Drag and terminal velocity

Drag is a resistive force exerted by a fluid on an object moving relative to that fluid. The drag force acts opposite to the object's velocity relative to the surrounding fluid. Unlike the simplest dry-friction model, drag generally depends on speed and on properties of both the object and the fluid.

$\displaystyle \mathbf v_{\rm rel}=\mathbf v_{\rm object}-\mathbf v_{\rm fluid},$

then the drag force has a direction opposite to $\mathbf v_{\rm rel}$ .

Two especially useful models are

$\displaystyle \mathbf F_D=-b\mathbf v_{\rm rel}$

for linear drag and

$\displaystyle \mathbf F_D = -\frac12\rho C_D A \vert\mathbf v_{\rm rel}\vert\mathbf v_{\rm rel}$

for quadratic drag.

Here $\rho$ is the fluid density, $C_D$ is a dimensionless drag coefficient, and $A$ is a reference area, usually the projected frontal area.

Why drag depends on speed

Drag arises from interaction between a moving body and the surrounding fluid. Depending on the geometry and flow regime, viscous shear, pressure differences, boundary-layer behavior, and wake formation all contribute.

A useful dimensionless quantity is the Reynolds number,

$\displaystyle \mathrm{Re}=\frac{\rho vL}{\mu},$

where

is a characteristic length and $\mu$ is the dynamic viscosity.

At very small Reynolds number, viscous effects dominate and a linear drag model is often appropriate. At larger Reynolds number, many bodies are better approximated over useful speed ranges by a force proportional to $v^2$ . This distinction is not a sharp universal boundary: the drag coefficient itself can vary with Reynolds number.

Free-body diagram for a falling body

Consider an object falling vertically through a fluid. The principal forces are its weight, the drag force, and the buoyant force.

Taking downward as positive,

$\displaystyle m\frac{dv}{dt} = mg-F_B-F_D.$

The buoyant force is

$\displaystyle F_B=\rho_f Vg,$

where $\rho_f$ is the fluid density and $V$

is the displaced volume.

In air, buoyancy is often small compared with the weight of a dense falling body and may be neglected. In liquids, or for low-density objects, it can be important.

Terminal velocity

As a falling object speeds up, its drag generally increases. A steady speed is reached when the net force vanishes:

$\displaystyle mg-F_B-F_D=0.$

At this point,

$\displaystyle \frac{dv}{dt}=0.$

The corresponding speed is the terminal speed. In a fixed vertical coordinate system its signed value may be called the terminal velocity.

Terminal speed is better understood as an asymptotic steady speed than as a universal maximum speed. If an object enters a fluid moving faster than its local terminal speed, drag can decelerate it toward the same terminal state from above.

Linear drag

For linear drag,

$\displaystyle F_D=bv,$

where

has units of $\mathrm{kg/s}$ .

Ignoring buoyancy and taking downward as positive,

$\displaystyle m\frac{dv}{dt}=mg-bv.$

The terminal speed follows from $dv/dt=0$ :

$\displaystyle \boxed{v_T=\frac{mg}{b}}.$

Define the time constant

$\displaystyle \tau=\frac{m}{b}.$

Then the solution for arbitrary initial velocity $v_0$

$\displaystyle \boxed{ v(t)=v_T+(v_0-v_T)e^{-t/\tau} }.$

For release from rest,

$\displaystyle \boxed{ v(t)=v_T\left(1-e^{-t/\tau}\right) }.$

The speed approaches $v_T$ exponentially.

Quadratic drag

For quadratic drag,

$\displaystyle F_D=cv^2,$

with

$\displaystyle c=\frac12\rho C_DA.$

Ignoring buoyancy and taking downward as positive,

$\displaystyle m\frac{dv}{dt}=mg-cv^2.$

At terminal speed,

$\displaystyle mg=cv_T^2,$

$\displaystyle \boxed{ v_T=\sqrt{\frac{mg}{c}} = \sqrt{\frac{2mg}{\rho C_DA}} }.$

For an object released from rest, the exact solution is

$\displaystyle \boxed{ v(t)=v_T\tanh\left(\frac{gt}{v_T}\right) }.$

Integrating once more gives the downward displacement from $x(0)=0$ :

$\displaystyle \boxed{ x(t) = \frac{v_T^2}{g} \ln\left[ \cosh\left(\frac{gt}{v_T}\right) \right] }.$

Both the linear and quadratic models approach a steady terminal speed asymptotically, but their time dependence is different.

Including buoyancy

For a body of volume $V$ and mass $m$ falling through a fluid of density $\rho_f$ , define

$\displaystyle F_{\rm eff}=mg-\rho_fVg.$

For quadratic drag,

$\displaystyle F_{\rm eff} = \frac12\rho_fC_DAv_T^2,$

$\displaystyle \boxed{ v_T = \sqrt{ \frac{2(m-\rho_fV)g} {\rho_fC_DA} } }$

for a body denser than the fluid.

