## Why is a standard g exactly 9.80665m/s/s?

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### Why is a standard g exactly 9.80665m/s/s?

I've wondered this for a while, but was reminded of it recently by this thread in Logic Puzzles. The strength of gravity varies over the surface of the Earth - not by a lot, but by as much as .04m/s2, or about one part in 200. So why is a standard gravity defined to be a precise value whose significant digits would indicate a precision of one part in a million? Is the quantity 9.80665m/s2 somehow more physically meaningful than 9.8 or 9.81 would be? If not, what quirk of history caused it to be this bizarre value rather than something rounder?

On a related note, it also bothers me when people use that exact number to compute a trajectory, since the added "precision" it imparts is wholly imagined.

In the thread that prompted this one, phlip and I had the following exchange:
skeptical scientist wrote:While we're on the subject, does anyone know why standard gravity is defined to be exactly 9.80665? Is this precise value in any way physically meaningful? If it isn't, why don't we use the much simpler value of 9.8 for standard gravity?

phlip wrote:Probably for the same reason that the speed of light isn't quite 3e8 m/s, and standard atmospheric pressure is over 1e5 Pa... and, for that matter, why an inch isn't 2.5cm... inertia, and the desire not to change a value too much when it gets redefined in terms of something else.

Presumably, when they decided to define standard gravity as a specific value in m/s/s, they just took the previous value for g, however that was defined, rounded it off to more digits than you'd ever really need, and used that.

I don't buy this argument, especially the comparison to the speed of light. Yes, the meter is now defined to be exactly 1/299,792,458 of the distance light travels in vacuum in one second, so one could say that the speed of light is 299,792,458 by definition, and one could rather have defined it to be 300,000,000. But before this was the definition of the meter, the length of a meter was defined by the length of a particular platinum bar held at the International Bureau of Weights and Measures. Also before the speed-of-light-definition of the meter, the speed of light in vacuum had been measured to 4 parts per billion, and was known to be 299,792,458±1 times the length of that platinum bar in a second. So when the meter was defined by a prototype meter bar, the speed of light really was 299,792,458 m/s, and that precise number had physical meaning. When the meter was redefined from the speed of light as opposed to from a prototype, it made sense to keep the same value, so as not to change the length of the meter.

I don't see how a similar argument can be made for a standard gravity. I understand how the number 299,792,458 used for the speed of light is physically meaningful, but I don't understand the physical significance of 9.80665 m/s2. Moreover, I don't see how it can be physically significant. It can't possibly be the exact force of gravity on the surface of the Earth, since that value varies between 9.78 m/s2 and 9.82 m/s2 based on location. So what, if anything, does it measure?
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### Re: Why is a standard g exactly 9.80665m/s/s?

Perhaps it's an average over a long time interval, over the surface of the earth.
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### Re: Why is a standard g exactly 9.80665m/s/s?

Wikipedia is, as usual, your friend:
The article on standard gravity wrote:The value of g0 defined above is a nominal midrange value on Earth, representing the acceleration of a body in free fall at sea level at a geodetic latitude of about 45.5°.

Apparently, though, the specifics are somewhat of a QWERTY-style phenomenon, where several countries happened to already have that value in their laws when it was made official.
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### Re: Why is a standard g exactly 9.80665m/s/s?

gmalivuk wrote:Wikipedia is, as usual, your friend:
The article on standard gravity wrote:The value of g0 defined above is a nominal midrange value on Earth, representing the acceleration of a body in free fall at sea level at a geodetic latitude of about 45.5°.

Apparently, though, the specifics are somewhat of a QWERTY-style phenomenon, where several countries happened to already have that value in their laws when it was made official.

Yeah, I saw that when I tried to google the answer for myself, but it doesn't really answer the question. It's a nominal midrange, but 9.8 would serve that purpose just as well, and I fail to see what's special about a latitude of "about 45.5º". I also found out when it was chosen: the 1901 meeting of the General Conference of Weights and Measures.
CGPM, 3rd meeting wrote:The value adopted in the International Service of Weights and Measures for the standard acceleration due to gravity is 980.665 cm/s2, value already stated in the laws of some countries.

However, this rather vague statement doesn't explain why it was chosen, just that it was "already stated in the laws of some countries" with no mention of which countries, or why they had adopted it.
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### Re: Why is a standard g exactly 9.80665m/s/s?

The 1930 International Gravity Formula gives 9.80635 as the average of its value at the equator and at the poles. So perhaps a prior formula gave .0003 more than this, and that's where they got the number?
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### Re: Why is a standard g exactly 9.80665m/s/s?

