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tony873004
2008-Sep-25, 07:24 AM
All units for this post are m/s2

On a recent exam, I was told to use 9.81 for g. The textbook I'm using, as well as several other textbooks and other sources I've seen also use 9.81.

My textbook was written by authors who live in San Francisco. According to Wikipedia, g in San Francisco is 9.793 which rounds off to 9.80. And throughout the world it varies. The high and low values in Wikipedia's table are 9.781 in Jakarta and 9.819 in Rome. Rounded off to 3 significant digits, this is 9.78 and 9.82. But rounded off to 2 significant digits, this is 9.8 and 9.8. In fact, everywhere on Earth rounds off to 9.8 in two significant digits, but there is discrepancy when rounded off to 3 significant digits.

Another Wikipedia page states Earth's "standard gravity" as 9.80665, which rounds to 9.81 in 3 significant digits. This might be why textbooks like this value. But they could just as easily like 9.807, or 9.8067. Why stop at 9.81? And as long as they're dropping digits, why not drop one more to 9.8, which is valid everywhere on Earth? So why do many Physics texts prefer 9.81?

http://en.wikipedia.org/wiki/Earth%27s_gravity
http://en.wikipedia.org/wiki/Standard_gravity

hhEb09'1
2008-Sep-25, 08:13 AM
So why do many Physics texts prefer 9.81?I think three digits are considered "just right", two is too few, four is too many. Kinda like pi = 3.14 instead of 3.1 or 3.142

And, as you say, once you decide on three digits, the rest more or less follows.

Ken G
2008-Sep-25, 10:34 AM
Why stop at 9.81? And as long as they're dropping digits, why not drop one more to 9.8, which is valid everywhere on Earth? So why do many Physics texts prefer 9.81?
I think you make a convincing case that a value of 9.8 is actually more scientifically honest. I agree with hhEb09'1 that three digits seems like an accuracy we can aspire to without going overboard, but the data you report certainly shows that 3 digits of precision is something of a fantasy, and actually promotes misconceptions about science.

Larry Jacks
2008-Sep-25, 01:01 PM
IIRC, gravity can vary a small amount depending on things like terrain*. If that's true, it might explain the different readings from around the world. When meteorologists report the local barometric pressure, they say things like "29.92 inches corrected to sea level." Perhaps those gravity readings need to be corrected to account for local factors as well.

*We had to account for this in geosynch satellite operations. These variations in gravity are one of the reasons why satellites up there have to perform periodic station-keeping maneuvers to stay in their assigned locations.

pghnative
2008-Sep-25, 01:17 PM
FWIW, when I took college physics (circa 1988), we used 9.8 m/s2 in our calcs. No longer have the textbook to reference it though...

AndreH
2008-Sep-25, 01:36 PM
All units for this post are m/s2

On a recent exam, I was told to use 9.81 for g. The textbook I'm using, as well as several other textbooks and other sources I've seen also use 9.81.

My textbook was written by authors who live in San Francisco. According to Wikipedia, g in San Francisco is 9.793 which rounds off to 9.80. And throughout the world it varies. The high and low values in Wikipedia's table are 9.781 in Jakarta and 9.819 in Rome. Rounded off to 3 significant digits, this is 9.78 and 9.82. But rounded off to 2 significant digits, this is 9.8 and 9.8. In fact, everywhere on Earth rounds off to 9.8 in two significant digits, but there is discrepancy when rounded off to 3 significant digits.

Another Wikipedia page states Earth's "standard gravity" as 9.80665, which rounds to 9.81 in 3 significant digits. This might be why textbooks like this value. But they could just as easily like 9.807, or 9.8067. Why stop at 9.81? And as long as they're dropping digits, why not drop one more to 9.8, which is valid everywhere on Earth? So why do many Physics texts prefer 9.81?

http://en.wikipedia.org/wiki/Earth%27s_gravity
http://en.wikipedia.org/wiki/Standard_gravity

Maybe this value was just agreed upon by one of these boards or conferences. I don't know. You have to agree on something to make calculations comparable, or in every calculation you would have include a list with the used values.
I guess the 9,80665 is obtained by a theoretical way. just take the Earth's mass as a point mass and claculate the theoretical value at the theoretical value of the Earth's radius (which again is something which was agreed upon).

I remember that back in school for some simplified calculations we used even 10.

hhEb09'1
2008-Sep-25, 01:44 PM
*We had to account for this in geosynch satellite operations. These variations in gravity are one of the reasons why satellites up there have to perform periodic station-keeping maneuvers to stay in their assigned locations.The gravity field of the earth is associated with the "geoid" which itself is described by spherical harmonic coefficients. I haven't looked them up but I'd imagine that the 9.80665 m/s/s figure mentioned in the OP is associated with the global degree zero spherical harmonic. There are no degree one coefficients, and the degree two variation involves the oblateness of the earth but also a term that represents the "pinching" of the equator. It is that pinching that causes the effect on the geosynch satellites, but it also means that there are two regions of geosynch orbit that are fairly stable. One of the degree three coefficients represents the "pear shape" of the earth.

