Hi Ronald Wieck.
The strenght of a gravitational field has the general form of:
$$g=\frac{GM}{R^2}$$
This is just the Newtonian gravitational force divided by mass of the object being accelerated, which incidentally means that it doesn't matter how massive an object is - all are accelerated at the same rate (but read on). It's the same as acceleration caused by the mass in the absence of any other forces.
The only actual constant in that equation is the gravitational constant G. The mass M and distance from its centre R depend on what is pulling you in, and how far from it you are standing.
Now, since most of the people that have ever lived are interested in acceleration on Earth's surface, we can just plug in Earth's mass and the distance from Earth's centre to the surface.
Earth's mass is the same for everybody, although denser mass concentrations below some areas of the surface have some tiny, yet measurable effect.
The radius varies depending on where you're standing - on the summit of Mt Everest or on the shores of Dead Sea, on one of the poles or on the equator.
However, the difference is not that great when compared to the average radius of the Earth. Looking at the equation above, it doesn't matter a whole lot if you divide GM by say, 6 370 000 m, squared, or by 6 378 000 m, squared (roughly, the height of Mt Everest compared to sea level). If you're thinking of differences in the range of a few dozen metres it becomes even more negligible.
That's why, for most intents and purposes, when one is talking about the acceleration from Earth's gravity on its surface, he can say it's everywhere the same and equals g (9.81m/s).
It's an approximate statement, and one needs to remember that it works only for the special case of the vicinity of the surface of the earth. In this sense it is constant. If you're a scientist whose work is to map the variation of g with very precise instruments, then of course, by definition, you won't treat it as constant. If you're plotting a trajectory of a space ship, or modelling its strength below the surface (what AT's graph shows), then you won't treat it as constant. If you're a regular shmoe, then it's as good as constant, and treating it otherwise without any good reason is pretty much pointless.
By the way, air resistance has no bearing on the strength of the gravitational field. It does have a bearing on the NET acceleration (like a parachutist at some point reaches 0 acceleration).
If you're asked to compute the acceleration due to gravity alone, you use the equation provided, or just use 9.81 if you're on Earth and not concerned with super-high accuracy. If you're asked to compute net acceleration (like with a projectile motion with air resistance - a standard exercise for college freshmen studying physics), then the acceleration will vary significantly. You'd still use the gravitational acceleration (treated as constant) in there, but that's a whole different kettle of fish.
