Editing Terminal velocity (section)

==Physics==
For terminal velocity in falling through air , where [[viscosity]] is negligible compared to the drag force, and without considering [[buoyancy]] effects, terminal velocity is given by
<math display="block">V_t= \sqrt\frac{2 m g}{\rho A C_d} </math>
where
*<math>V_t</math> represents terminal velocity,
*<math>m</math> is the [[mass]] of the falling object,
*<math>g</math> is the [[Earth's gravity|acceleration due to gravity]],
*<math>C_d</math> is the [[drag coefficient]],
*<math>\rho</math> is the [[density]] of the fluid through which the object is falling, and
*<math>A</math> is the [[projected area]] of the object.<ref>{{cite book|url=https://books.google.com/books?id=OPsTDAAAQBAJ&pg=PA26|title=Dispersal in Plants: A Population Perspective|last1=Cousens|first1=Roger|last2=Dytham|first2=Calvin|last3=Law|first3=Richard|publisher=[[Oxford University Press]]|date=2008|pages=26–27|isbn=978-0-19-929911-9}}</ref>

In reality, an object approaches its terminal speed [[asymptotically]].

Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using [[buoyancy|Archimedes' principle]]: the mass <math>m</math> has to be reduced by the displaced fluid mass <math>\rho V</math>, with <math>V</math> the [[volume]] of the object. So instead of <math>m</math> use the reduced mass <math>m_r = m-\rho V</math> in this and subsequent formulas.

The terminal speed of an object changes due to the properties of the fluid, the mass of the object and its projected cross-sectional [[surface area]].

Air density increases with decreasing altitude, at about 1% per {{convert|80|m|ft}} (see [[barometric formula]]). For objects falling through the atmosphere, for every {{convert|160|m|ft}} of fall, the terminal speed decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed ''decreases'' to change with the local terminal speed.

Using mathematical terms, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the [[drag equation]]):

<math display="block">F_\text{net} = m a = m g - \frac{1}{2} \rho v^2 A C_d,</math>

with ''v''(''t'') the velocity of the object as a function of time ''t''.

At [[List of types of equilibrium|equilibrium]], the [[net force]] is zero (''F''<sub>net</sub> = 0)<ref>{{cite book|url=https://books.google.com/books?id=ouzxCAAAQBAJ&pg=PA22|title=Fluid Mechanics for Marine Ecologists|last=Massel|first=Stanisław R.|publisher=[[Springer Science+Business Media]]|date=1999|page=22|doi=10.1007/978-3-642-60209-2|isbn=978-3-642-60209-2}}</ref> and the velocity becomes the terminal velocity {{math|1=lim{{sub|''t''→∞}} ''v''(''t'') = ''V''<sub>''t''</sub>}}:

<math display="block">m g - {1 \over 2} \rho V_t^2 A C_d = 0.</math>

Solving for ''V''<sub>''t''</sub> yields:

{{NumBlk|:|<math>V_t = \sqrt\frac{2mg}{\rho A C_d}.</math>|{{EquationRef|5}}}}

The drag equation is—assuming ''ρ'', ''g'' and ''C''<sub>''d''</sub> to be constants:
<math display="block"> m a = m \frac{\mathrm{d}v}{\mathrm{d}t} = m g - \frac{1}{2} \rho v^2 A C_d.</math>

Although this is a [[Riccati equation]] that can be solved by reduction to a second-order linear differential equation, it is easier to [[Separation of variables|separate variables]].

A more practical form of this equation can be obtained by making the substitution {{math|1=''α''<sup>2</sup> = {{sfrac|''ρAC''<sub>''d''</sub>|2''mg''}} }}.

