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Say we have a fall while considering air resistance

$$ma=mg-Bv^2.$$

It is a standard differential equation.

But then I was asked what would happen, if we ignore the gravitational force $mg$. Now I am confused, why or when can we ignore the gravitational force? I assume when the object is very very small so that it doesn't matter anymore? But then

$$m\frac{d^2 s}{dt^2}=-B\left(\frac{ds}{dt}\right)^2.$$

What exactly does this say about the situation?

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    $\begingroup$ I am not sure what the person asking you the question meant by "what would happen". Do they mean what would the motion be if gravity was suddenly 'switched off'; or do they mean what would be the change in the equation describing the motion? $\endgroup$
    – Martin CR
    Jan 10 at 11:19
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    $\begingroup$ A circumstance where the gravitational force becomes irrelevant, might be when the aerodynamic drag is massively higher than the weight... e.g. soon after a high-velocity bullet leaves the barrel of a gun, fired vertically; or a meteorite enters the earth's atmosphere at transonic speeds $\endgroup$
    – Martin CR
    Jan 10 at 11:21
  • $\begingroup$ It's not clear exact situation what is asked. If gravity would be switched off in the process of free fall, then obviously body will stop at some point due to air resistance (or it would hit the surface due to inertia if drag force not big enough to stop the body at $h=0$). If gravity is switched-off right before body is released for falling - it would simply stay at the same height $h=h_0=\text{const}$ at rest. Depends... $\endgroup$ Jan 10 at 11:51
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    $\begingroup$ Comment about terminology: A fall with air resistance is by definition not a free fall. $\endgroup$
    – Qmechanic
    Jan 10 at 12:00
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    $\begingroup$ I assume when the object is very very small so that it doesn't matter anymore? This is wrong. Gravity affects the object regardless of it's size. Aerodynamic drag does depend on size - basically the bigger it's cross section the more drag. $\endgroup$ Jan 10 at 16:05

2 Answers 2

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This is a simplification when drag is much larger than the gravitational force. In order for this to work the drag must be at least order of magnitude larger. When initial velocity is (close to) zero, the simplification obviously does not work as it predicts the object to remain at rest.

The free-fall example is little bit difficult to imagine. Much better example would be a moving car with no force from the motor, assuming the car moves on a horizontal surface (road).


If the equation of motion is given as:

$$m \frac{d^2}{dt^2} x(t) = -k \cdot v(t)^2$$

it can be written as

$$m \frac{d}{dt} v(t) = -k \cdot v(t)^2$$

which equals to

$$\int \frac{dv(t)}{v(t)^2} = -\int \frac{k}{m} dt$$

This finally leads to

$$\frac{1}{v(t)} = \kappa t + C_1$$

where $\kappa = \frac{k}{m}$, $C_1 = \frac{1}{v_0}$, and $v_0$ is the initial velocity.

The equation for velocity is

$$\boxed{v(t) = \frac{v_0}{1 + \kappa v_0 t}}$$

From this it is trivial to find the equation for position:

$$\boxed{x(t) = \frac{1}{\kappa} \ln(1 + \kappa v_0 t)}$$

Note that the velocity approaches zero with time, i.e. the drag is slowing the object.

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  • $\begingroup$ Hi question on initial velocity. What shall we do if it’s 0? $\endgroup$ Jan 10 at 12:23
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    $\begingroup$ The equations I derived do not put any constraints to initial velocity. For $v_0 = 0$ the object remains at rest, i.e. $x(t) = 0$ and $v(t) = 0$. $\endgroup$ Jan 10 at 12:36
  • $\begingroup$ Which means the object would float in air at whatever height? $\endgroup$ Jan 10 at 13:33
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    $\begingroup$ That would be reasonable to assume, but the thing is that this simplification works only while drag is much higher than the gravitational force, I'd say at least order of magnitude higher. When velocity is sufficiently small, you can no longer neglect gravitational force. $\endgroup$ Jan 10 at 13:35
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    $\begingroup$ I find the phrasing "the drag is stopping the object" a bit misleading since the object will travel arbitrarily far given enough time. Maybe "slowing" would be better? $\endgroup$
    – schtandard
    Jan 10 at 23:07
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@MarkoGulin has fully addressed the mathematical aspect of the problem. I thus add a remark regarding the physical situations where the equation in question can be relevant:

  • Drag force is ubiquitous when we talk about movement of macroscopic objects in a gas or liquid. It is not hard to think of examples that do not involve free fall: e.g.,
    • a boat or a submarine that has turned off its engines,
    • a glider aircraft,
    • an immobile helicopter or a baloon carried by a gust of wind,
    • a meteorite or a satellite entering the Earth's atmosphere, etc.
  • Projectile motion: here we are dealing with a bit more complicated situation, since the drag force is proportional to the square of the full velocity: $$m\dot{\mathbf{v}}=m\mathbf{g}-B\frac{\mathbf{v}}{|\mathbf{v}|}\mathbf{v}^2.$$ However, when the horizontal velocity is very big, we can approximate it by $$m\dot{v}_x=-Bv_x^2. $$
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