How many soda bottles in Brian's raft?

The question from Dec. 14:

Mr. Jacobs’ friend Brian Jackson saved two-liter soda bottles throughout his senior year of college.  During “Haverfest," he duct taped the bottles together to form a raft.  He then successfully floated himself out onto the duck pond.

Estimate how many bottles Brian used.  Explain your reasoning thoroughly and show all calculations for full credit.

While the majority does not always rule in physics, in this case "they" were right on.  My reasoning:

Call Brian 80 kg or so.  His weight is then 800 N.  That weight must be supported by the buoyant force, which is equal to the density of water times the displaced volume times g.  If each bottle is fully submerged, it displaces 2 L, or 0.002 cubic meters.  The buoyant force created by one bottle is then (1000 kg/m^3)(0.002 m^3)(10 N/kg) = 20 N.  To get to 800 N at 20 N per bottle, you'd need about 40 bottles.

What if Brian's not 80 kg?  Well, as I have to point out to people, 80 kg is a reasonable estimate for Brian, but college guys who drink soda are often heavier; and, in a recent development of Haverlore, I have discovered that Brian supported a second person on the raft as well.  Furthermore, even if Brian were 75 kg, 40 bottles would have to be nearly fully submerged, leaving essentially no safety margin, and getting Brian's feet* wet.  This is an order of magnitude estimate... why not double the estimate to 80 bottles or so?  Then the bottles are in the neighborhood of halfway underwater.  Brian can sit dry, he can bring a friend, he can eat at the COOP** all he wants; 80 bottles will support him.

* Or more likely, his tuckus

** The yummy snack bar... it used to be too expensive for me, but now I find out that students can make their parents pay for the COOP as part of these newfangled meal plans.  Ach, and nowadays students can access email from their rooms, too.

What about the weight of the soda bottles themselves?  Some students will tell me "the soda bottles are of negligible weight."  Okay, but are they?  What's the evidence?  

Some students found that an empty bottle has mass about 40-50 g, for a weight of about half a newton or so. That means that each bottle will only support 19.5 N of Brian rather than the 20 N previously conjectured.  

Does that mean, as some say, that the proper answer is "41 bottles?"  No, certainly not.  As discussed in the previous paragraph, the uncertainty in Brian's mass, and in just how much of the raft is submerged, far outweighs this 3% change due to the weight of the bottle.  The answer is still somewhere around 80 bottles, or better yet, some dozens of bottles. 

Pole vaulting and the apparent weight at the equator

http://xkcd.com/852/

My first reaction when my colleague El Mole showed me the cartoon at the right was that 2 cm is way too significant a difference.

[Pause while you read the comic.  Good, ain't it?]

The fundamental principle is correct.  Because the linear speed of a point on the earth's surface is larger at the equator than at the poles, the "apparent weight" of a person, and thus the "apparent gravitational field g," will be smaller.  In advanced mechanics classes, Newton's Second Law is formulated in the rotating reference frame of the earth, and the effective g is reduced by a  centrifugal acceleration term equal to v²/r.

Why centrifugal and not centripetal?  In an INERTIAL reference frame, acceleration in circular motion is toward the center, i.e. centripetal.  "Inertial reference frame" means, in a sense, imagine that we observe the universe from a stationary camera placed above the rotating object.  Then the net force on the object is continually changing direction so as to push the object toward the circle's center.  However, if we instead observe the world from the eyes of the rotating object itself, then it seems like we are being pushed away from the center of the circle, i.e. in a centrifugal direction.  And if we consider a person rotating at the equator, it makes sense to consider the rotating reference frame; it's more interesting and useful to figure out what the rotating person feels than to figure out what would be observed by a stationary flying saucer over the north pole..

