[Orange (OP)]

Hello, the commonly accepted value of the Earth's gravitational acceleration at the poles is 9.83 m/s2, however when I calculate it from the Earth's polar radius (6356.752 km) I arrive at 9.864 m/s2.

What explains this difference?

Thanks !

[Janus]

The formula for Gravitational acceleration is GM/r^2 
G is the universal gravitational constant
M is the mass of the body in question (Earth for example)
r is the distance from the center.
The 9.83 m/s^2 is the accepted mean gravitational acceleration at the surface of the Earth(sea level)
Ergo, gravity will slightly weaker than that at the top of a mountain high above sea level, and slightly stronger in a valley below sea level.
As far as the pole goes, the Earth is not a perfect sphere, It is an oblate spheroid, and thus the Poles are closer to the center of the Earth than the average.  And  the actual value of g will be higher there. 
Then you have the Earth's spin, which makes the local "effective"(what you'd measure if you dropped an object) value of gravitational acceleration vary between poles and equator.

[Kryptid]

Perhaps you can show your math?

[Orange (OP)]

I indeed took this formula to calculate the gravitational acceleration GM/r^2 with the polar radius 6356.752 km and that gives me 9.864 m/s2.

I also did it for the equator by taking the equatorial radius (6378.137 km) and that gives me 9.798 m/s2.

On the internet it says:
Average gravity: 9.806 m/s2
Gravity poles: 9.832 m/s2
Gravity equator: 9,780 m/s2

I imagine that the difference between the gravity that I calculated for the equator and that given for the equator on the internet could be explained by the centrifugal effect of the Earth on the equator but I am not sure .
On the other hand, for the difference between the value that I calculated for the poles and that given for the poles on the internet, I don't really have an explanation.
I'm trying to figure out what could explain this difference?

[Halc]

I indeed took this formula to calculate the gravitational acceleration GM/r^2 with the polar radius 6356.752 km and that gives me 9.864 m/s2.

The equation GM/r² works only for spherically symmetric masses, which Earth isn't. So there's mass near the equator that is outside that 6356 radius and thus contributes less than it would were it packed into a smaller space of a r=6356 sphere.

I also did it for the equator by taking the equatorial radius (6378.137 km) and that gives me 9.798 m/s2.
...
Gravity equator: 9,780 m/s2

I don't know where you got those figures, but it seems like something as measured as weight of a known mass object stationary on the equator as opposed to acceleration relative to some inertial frame. Yes, the difference is the subtraction of the centripetal acceleration you get from Earth's spin.
Also, once again, the Earth under you is not a spherically symmetric distribution of mass, so the GM/r² formula doesn't work directly. With Earth, gravity goes up as you measure it in a hole dug considerably below sea level. Given a spherical mass with uniform mass density, it would immediately go down if measured below the surface like that. But the density/depth curve has little to do with surface gravity, at least for a spherical object that doesn't spin, but Earth is neither.

[alancalverd]

Where did you get your value for M? Is it the same as what you would calculate if you began with the values for r and g used by whoever produced the internet values?   

[Orange (OP)]

Where did you get your value for M? Is it the same as what you would calculate if you began with the values for r and g used by whoever produced the internet values?

I used this value for M: 5.97217×10^24 (Wikipédia) but I don't know what value was used to calculate the results I saw on the internet.

[Orange (OP)]

I don't know where you got those figures

The values I saw on the internet come from here: I can't send a link but it's on Wikipedia, Theoretical gravity.

[alancalverd]

The problem you have is that M is only calculated by measuring the weight of a standard mass at a point where you have also measured g. Since g varies all over the place, and the barycenter of M is not the geometric center of a symmetrical shape, any subsequent estimates of g at any point are subject to very significant systematic errors. Which is why, when it matters, we measure g rather than calculate it.

[Orange (OP)]

The problem you have is that M is only calculated by measuring the weight of a standard mass at a point where you have also measured g. Since g varies all over the place, and the barycenter of M is not the geometric center of a symmetrical shape, any subsequent estimates of g at any point are subject to very significant systematic errors. Which is why, when it matters, we measure g rather than calculate it.

But then, is there another way to measure the mass of the Earth (M) which does not require measuring g at the beginning and which would allow us to know the real total mass of the Earth?

[evan_au]

is there another way to measure the mass of the Earth (M)

The standard way used by astronomers is to measure the orbital period of a small satellite.

The Moon is far enough away so that it won't be affected very much by centripetal bulges.
- Unfortunately, at 1/80 the Earth's mass, it doesn't really qualify as "small"
- You could look at Geostationary satellites, which qualify as "small" (compared to the Earth), and are still a fair distance away from the Earth's midline bulge

Unfortunately, this still involves the use of "G", which remains one of the poorest measured fundamental constants, certain to to only 4 significant figures.

[alancalverd]

So we assume G is constant, if not well-known, and use the wobbles in satellite altitude (i.e. g) to identify departures of the geoid from a symmetric oblate spheroid. The departures are sufficiently important and both spatially and temporally variable that GPS-based instrument approaches to runways are revised at regular intervals when the geoid is recalculated.

[Orange (OP)]

If I understood correctly, the difference between my values and those I have seen regularly on the internet is explained because the GM/r² formula works for a spherical symmetric mass and does not work for a planet which is oblate like the Earth and also because the value of M does not represent the real value of the mass of the Earth because it is calculated from the value of g which varies at different places on the surface of the Earth.

On the other hand, concerning the last reason, I think but I am not sure, the uncertainty of the mass of the Earth could be reduced if it were calculated with the orbital period of a geostationary satellite as explained by evan_au (instead of the weight of a mass on the surface of the Earth) and then in the formula we would no longer have the value of g which varies everywhere on the surface of the Earth, in the formula, there would remain only the uncertainty of the value of G. So we would reduce the uncertainty of the value of the real mass of the Earth (the value of M would be more accurate) ?