If the body is less dense than the fluid, the steady motion may be upward rather than downward, as for a rising bubble.

Stokes drag

For a small sphere moving slowly through a viscous fluid in the creeping-flow regime,

$\displaystyle \mathrm{Re}\ll1,$

Stokes' law gives

$\displaystyle \boxed{ F_D=6\pi\mu Rv },$

where

is the sphere radius.

For a sphere of density $\rho_s$ falling in a fluid of density $\rho_f$ ,

$\displaystyle \frac43\pi R^3(\rho_s-\rho_f)g = 6\pi\mu Rv_T.$

Therefore

$\displaystyle \boxed{ v_T = \frac{2R^2(\rho_s-\rho_f)g}{9\mu} }.$

In terms of sphere diameter $d=2R$ ,

$\displaystyle \boxed{ v_T = \frac{gd^2(\rho_s-\rho_f)}{18\mu} }.$

Drag coefficient

The drag coefficient $C_D$ is not a universal constant for a particular material. It depends on body shape, orientation, surface condition, and flow regime.

For quadratic drag,

$\displaystyle F_D=\frac12\rho C_DAv^2.$

The reference area must be stated consistently with the definition of $C_D$ . For a sphere it is normally the projected circular area,

$\displaystyle A=\pi R^2.$

Because $C_D$ can vary with Reynolds number, the constant- $C_D$ quadratic model is an approximation over a chosen operating range.

Scaling of terminal speed

For quadratic drag without buoyancy,

$\displaystyle v_T=\sqrt{\frac{2mg}{\rho C_DA}}.$

Thus,

$\displaystyle v_T\propto\sqrt{m}, \qquad v_T\propto\frac{1}{\sqrt{A}}, \qquad v_T\propto\frac{1}{\sqrt{\rho}}, \qquad v_T\propto\frac{1}{\sqrt{C_D}}.$

Increasing frontal area or drag coefficient lowers terminal speed, which is the central idea behind a parachute.

Power dissipated by drag

The instantaneous mechanical power associated with drag is

$\displaystyle P_D=\mathbf F_D\cdot\mathbf v.$

Because drag opposes the velocity,

$\displaystyle P_D<0.$

For linear drag,

$\displaystyle P_D=-bv^2.$

For quadratic drag,

$\displaystyle P_D=-cv^3.$

At terminal speed, kinetic energy is no longer changing even though energy is continuously being transferred to the surrounding fluid. Gravitational potential energy is then converted into thermal energy, wake motion, sound, and other fluid disturbances.

Worked examples

Example 1: linear drag

A $0.20\,\mathrm{kg}$ object falls with linear drag coefficient $b=0.50\,\mathrm{kg/s}$ . Neglect buoyancy.

The terminal speed is

$\displaystyle v_T=\frac{mg}{b} = \frac{(0.20)(9.81)}{0.50} = 3.92\,\mathrm{m/s}.$

The time constant is

$\displaystyle \tau=\frac{m}{b}=0.40\,\mathrm{s}.$

Released from rest,

$\displaystyle v(t) = 3.92\left(1-e^{-t/0.40}\right)\,\mathrm{m/s}.$

Example 2: speed after one time constant

For the preceding object,

$\displaystyle v(\tau)=v_T(1-e^{-1})\approx0.632v_T.$

Thus,

$\displaystyle v(\tau)\approx2.48\,\mathrm{m/s}.$

Example 3: quadratic terminal speed of a falling person

Consider a person of mass

$\displaystyle m=80\,\mathrm{kg},$

with

$\displaystyle \rho=1.225\,\mathrm{kg/m^3}, \qquad C_D=1.0, \qquad A=0.70\,\mathrm{m^2}.$

Then

$\displaystyle v_T = \sqrt{ \frac{2(80)(9.81)} {(1.225)(1.0)(0.70)} } \approx42.8\,\mathrm{m/s}.$

This is about

$\displaystyle 154\,\mathrm{km/h}.$

The number is only an estimate because $C_D$ , area, orientation, and air density can vary.

