By user13107


2012-09-07 18:00:25 8 Comments

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

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

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

1 comments

@dmckee 2012-09-07 18:13:31

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".

@user13107 2012-09-07 18:16:50

Thanks. Shouldn't this be included in textbooks ?

@dmckee 2012-09-07 18:21:17

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.

@David Z 2012-09-07 18:21:46

@user13107 It is, but perhaps not the ones you've been reading.

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