But, if the textbook only uses it in the fashion, "assume gravity to be 9.81 m/s/s" or "use 9.81 m/s/s as the acceleration of gravity", that seems to be OK. After all, gravity does vary, why not use a value that represents the average if you have to use one value? What if gravity varied even more? Wouldn't the problem be justified in saying something like, "for this problem, use 11 m/s/s as the acceleration of gravity"?

Ken G
2008-Sep-25, 01:59 PM
But, if the textbook only uses it in the fashion, "assume gravity to be 9.81 m/s/s" or "use 9.81 m/s/s as the acceleration of gravity", that seems to be OK. After all, gravity does vary, why not use a value that represents the average if you have to use one value?Well, for one thing, for satellite orbits you would want the gravitational acceleration only, whereas if you are dropping blocks in a laboratory, you need the value corrected for the local rate of surface rotation. I think the OPer's point is that if there are important factors like that going on, then even a conventional choice should not overstate the useful level of precision-- because so doing tends to obscure those other effects and gives people a naive sense of the issues scientists must weigh (no pun intended).

What if gravity varied even more? Wouldn't the problem be justified in saying something like, "for this problem, use 11 m/s/s as the acceleration of gravity"?Certainly the problem can stipulate that the experiment is happening in any gravity it wants, even on the Moon, but the implication is that the value you are taking is somehow applicable to what you would actually be doing in, say, physics lab. As such, a value that does not over-report the precision is all part of responsible error management, a topic that is in other ways viewed as quite important when counting significant figures. At least that's how I interpret the question in the OP, which I think has a valid beef.

hhEb09'1
2008-Sep-25, 02:08 PM
Certainly the problem can stipulate that the experiment is happening in any gravity it wants, even on the Moon, but the implication is that the value you are taking is somehow applicable to what you would actually be doing in, say, physics lab. As such, a value that does not over-report the precision is all part of responsible error managemennt, a topic that is in other ways viewed as quite important when counting significant figures. At least that's how I interpret the question in the OP, which I think makes a valid point.Judging by the values reported in the OP, it looks more like the precision is under-reported, if anything. The OP has all the local values to another decimal place yet.

But, still, it seems like the values are presented in the fashion, "use this value". Because there is not a single value. In my experience, that's just to forestall the rising of a dozen arms asking which value of g should be used. :)

George
2008-Sep-25, 02:28 PM
Could it be that it is more common to have two decimal places used in the other values of the terms found in many related equations?

For San Fransicso, the tasty food alone would quickly cause someone going their with a known given mass to suddenly, after only one meal, weigh a 1/3 lb. more, which produces and effective g of 9.81 (assuming they choose not to admit to a gain in mass). :)

nauthiz
2008-Sep-25, 03:59 PM
I just took a look on the Wiki page for Earth's gravity (http://en.wikipedia.org/wiki/Earth%27s_gravity), and it mentions that the standard gravity (9.80665) is a 'nominal' average. Curious, I looked into it a little bit more and it turns out that value was set somewhat arbitrarily at a 1901 convention and apparently wasn't drawn from anything along the lines of a global data set, so it would seem to be more of a convention than an actual average.

Jeff Root
2008-Sep-25, 04:58 PM
My high school physics textbook, printed in 1965, has a table of 12 cities
and other places around the globe with magnitude of the gravitational field
given to four significant digits. The altitude of each location is also given.
The table is very close to the beginning of the section on motion at the
Earth's surface.

At the end of the section are 39 problems. Many of them involve gravity,
and some specify g, as follows:

Problem #1 specifies 9.80 N/kg
Problem #4 specifies 9.81 m/s^2
Problem #7 specifies 9.80 N/kg
Problem #30 specifies 9.8 m/s^2
Problem #31 specifies 9.8 N/kg

-- Jeff, in Minneapolis

tony873004
2008-Sep-25, 05:56 PM
I think three digits are considered "just right", two is too few, four is too many. Kinda like pi = 3.14 instead of 3.1 or 3.142

And, as you say, once you decide on three digits, the rest more or less follows.

That's a good point. I always assumed pi was usually expressed to 3 digits because this is a convienent place to stop since the next digit is a 1. (If pi were 3.144917, would 3.14 still be popular?)



...As such, a value that does not over-report the precision is all part of responsible error management, a topic that is in other ways viewed as quite important when counting significant figures. At least that's how I interpret the question in the OP, which I think has a valid beef.
Your interpretation is correct. Thanks everyone for the great replies.

nauthiz
2008-Sep-25, 06:05 PM
That's a good point. I always assumed pi was usually expressed to 3 digits because this is a convienent place to stop since the next digit is a 1. (If pi were 3.144917, would 3.14 still be popular?
If pi were 3.144917, folks would round it to 4 digits as 3.145. Other folks would round that to 3.15, and still more folks would round it to 3.2.

