I just learned of Newton's law of gravitation and that distance between two bodies is a factor in the gravitational force. My question is if that's true why is the Earth's gravitational acceleration a constant $9.8\mathrm{ms^{-2}}$ as change in force mathematically should mean change in acceleration if you refer to the equation $F=ma$.
Answer
Although @JamalS's answer covers the variation of $g$ due to latitude and terrain variations, here is a simpler reason why $g$ is taken to be constant close to the Earth's surface.
If the Earth has a radius $R$, and there is a mass $m$ at a distance $r$ above the surface, the force is given by: $$F=\frac{GMm}{(R+r)^2}$$
This formula can be expanded as a series for small values of $r$, and this gives (upto three terms): $$F=\frac{GMm}{R^2}-\frac{2GMmr}{R^3}+\frac{3GMmr^2}{R^4} + \mathcal O (r^3)$$
The first term is a constant, but the second term and so on have ratios of the form $\frac{r}{R^3}$ which are very small when $r$ is small compared to the radius. (for example if $r$ is in the order of a few metres)
For most examples used in high school physics or introductory classical mechanics, $r$ usually is very small compared to the radius of the Earth, which is 6371 kilometers.
In cases like these, the other terms are as good as nothing, and the first term is the most significant contribution. This first term happens to be $mg$, where: $$g=\frac{GM}{R^2}$$
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