Falling body with air resistance

[chengbin]

What is the difference between these two equations.

m dv/dt = mg - kv

and

m dv/dt = mg - kv^2

as the equation modeling a falling body with air resistance?

 

[AlephZero]

The first equation is only accurate for very low Reynolds numbers (Re < 1). For bodies falling in the Earth's atmosphere, it only applies to very small objects like dust particles, tiny insects, etc.

The second equation is a fairly good model for larger Reynolds numbers, at least for speeds below about Mach no 0.2 (about 70 m/s) Above that speed, the compressibility of the air begins to affect the situation and "k" is no longer a constant.

 

[RandomGuy88]

From a mathematics point of view, the second equation is nonlinear making it much more difficult to solve.

 

[Metaleer]

A bit of theory first.

The function f(v) that gives the magnitude of the air resistance when the object's velocity is less than the speed of sound may be written, with the Taylor series expansion in mind, as

f(v) = bv + c v^2 = f_\text{lin} + f_\text{quad}.​

Now, the physical origin of these two terms are different. The first linear term is due to the viscous drag of the medium, and it's usually proportional to the viscosity of the medium and the linear dimensions of the object. The second quadratic term arises from the projectile having to accelerate the mass of air with which it is continually colliding; it is proportional to the density of the medium and the cross-sectional area of the projectile. For spherical particles, we have

b = \beta D

c = \gamma D^2​

where D is the diameter of the sphere and \beta and \gamma are constants that depende on the nature of the medium. For spherical projectiles in air and for standard conditions, we have approximate values of

\beta = 1.6 \cdot 10^{-4} ~\text{N}\cdot\text{s}/\text{m}^2​

and

\gamma = 0.25 ~\text{N}\cdot\text{s}^2/\text{m}^4.​

Now, to what you asked: often, one can neglect one of the terms compared to the other. To decide which can be neglected, we need to compare the size of the two terms:

\frac{f_\text{quad}}{f_\text{lin}} = \frac{\gamma D}{\beta}v = \left(1.6 \cdot 10^3 \frac{\text{s}}{\text{m}^2}\right) D v.​

So, in a given problem, you have to substitute the values of the diameter and the velocity into this equation to figure if any of the two terms can be neglected. If \frac{f_\tex{quad}}{f_\tex{lin}} \approx 1 then unfortunately you have to keep both terms to have a more accurate mathematical model.

EDIT: (Further information) As a rule of thumb, for very small liquid drops in air, but also slightly larger objects in a very viscous fluid, the drag force can usually be taken as linear. For most other projectiles, such as golf balls, cannonballs, humans in free fall, the dominant drag force is the quadratic one.

As AlephZero has mentioned, this discussion can be related to the Reynolds number. As I have mentioned previously, the linear drag can be related to the viscosity of the fluid, and the quadratic term is related to the inertia (density) of the fluid. Thus, the ratio \frac{f_\text{quad}}{f_\text{lin}} can be related to the fundamental parameters of viscosity and density. The result is that this ratio is of roughly the same order of magnitude as the dimensional number R = \frac{D v \rho}{\eta} we call the Reynolds number. This means that, in general, the quadratic drag is dominant when R is large and that linear drag is dominant when R is small.

Reference: Classical Mechanics, John R. Taylor.

Hope this helps.

 

[HallsofIvy]

From a mathematics point of view, the second equation is nonlinear making it much more difficult to solve.

I disagree with that. For first order equations, the distinction between "linear" and "non-linear" is not as important as for higher order equations. And certainly not in the case of this particular equation which is "separable" and the general solution is easily written.

 

[chengbin]

Thank you guys.

@RandomGuy88, this is so true. This is why I'm asking this equation in my other thread here

https://www.physicsforums.com/showthread.php?p=3197469#post3197469

@HallsofIvy, The non-linear equation ends up involving tanh and stuff, which I didn't learn yet. From Google searches, I have to do partial fractions with square roots. Sounds hard and a lot of work.

 

[Metaleer]

Thank you guys.

@RandomGuy88, this is so true. This is why I'm asking this equation in my other thread here

https://www.physicsforums.com/showthread.php?p=3197469#post3197469

@HallsofIvy, The non-linear equation ends up involving tanh and stuff, which I didn't learn yet. From Google searches, I have to do partial fractions with square roots. Sounds hard and a lot of work.

You could leave the result as a function of logarithms, if hyperbolic tangents don't make you feel comfortable, but in any case, this is a good example so that you can see where hyperbolic functions appear in real life.

I disagree with that. For first order equations, the distinction between "linear" and "non-linear" is not as important as for higher order equations. And certainly not in the case of this particular equation which is "separable" and the general solution is easily written.

This may be true for a simple free fall problem, but in the linear case when the motion is not restricted to a line (that is, assume you throw something), the equations for both components aren't coupled, so they can be easily solved separately. However, when the drag is quadratic, each component is coupled, and the equations aren't linear to boot, so mathematically, assuming the problem you're treating isn't a simple free fall but rather a general 2 dimensional motion, the problem is more complicated.