I know that free fall in vacuum is independent of mass. But what happens in air?

Question

Pardon me if this sound somewhat trivial to some of the highly qualified users here.

So I know two objects will fall at the same rate if they are in absolute vacuum, so what I'm assuming is that the free fall formula is only correct if there is no air, right?

I'm a programmer and I want to implement a little physics in some code I'm creating. Let's say two objects, weighing 1 pound and 10 pounds respectively, fall at the same speed only if there is no air. But what if I want to consider air, and mass, the heavier object from my logic will fall faster and will touch the ground faster.

So the free fall formula is not what I'm looking for? What formula should I consider?

Answer 1

You need to research the mechanics of drag and the Drag Co-efficient (start with this wiki page).

A simple model (where drag is a Ram Pressure) holds the drag to be proportional to the square of the speed. This can be justified on simple momentum conservation grounds, in the case of pure ram pressure. The drag co-efficient is an empirically-found "fudge factor" that multiplies the pure ram pressure to get the drag.

So, if the body is falling straight down, the nett force on it downwards is

$$F_\downarrow = m\,g - \frac{1}{2}\,\rho\,A\,c_d\,v^2\tag{1}$$

where $\rho$ is the fluid (air) density, $A$ the body's horizontal cross-sectional area (the face area presented to the flow), $v$ the body's downwards speed and $c_d$ the drag co-efficient. You can look these co-efficients up.

So if we put (1) into Newton's second law, you get a differential equation for the downward speed:

$$\frac{\mathrm{d}\,v}{\mathrm{d}t} = v\,\frac{\mathrm{d\,v}}{\mathrm{d}x} = g - \frac{\rho\,A\,c_d}{2\,m} v^2\tag{2}$$

where $x$ is the distance fallen. Something like Mathematica will give you explicit solutions to this equation. If you've studied differential equations, (2) is straightforward to solve by hand to get an analytic solution, too. Alternatively, you could integrate these equations numerically in your program, but I suspect the explicit solutions will be easier to get going.

The same ideas should give you enough to derive a vector description of dynamics if there is an initial horizontal component of velocity.

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R. Rankin's answer includes a term proportional to the speed as well as the quadratic term. This proportional term is viscous drag as described in the Wiki article on the Stokes's Law (not to be confused with Stoke's theorem on differential forms). I believe this is a fairly insignificant term for air, but ultimately you may want to consider differential equations like

$$\frac{\mathrm{d}\,v}{\mathrm{d}t} = v\,\frac{\mathrm{d\,v}}{\mathrm{d}x} = g - \frac{\beta}{m}\,v-\frac{\rho\,A\,c_d}{2\,m} v^2\tag{3}$$

This one you can solve analytically by hand too, or shove it into Mathematica.

Answer 2

Your want to add a drag term to your force equation. It will be more complicated and involve the geometry of the objects in question. Note also your supposition will not always hold, an adult in a parachute will not fall faster than a child without a parachute. {disclaimer: Not tested empirically!!!!} You will have drag force term like:

$$\vec{F}_{drag}\propto-\alpha\vec{V}_{elocity}-\beta\mid\vec{V}_{elocity}\mid^{2}\hat{V}_{elocity}$$

Where $\alpha$,$\beta$ are functions of the geometry, the fluid and such (pressure and whatnot). You can look this type of thing up. If you want to get into more depth (and accuracy) you will want to look at fluid flow equations such as those of Euler and Bernoulli.