Ken G
2008-Sep-25, 06:53 PM
But, still, it seems like the values are presented in the fashion, "use this value". Because there is not a single value. In my experience, that's just to forestall the rising of a dozen arms asking which value of g should be used. I see what you are saying, but my reaction would be "use any value you want that you can justify, because if you think I'm going to grade the difference between 9.81 and 9.8, you must think this is a class in calculator dexterity".

hhEb09'1
2008-Sep-25, 07:29 PM
I see what you are saying, but my reaction would be "use any value you want that you can justify, because if you think I'm going to grade the difference between 9.81 and 9.8, you must think this is a class in calculator dexterity".You require them to justify it on the test? You mean, like documentation of a local value?

Ken G
2008-Sep-25, 08:47 PM
I mean like, they better not use 15 or 1, but I'm certainly never going to mark an answer wrong that is the difference between an experiment in Denver and an experiment in Buffalo. In fact, I'd rather they understand the limitations of the numbers that come out of their calculators.

Spaceman Spiff
2008-Sep-26, 12:47 AM
All units for this post are m/s2

On a recent exam, I was told to use 9.81 for g. The textbook I'm using, as well as several other textbooks and other sources I've seen also use 9.81.

My textbook was written by authors who live in San Francisco. According to Wikipedia, g in San Francisco is 9.793 which rounds off to 9.80. And throughout the world it varies. The high and low values in Wikipedia's table are 9.781 in Jakarta and 9.819 in Rome. Rounded off to 3 significant digits, this is 9.78 and 9.82. But rounded off to 2 significant digits, this is 9.8 and 9.8. In fact, everywhere on Earth rounds off to 9.8 in two significant digits, but there is discrepancy when rounded off to 3 significant digits.

Another Wikipedia page states Earth's "standard gravity" as 9.80665, which rounds to 9.81 in 3 significant digits. This might be why textbooks like this value. But they could just as easily like 9.807, or 9.8067. Why stop at 9.81? And as long as they're dropping digits, why not drop one more to 9.8, which is valid everywhere on Earth? So why do many Physics texts prefer 9.81?

http://en.wikipedia.org/wiki/Earth%27s_gravity
http://en.wikipedia.org/wiki/Standard_gravity

Am I missing something, here? I am almost certain that the quoted precisions (4 sig figs) at these locations are significant. The local variations are due to local continental densities and elevations different from sea level, the earth's rotation for gosh sakes, and even the latitude on an oblate spheroid (and Earth isn't actually an oblate spheroid). All of this explained in the first link. So what's the mystery?

Ken G
2008-Sep-26, 12:50 AM
The quoted figures are significant-- the issue is, why do we choose to quote g=9.81 when we do physics problems? In most places that would apply such problems to experiment, that would be an incorrect value, to that many sig figs. So why not just back on the sig figs until everyone can actually use the value meaningfully? Note I don't even know if that is corrected for surface rotation, and I'll bet most others don't either, off hand.

hhEb09'1
2008-Sep-26, 12:55 AM
We could always go to 10, I'd like that :)

PS: Here's a coincidence, from the Earth Gravity wiki page (http://en.wikipedia.org/wiki/Earth's_gravity): The effective gravity (factoring in the centrifictional force and its bulge, as well as maybe other things) at the poles is 9.832 m/s/s, and that at the equator is 9.789 m/s/s. Of course, ironically, the exact average of these two values is 9.8105 m/s/s

Ken G
2008-Sep-26, 01:39 AM
Introducing too many caveats and exceptions at that level tends to turn the light over the students' heads off rather than on.
It could be argued that what it would actually do is teach them to do science, rather than rote calculations. What I see in science courses is a real lack of understanding of any concept of error or uncertainty. We want everything to be so clean, to avoid muddling our young minds, and we end up with unmuddled young minds that have little concept at all of what science is beyond a series of recipes they must apply, like jumping through hoops, to get to what they really want to do.

nauthiz
2008-Sep-26, 02:29 AM
It could be argued that what it would actually do is teach them to do science, rather than rote calculations.
Yah. As far as teaching science goes, I imagine a class period spent explaining that the strength of the planet's gravitational pull varies quite a bit depending on where you are would have more value than a week's worth of the usual grinding through calculations on worksheets.

Leave the math problem sets for math class.

Ken G
2008-Sep-26, 03:17 AM
Hmm.. Perhaps I'm showing my age, then, but I tend to believe that it's best to teach elementary principles first and then add complexity as the student progresses and shows increasing interest and/or ability.
The vast majority of students using g=9.81 m/s2 will likely never take another mechanics class. And if you are talking about physics majors, all the more reason to treat them like adults that can handle a little sophistication. If we are talking high school though, I suppose one can only do so much.


This is my beef with "new math" and the "new new math" where concepts and set theory are introduced without a thorough grounding in "math facts". That's at the early educational stages, and I confess I don't know much about that. There is a whole different set of challenges there, I imagine it's pretty tough sledding and there is a role for simple drilling. But the students I see in college seem to have much more trouble with understanding what math is and what it is for, than they do going through the rote motions of math (which they also don't do well, but it's less debilitating than not even knowing the what or why).


