results in the length of the day changing through out the year.

Distance from the Sun would lose you points: the Earth's orbit is so close to circular that the distance does not make any measurable difference in the weather on Earth. Earth is actually closest to the Sun on Jan 3 or so, Northern Hemisphere Winter, Southern Hemisphere Summer - if it were simply distance from the orbit then both hemispheres would have the same season. Saying the tilt makes one part closer to or farther from the Sun is a little more creative (and would gain partial credit from the tilt), but also doesn't make a big enough difference.


2) Is Pluto currently considered a planet by astronomers? Why or why not?

The current definition of planet require an object to orbit the Sun, be small enough to not be undergoing nuclear fusion, be large enough to be round from self-gravitation, and have cleared its orbit. Pluto fails the last criterion, as there are many other objects orbiting just outside Neptune's orbit, variously called plutinos, plutiods, trans-Neptunian objects, and a larger group called Kuiper Belt Objects.


3) Compare and contrast the Greenhouse Effect and the Ozone Layer.

The greenhouse effect is caused by greenhouse gases (such as CO2, H20, CH4, etc.) distributed throughout the atmosphere (or closer to the ground) while the ozone layer is one region of the stratosphere containing O3. The ozone layer acts simply to absorb much of the incoming UV and prevent it from reaching the ground. The greenhouse effect is more complex, as (nearly) full spectrum light hits the ground, some is reradiated as IR, and then the greenhouse gases trap much of the IR light within the atmosphere.

FWIW I know the wording of this question needs improvement. A better wording might perhaps be "Greenhouse Gases and the Greenhouse Effect, and the Ozone Layer and its effects." I typically explain what I'm looking for before giving this quiz question - I usually tell the students this will be the next quiz, and explain that the two are "opposite and complementary." The main thing I'm looking for here is for the students to distinguish that they are two entirely separate phenomena, since the media always conflates them.


4) Why do astronauts in the Space Shuttle experience weightlessness?

As the Space Shuttle orbits the Earth it is moving just fast enough that as the Shuttle (and everything within) falls towards the Earth due to gravity, the surface of the Earth (and the path of their orbit) curves away underneath them by the same amount that they fall due to the gravity. Therefore they are in free fall, or experience weightlessness even though they are still influenced by gravity. (Compare this situation to Newton's synthesis of the apple and the Moon, and the pictures of the cannonball around the Earth as it is launched faster and faster.)

Lose points for distance from Earth, or microgravity if the term is not accompanied by the above explanation. The distance of the Space Shuttle from the Earth is not significant enough to result in a low gravity environment by itself - if we had a platform rising from the Earth and attached to the ground, you could stand on it and still experience gravity. It's only b/c the Shuttle is in orbit that it appears gravity is not affecting you.




Once again, thanks for the help!

X-posted

Comments

[galbinus-caeli.livejournal.com]

Question 1: Ahem. You never said "on earth", nor did you say "winter, spring, summer, fall" as opposed to "monsoon, dry" as experienced in much of South Asia.

Don't expect your students to give a specific answer to a general question. Some of us get confused by that sort of thing.

Question 4: You never said "in orbit". Most of the time the space shuttles are sitting on the ground and the only time someone on board is going to experience freefall is if they trip.

[zandperl.livejournal.com]

1) Oh, I thought your tidal locking was amusing. :) The wording was a bit confusing I thought. A tidally locked planet will experience seasons - the side towards the star will always be summer, and the side away always winter - but a particular location on the planet will not.

4) yeah yeah yeah. I'll probably say ISS next time I give this question.

[jrtom.livejournal.com]

I'm not the one with the astronomy degree, nor yet one with a physics degree. But I have a couple of quibbles with your answers; I'd be interested in finding out if I missed something. :)

(1) Does the _directness_ (i.e., presumably, how much atmosphere the sunlight has to go through in order to reach the planet's surface) of the light matter nearly as much as the number of hours during which the sun is heating that part of the globe? I'd bet that the primary thermodynamic effect is from the latter, especially because much of what's important (I assume) is how much the atmosphere itself is getting warmed.

