8
$\begingroup$

I've tried doing research on this and there are some similar questions. However, they have slightly different scenarios and to make sure I understand things perfectly, I wanted to word it into my own words and confirm that my thought process is correct. I am just getting back into physics after a decade of not doing anything remotely related, so I'd like to make sure I'm not confusing things.

We have two objects that are the exact same size and shape. Let's call them A and B.
A is twice the mass of B.

Both are dropped from an airplane at the same time.

Am I correct to say that since B is half the mass of A, it will reach its terminal velocity much sooner, producing 0 acceleration from there on.

A, on the other hand, will continue accelerating until the positive force from the air resistance equals its weight, at which point it reaches its terminal velocity and 0 acceleration.

A will have a higher terminal velocity and reach the ground sooner than B, because A was accelerating for a longer period of time.

$\endgroup$
  • $\begingroup$ The edit was not really necessary - when applied to 2 objects in the same reference frame, mass and weight are pretty much interchangeable. $\endgroup$ – Pranav Hosangadi Sep 2 '13 at 3:29
  • $\begingroup$ Also, I don't really see your question. If you meant to ask if your reasoning is correct, then my answer would be that it is. This is why a person with a parachute reaches the ground much later than a person without one. $\endgroup$ – Pranav Hosangadi Sep 2 '13 at 3:37
  • $\begingroup$ @PranavHosangadi Yes, that's actually exactly what I asked. In the first paragraph I'm asking to confirm that my reasoning is correct. I ask, "Am I correct to..." $\endgroup$ – B.K. Sep 2 '13 at 3:40
11
$\begingroup$

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

enter image description here

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.

$\endgroup$
  • 1
    $\begingroup$ You do have to keep in mind that unless other forces are involved or if you would include the decrease of air density with increase of altitude, such that the solution you gave will be a good approximation, than terminal velocity will actually never be reached. The velocity approaches terminal velocity when time goes to infinity, however in real-life any turbulence will disturb this when the velocity is very close to terminal velocity. $\endgroup$ – fibonatic Aug 16 '14 at 19:21
  • $\begingroup$ Shouldn't those "horizonta lines" in reality be slightly inclined towards down-positive x? (air density increase by decreasing altitude) $\endgroup$ – GameDeveloper Aug 9 '16 at 8:31
  • 1
    $\begingroup$ @DarioOO Quite possibly yes. I certainly make an assumption of constant density in my answer. If I have time in the coming days, I may compute a numerical solution to the ODE for $v$ with more realistic density as a function of height and see what pops out. It's not obvious to me what the result would be without having run a simulation. One can't simply put spatially-varying density into the expression I derived. $\endgroup$ – joshphysics Aug 9 '16 at 20:00

protected by ACuriousMind May 28 '17 at 11:31

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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