I can't seem to find a straight forward answer to this. I really just want to know if changing mass of an object affects the terminal velocity. If two objects of the same dimensions except one had twice the mass, fell from a plane, would the one with higher mass reach a faster terminal velocity, therefore making it hit the ground before the one with less mass? I know all objects have the same gravitational pull which makes a marble and a bowling ball hit the ground at the same time if you drop them. But if they were both dropped from a plane would the marble max out at terminal velocity slower than the bowling ball, making the bowling ball hit first?


Suppose your object is a sphere with a radius $r$ and mass $m$. The aerodynamic drag on a sphere is given by:

$$ F_{drag} = \tfrac{1}{2}C_d \rho \,\pi r^2 \,v^2 \tag{1} $$

where $\rho$ is the density of the air and $C_d$ is the drag coefficient. The drag coefficient varies with speed (the NASA article I linked shows how $C_d$ changes with speed) but over a limited range of speeds it can usefully be taken as constant.

The downward force on the object is simply:

$$ F_{grav} = mg \tag{2} $$

and terminal velocity is reached when the two forces are in balance i.e. when $F_{drag} = F_{grav}$. If we equate equations (1) and (2) we get:

$$ \tfrac{1}{2}C_d \rho \,\pi r^2 \,v^2 = mg $$

and rearranging gives:

$$ v_{term} = \sqrt{\frac{2mg}{C_d \rho\pi r^2}} $$

In your case you keep the size of the spheres constant, in which case we get:

$$ v_{term} \propto \sqrt{m} $$

So terminal velocity does increase with mass. The heavier sphere will have a higher terminal velocity.

  • $\begingroup$ Thank you for this! So in conclusion, the sphere with the higher mass would hit the ground first, given they are dropped high enough to reach their terminal velocities? $\endgroup$ Jul 17 '15 at 6:41
  • $\begingroup$ @NathanMcMasters: The higher mass will always hit the ground first, even if the height isn't great enough to reach terminal velocity. See Different density objects falling on Earth's atmosphere? $\endgroup$ Jul 17 '15 at 7:25
  • $\begingroup$ Ok that makes sense, I'm just confused why so many people say if you dropped them at the same time they hit at the same time. Do they just mean approximately, but in reality the higher mass object hits slightly before? $\endgroup$ Jul 17 '15 at 18:04
  • $\begingroup$ @NathanMcMasters: in the absence of air resistance, or if air resistance is not a significant factor, they will hit the ground together. Galileo did his experiments by rolling balls down an inclined plane at low speeds where air resistance could be ignored. $\endgroup$ Jul 17 '15 at 19:33

Imagine 3 objects. One is a flat piece of paper. The second is an identical piece of paper rolled into a ball. The third object is the exact same shape and size as the rolled up piece of paper but it's made of iron.

If you drop the flat paper and the rolled up paper at the same time, the rolled up paper hits first because it has less air resistance due to having less surface area against incoming wind. (I'm sure you are of aware of that). Now drop the rolled up paper and the piece of iron at the same time. The iron will hit first. Air resistance depends on the velocity of the object and its surface area, which are the same in this case. So what gives? Well just because the wind resistance is the same doesn't mean they fall at the same rate. The iron is more massive so a wind resistance force slows it down less than it does the rolled up piece of paper.

  • 1
    $\begingroup$ Are we talking about dropping the piece of rolled up paper and the iron ball from 5 feet up (Like my eye level to my feet) or 500 feet (like off a building?) I would think the iron and paper ball would hit at the same time if it was 5 feet. $\endgroup$ Jul 17 '15 at 6:11
  • 1
    $\begingroup$ You're right. If the drop is from a low height the objects don't have much times to accelerate to high velocity and therefore wind resistance will be tiny and almost negligible. $\endgroup$
    – Alex
    Jul 17 '15 at 6:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.