How to arrive to / calculate gravitational acceleration on Earth? What are the exact steps and figures to calculate gravitational acceleration on Earth (9.8 m/s^2) and where do these figures come from?
(Please use an algebraic explanation, but assume understanding of the principles of calculus without being handy with the notation.)
I'm assuming (getting some figures from WolframAlpha) we're looking at something along the lines of these for values:

G = 6.674×10^-11 
m(earth) = 5.9721986×10^24 kg
d(r of earth) = 6371.0088 km

As far as the simplest answer, I'd be happy with some pointers to solid reference material in addition to a simple walkthrough of the equation.
 A: From a Newtonian perspective, every particle of mass ($m_j$) produces a gravitational field in the space surrounding it:
$$\vec{g}_j=\dfrac{-Gm_j}{r^2}\hat{r},$$
where $r$ is the distance from the mass to the point in space you're interested in, and $\hat{r}$ is a unit vector pointing from the mass toward the point in space.
If you have several masses, the gravitational field at a point is the sum of all the individual fields from the individual masses. If you place a separate mass, $M$, at that point in space, it will experience a force:
$$\vec{F}_{\mathrm{on }M}=M \sum_j \vec{g}_j=M\vec{g}_{\mathrm{net}}.$$
As a consequence of applying Newton's 2nd Law to this force, we find that the acceleration is equal to the gravitational field:$$\vec{a}=\vec{g}_{\mathrm{net}}.$$
For spherically symmetric mass distributions it turns out (see Gauss's law for gravity) that the net gravitational field (that sum) is equal to the field of a single, large mass as the same location as the center of the distribution. In other words, for a planet or a star, the gravitational field at a distance $r$ from the center of that object is simply
$$\vec{g}=\dfrac{-Gm_{\mathrm{total}}}{r^2}\hat{r}.$$
This holds for any distance $r$ larger than the radius of the sphere, so not only can you use it to find the gravitational field at the surface of Earth, but also 2000 km above the surface (where it will be less).
A: One straightforward way is the use the Universal Gravitation Law, where M1 = mass of Earth and M2 = mass of object being accelerated by the earth.
F = (G)(M1)(M2) / r^2
Here we divide M2 by both sides:
F / M2 = (G)(M1) / r^2
Since we know F = Ma, F / M2 = a, which is the acceleration due to gravity by Earth:
a = (G)(M1) / r^2
Now you simply plug in your values for G, M1, and r.
G is the universal gravitation constant, which is just a number that was derived to solve gravitational attraction problems.
r is the distance from both masses' centres.  If your object was falling at the surface of the earth, r = Earth's radius (which is the distance from Earth's surface to the centre).  If your object was, let's say, 5000 km above the surface of Earth, r = 5000 km + radius of Earth.
Hope that helps.