Example 4: opening a parachute

Suppose the effective product $C_DA$ increases by a factor of $16$ when a parachute opens. Since

$\displaystyle v_T\propto\frac{1}{\sqrt{C_DA}},$

the new terminal speed is

$\displaystyle v_{T,\mathrm{new}}=\frac14v_{T,\mathrm{old}}.$

Example 5: small sphere in a viscous fluid

A small sphere has

$\displaystyle R=0.50\,\mathrm{mm}, \qquad \rho_s=2500\,\mathrm{kg/m^3}, \qquad \rho_f=1000\,\mathrm{kg/m^3},$

and falls through a fluid with

$\displaystyle \mu=1.0\,\mathrm{Pa\,s}.$

Using Stokes' terminal-speed result,

$\displaystyle v_T = \frac{2R^2(\rho_s-\rho_f)g}{9\mu},$

we obtain

$\displaystyle v_T\approx8.18\times10^{-4}\,\mathrm{m/s}.$

Example 6: force balance at terminal speed

A falling object has weight

$\displaystyle W=120\,\mathrm{N}$

and buoyant force

$\displaystyle F_B=8\,\mathrm{N}.$

At terminal speed,

$\displaystyle F_D+F_B=W,$

$\displaystyle F_D=120-8=112\,\mathrm{N}.$

Common mistakes

Drawing drag in the wrong direction. Drag opposes motion relative to the fluid, not necessarily motion relative to the ground.
Treating as a universal constant independent of Reynolds number.
Assuming all drag is proportional to .
Forgetting buoyancy when the displaced-fluid weight is significant.
Calling terminal speed the greatest speed an object can ever possess.
Using total surface area when the drag formula requires projected reference area.
Forgetting the absolute-value form in the vector quadratic-drag law when motion can reverse.
Applying Stokes' law outside the low-Reynolds-number regime.

Practice exercises

An object moves east through still air. In what direction does its drag force act?
A body is carried east by a wind faster than the body moves east relative to the ground. What direction can the drag force have?
A $0.50\,\mathrm{kg}$ body has linear drag coefficient $b=0.20\,\mathrm{kg/s}$ . Find its terminal speed.
For the previous body, find the time constant.
Show that an object released from rest with linear drag reaches after one time constant.
Derive the quadratic-drag terminal speed $v_T=\sqrt{2mg/(\rho C_DA)}$ .
A skydiver doubles effective frontal area without changing . By what factor does the quadratic-drag terminal speed change?
A body enters a fluid at a downward speed greater than its terminal speed. State the initial direction of its acceleration.
Include buoyancy and derive the quadratic terminal speed of a body of volume .
Derive the Stokes terminal speed of a small sphere.
For a sphere obeying Stokes' law, how does terminal speed scale with radius?
Estimate the drag power dissipated at terminal speed for an $80\,\mathrm{kg}$ falling body when buoyancy is negligible and $v_T=40\,\mathrm{m/s}$ .
Explain why a single constant drag coefficient may fail over a very large range of speeds.
Starting from

$\displaystyle m\dot v=mg-cv^2,$
verify directly that

$\displaystyle v(t)=v_T\tanh(gt/v_T)$
satisfies the equation for release from rest.

GRE-style speed checks

For linear drag , the SI units of are (A) $\mathrm{kg}$ (B) $\mathrm{kg/s}$ (C) $\mathrm{N\,s^2/m}$ (D) $\mathrm{m/s}$ . Answer: B.
A falling object reaches terminal speed when (A) gravity vanishes (B) velocity vanishes (C) the net force vanishes (D) drag vanishes. Answer: C.
With quadratic drag and negligible buoyancy,

$\displaystyle v_T\propto$
(A) (B) (C) $\sqrt m$ (D) . Answer: C.
If effective area is multiplied by four while all other quadratic-drag parameters remain unchanged, terminal speed is multiplied by (A) (B) (C) (D) . Answer: C.
At very small Reynolds number around a sphere, the standard drag law is (A) Coulomb friction (B) Stokes drag (C) Hooke's law (D) inverse-square drag. Answer: B.
An object moving faster than its terminal falling speed can (A) never slow down (B) accelerate downward indefinitely (C) decelerate toward terminal speed (D) have zero drag. Answer: C.

Source and licensing note

The general definitions of drag and terminal velocity were informed by the Wikipedia articles Drag (physics) and Terminal velocity, whose text is available under a Creative Commons Attribution–ShareAlike license.

The development here has been reorganized and expanded for PhysicsLibrary, including separate linear- and quadratic-drag solutions, buoyancy, Stokes drag, energy dissipation, worked examples, practice exercises, and GRE-style checks.

Standard Newtonian drag and terminal-speed formulas were cross-checked against openly available Physics LibreTexts material.

All figures included with this entry were generated specifically for this PhysicsLibrary article.

Bibliography

1: Wikipedia contributors, “Drag (physics),” Wikipedia, The Free Encyclopedia. Drag (physics)
2: Wikipedia contributors, “Terminal velocity,” Wikipedia, The Free Encyclopedia. Terminal velocity
3: Physics LibreTexts, “Drag Force and Terminal Speed.” Physics LibreTexts

Unless otherwise noted, this PhysicsLibrary entry is intended for release under the Creative Commons Attribution–ShareAlike 4.0 International license.

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