Way back in the dinosaur era when I was taught high school physics, we learned the very basics of gravitation and we used 9.81 m/s^2 as the value at sea level while we got comfortable with computing the acceleration due to gravity on the objects in our problems.OK, then let me demonstrate my point by asking you, off the cuff, when you were taught that value of g, do you think it did or did not account for the surface rotation of the Earth? If you don't know off hand, then what did that number even mean?


I was the only student who knew better, and my teacher knew that I knew, and very politely and gently asked me to shut [...] up. The other students, who had no intention of ever working in physical science, were perfectly content to pretend that Earth's gravitational field is perfectly uniform and did all their computations just like the book asked them to. But that's just the kind of teaching I think kills the interest in physics. It's just not that important that those students can plug-and-chug into kinematics formulas, it really isn't. I'll bet the "shut up and follow the recipe" approach earned few converts into the field of physics from your classmates. Admit it-- what did they say about physics behind that teacher's back?

I feel that he should have encouraged you to explore the reasons why g varies-- that would have been a more important lesson that could have left an imprint on those students long after they had forgotten the formulas of kinematics. They would have learned quantitative critical thinking skills, which is really what physics is all about-- not formulas.

Ken G
2008-Sep-26, 03:24 AM
As far as teaching science goes, I imagine a class period spent explaining that the strength of the planet's gravitational pull varies quite a bit depending on where you are would have more value than a week's worth of the usual grinding through calculations on worksheets.I tend to agree, though there's certainly a trade-off and we'd like a balance of both basic competence and an appreciation for the difference between critical thinking and sheepish adherence to what one is told.
Well, were it not for basic high school physics, I would never have grasped trigonometry.That is certainly as valid today as ever-- many people find that math is easier to understand when it has a physics application.

nauthiz
2008-Sep-26, 04:05 AM
I tend to agree, though there's certainly a trade-off and we'd like a balance of both basic competence and an appreciation for the difference between critical thinking and sheepish adherence to what one is told.
Sure, and I think the students should work through a few applications of the equations themselves. But I still remember high school physics class where we got these worksheets with 30 identical problems where the goal wasn't do do anything that required much of a knowledge of physics, it was just to repeatedly fill in a couple variables and maybe grind through a little basic algebra.

I don't think the kids who did poorly on those worksheets did so because they couldn't grasp science or the basic principles of physics, it was because they weren't so great at math. (Yeah, don't even get me started on how math in my school system was taught past about grade 5.)

AndreH
2008-Sep-26, 09:27 AM
I see what you are saying, but my reaction would be "use any value you want that you can justify, because if you think I'm going to grade the difference between 9.81 and 9.8, you must think this is a class in calculator dexterity".

I am with hheb09'1 here. And my first answer why to use 9,81 is simply "because this is the value agreed upon" including a remark that it can ofcourse vary locally and that there are other "constants" (like radius of the earth) which also may vary locally. And for the sake of making calculations a little less confusing there is a table with internationally agreed values which will be used.

So in short terms: use this, but be aware in reality it could vary for different reasons

Also I would let students make a calculation of the difference in g due to Earth rotation at the poles and equator. (But then what value for R?:)).
And maybe conduct the "real pendulum" experiment to see if they could find (and to what accuracy) the local value of g.
And of course additionally if it comes to calculation I would train them always to put down or derive the formula before plugging in the numbers in the end.

grant hutchison
2008-Sep-26, 10:34 AM
The quoted figures are significant-- the issue is, why do we choose to quote g=9.81 when we do physics problems? In most places that would apply such problems to experiment, that would be an incorrect value, to that many sig figs. So why not just back on the sig figs until everyone can actually use the value meaningfully? Note I don't even know if that is corrected for surface rotation, and I'll bet most others don't either, off hand.The 9.80665 m.s-2 is neither corrected nor uncorrected for rotation, really. It's just the exact figure to be used when converting acceleration in m.s-2 to acceleration in "g"; agreed by a committee at a time when the measurement precision for surface gravity didn't actually merit the precision of the agreed decimals, and correctly designated gn (I think the "n" is for "nominal") to differentiate it from ge, gp and gΦ.

Without having seen the problem in Tony's exam paper, it seems to me that the most likely reason to provide a figure of 9.81 for gn would be just because the problem was set up using three significant figures for the other data, too. Tony?

Grant Hutchison

George
2008-Sep-26, 01:35 PM
Without having seen the problem in Tony's exam paper, it seems to me that the most likely reason to provide a figure of 9.81 for gn would be just because the problem was set up using three significant figures for the other data, too. Tony?
That would make sense for us old-timers, too. We almost always used 32.2 fps2, so it doesn't feel right without 3 significant figures. It was very rare, in engineering at least, that we used any other value for g, IIRC.

Ken G
2008-Sep-26, 01:39 PM
The 9.80665 m.s-2 is neither corrected nor uncorrected for rotation, really. It's just the exact figure to be used when converting acceleration in m.s-2 to acceleration in "g"That explains the meaning of that designation, but note that is not the acceleration of gravity that applies to the problem being solved. Indeed, I would say that means the acceleration used in the problem is not corrected for surface rotation, so will be inaccurate for that reason as well.

