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?
[Physics] If two objects have all the same conditions except different masses. Will their terminal velocity be different
gravitymassvelocity
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Terminal velocity is reached when the drag force due to moving through air is equal (but opposite) to the gravitational force. Now, the gravitational force is proportional to the mass, while the drag force has nothing to do with mass, but everything to do with how large and "streamlined" the object is. Suppose object A is twice as heavy as object B. If object A also experiences twice the drag force as object B (at a given speed), then their terminal velocities will be the same.
To put it another way, let's suppose that the two objects have the same masses, and therefore the same weights; they have the same gravitational forces. The question becomes: do they have the same drag force?
Drag comes from the resistance of the air to an object's movement, so – all else being equal – something that's more streamlined will have less resistance. If one of this is shaped like a bullet, and one is shaped like a big hollow ball, the big ball will have the same amount of drag at low speeds as the bullet does at high speeds. So the ball's terminal velocity will be a lot lower.
(I've simplified a little; the force of buoyancy should be added to the drag force. But usually this is relatively small, so we can ignore it just for simplicity.)
Yes, although I don't think it's totally obvious that your statements are true. Let's assume that the drag force on a given object is \begin{align} \mathbf F_\mathrm{drag} = -\frac{1}{2}\rho AC_dv\mathbf v \end{align} where $\rho$ is the mass density of the fluid in which it moves, $A$ is its cross-sectional area, $C_d$ is its drag coefficient, $\mathbf v$ is its velocity, and $v=|\mathbf v|$ is its speed. Then Newton's Second Law gives the following equation of motion for an object falling near the surface of the Earth under the influence of gravity: \begin{align} ma = m g -\frac{1}{2}\rho AC_d v^2 \end{align} So that the acceleration of the object is \begin{align} a = g - \frac{1}{2}\frac{\rho AC_d}{m}v^2 \end{align} In particular, for a fixed cross-sectional area, increasing the mass of the object will increase its acceleration because the second term will be smaller in magnitude. But that also means that the object's speed with increase faster, so that the second term will grow faster; there are competing affects. So which one wins out? Well, the equation of motion can be looked upon as a differential equation for the velocity $v(t)$ as a function of time; \begin{align} \dot v(t) = g-\frac{1}{2}\frac{\rho AC_d}{m} v(t)^2 \end{align} With the initial condition $v(0) = 0$, namely if you just drop the object, the solution (thanks to to Stephen Wolfram) is \begin{align} v(t) = \sqrt{\frac{2gm}{AC_d\rho}}\tanh\left(\sqrt\frac{AC_dg\rho}{2m}t\right) \end{align} Let's plot this function for some different mass values but keeping all other parameters the same
The lowest blue curve corresponds to the lowest mass, and each successive curve above it corresponds to a mass twice as large as for the last curve.
It's clear that the terminal velocities of the more massive objects are greater and that these velocities are achieved at a later time. Moreover, after terminal velocity is reached, the object no longer accelerates.
Best Answer
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.