Question: There are plenty of formulas that use gravity acceleration of Earth. This is represented with the symbol $g$. In my school work (I am a high school student) we usually take it as $g= 9,8 \,\text m/\text s^2$. 

This thing is obviously a number that is only usable on Earth. What I want to know is that, what if I want to make my calculations according to another planet? How the number is going to change?

Answer: 
The gravitational acceleration constant defined as $g$ for earth depends on the mass of the earth and the distance from it. The formula is $g(r) = \frac{GM(r)}{r^2}$. (See [Newtons Law of Universal Gravitation](http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation) for more detail). So $g$ is not a constant even on earth but depends on your altitude, albeit rather slowly.
If you are in the moon, the mass of the moon $(~10^{22} kg)$ is less than that of earth $(~10^{24} kg)$ and thus the gravitational force you would feel, $mg$ would be far less due to the $g$ being smaller, about $1.62 m/s^2$. 

Also, the units of $g$ are $m/s^2$ and not $N/s^2$
An easy way to think about this is to consider that the acceleration of gravity, at the surface of, say, a planetary body, essentially depends on two quantities: the mass of the body and the radius.

The surface acceleration increases with the mass of the body (if you double the mass, you double the acceleration) and decreases with square of the radius (if you double the radius , the acceleration is quartered).

So, for example, the Moon's radius is about 0.273 times the Earth's radius but the Moon's mass is about 0.0123 the Earth's mass. So, we would expect the acceleration at the surface of the Moon to be

$g\_m = g\_e (.0123)\dfrac{1}{(.273)^2} \approx \dfrac{g\_e}{6}$

and, sure enough, the surface gravity of the Moon is about $1.62\frac{m}{s^2}$

So, if you know the mass and radius of, say, Mars, you can determine the surface gravity of Mars as follows:

$g\_M = g\_e\dfrac{M\_M}{M\_e} \cdot \dfrac{R^2\_e}{R^2\_M}$