Without having seen the problem in Tony's exam paper, it seems to me that the most likely reason to provide a figure of 9.81 for gn would be just because the problem was set up using three significant figures for the other data, too. Tony?
I would wager that all the other numbers in the problem were integers. That is certainly the norm for such problems, further underscoring the absence of any effort to communicate how science actually works.

Ken G
2008-Sep-26, 01:40 PM
That would make sense for us old-timers, too. We almost always used 32.2 fps2, so it doesn't feel right without 3 significant figures. It was very rare, in engineering at least, that we used any other value for g, IIRC.
I count that as pretty scary in engineering, of all places, where the answers are supposed to actually mean something quantitative.

hhEb09'1
2008-Sep-26, 02:13 PM
That explains the meaning of that designation, but note that is not the acceleration of gravity that applies to the problem being solved. Indeed, I would say that means the acceleration used in the problem is not corrected for surface rotation, so will be inaccurate for that reason as well.But there is a place on the earth where the value in the OP would represent that, so how would we know?

Personally, I think g should be defined as 9.86960 m/s/s :)

I would wager that all the other numbers in the problem were integers. That is certainly the norm for such problems, further underscoring the absence of any effort to communicate how science actually works.Probably more than one problem.

grant hutchison
2008-Sep-26, 02:37 PM
That explains the meaning of that designation, but note that is not the acceleration of gravity that applies to the problem being solved. Indeed, I would say that means the acceleration used in the problem is not corrected for surface rotation, so will be inaccurate for that reason as well.I believe I could find a place where the given figure is exact, neglecting rotation, and another where the figure is exact, including rotation. So I stand by my contention that it is neither corrected nor uncorrected, unless Tony tells me his problem included a stipulated latitude.

Grant Hutchison

George
2008-Sep-26, 03:30 PM
I count that as pretty scary in engineering, of all places, where the answers are supposed to actually mean something quantitative.
What would have been much scaryier would have been to claim 3 or 4 decimal places sense we only had slide rules. :) I suspect the 32.2 value was used simply because of the slide rule issue; it was a slide rule rule. ;)

[Then there was "Fast Eddie", one of our thermo profs., who claimed he could get 5 place accuracy on his slide rule and that there were only 5 others in the world who could do so, and he taught them (or something like that). He was one of the three profs. -- others were "Rocket Man" (vibrations prof.) and Tulley (fluid mechanics prof.) -- of the brutal triad where any student who managed to pass all three were installed into an elite and small membership. I never asked this club what they did. :)]

Ken G
2008-Sep-26, 03:59 PM
I believe I could find a place where the given figure is exact, neglecting rotation, and another where the figure is exact, including rotation. So I stand by my contention that it is neither corrected nor uncorrected, unless Tony tells me his problem included a stipulated latitude.But as I said, a problem is welcome to say it is on Mars, or some hypothetical place. What difference does it make if you can find a place on Earth where that number holds? The real issue is, is that really "the acceleration of gravity", as students would be able to experience in the laboratory, and of course it is not. So why not ask them to use a value that is, by stripping as many significant figures as necessary to do that? Why are we complicating the number if there is no corresponding increase in the accuracy of the result in practice, are we teaching "calculator dexterity" that we want them to push that extra button? The issue is not whether 3 significant figures, or 10, can in principle apply in some situation. It is, what is the relevant accuracy of the quantities we are asking the students to use, and are we asking them to even understand the concept of accuracy. That's what I mean about teaching them science while they are being taught science-- is g a fundamental constant of nature, like c?

Ken G
2008-Sep-26, 04:06 PM
We want to give them a taste, and the basic understanding of what's involved in doing the work to learn that subject matter. But this is exactly my point. What "taste" is more important-- how to push buttons on a calculator, or what scientific accuracy even means?
Wax too esoteric on the details presented, and IMO you risk losing the students before the light bulbs go on.But I would say that teaching rote kinematic formulas that students will never use the day they walk out of the course, while not teaching what science is, is precisely "waxing too esoteric on the details".



To sit in the ivory tower and pooh pooh the very idea of "someone so naive to actually USE the value 9.81 m/s^2 for the acceleration of Earth's gravity" accomplishes nothing more than to reinforce the idea that scientists and engineers are a bunch of elitists. Were that anything close to the argument I am presenting, I would agree.

