# Is acceleration due to gravity constant? I was taught in school that acceleration due to gravity i

Is acceleration due to gravity constant?
I was taught in school that acceleration due to gravity is constant. But recently, when I checked Physics textbook, I noted that
$F=\frac{G{m}_{1}{m}_{2}}{{r}^{2}}.$
So, as the body falls down, $r$ must be changing, so should acceleration due to gravity.
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Gia Schaefer
This is a first introduction to the issue of the relative changes in physics.
Consider the motion of objects near the Earth's surface. Call the nominal radius of the Earth , and the height of the object $h$
Now the acceleration due to gravity at $h$ is
$g=\frac{{F}_{g}}{m}=G\frac{{M}_{e}m}{\left(R+h{\right)}^{2}m}=G\frac{{M}_{e}}{\left(R+h{\right)}^{2}}$
and lets manipulate this a little
$g=G\frac{{M}_{e}}{\left(R+h{\right)}^{2}}=G\frac{{M}_{e}}{{R}^{2}\left(1+h/R{\right)}^{2}}\approx G\frac{{M}_{e}}{{R}^{2}}\left(1-2\frac{h}{R}\right).$
The last approximation there is dropping higher order terms in $\frac{h}{R}$ which will shortly be seen to be justified.
So, ask yourself how big is $\frac{h}{R}$ for the situations you encounter in your life. A few meters or a few tens of meters at most, right? So $\frac{h}{R}$ is of order ${10}^{-5}$ or smaller over human scales or ${10}^{-3}$ even over the whole height range that we use including airplane elevations.
So, for almost all calculation that you want to make the variation of $g$ negligible.
Physicists get a lot of millage out of these kinds of considerations to the point that you there is a fair amount of shorthand devoted to discussing fractional changes. People say things like "Yeah, but it's down by two orders of magnitude, so we can neglect it".