Why is Earth's gravitational acceleration $9.8 \frac{m}{s^2}$? How was the value of $g$ determined as 9.8 $\frac{m}{s^2}$?
I am not requesting the derivation but the factors/parameters that influence this value.
 A: $$g=\frac{GM}{r^2}$$ so the factors/parameters are Newton’s universal gravitational parameter, the mass of the earth, and the radius of the earth. 
A: One thing I'd like to add to these excellent answers is that although you didn't ask for a derivation, I think it is necessary in understanding the $g = \frac{GM}{r^2}$ equation.
The force due to gravity between two masses is given by Newton's law of universal gravitation;
$F=G\frac{m_{1} m_2}{r^2}$, where;
$F$ = Force between the two masses
$G$ = Gravitational constant
$m_1$ = Mass of object 1 (In this example, this can be the mass of the Earth)
$m_2$ = Mass of object 2 (In this example, this can be the object on the surface of the Earth)
$r$ = distance between the centre of the two masses (In this example, this will be the radius of the Earth - assuming the object is on the surface of the Earth)
Therefore, the acceleration of the object on the surface of the Earth can be found using Newton's second law;
$F=m_2a_2$. Therefore,
$m_2a_2 = G\frac{m_{1} m_2}{r^2}$, and this simplifies to;
$a_2 = G\frac{m_{1}}{r^2}$.
Then, the acceleration can be replaced with $g$, the mass of the Earth can be replaced with $M$ and this forms the equation;
$g=\frac{GM}{r^2}$.
This is the reason why the acceleration at the Earth's surface is always 9.81 $m{s^{-2}}$, regardless of the mass of the other object. Therefore, in answer to your question, the only factors affecting the acceleration due to gravity at the Earth's surface are;


*

*The gravitational constant

*The mass of the Earth

*The radius of the Earth (if we assume the centre of mass is exactly in the centre of the Earth)

A: 9.8 m/s^2 is not 'the Earth's gravity', it's the at-mean-sea-level acceleration
due to Earth's gravity.   The acceleration would be quite different
if measured elsewhere, like at the lunar orbit.
The effective gravity constant also varies due to local mineral density,
and latitude.
Earth's gravity, in the universal sense, is entirely characterized
by the mass of the planet, roughly 5.97 *10^(24) kg,
To calculate acceleration, multiply that by the universal gravity constant G and divide by the
square of the distance from the center of the
planet.   Only if you pick Earth's radius does that give the 9.8 m/s^2 value.