Its kind of ridiculous. With slightly varying g, what you do is not to designate a standard g with 6 significant digits, you express g with the appropriate uncertainty for the value to be correct everywhere. That is, g = 9.8(2) m/s2.
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### Re: Why is a standard g exactly 9.80665m/s/s?

Does NASA have to use a particular value of g, calculated/measured for their launch site for launches?

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### Re: Why is a standard g exactly 9.80665m/s/s?

Probably, but the 1901 official value likely didn't take that into consideration...
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### Re: Why is a standard g exactly 9.80665m/s/s?

BlackSails wrote:Does NASA have to use a particular value of g, calculated/measured for their launch site for launches?

Since they travel far enough from the surface for error to be small albeit significant, I doubt they use small g at all.
Last edited by You, sir, name? on Wed Mar 17, 2010 11:05 pm UTC, edited 1 time in total.
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### Re: Why is a standard g exactly 9.80665m/s/s?

It does make sense to define a standard gravity, for the same reason as you define standard pressure - it is important to have a reference. It is often used as a unit of acceleration, so it is obviously important that it has a standard value. But giving the standard value 6 digits when none but the first two are physically meaningful makes no sense to me.
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Incidentally, this is wrong:
In the other thread, skeptical scientist wrote:the centrifugal acceleration at the equator...is less than .001 m/s2.

Acceleration around a circle is not radius/period^2. It's v^2/r, which works out to 4pi^2 times the figure you quoted.
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### Re: Why is a standard g exactly 9.80665m/s/s?

Doh! Actually the formula I used was [imath]\omega^2r[/imath]; I just forgot that angular velocity has units of radians per second, not rotations per second.
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### Re: Why is a standard g exactly 9.80665m/s/s?

skeptical scientist wrote:It does make sense to define a standard gravity, for the same reason as you define standard pressure - it is important to have a reference. It is often used as a unit of acceleration, so it is obviously important that it has a standard value. But giving the standard value 6 digits when none but the first two are physically meaningful makes no sense to me.

http://discovermagazine.com/2007/mar/gr ... ld-700.jpg

1 mGal = 1x10-5 m/s2

Most areas on Earth (at a given elevation) seem to have gravitational anomalies of -40 to +40 mGal, or on the order of 1x10-4 m/s2... so precision out to five decimals (or at least four) seems pretty reasonable to me.

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### Re: Why is a standard g exactly 9.80665m/s/s?

You, sir, name? wrote:Its kind of ridiculous. With slightly varying g, what you do is not to designate a standard g with 6 significant digits, you express g with the appropriate uncertainty for the value to be correct everywhere. That is, g = 9.8(2) m/s2.

No. Standard gravity is not a measured quantity, it is a defined quantity. Expressing it to six digits is simply a shorthand; the actual value is exactly 9.8066500000000000000...m/s/s.

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### Re: Why is a standard g exactly 9.80665m/s/s?

Carnildo wrote:No. Standard gravity is not a measured quantity, it is a defined quantity. Expressing it to six digits is simply a shorthand; the actual value is exactly 9.8066500000000000000...m/s/s.

Sure, but the question is why it was defined as that, and not 9.800000000000... m/s/s. Since that value is much easier to remember, and is just as good an approximation for the actual gravity in any given place on the Earth as the actual value.

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### Re: Why is a standard g exactly 9.80665m/s/s?

I guess because the "standard" it was based on was the average between the poles and the equator - albeit a bit out by modern reckoning.

Rounding it off wouldn't make it relevant to any particular physical phenomenon, and seems rather unscientific.

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### Re: Why is a standard g exactly 9.80665m/s/s?

Fume Troll wrote:Rounding it off wouldn't make it relevant to any particular physical phenomenon, and seems rather unscientific.
But leaving it at the rather random number it's at doesn't make it any more relevant to any physical phenomena, and isn't any more scientific...

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### Re: Why is a standard g exactly 9.80665m/s/s?

Fume Troll wrote:I guess because the "standard" it was based on was the average between the poles and the equator - albeit a bit out by modern reckoning.

That's still only a guess as far as this thread is concerned, unless someone else has come across actual references to that fact.

phlip wrote:But leaving it at the rather random number it's at doesn't make it any more relevant to any physical phenomena, and isn't any more scientific...

Well sure, but I don't see how rounding it off outweighs the disadvantages of fucking with an international standard that's been in place for over a century.
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### Re: Why is a standard g exactly 9.80665m/s/s?

Just a thought: an exactly defined g is important for laws and standards (and contracts!) that for some reason define quantities of matter as a weight. You want to be able to buy so many tonnes or pounds without having to worry whether those are polar tonnes or equatorial tonnes, or going into discussing whether those tonnes were intended really as a mass or a weight. So under those circumstances I can imagine that countries defined a fixed g before an international standard was chosen.