Ken G
2008-Sep-26, 04:07 PM
What would have been much scaryier would have been to claim 3 or 4 decimal places sense we only had slide rules. :) I suspect the 32.2 value was used simply because of the slide rule issue; it was a slide rule rule.Sadly, I must confess I don't even know what that means!

grant hutchison
2008-Sep-26, 04:17 PM
What difference does it make if you can find a place on Earth where that number holds? The real issue is, is that really "the acceleration of gravity", as students would be able to experience in the laboratory, and of course it is not.It might well be. The number, as quoted, holds over a significant swathe of the inhabited area of the Earth.
But my point had nothing to do with the teaching of science; I was simply defending my original statement that gn is neither corrected nor uncorrected for rotation. You may or may not think it's a good value to use for a specific problem. :)


Sadly, I must confess I don't even know what that means!Three sig figs easily at the left end of the slide-rule scale, and just about, with a hawk-sharp eyeball, at the right. Problems were not presented with more sig figs, and the answer was likewise limited.
I remember I sometimes used to juggle the order of division and multiplication in an attempt to keep my intermediate values as close to the left end of the rule as I could. :)

Grant Hutchison

nauthiz
2008-Sep-26, 04:49 PM
Why even bother with two significant figures? If a teacher wants to make my life easier by saying the acceleration due to gravity is 10m/s2, I'd be the last to complain.

Ken G
2008-Sep-26, 04:54 PM
I was simply defending my original statement that gn is neither corrected nor uncorrected for rotation.I accepted that statement holds for the constant gn, but I am talking about the appropriateness of blindly using that constant, with that many sig figs, as if it meant something in the physics problems it is being applied to. That would be overlooking the big difference between that "constant" and the things we view as fundamental constants, which is the lost lesson there.

You may or may not think it's a good value to use for a specific problem. Right, that is the issue-- what lessons are in the presence or absence, in the general education of introductory physics, of that seemingly innocuous extra "1".

Ken G
2008-Sep-26, 05:01 PM
Why even bother with two significant figures? If a teacher wants to make my life easier by saying the acceleration due to gravity is 10m/s2, I'd be the last to complain.But there is a reason for using 9.8-- that is the number that will apply in a generic lab experiment. Using 10 is easier, but loses accuracy for no real reason, and might cause us to look for explanations of why our answer did not agree with experiment if we anticipated agreement at a higher accuracy level. That's a key aspect of science, and is part of what I'm saying we should teach. When is a theory "good enough", and when should we look to modify it? What is a theory, in the first place? These are important questions, that if we don't address in the introductory classes that most educated people take, then when were we planning on addressing them? The practitioners don't really need to be told what a theory is, they will have ample opportunity to figure that out.

NEOWatcher
2008-Sep-26, 05:05 PM
It's amazing how this is turning into an argument of generalizations.
It's obvious we learn in different ways.
It's obvious that the context is important.
Now; I've been following the discussion, and I don't even know what class Tony is talking about. Everyone seems to be assuming physics, or at least some kind of science.

It could be calculus. In calculus we used the value of 32 f/s/s. Did it matter to us? No. Did the physics types say anything about the accuracy? Heck no. It just didn't matter in that context.

We learned that the application of the formulas and interpretation of the problem was a big part of solving the problem. If that were not done correctly, then it didn't matter how accurate the rest of it was.

hhEb09'1
2008-Sep-26, 05:16 PM
Trying to itemize the effects of topography and rotation is going to be problematic--is the extra distance to the center of the earth due to the centrifictional bulge, or to the extra mass between the surface and the center? But maybe no one is going down that direction anyway.

Now; I've been following the discussion, and I don't even know what class Tony is talking about. Everyone seems to be assuming physics, or at least some kind of science. :) The OP mentions Physics texts, but at the bottom.

mugaliens
2008-Sep-26, 05:19 PM
If pi were 3.144917, folks would round it to 4 digits as 3.145. Other folks would round that to 3.15, and still more folks would round it to 3.2.

If they did, they'd be violaing the rules of rounding. You cannot round a rounded number.

Thus, 3.144917 rounded to 4 digits is indeed 3.1449.

Rounded to 3 digits it's 3.145

Rounded to 2 digits it's 3.14

Rounded to 1 digit it's 3.1.

You must always round numbers based on the most accurate value available.

By the way, 3.2 isn't due to rounding error (http://en.wikipedia.org/wiki/Rounding_error), which is the different between the rounded number and it's exact mathematical value. Rather, because it's been improperly rounded, it's due to bone-headed error.

hhEb09'1
2008-Sep-26, 05:20 PM
I remain convinced that is precisely the argument you are presenting. You would seem to be stipulating that a simple explanation that doesn't consider every possible mitigating factor is useless to the teaching of science. I'm not a professional educator, but I do have classroom experience in providing both corporate and military training. I can't imagine an approach more likely to convince students the subject is impossible to learn than demanding they first find the local value for g to 4 or more digits of precision, then correct for...[snip]But Ken G is advocating using 9.8 m/s/s :)

nauthiz
2008-Sep-26, 05:23 PM
But there is a reason for using 9.8-- that is the number that will apply in a generic lab experiment. Using 10 is easier, but loses accuracy for no real reason, and might cause us to look for explanations of why our answer did not agree with experiment if we anticipated agreement at a higher accuracy level. . .

If they did, they'd be violaing the rules of rounding. You cannot round a rounded number. . .

Y'all have no love for lame humor, do you? :cry:

NEOWatcher
2008-Sep-26, 05:27 PM
The OP mentions Physics texts, but at the bottom.
Well; the OP doesn't exactly say that the test and text he has fits into the subgroup of "many phyiscs text". But; I will accept the assumption since any other discussion might concentrate on the wording instead of the concept.

