No uncertainty for standard gravitational acceleration?

Question

The other day I asked about the uncertainty of light, and this issue triggered me to start looking into other physical constants and try to understand why other constants have no uncertainty.

One of those constants is the standard gravitational acceleration, which is the acceleration due to Earth's gravity ($\approx 9.8\: \mathrm{m/s^2}$)... which is also a physical quantity for which I could easily suggest a lot of reasons why it should have uncertainties, like:

• The meter itself has uncertainty,
• The radius of the Earth is different everywhere

Why doesn't $g$ have uncertainty?

I would like to point out there are other constants with no uncertainty (like permittivity of free space) that I don't understand. Isn't there a collective reference to explain those conventions?

EDIT:

Maybe I wasn't clear. Check the latest review from Physical Review D, and you'll see that the uncertainty is given as zero at page 108. And also from CODATA NIST:

Answer 1

The typical gravitational acceleration on the surface of the Earth, $g \approx 9.8\: \mathrm{m/s^2}$, has uncertainty. That's one of the reasons why the $\approx$ symbol is used.

The Earth's gravitational field varies a lot due to oceans, the thickness of the crust, mountains, non-uniform density in the crust and mantel, etc.

A pair of satellites was launched for the Gravity Recovery and Climate Experiment (GRACE) and based on that data a map was made showing the variation:

From Wikipedia:

Apparent gravity on the earth's surface varies by around $0.7\%$, from $9.7639\: \mathrm{m/s^2}$ on the Nevado Huascarán mountain in Peru to $9.8337\: \mathrm{m/s^2}$ at the surface of the Arctic Ocean.

As others have mentioned, the constant of gravitational acceleration, $g_0$ that is defined exactly as $9.80665\: \mathrm{m/s^2}$ is used for the standardization of weight like the pound against units of mass like the kilogram.

Answer 2

The table cited is giving the definition of a unit called the standard acceleration of gravity. It happens that the acceleration of gravity on various places on earth is close to 1 when measured in these units, but that does not change the exactness of the definition.

By analogy. One foot is exactly 12 inches, despite the varying size of various human feet found in the population...

Answer 3

As a first order approximation the Earth can be considered a sphere, if you'd like a little bit more accurate approximation you can represent it an ellipsoid of revolution (meaning two of the three axis have equal length and it rotates about the third, unequal axis). Treating the Earth as an ellipsoid -- and therefore it's gravity field as an ellipsoid -- makes the math a lot more tractable and the differences between this ellipsoidal gravity field and the actual gravity field are actually so small that they can be treated as linear. This treatment leaves you with a "normal" gravity field and a "disturbing" gravity field and makes determining it through observations a lot more simple.

The current most commonly used reference ellipsoid is WGS 84 and is completely described by four parameters:

Parameter and Value	Description
$a = 6,378,137 \: m$	semi-major axis of the ellipsoid
$f = 1/298.257223563$	ellipsoidal flattening
$GM = 3,986,004.418 $x$ 10^{8}m^{3}s^{-2}$	geocentric gravitational constant of the Earth (including atmosphere)
$\omega = 7,292,115$x$10^{-11}rad\:s^{-1} $	Earth's angular velocity

These parameters are constrained through fitting the ellipsoid to empirical observation. In the satellite era where we have global data coverage from direct observations made by missions like GRACE, GOCE, etc., along with decades of terrestrial measurements, and satellite altimetry, these parameters are considered known. Note, this is for the reference ellipsoid not Earth. The value for Earth's acceleration of gravity $g \approx 9.8\: m\:s^{2}$ you referenced (and that is commonly referenced) is the average acceleration of normal gravity, or $g_{n} = 9.806\: m\:s^{2}$. This is a value that is referenced to the ellipsoid and is the theoretical value at 45 degrees latitude on the reference ellipsoid. This accuracy is sufficient for many applications.

So, in short, $g_{n}$ is known and has no uncertainty but $g_{n} \neq g_{actual}$ on Earth's actual surface. Typically $g_{actual}$ has many sources of uncertainty.

Another interesting thing to note is that we know $GM$ very well, but not so much $G$ or $M$.

Answer 4

You're actually looking at the "standard gravitational acceleration," $g$, which is not the same thing as the actual acceleration of gravity. This constant $g$ is used to provide a consistent definition of certain units of force, for example the pound force (lbf), so it needs to be a defined constant, not a measurement. $g$ was chosen to be close to the acceleration of gravity on the Earth's surface (for obvious reasons), but other than that it has nothing to do with the actual strength of the Earth's gravitational field.

In particular, $g$ does not reflect the most up-to-date measurements of the strength of Earth's gravity. Don't let its appearance in the PDG Review (your PRD reference) fool you: that number has been the same for decades. If you go back and check it in previous versions of the review, you will find the same value.