Also, "9.8" is only simple in a metric system. I can imagine that our beloved inchy friends across the waters thought that if they were going to get a weird number, so was everyone else.

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### Re: Why is a standard g exactly 9.80665m/s/s?

No, because it's still only a terminating value in metric, and not in inches or feet.
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### Re: Why is a standard g exactly 9.80665m/s/s?

phlip wrote:Sure, but the question is why it was defined as that, and not 9.800000000000... m/s/s. Since that value is much easier to remember, and is just as good an approximation for the actual gravity in any given place on the Earth as the actual value.

And the answer is that it's not just as good an approximation.

The OP was incorrect about the variability of gravity at a given elevation over the Earth's surface.

MallowTheCloud wrote:http://discovermagazine.com/2007/mar/grace-in-space/world-700.jpg

1 mGal = 1x10-5 m/s2

Most areas on Earth (at a given elevation) seem to have gravitational anomalies of -40 to +40 mGal, or on the order of 1x10-4 m/s2... so precision out to five decimals (or at least four) seems pretty reasonable to me.

Yes, elevation plays a much larger role, but there's a precise equation to get the force of gravity given standard g and your elevation, so that doesn't fly either. Standard g should have at least 5 significant digits, and 6 seems reasonable as well.

EDIT: Latitude also plays a significant role (due to "centrifugal force"), but again, this can be calculated precisely.

So, basically, a "standard gravity" will generally be accurate to at least four decimal places after you take latitude and altitude into consideration.

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### Re: Why is a standard g exactly 9.80665m/s/s?

gmalivuk wrote:No, because it's still only a terminating value in metric, and not in inches or feet.

That's not actually correct. 1 inch is exactly 2.54 cm, and I get this in clisp:

Code: Select all

[20]> (format t "~10,100F" (/ (/ (* 980665/100000 100) 254/100) 12))32.1740500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000NIL[21]>

fractions in lisp are exact, and I only converted to floats during printing.

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### Re: Why is a standard g exactly 9.80665m/s/s?

MallowTheCloud wrote:
Spoiler:
phlip wrote:Sure, but the question is why it was defined as that, and not 9.800000000000... m/s/s. Since that value is much easier to remember, and is just as good an approximation for the actual gravity in any given place on the Earth as the actual value.

And the answer is that it's not just as good an approximation.

The OP was incorrect about the variability of gravity at a given elevation over the Earth's surface.

MallowTheCloud wrote:http://discovermagazine.com/2007/mar/grace-in-space/world-700.jpg

1 mGal = 1x10-5 m/s2

Most areas on Earth (at a given elevation) seem to have gravitational anomalies of -40 to +40 mGal, or on the order of 1x10-4 m/s2... so precision out to five decimals (or at least four) seems pretty reasonable to me.

Yes, elevation plays a much larger role, but there's a precise equation to get the force of gravity given standard g and your elevation, so that doesn't fly either. Standard g should have at least 5 significant digits, and 6 seems reasonable as well.

EDIT: Latitude also plays a significant role (due to "centrifugal force"), but again, this can be calculated precisely.

So, basically, a "standard gravity" will generally be accurate to at least four decimal places after you take latitude and altitude into consideration.

If you take latitude and altitude into consideration, you are no longer using standard gravity.
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### Re: Why is a standard g exactly 9.80665m/s/s?

scarecrovv wrote:
gmalivuk wrote:No, because it's still only a terminating value in metric, and not in inches or feet.

That's not actually correct. 1 inch is exactly 2.54 cm, and I get this in clisp:

Code: Select all

[20]> (format t "~10,100F" (/ (/ (* 980665/100000 100) 254/100) 12))32.1740500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000NIL[21]>

fractions in lisp are exact, and I only converted to floats during printing.

What now?

Are you claiming that 980665/254 or 980665/3048 are terminating decimals? Because they're definitely not, and I'm furthermore not sure why you obfuscated that with your bit of code there when WolframAlpha would have told you quite easily.
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### Re: Why is a standard g exactly 9.80665m/s/s?

You, sir, name? wrote:
MallowTheCloud wrote:
Spoiler:
phlip wrote:Sure, but the question is why it was defined as that, and not 9.800000000000... m/s/s. Since that value is much easier to remember, and is just as good an approximation for the actual gravity in any given place on the Earth as the actual value.

And the answer is that it's not just as good an approximation.

The OP was incorrect about the variability of gravity at a given elevation over the Earth's surface.

MallowTheCloud wrote:http://discovermagazine.com/2007/mar/grace-in-space/world-700.jpg

1 mGal = 1x10-5 m/s2

Most areas on Earth (at a given elevation) seem to have gravitational anomalies of -40 to +40 mGal, or on the order of 1x10-4 m/s2... so precision out to five decimals (or at least four) seems pretty reasonable to me.