(4) I think that the answer you wrote is to the question "why do astronauts orbit the earth instead of falling down to it", not "why do astronauts experience weightlessness" (to which the answer is more like what I wrote, I think, i.e., it's because there are no significant accelerations on them, which is true because they're in free fall, which is true because they're orbiting the planet). Skipping right to the orbiting thing doesn't really (to my mind) address the original question.

[zandperl.livejournal.com]

1) I was taught that the angle of incidence is the primary factor, with length of day as a secondary correction on top of that.

If you wish to calculate it, the position of the Sun in the sky at noon on the equinoxes will be at an altitude (angle above the horizon) of 90º - your latitude. I'm in New England so that's around 43º latitude, so on the equinoxes the Sun's altitude is 47º. On the solstices it will be 23.5º higher or lower, so the range is roughly 23º-71º. The intensity of light you receive (sometimes called insolation) will be equal to the Sun's irradiance * cos(θ), where θ is the angle between the Sun and the zenith, or 90º-altitude, so θ ranges from 19º-87º. (It turns out this effect is called Lambert's cosine law, though I was never taught the name of it, just told it was an obvious truth.)

The cosine of 19º-87º ranges from 0.05-0.95. That is, here in New England, in winter at noon I'm receiving 5% of the maximum possible light from the Sun (W/m^2), while in summer at noon I'm receiving 95% of the maximum possible. (Technically this is how much is incident on the upper atmosphere, but that's a third order correction.) From winter to summer the incident radiation increases a whopping 19 times. Here in New England, on Winter Solstice the Sun is up for approximately 9 hours, and on Summer Solstice it's up for around 15 hours [source: USNO], an increase of only 1.7 times. It is clear to me from this data that the angle will make a much bigger difference than the length of time, since it is such a drastic change.

An even more detailed analysis (i.e., how do the insolation, length of time, and atmospheric absorption affect things?) would require a degree in meteorology, or at least much more effort than I am willing to put in.

4) They are too accelerating significantly: when they're orbiting, they're moving in a circle. A circle is not a straight line. Therefore by Newton's laws the body is accelerating. Gravity provides that acceleration, so gravity is experienced as moving in a circle rather than experiencing it as your weight on the ground (technically a normal force acting back up from the ground).

I was actually surprised how many people had their heads stuck in the non-inertial reference frame of the astronauts when looking at this question - more people in the physics community actually approached it that way. While it's possible to get a partial understanding from that point of view, to get a full understanding from that point of view is really difficult (since it requires fictitious forces), and I didn't see anyone do it successfully.

[jrtom.livejournal.com]

(1) I'll have to read the references you provided on insolation, etc.; thanks. There's still something counterintuitive about that--I think it's the notion that the angle is significant that's bothering me. (I'm not saying you're wrong, you understand, I'm just explaining the specific nature of my confusion/doubt.) In essence, it's not obvious to me why the angle at which a photon hits the atmosphere should be significant in terms of the amount of energy transferred.

(4) You're correct that there's an acceleration; I expressed myself poorly. However, I'm not sure that I'm totally wrong here. If I jump off a bridge, I will (briefly) 'experience weightlessness'. In that case there's one significant force operating on me (i.e., gravity--neglecting air resistance and other gravitational effects) and it's doing so uniformly across my body. So, thinking through this, I'm guessing that I feel weight (as opposed to weightlessness) when there's more than one perceptible physical force acting on me (typically a normal force, as you said, acting up from the ground, but it need not be the ground--I still feel weight in an airplane as long as the airplane is not itself free-falling).
So, never mind acceleration and reference frames; I don't think they're relevant here. Or am I totally off base?

(Incidentally, I realize that you get paid for this, and not by me...so feel free to punt; I have other folks with physics backgrounds that I can work through this with if you don't have the time or inclination to take this further. I shall not be offended if you so choose.)

[zandperl.livejournal.com]

1) it's not obvious to me why the angle at which a photon hits the atmosphere should be significant in terms of the amount of energy transferred.

