What do you mean by acceleration due to gravity?

Hint: The acceleration due to gravity is the acceleration gained by an object due to the attractive gravitational force between two objects. The S.I unit of gravity is the same as that of acceleration, hence $m{{s}^{-2}}$. The acceleration due to gravity of earth is a standardized value denoted by (g). The standard value of g at mean sea level is $9.8\ m{{s}^{-2}}$.


Complete step by step answer:
As per the definition, acceleration due to gravity is the acceleration gained by an object due to the attractive gravitational force between two objects. The S.I. unit of gravity is the same as that of acceleration, hence $m{{s}^{-2}}$.
Further on, acceleration due to gravity of earth is a standardized value denoted by (g). The standard value of g at mean sea level is $9.8\ m{{s}^{-2}}$.
Let us understand how, do we find the value of acceleration due to gravity of earth and in general.
Let us consider a heavy body of mass (M) and another lighter body of mass (m). These two bodies are separated by a distance (R) in between them.
The gravitational force acting on both the bodies will be given by: ${{F}_{G}}=\dfrac{GMm}{{{R}^{2}}}$, where (G) is the universal gravitational constant, given by: \[G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}\]. The gravitational force is of attractive nature, hence the gravitational force will act on the bodies such that the distance between these two bodies will reduce.
We will equate the gravitational force with the general force formula, which states that force is equal to mass times the acceleration of the body. That is: $F=ma$.
Hence, the acceleration of the lighter body (m) towards the heavier body of mass (M) is given by: ${{F}_{G}}=F\Rightarrow \dfrac{GMm}{{{R}^{2}}}=m{{a}_{m}}\Rightarrow {{a}_{m}}=\dfrac{GM}{{{R}^{2}}}$.
Similarly, the acceleration of the heavier mass (M) will be: ${{F}_{G}}=F\Rightarrow \dfrac{GMm}{{{R}^{2}}}=M{{a}_{M}}\Rightarrow {{a}_{M}}=\dfrac{Gm}{{{R}^{2}}}$.
If the mass of the lighter object is much less than the heavier object, then the acceleration of the heavier mass object can be approximated to be zero with respect to the lighter mass object.
To understand it better, let’s consider, the heavier object to be Earth, having a mass of \[{{M}_{e}}=5.972\times {{10}^{24}}kg\] and the lighter mass to be us having a mass of m=70kg. The radius of earth is ${{R}_{e}}=6400km=6.4\times {{10}^{6}}m$.
The acceleration of the lighter mass (m) would be: ${{a}_{m}}=\dfrac{GM}{{{R}^{2}}}\Rightarrow {{a}_{m}}=\dfrac{\left( 6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}} \right)\left( 5.972\times {{10}^{24}}kg \right)}{{{\left( 6.4\times {{10}^{6}}m \right)}^{2}}}\Rightarrow {{a}_{m}}\approx 9.8m{{s}^{-2}}$. Similarly, the corresponding acceleration of earth is $\approx {{10}^{23}}$ times lesser than the value of ${{a}_{m}}$, hence approximately equal to zero.
Hence, the acceleration due to gravity of earth is $g={{a}_{m}}=9.8m{{s}^{-2}}$.

Note:
The value of gravity changes from place to place on the surface of earth, since earth isn’t a complete sphere. The radius of earth near the equator is more than the radius of earth towards the poles. Hence, the standardized value of gravity is considered to be (g=$9.8\ m{{s}^{-2}}$) changes. Thus, the value of acceleration due to gravity increases gradually from the equator towards the poles.