1. How does the presence of air affect the rate at which a body falls?
  2. What is the rule for falling objects, when there is little or no air resistance?
  • 3
    $\begingroup$ That's drag for you. $\endgroup$
    – ACuriousMind
    Apr 26, 2015 at 3:54
  • $\begingroup$ Can you please explain it further? $\endgroup$
    – Leil
    Apr 26, 2015 at 4:04
  • 1
    $\begingroup$ In most cases(depending on the shape of the body, the nature of fluid and the precision of answer required this changes) we just assume that the presence of air is modelled as a velocity dependent retarding force ($\mathbf{F}_{retarding}=-c\mathbf{v}$) where c is some constant. When there is no air you don't have this force. Simple as that. $\endgroup$
    – gautam1168
    Apr 26, 2015 at 4:23

2 Answers 2


We'll start with your second question, as that's (much) simpler.

Close to the surface of the earth, it's safe to assume that the force of gravity is proportional to the mass of your object:

$$F_G = mg$$

where $m$ is the mass, $F_G$ is the force of gravity, and $g$ is a constant (for the earth $g \approx 9.8 m/s^2$.) Then Newton's second law tells us that the object's acceleration will be:

$$\ddot{y} = -\frac{F_G}{m} = -\frac{mg}{m} = -g$$

where $y$ is your height; we use a negative sign because we are accelerating downwards.

We now have to integrate this differential equation. If we assume that we started at a height $y_0$ with no initial velocity, then after time $t$ our height would be:

$$y(t) = y_0-\frac{gt^2}{2}$$

so you see our height decreases quicker and quicker with time. This is just standard high school physics.

That's the case of no air resistance. But as bodies move faster, the air-resistance can become significant. As already mentioned in a comment, the simplest thing to do is to add a new force on the body: we've already got a gravitational force $F_G$, but now we'll consider the air-resistance force $F_R$. A natural first guess is to say that this force increases in proportion to your velocity - if you double your speed, you get twice as much air resistance. In this case,

$$F_R = -\alpha \dot{y}$$

Where $\alpha$ is just a constant telling you how strong the air resistance is. And Newton's second law now says:

$$\ddot{y} = -\frac{mg+\alpha \dot{y}}{m}$$

And again this is just a linear differential equation that can be solved for any initial position and velocity by standard techniques, taught in a first course on differential equations.

Now it must be remembered in all this that we assumed the equation for $F_R$ is of this particular form. Often this isn't the case. For different shaped objects, moving in different ways, we may have different air resistances. In some cases, we might be able to use some experimental data. But in many cases we may have to resort to modelling the air-flow itself, and this is a much more difficult computational task.


When the body is falling down without the affect of air resistance,it will experience only one force,I.e.the gravitational force which pulls out downwards towards the earth..and the object will fall to the surface of the earth will an acceleration 'g'. Coming to the first case...when the object is falling down in the presence of a drag force (friction due to the air,or air resistance) it would take little more time to fall down since the friction force acts opposite to the gravitational force (in this case) and it reduces the acceleration of the object. Hope this answer is helpful to you. Cheers!


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