Yes, elevation plays a much larger role, but there's a precise equation to get the force of gravity given standard g and your elevation, so that doesn't fly either. Standard g should have at least 5 significant digits, and 6 seems reasonable as well.

EDIT: Latitude also plays a significant role (due to "centrifugal force"), but again, this can be calculated precisely.

So, basically, a "standard gravity" will generally be accurate to at least four decimal places after you take latitude and altitude into consideration.

If you take latitude and altitude into consideration, you are no longer using standard gravity.

True. If there were equations which allowed you to use g to more simply calculate acceleration due to gravity at any latitude or altitude, then it would be worthwhile to keep the precision. However, it seems the equations to approximate the effects of latitude and altitude don't even involve "standard gravity", so I can't really defend the precision.

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### Re: Why is a standard g exactly 9.80665m/s/s?

gmalivuk wrote:
scarecrovv wrote:
gmalivuk wrote:No, because it's still only a terminating value in metric, and not in inches or feet.

That's not actually correct. 1 inch is exactly 2.54 cm, and I get this in clisp:

Code: Select all

[20]> (format t "~10,100F" (/ (/ (* 980665/100000 100) 254/100) 12))32.1740500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000NIL[21]>

fractions in lisp are exact, and I only converted to floats during printing.

What now?

Are you claiming that 980665/254 or 980665/3048 are terminating decimals? Because they're definitely not, and I'm furthermore not sure why you obfuscated that with your bit of code there when WolframAlpha would have told you quite easily.

Oh dear, it seems you're correct! I didn't think to use Wolfram Alpha, and I assumed format would print fractions as decimals with as much precision as I asked for. Oh well.

Never mind me.

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### Re: Why is a standard g exactly 9.80665m/s/s?

gmalivuk wrote:
Fume Troll wrote:I guess because the "standard" it was based on was the average bewteen the poles and the equator - albeit a bit out by modern reckoning.

That's still only a guess as far as this thread is concerned, unless someone else has come across actual references to that fact.

Fair enough.

That's what's suggested here: http://en.wikipedia.org/wiki/Kilogram_force

and is here: http://mtp.jpl.nasa.gov/notes/altitude/altitude.html
edited for quotes.
Last edited by Fume Troll on Mon Mar 22, 2010 2:44 pm UTC, edited 1 time in total.

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### Re: Why is a standard g exactly 9.80665m/s/s?

Fume Troll wrote:That's what's suggested here: http://en.wikipedia.org/wiki/Kilogram_force

To be fair, that only says that standard gravity is "a conventional value approximating the average magnitude of gravity on Earth". This is true of lots of other values, including 9.8.

Interestingly, that says "The original definition of $$\gamma_{45}$$ was 9.8 m/s2; however, this was changed in the US in 1935 to the current value: $$\gamma_{45}$$=9.80665 m/s2. Finally, the earth's gravity model has been modified slightly since this most recent definition of $$\gamma_{45}$$, so $$\gamma_{45}$$ is now the gravity at 45.542 degrees instead of 45.0 degrees." So originally the US had the value which makes more sense to me, and then changed it to the one which makes less sense (presumably to bring it in line with the international standard, which leaves open the question of why the international standard was the strange value in the first place. It also suggests that even though 9.80665 m/s is now the gravity at 45.542º, at one point it may have been the gravity at 45º exactly. Unfortunately, it doesn't actually say this.
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### Re: Why is a standard g exactly 9.80665m/s/s?

More here: http://www.numericana.com/answer/units.htm#g
Spoiler:
To an actual measurement of 9.80991 m/s2 in Paris, a theoretical correction factor of 1.0003322 was applied which gives a sea-level equivalent at 45° of latitude. The result (9.80665223...) was rounded to five decimals to obtain the value officially enacted by the third CGPM, in 1901.

The above includes a centrifugal component due to the rotation of the Earth, whereas the gravitational field at altitude zero has a slightly larger value, used when tracking satellites outside the atmosphere in nonrotating coordinates (9.82025048(2) m/s2 ) which is the ratio of the Earth's gravitational constant (3.986004415(8) 1014 m3/s2 ) to the square of the conventional radius of the Earth (R = 6371000 m).

Shame I can't find the conference proceedings.

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### Re: Why is a standard g exactly 9.80665m/s/s?

Well even without conference proceedings, you did at least finally find something that said explicitly how the number was arrived at. So thanks for that.

I wonder why the Internet made this particular question so difficult to actually pin down...
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### Re: Why is a standard g exactly 9.80665m/s/s?

Thanks, fume troll.
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