But to have the earth's rotation make a difference of nearly an inch?  An inch is significant in pole vaulting!  The last time the pole vaulting world record was broken, it was by Ukrainian Sergey Bubka over a ten year stretch from 1984-1994.  Each time he broke the previous record, he did so by just one or two centimeters.  The question that the comic begs is, should someone aspire to break records, should he compete exclusively in Ecuador rather than in, say, London?  Would the location make any difference at all?

I made my own order-of-magnitude estimate to check the comic.*  The gravitational field due to earth, without reference to rotation, is about 10 m/s². That term will be lessened by the "centrifugal" acceleration** v²/r. 

First, find v.  The radius of the earth is about 6000 km.  Multiply by 2π to get the circumference at the equator to be about 40,000 km, which is 40 million m.  We go around this circumference in 24 hrs = 80,000 s or so.  This gives a speed in the neighborhood of 500 m/s.

The "centrifugal" acceleration is then (500 m/s)²/(6,000,000 m) = 0.04 m/s². Compared to the gravitational g of 10 m/s², the centrifugal term is, say, four tenths of a percent.

Now, in the absolutlely simplest model, we might consider a pole vaulter as running at a fast horizontal speed, then launching himself as a projectile with that same speed.  The maximum vertical height the vaulter obtains is governed by vertical kinematics, with a known vertical launch velocity v_oy, final vertical velocity of zero, and acceleration of g downward.  This max height can be shown to be v_oy² / 2g.  Point is, the maximum height depends inversely on the first power of g.

So now we reach the end of the story:  what happens when we reduce g by a few tenths of a percent?  We increase the pole vaulter's maximum height by a few tenths of a percent as well. 

Bubka's record vault is 6.15 m.  Increasing that jump by four tenths of a percent would increase his vault height by... a couple of centimeters.  The comic is right. 

In practice, could Bubka have just gone to Indonesia to add two centimeters to his record?  Not exactly.  Four tenths of a percent is the difference in the apparent g between the pole and the equator.  If Bubka set his record at the 1994 Santa Claus's Merry Elves Invitational, then our analysis is sound.  But the farthest north city I can envision holding a major international track meet is, say, Oslo, Norway, at 60 degrees north latitude.  In Oslo, the effective g will be less than at the pole, but not an entire 0.4% less.  Since the linear speed of someone rotating on the globe drops off from equator to pole as the cosine of the latitude***, in Oslo the effective g is reduced by only 0.1%. 

The Oslo-Jakarta pole vault differential is more like 1.5 cm, not a full 2 cm.  Close enough.

Having read all this, my question for you is, who is the more complete nerd?  The xkcd author for carrying out this calculation and basing a comic strip on it, or me for checking the accuracy of the calculation?

GCJ
*This particular comic is generally quite good about its physics.  In fact, I'd be far more comfortable asking the xkcd writers to check me than vice versa.

** xkcd can explain this better than I ever could:  http://xkcd.com/123/

*** At the equator, cos (0) = 1, so his speed relative to Earth's center would be 500 m/s; at the pole, cos (90) = 0, so his speed is 0 m/s.  In Oslo, his speed is 500 m/s cos (60), or half his speed at the equator, and by the calculation above, the correction to g is one-fourth of the correction at the equator.

Universal Gravitation -- instructions and strategy

It's time to discuss universal gravitation.  Since quantitative demonstrations won't work here -- what, you want to hook a spring scale to the moon to verify the force the earth is exerting on it? -- I make some calculations with interesting results.

First, I calculate the gravitational field g, and show that g is still the same value at the top of Mount Everest. 

Next, I calculate the force of the earth on the moon using universal gravitation.  We get 10¹⁷ N.  As a check on that calculation, I use circular motion (which we have just finished covering):  I find the centripetal force required to keep the moon in its orbit.  We get the same answer... and then we discuss how the combination of circular motion and universal gravitation allows calculation of all sorts of astronomical quantities. 

A quiz a few days in asks to find the mass of the Milky Way galaxy, given the sun's orbital period around the center, assuming a two-body problem and circular motion. This calculation gives an accurate result, despite the crazy assumptions.