And; since I really don't know what kind of labs are done in this class, and the inherent accuracy of these. I really don't know if that kind of accuracy is really called for.

Just like pi. I've never seen a need throughout any of my education to use anything more accurate than 3.14, yet, almost since I first learned of pi, I learned many more digits, and the issues of rounding.

Jeff Root
2008-Sep-26, 05:54 PM
The 9.80665 m.s-2 is neither corrected nor uncorrected for
rotation, really. It's just the exact figure to be used when
converting acceleration in m.s-2 to acceleration in "g"
That explains the meaning of that designation, but note that is not
the acceleration of gravity that applies to the problem being solved.
Indeed, I would say that means the acceleration used in the problem
is not corrected for surface rotation, so will be inaccurate for that
reason as well.
Can you explain what you mean by the terms "corrected for" and
"inaccurate", above?

If I fail to "correct for" rotation in my report of a measurement,
does that mean I forgot to remove the effect of Earth's rotation
from my measurement, or does it mean that I forgot to add the
effect of Earth's rotation to my measurement? Why should my
report of the measurement be "corrected"? In what way is it
"inaccurate" if I don't "correct" it?

-- Jeff, in Minneapolis

Ken G
2008-Sep-26, 06:01 PM
At what point were we talking about teaching scientific accuracy? The original question was "why are we taught this value for g?" I think you are interpreting that question differently. What it was about was very much about teaching scientific accuracy, or more to the point, not teaching it, by using a generic g value with an inappropriate precision.

At what point did we veer aside from teaching the basics of kinematics to failing to teach "what science is"?The instant those students walked into what is likely the only physics course they will ever take. Or are the students at your school all jazzed about learning kinematics? Perhaps yours was the one place on the planet where the students were buzzing in the halls "yippee, now is when I finally get to learn kinematics."

When the students express an interest in knowing why their results deviate from the predicted values, then the teacher can introduce the other factors that must be included when scientists and engineers solve real problems in orbital calculations, structural loading, etc.Did that ever happen?


How is it somehow cheating the students out of their education to wait for the "aha" moment before discussing that? It isn't-- if it ever happened.
You would seem to be stipulating that a simple explanation that doesn't consider every possible mitigating factor is useless to the teaching of science.You already mentioned that interpretation of my point, but I still view it as bizarre. As hhEb09'1 said, I am advocating matching the simplicity of the calculation with the factors that are likely to be actually relevant to that calculation. A "wrong" answer in science is nothing other than one that does not "meet the desired accuracy criteria". Now, how many science classes are delivering that lesson, when their own numbers don't underscore what the desired accuracy criteria even is?

Ken G
2008-Sep-26, 06:02 PM
Y'all have no love for lame humor, do you? :cry:
Not when it comes to science, baby! :)

Ken G
2008-Sep-26, 06:08 PM
Can you explain what you mean by the terms "corrected for" and
"inaccurate", above?I mean that if we treat g like it is a fundamental constant of nature, rather than make the point that it is not, we are introducing error into the interpretation of our own calculations. Of course, the error itself is irrelevant, what matters is the understanding of error. That is a far more important aspect of physics, for most people, than kinematics formulae. If you don't believe me, go into a crowded mall and ask "what is the distance covered in time t with acceleration a" and see how many people know the answer. But had they been taught how science deals with error, they would know something they would never forget.



If I fail to "correct for" rotation in my report of a measurement,
does that mean I forgot to remove the effect of Earth's rotation
from my measurement, or does it mean that I forgot to add the
effect of Earth's rotation to my measurement?One does not correct for things in measurements, one does it in predictions of measurements. I took that as self-apparent-- one does not need the number 9.81 to do a measurement.

tony873004
2008-Sep-26, 07:03 PM
My professor would probably think I'm overanalyzing this by even asking, as this is only a lower division course. So my OP is merely a curiosity.


Without having seen the problem in Tony's exam paper, it seems to me that the most likely reason to provide a figure of 9.81 for gn would be just because the problem was set up using three significant figures for the other data, too. Tony?

Grant Hutchison
The exam is still being graded, so I don't remember the other figures. They were fluid problems. The formula page the professor gave us gives the density of water to be 1.000E3 kg/m^3, and 1.013E5 Pa for atmospheric pressure. These values were needed in the problem, so he's mixing 3 & 4 sig figs, and possibly 1 & 2 sig figures as well. If I remember correctly, one of the pieces of data was 30 m (1 sig figure), but was irrelevant, as I didn't need it for my answer (assuming I did it properly). Regardless, they could have used this as a nit-pick, giving all figures to 3 digits except for g, and then deducted a point for any answer expressed to 3 digits. I know the rules for significant digits are more complicated than this, but I've never had points deducted in any class for ignoring those rules and simply rounding the answer to the precision of the least precise input. Most teachers just don't want to see an answer straight off the calculator like 17.3745857483.

