# The measurement of $g$ and air resistance

Suppose I am trying to determine the value of the acceleration due to gravity by simply dropping objects from a ledge. If I did not account for air resistance, would the acceleration due to gravity measured be less than if I did account for air resistance?

This is what a thought: Suppose I had an object of mass $m$ that was being dropped and let $F_{drag}$ be the force due to air resistance. Then

$mg - F_{drag} = ma$. Then with some algebra I get $a = g - \frac{F_{drag}}{m}$. If I do not account for air resistance, I get $a = g$. But if I do account for air resistance $a < g$. So I get that the value of $g$ is greater if I do not account for air resistance.

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No, g is constant in this situation (assuming the gravitational field doesn't change). It's a that actually changes. So a1 < g with air resistance (correct) but a2 = g without air resistance (still correct). But that just means that a1 < a2 since g is the same in both cases. – user758556 Jun 7 '12 at 3:41

You are attempting to measure what is the acceleration due to gravity $g$, at earths surface.

So lets say you dropped something and measured the acceleration it had. Then you want to find out how this relates to the acceleration due to gravity.

If you do not include air resistance in your equations, you will say, $$ma = mg \implies g = a$$

So you will claim that $g$ is whatever the acceleration you measured is.

If you do include air resistance, you will say

$$ma = mg - F_{air} \implies g = a + F_{air}/m$$

So indeed you are underestimating $g$ if you do not include air resistance, because air resistence is (as its name implies) resisting the motion of your ball, yielding a smaller acceleration than if it wasn't there!.

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@Daven Ware: Thanks very much! – Student Jun 7 '12 at 16:30

No, the force of gravity is constant. The acceleration will reduce due to drag, but the same force of gravity (9.8 assuming your talking about earth) will be acting on it.

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