Throughout the unit I am modeling ORDER OF MAGNITUDE ESTIMATION.  I don't ever pound the calculator, and students are forbidden from having calculators on their desks.  Instead, I just write every calculation in scientific notation in standard units.  From that, I cancel powers of ten in my head, and write the answer to zero or one significant figures.  And, we check the reasonability of all answers where possible by direct comparison to known quantitites -- for example, if we calculate the mass of a star, we compare that to the mass of the sun.

Below are the instructions I write out on my problem sets for universal gravitation.  I follow through on my demands, too -- a problem without an order-of-magnitude estimate written out loses at least 3 out of 10 points.  The nebulous and meaningless statement "that's a really huge mass" rather than "that's 1/10 the mass of the sun, so reasonable for a star" loses 3 of 10 points. 

Topics to be discussed: Newton’s law of universal gravitation. I find that, amazingly, the biggest trouble that students have when solving universal gravitation problems is plugging correctly into their calculators. To be sure you don’t screw up, follow this advice:

• Solve in variables as far as possible in each problem. Only plug in values at the end.

• Do an order of magnitude estimate of your answer without the calculator, to be sure you’re not way, way off.

• Check the reasonability of the answer. When you’re asked to make a comparison, do this right – don’t just say “that’s a big mass”, say “that’s twice the mass of the earth” or “that’s close to the mass of the sun”. Astronomical data is easily available online, and there are astronomical tables in all physics texts.

• I will be looking at comparisons VERY CAREFULLY on gravitation problems. Do these right.

The Milk Problem

Finally, yesterday, I got to teach. On the first day, as discussed previously, I get into real physics: equilibrium situations in AP, position-time graphs in general.

I found a new problem appropriate for the first night’s assignment. I want to give students a sense of how physics will be different from math class, but I’m not yet ready to assign conceptual questions about the current topics – we haven’t gotten far enough. I use problems that require serious reasoning, especially those which lend themselves to order-of-magnitude estimates. The following is based on a problem from the Young and Freedman text:

Milk is often sold by the gallon in plastic containers. You are to solve by calculation and reasoning, not through research.

(a) Estimate the number of gallons of milk that are purchased in the United States each year. (Obviously, your answer should include both verbal and mathematical reasoning.)
(b) What approximate weight of plastic does this represent? Compare this weight to something with which you are familiar.

What an excellent question! Even if students WANTED to try to answer through library or google research, that’s a daunting task… as I found out.

My own reasoning: with 300 million people in the USA, figure about a gallon per week for a family of four. That’s around 70 million gallons per week, times 52 weeks, or somewhere near 4 billion gallons of milk per year sold in the US.

A bunch of googling produced

this article from the US General Accounting Office, suggesting that about 7 billion gallons of milk per year were sold in 2001. Hey! I’m well within a factor of 10, which is the goal of such a problem, anyway.

As for the weight of plastic… I originally guessed about 20 g from an empty milk jug, based on my experience with my hanging weights. This gives 108 kg of plastic for the billion gallons sold each year, or about 100,000 tons.

My chemistry colleague, the Atlanta Cracker, Mr. Paul Vickers weighed an empty half-gallon milk carton, getting 47 g. Woodberry Forest Librarian Phoebe Warmack turned up a claim that nowadays gallon milk jugs are less than 60 g. So my estimate was definitely good enough, because it gives the same ~100,000 tons of plastic as does a weight of 50 g or so.

And this is the whole point of the exercise. Not only do I want my students to gain their first exposure to “Fermi problems” and order-of-magnitude estimation, I also want to make a preemptive strike against the arguments I inevitably hear in class: “You said the answer was 5.9 N, but I got 5.8 N. What did I do wrong?” Or, as always, “isn’t g 9.8, not 10?” Hopefully they will see that a 2% difference is meaningless when making everyday measurements.