Let $o$ be a spherical object with mass $m$ and surface $s$.

Let $g$ be the gravitational acceleration and $h$ the height.

Let the gas where we drop $o$ in have density $d$ and pressure $p$ at height $h$.

If we drop $o$ at height $h$ what will be its speed $v(t)$ and height $H(t)$ as a function of time $t$ ?

Note that $g$ is a parameter here, not ness 9,81 as here on earth.

I know here on earth there is a speed limit to falling objects in the air. Do we always get a speed limit?

What if we add an initial velocity to $o$ paralell to the falling direction $v$ , where the velocity $v$ can be both positive or negative?

What is the logic and justification behind the equations and solution? Is this analogous to sliding on a ramp with friction, or is it more like collosions with particles? Or maybe both are valid and are possible origins of the solution equations?


1 Answer 1


Terminal velocity of a free falling object is obtained at the moment its acceleration vanishes \begin{equation} \Sigma\mathbf{F}=0. \end{equation} The forces that act on the object while falling are the gravity force, \begin{equation} F_g=-mg, \end{equation} and the drag force \begin{equation} F_d=\frac{1}{2}v_t^2dC_dA \end{equation} where $v_t$ is the terminal velocity, d is the density, $C_d$ is the drag coefficient and A is the cross section of the object. Substituting the expression of the forces into the force equilibrium equation we obtain \begin{equation} v_T=\sqrt{\frac{2mg}{d C_dA}}. \end{equation} This solution is written generally for any kind of falling object in inviscid fluid. For the particular case of a sphere, $A=\pi r^2$ where $r$ is the radius of the sphere. The drag coefficient value, $C_d$, can be found in the literature.


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