I don't have a problem with them telling me to use 9.81 for a specific question. I'll just pretend the experiment took place in Germany instead of San Francisco. As hhEb pointed out, they could have told me to use 11 m/s/s, in which case, I'd just assume I was on another planet. It's just merely a curiosity as to why 9.81 seems to be preferred over (or at least as often as) the more reasonable (IMHO) 9.8.


...unless Tony tells me his problem included a stipulated latitude.

Grant Hutchison

No stipulated latitude. This is a lower division class :)

I've asked a professor a few years ago if g was acceleration due to gravity, or net downward acceleration, in which case centripetal acceleration would have to be subtracted. I was told the latter. The first Wikipedia page I linked to gives a formula for computing g that takes latitude (and hence centripetal acceleration and distance from center of Earth) into account when computing g.

** In my OP I said that Jakarta was the lowest value in the table. Actually, its Mexico City. That makes sense as it's got appreciable elevation, as well as being tropical where it sits high on Earth's equatorial bulge, and experiences close to max centripetal acceleration.) Although not on the table, I'd bet Lake Titicaca in Peru beats Mexico City.

grant hutchison
2008-Sep-26, 07:33 PM
The formula page the professor gave us gives the density of water to be 1.000E3 kg/m^3, and 1.013E5 Pa for atmospheric pressure. These values were needed in the problem, so he's mixing 3 & 4 sig figs, and possibly 1 & 2 sig figures as well.Okay, there goes my best guess. :)


I've asked a professor a few years ago if g was acceleration due to gravity, or net downward acceleration, in which case centripetal acceleration would have to be subtracted. I was told the latter.Well, that's the difference between gn and the other values. The values for ge, gp and gφ all include the centrifugal component, as you say, whereas gn is just an agreed convention for unit conversion.

Grant Hutchison

Ken G
2008-Sep-26, 07:46 PM
"The approximate value of g that we're going to use for the purpose of this lab is 9.81 m/s^2" Yes, I'd say that's a fair statement of the problematic stance. The problem is twofold:
1) the approximate value adopts a third significant figure that is downright meaningless in most of the places where such a lab could be carried out. For labs intended to be used in the region where that value is actually defensible, that's another story-- the OP is about generic values.
2) the stance can quite naturally be interpreted as implying that the value of g is approximate in the way that further decimals have been truncated, but in fact that is not the sense to which that number is approximate.

Both of these issues are problems because if a student went into lab and attempted to verify this "law", and if they could achieve 3 sig figs of precision in their experiment, they could very well be led to believe falsely that there was a problem with the law itself, and not just the arbitrary over-reporting of precision relevant to the experiment in one of its parameters.


Yes, we did have the "aha" moment after the lab was completed where we discussed many of the factors that limited the accuracy and precision of our actual results vs. calculated predictions.Then the result is achieved by other means, which is fine. There's nothing wrong with a "strawman" approach to education, whatever works. The OP is reporting a different experience.
There was never any point that my teacher, at any rate, ever suggested g was a universal constant and certainly not that 9.81 m/s^2 was a precise value.Fine, but the thread is about the OP-- not about your teacher. The only part relevant to your teacher was the "shut up about precision issues" conversation, and you already clarified what that was really about. It is not my goal to besmirch your teacher, but rather to point to reasons why g should be generically taken as 9.8 and not 9.81.

George
2008-Sep-26, 11:49 PM
Sadly, I must confess I don't even know what that means! Ug. Kids, I love 'em! ;)

Prior to hand calculators was an ancient device somewhat superior to the Chinese abacus -- the slide rule (http://en.wikipedia.org/wiki/Slide_rule). In my day, all freshman engineers, regardless of species, were required to take Engineering 101 that spent the majority of the semester learning the use of this slidding apparatus. :)

HP hit the market in the early 70's, while I was in college, but it was horribly expensive (> $ 400 in 1970 dollars). Then came TI and prices dropped, and the rest is history. The slide rule now sits next to buggie whips.

What is significant about the slide rule is that 32.2 was the number to use for text problems because of the slide rule's limitations to the number of decimal places that could be observed on the stick. As Grant has pointed out, we have 3 significant places in the 9.81, which was the slide rule equivalent to 32.2. I would not be surprised if there lies the answer to the O.P. question (ignoring the other nuances).

Spaceman Spiff
2008-Sep-27, 12:53 AM
The quoted figures are significant-- the issue is, why do we choose to quote g=9.81 when we do physics problems? In most places that would apply such problems to experiment, that would be an incorrect value, to that many sig figs. So why not just back on the sig figs until everyone can actually use the value meaningfully? Note I don't even know if that is corrected for surface rotation, and I'll bet most others don't either, off hand.


Ok.
To the best of my knowledge the quoted values of g do include rotation, and various local effects (such as I mentioned, above).

As mentioned above, 9.81 m/s^2 is likely to be some sort of average, although it would be nice to know how that average was determined.

Spaceman Spiff
2008-Sep-27, 01:05 AM
This website (http://geophysics.ou.edu/solid_earth/notes/potential/igf.htm) is interesting, and has a link to useful excel spreadsheet for computing g.