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Explanation of proportionality with falling objects.

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zeralda21
#1
May1-12, 02:28 PM
P: 119
A falling object with no initial velocity with mass m is influenced by a gravitational force g and the air resistance which is proportional to the object´s speed. By Newton´s laws this can be written as:

(1) mg-kv=ma or (2) mg-kv^2=ma (for large velocities).

I assume that k is a positive constant that depends on the geometry of the object and the viscosity. But how can one explain that the air resistance is proportional to the velocity? And to the velocity squared in the second equation?
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Khashishi
#2
May1-12, 02:42 PM
P: 886
I think those are phenomenological laws. I mean, they are ansatz forms for the air resistance which aren't based on detailed physics, but rather generic expressions with adjustable parameters to give it flexibility to fit the experiment. We can expect the air resistance to increase monotonically with velocity, but not necessarily linear.

We can approximate many (analytic) functions as linear for small input using a Taylor series expansion. For small enough input, assuming 0 constant term, the linear term (if non-zero) will dominate. This is the basis of a great many laws like Hooke's law or Ohm's law, which don't really have a derivation from fundamental physics. For large input values, the higher order terms will tend to dominate over the linear.

In the case of wind, by symmetry, I expect the squared term to be zero, and the first non-linear term to be velocity cubed.
Redbelly98
#3
May1-12, 06:41 PM
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There are two effects at play, both pretty well grounded in basic physics concepts.

One effect is from simple collisions between the object and the air molecules. This results in a force proportional to v2.

The linear-in-v term is due to the tendency of the object to drag the air along with it, and the viscosity of the air.


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