Dividing both sides by ''m'' gives
<math display="block">\frac{\mathrm{d}v}{\mathrm{d}t} = g \left( 1 - \alpha^2 v^2 \right).</math>

The equation can be re-arranged into
<math display="block">\mathrm{d}t = \frac{\mathrm{d}v}{g ( 1 - \alpha^2 v^2)}.</math>

Taking the integral of both sides yields
<math display="block">\int_0^t {\mathrm{d}t'} = {1 \over g}\int_0^v \frac{\mathrm{d}v'}{1-\alpha^2 v^{\prime 2}}.</math>

After integration, this becomes
<math display="block">t - 0 = {1 \over g}\left[{\ln(1 + \alpha v') \over 2\alpha} - \frac{\ln(1 - \alpha v')}{2\alpha} + C \right]_{v'=0}^{v'=v} ={1 \over g} \left[{\ln \frac{1 + \alpha v'}{1 - \alpha v'} \over 2\alpha} + C \right]_{v'=0}^{v'=v}</math>

or in a simpler form
<math display="block">t = {1 \over 2\alpha g} \ln \frac{1 + \alpha v}{1 - \alpha v} = \frac{\mathrm{artanh}(\alpha v)}{\alpha g},</math>
with artanh the [[inverse hyperbolic tangent]] function.

Alternatively,
<math display="block">\frac{1}{\alpha}\tanh(\alpha g t) = v,</math>
with tanh the [[hyperbolic tangent]] function. Assuming that ''g'' is positive (which it was defined to be), and substituting ''α'' back in, the speed ''v'' becomes
<math display="block">v = \sqrt\frac{2mg}{\rho A C_d} \tanh \left(t \sqrt{\frac{g \rho A C_d }{2m}}\right).</math>

Using the formula for terminal velocity
<math display="block">V_t = \sqrt\frac{2mg}{\rho A C_d}</math> 
the equation can be rewritten as
<math display="block"> v = V_t \tanh \left(t \frac{g}{V_t}\right).</math>

As time tends to infinity (''t'' → ∞), the hyperbolic tangent tends to 1, resulting in the terminal speed
<math display="block"> V_t = \lim_{t \to \infty} v(t) = \sqrt\frac{2mg}{\rho A C_d}.</math>

[[Image:Stokes sphere.svg|thumb|upright|Creeping flow past a sphere: [[Streamlines, streaklines, and pathlines|streamlines]], drag force ''F''<sub>d</sub> and force by gravity ''F''<sub>g</sub>]]

For very slow motion of the fluid, the inertia forces of the fluid are negligible (assumption of massless fluid) in comparison to other forces. Such flows are called [[Stokes flow|creeping or Stokes flows]] and the condition to be satisfied for the flows to be creeping flows is the [[Reynolds number]], <math>Re \ll 1</math>. The equation of motion for creeping flow (simplified [[Navier–Stokes equation]]) is given by:

<math display="block">{\mathbf \nabla} p = \mu \nabla^2 {\mathbf v} </math>

where:

* <math>\mathbf v</math> is the fluid velocity vector field,
* <math>p</math> is the fluid pressure field,
* <math>\mu</math> is the liquid/fluid [[viscosity]].

The analytical solution for the creeping flow around a sphere was first given by [[George Gabriel Stokes|Stokes]] in 1851.<ref>{{cite journal |last1=Stokes |first1=G. G. |title=On the effect of internal friction of fluids on the motion of pendulums |journal=Transactions of the Cambridge Philosophical Society |date=1851 |volume=9, part ii |pages=8–106 |bibcode=1851TCaPS...9....8S |url=https://babel.hathitrust.org/cgi/pt?id=mdp.39015012112531;view=1up;seq=208}}  The formula for terminal velocity (''V'')] appears on p. [52], equation (127).</ref>
From Stokes' solution, the drag force acting on the sphere of diameter <math>d</math> can be obtained as
{{NumBlk|:|<math> D = 3\pi \mu d V \qquad \text{or} \qquad C_d = \frac{24}{Re} </math>|{{EquationRef|6}}}}

where the Reynolds number, <math>Re = \frac{\rho d}{\mu}  V</math>. The expression for the drag force given by equation ({{EquationNote|6}}) is called [[Stokes' law]].

When the value of <math>C_d</math> is substituted in the equation ({{EquationNote|5}}), we obtain the expression for terminal speed of a spherical object moving under creeping flow conditions:<ref>{{cite book | first=H. | last=Lamb | author-link=Horace Lamb | year=1994 | title=Hydrodynamics | publisher=Cambridge University Press | edition=6th| isbn=978-0-521-45868-9  |pages=599}} Originally published in 1879, the 6th extended edition appeared first in 1932.</ref>

<math display="block">V_t = \frac{g d^2}{18 \mu} \left(\rho_s - \rho \right),</math>
where <math>\rho_s</math> is the density of the object.