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The greater the mass, the higher the terminal velocity. But what if the object's size (area) also increase, will the air resistance cause the heavier object to become slower than the lighter object?

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  • $\begingroup$ Is there a difference in terminal velocity of a parachute and a stone, if they weigh the same? $\endgroup$ – Steeven May 25 '17 at 17:47
  • $\begingroup$ Yes. Drag coefficients are usually drastically different for a parachute vs. a stone. $\endgroup$ – David White May 25 '17 at 18:03
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Weight is $w=mg$ and air drag $D=\frac12 \rho C A v^2$. (Here $g$ is the gravitational acceleration, $C$ the drag coefficient, $A$ the cross section area seen from the flying direction, and $v$ the speed.)

Let's put both unto Newton's 1st law, which counts when there is terminal velocity (no more acceleration):

$$D-w=0\\ \frac12 \rho C A v^2-mg=0\\ v=\sqrt{\frac{mg}{\frac12 \rho C A}}=\sqrt{\frac{Vg}{\frac12 C A}}$$

This is the terminal speed. We have inputted $m=\rho V$.

  • Here it clearly shows that density doesn't really make a difference. But size does. Size in the form of volume, because that means more matter.

So you cannot simply say that mass has influence; two different types of stones can have same size and volume but different masses.

  • Shape also has an influence. Because of the cross section area. That is why a parachute falls slower but a spear faster.

So, the answer to your question is that yes, size can have an influence, but you need to know more. You need to know shape as well.

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    $\begingroup$ More over, you can cheat. A demo I do for gen ed classes and public demonstrations involves a variant on the Galileo drop. Drop and book and a basket style coffee filer one next to the other, first. Obviously the book gets to the ground first. Then repeat the experiment with the coffee filter on the book. They drop together, often with the filter remaining in contact with the book all the way down. $\endgroup$ – dmckee --- ex-moderator kitten May 25 '17 at 18:31
  • $\begingroup$ Well, thank you for you all.I have still one more question.I don't understand why object will fall with same acceleration before they reach their terminal velocity.Can't I use mg-drag=net force and divide by the mass and get the acceleration of the object? $\endgroup$ – Tlx 2208 May 26 '17 at 15:04
  • $\begingroup$ @Tlx2208 "I don't understand why object will fall with same acceleration before they reach their terminal velocity." This is wrong. They don't fall with the same accelerations (because the net pull in them is different). And furthermore one object does not have the same acceleration the whole way - because the air drag grows with speed. $\endgroup$ – Steeven May 26 '17 at 15:52
  • $\begingroup$ @Tlx2208 "Can't I use mg-drag=net force and divide by the mass and get the acceleration of the object?" Yes you can. But the acceleration you get is only for that specific moment in time. Only for that specific speed $v$ that you plug into the drag-formula. That speed changes all the time (until terminal speed is reached) $\endgroup$ – Steeven May 26 '17 at 15:56
  • $\begingroup$ Oh ok.So two objects will drop with same acceleration with another until they reach their terminal velocity? $\endgroup$ – Tlx 2208 May 26 '17 at 16:06
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So the two objects have the same density. Assume that they have the same shape as well, maybe spheres. Then the weight increases as $R^3$ and the air resistance as the area, or $R^2$ where R is the radius of the sphere. The terminal velocity depends on the ration W/Fr where W is the weight and Fr the air resistance. This ratio increase with R (not necessarily in a linear mode). So the terminal velocity will be larger for the larger ball.

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