Motion equation of a free falling object, with air resistance

**arbolis** · February 25th 2010, 06:43 PM

I solved it for fun, but a question raised.
$m \dot v =mg-\gamma v$ . With the integrating factor method, I reached $v=\frac{gm}{\gamma}+e^{-\frac{\gamma t}{m}}C$ .
If I set $v(0)=0$ , I get $v=\frac{gm}{\gamma} \left [ 1- e^{-\frac{\gamma t}{m}} \right ]$ .
So when $t\to \infty$ , $v\to \frac{gm}{\gamma}$ .
It surprises me a bit, it means that if one let fall 2 balls of different masses, the heavier one will fall faster than the less heavy one. I'm not sure this result is true, can you confirm it please?

Also it means that the terminal velocity is $\frac{gm}{\gamma}$ , in other words, weight over the drag coefficient. This could make sense, the lesser the drag coefficient, the faster the object falls.

**CaptainBlack** · February 25th 2010, 08:23 PM

Originally Posted by arbolis

I solved it for fun, but a question raised.
$m \dot v =mg-\gamma v$ . With the integrating factor method, I reached $v=\frac{gm}{\gamma}+e^{-\frac{\gamma t}{m}}C$ .
If I set $v(0)=0$ , I get $v=\frac{gm}{\gamma} \left [ 1- e^{-\frac{\gamma t}{m}} \right ]$ .
So when $t\to \infty$ , $v\to \frac{gm}{\gamma}$ .
It surprises me a bit, it means that if one let fall 2 balls of different masses, the heavier one will fall faster than the less heavy one. I'm not sure this result is true, can you confirm it please?

Also it means that the terminal velocity is $\frac{gm}{\gamma}$ , in other words, weight over the drag coefficient. This could make sense, the lesser the drag coefficient, the faster the object falls.

Yes, in a resisting medium a falling object does reach a terminal velocity, and yes it should be greater for a heavier object than for a lighter object.

Since for objects of the same shape but different weights, at the same speed the net downward force is greater on the heavier than the lighter (as the drag will be equal). At the terminal velocity of the lighter there will still be a net downward force on the heavier and so it will still be accelerating.

Note the form of drag you used (drag proportional and in the opposite direction to speed) is appropriate at small Reynolds' number. In practice this means a small or large surface area object falling in air, or a medium more viscous than air.

This type of result is why Galileo and Aristotle disagree about falling objects. Aristotle's version that heavier objects fall faster then light is obvious from everyday experience. It takes a Galileo to realise why this is not useful and try to decouple drag forces from gravitational.

CB

**arbolis** · February 26th 2010, 05:20 AM

Originally Posted by CaptainBlack

Yes, in a resisting medium a falling object does reach a terminal velocity, and yes it should be greater for a heavier object than for a lighter object.

Since for objects of the same shape but different weights, at the same speed the net downward force is greater on the heavier than the lighter (as the drag will be equal). At the terminal velocity of the lighter there will still be a net downward force on the heavier and so it will still be accelerating.

Note the form of drag you used (drag proportional and in the opposite direction to speed) is appropriate at small Reynolds' number. In practice this means a small or large surface area object falling in air, or a medium more viscous than air.

This type of result is why Galileo and Aristotle disagree about falling objects. Aristotle's version that heavier objects fall faster then light is obvious from everyday experience. It takes a Galileo to realise why this is not useful and try to decouple drag forces from gravitational.

CB

Thanks a lot for your reply. Very well explained.

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