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I was taught in school that acceleration due to gravity is constant. But recently, when I checked Physics textbook, I noted that

$$F = \dfrac{G \cdot m_1 \cdot m_2}{r^2} $$

So, as the body falls down, $r$ must be changing, so should acceleration due to gravity.

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This inverse-square law is actually meant for "Gravitational attraction between two bodies" $F=mg$ is the actual one... $r$ doesn't come there... – Waffle's Crazy Peanut Sep 7 '12 at 18:05
Ok, but didn't $F = mg$ originate from the inverse-square law ? – user13107 Sep 7 '12 at 18:07
@CrazyBuddy that sort of information would be better provided in an answer. – David Z Sep 7 '12 at 18:20
@CrazyBuddy: In $F=mg$, $g$ varies with height. – Ben Crowell Sep 8 '12 at 0:10
up vote 9 down vote accepted

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 $R \approx 6400\text{ km}$, and the height of the object $h$.

Now the acceleration due to gravity at $h$ is $$ g = \frac{F_g}{m} = G\frac{M_e m}{(R + h)^2 m} = G\frac{M_e}{(R + h)^2} $$and lets manipulate this a little $$ g = G\frac{M_e}{(R + h)^2} = G\frac{M_e}{R^2(1 + h/R)^2} \approx G\frac{M_e}{R^2}\left(1 -2\frac{h}{R} \right).$$ 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".

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Thanks. Shouldn't this be included in textbooks ? – user13107 Sep 7 '12 at 18:16
No. Your instructor should have pointed you in the right direction if you showed an interest and not disturbed those who didn't ask with the extra math. – dmckee Sep 7 '12 at 18:21
@user13107 It is, but perhaps not the ones you've been reading. – David Z Sep 7 '12 at 18:21

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