It's not the angle at which one individual photon hits that matters. It's the resulting photons per square meter that matters. If you've got a Mag light or some other good flashlight at home, you can experiment with this. Go into a dark room and shine the light straight on a wall so it makes a nice circle. Now keep the light shining towards the same spot on the wall, but arc your hand around to the right or left so the circle smears out to make an ellipse. The same amount of light is now spread out over a larger area, and in that area it is dimmer. If you looked at the same circular area on the wall as you originally had, there's less light hitting that circular area.

I'm slapping a couple image links down here in case they help. In this one, notice how the same amount of light from the Sun is spread out over a larger area of the Earth. This one shows how to calculate that mathematically. Also consider the extremes: let's look at one of the equinoxes. The Sun is shining straight down on the equator, it's straight overhead and thus will appear very strong and bright. At the poles however, the sunlight just strikes with a "glancing blow," and the Sun will appear directly on the horizon - and it's not just that the sunlight has to go through more atmosphere that makes it seem dimmer when it's on the horizon, it really does shine less light on the ground and every little tiny pebble casts a shadow.

4) Ah hah, now we are getting to the heart of the problem with the word "weight," a problem I have yet to solve to my satisfaction. If you draw a free body diagram of a person standing in a room, the force of gravity points downwards, and the normal force (the ground pushing back up on the person's feet) points up. Which is the weight? It depends who you ask, or what the circumstances are. Do you still have weight if the floor is removed and thus the only force is the gravity downwards (causing free fall)? That's debatable too.

If you really want to make your head hurt, jethereal and I have been arguing over what an object's weight is at the bottom of the ocean, where there's gravity acting down, normal force from the ocean floor acting up, a "buoyancy force" acting up, and water pressure on all surfaces of the object. Add to that the question of whether there's water pressure on the bottom surface and I've now asked three physicists what the object's weight was and gotten nine answers. :)

[jrtom.livejournal.com]

(1) You're right; that's an excellent point that I'd overlooked. Good explanation. Had you considered going into teaching as a profession? ;)

(4) "the heart of the problem": precisely. How about this: "weightlessness" = _net_ _perceptible_ physical force is 0 (vector). It seems to me that any definition of 'weight' that means that you can have weight and still be experiencing weightlessness (which might be viewed as a generalization of 'free fall') is not a very useful one. Clearly I still have mass in free fall, but suggesting that I have 'weight' seems like an abuse of terminology.

One possible weasel-wording: you said 'experience weightlessness'. People are not perfect detectors of gravity fields, so minor effects can be ignored for this question.

"...bottom of the ocean": this is actually not materially different from the problem of weight when standing on dry land (assuming that the planet has an atmosphere) from a strict physics standpoint, as the air supplies a 'buoyancy force' just as water does, just about 3 orders of magnitude smaller (assuming it's proportional to density, which I assume it is).

I'll ask a few physicists of my acquaintance what they think about this; if you don't hear back from me on this, ping me. :)

[zandperl.livejournal.com]

1) Har har, that's a thigh slapper. I'll admit the cosine of the angle thing took me a while to grasp myself, and then even when I did grasp it (in the context of light emitted from a layer within a star) it took me a while to connect it to the phenomenon of the seasons.

2) Problem is in intro mechanics the terms "force of gravity" (written F_g) and "weight" (written W) are often used interchangeably. According to that usage, an object in freefall above the surface of a planet does have weight. One definition that I like (sometimes, I change my mind often) is that weight's the normal force applied by the object onto the surface it is resting upon. By this definition, weight is what your household scale reads, even if you're in an accelerating elevator.

The reason I said "bottom of the ocean" is b/c it's easier for us to think of pressure forces in that circumstance, b/c many of us are used to thinking of air as not having any mass. jethereal and I argued over whether there's pressure on the bottom surface acting upwards. Some more things to throw into the mix: it's my understanding that it's the pressure differential between the top and bottom surfaces that causes buoyancy; it's jethereal's understanding that it's a simple Archimedian displacement of water. And also consider the difference between a book lying flat on a table, and a half-shell from which the air is sucked out (or a device like this) - which of these objects will "weigh" more? and if air pressure is what's holding it onto the table, then is there extra air pressure on the underside of the table?

And all this is still simple intro Mechanics. To think I'm teaching Calc Physics II (Thermo and more) in the fall for the first time...