drag | Physics Library

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.

v = v − v , rel object fluid

then the drag force has a direction opposite to v_rel.

Two especially useful models are

FD = − bvrel

for linear drag and

1 FD = − -ρCDA |vrel|vrel 2

for quadratic drag.

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

1 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,

ρvL Re = ----, μ

where L is a characteristic length and μ 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 v². This distinction is not a sharp universal boundary: the drag coefficient itself can vary with Reynolds number.

2 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,

m dv- = mg − F − F . dt B D

The buoyant force is

FB = ρfV g,

where ρ_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.

3 Terminal velocity

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

mg − FB − FD = 0.

At this point,

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.

4 Linear drag

For linear drag,

FD = bv,

where b has units of kg∕s.

Ignoring buoyancy and taking downward as positive,

dv m ---= mg − bv. dt

The terminal speed follows from dv∕dt = 0:

|-----mg--| |vT = ----. --------b--

Define the time constant

m- τ = b .

Then the solution for arbitrary initial velocity v₀ is

|--------------------------| |v(t) = vT + (v0 − vT)e− t∕τ . ---------------------------

For release from rest,

|--------------------| | ( −t∕τ) | v(t)-=-vT--1 −-e-----.

The speed approaches v_T exponentially.

5 Quadratic drag

For quadratic drag,

F = cv2, D

with

1- c = 2ρCDA.

Ignoring buoyancy and taking downward as positive,

dv- 2 m dt = mg − cv .

At terminal speed,

mg = cv2, T

|-----------------------| | ∘ ---- ∘ -------| |vT = mg--= 2mg---. ----------c-------ρCDA---

For an object released from rest, the exact solution is

|---------------------| | ( gt ) | |v(t) = vT tanh --- . -----------------vT---

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

|--------------------------| | v2 [ ( gt) ] | |x(t) = -T-ln cosh --- . --------g------------vT----

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

6 Including buoyancy

For a body of volume V and mass m falling through a fluid of density ρ_f, define

Feff = mg − ρfV g.

For quadratic drag,

Feff = 1-ρfCDAv2 , 2 T

|-----∘---------------| | 2(m − ρfV )g | vT = ------------- | -----------ρfCDA-------

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.

7 Stokes drag

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

Re ≪ 1,

Stokes’ law gives

|-------------| -FD--=-6πμRv--,

where R is the sphere radius.

For a sphere of density ρ_s falling in a fluid of density ρ_f,

4 3 --πR (ρs − ρf )g = 6πμRvT . 3

Therefore

|--------------------| | 2R2 (ρs − ρf)g | |vT = -------------- . ------------9μ-------

In terms of sphere diameter d = 2R,

|------------------| | gd2(ρs-−-ρf) | vT = 18μ . --------------------

8 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,

F = 1ρC Av2. D 2 D

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

A = πR2.

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

9 Scaling of terminal speed

For quadratic drag without buoyancy,

∘ ------- 2mg vT = ------. ρCDA

Thus,

√ -- 1 1 1 vT ∝ m, vT ∝ √---, vT ∝ √--, vT ∝ √----. A ρ CD

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

10 Power dissipated by drag

The instantaneous mechanical power associated with drag is

PD = FD ⋅ v.

Because drag opposes the velocity,

PD < 0.

For linear drag,

2 PD = − bv .

For quadratic drag,

3 PD = − cv .

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.

11 Worked examples

Example 1: linear drag

A 0.20 kg object falls with linear drag coefficient b = 0.50 kg∕s. Neglect buoyancy.

The terminal speed is

mg (0.20)(9.81) vT = ----= ------------= 3.92 m ∕s. b 0.50

The time constant is

τ = m- = 0.40 s. b

Released from rest,

( −t∕0.40) v(t) = 3.92 1 − e m∕s.

Example 2: speed after one time constant

For the preceding object,

v(τ) = vT(1 − e−1) ≈ 0.632vT .

Thus,

v(τ) ≈ 2.48m ∕s.

Example 3: quadratic terminal speed of a falling person

Consider a person of mass

m = 80 kg,

with

3 2 ρ = 1.225 kg∕m , CD = 1.0, A = 0.70m .

Then

∘ ------------------ 2(80)(9.81) vT = ------------------≈ 42.8 m ∕s. (1.225)(1.0)(0.70)

This is about

154km ∕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

--1---- vT ∝ √C---A-, D

the new terminal speed is

1- vT,new = 4vT,old.

Example 5: small sphere in a viscous fluid

A small sphere has

R = 0.50mm, ρ = 2500kg ∕m3, ρ = 1000 kg∕m3, s f

and falls through a fluid with

μ = 1.0Pa s.

Using Stokes’ terminal-speed result,

2R2-(ρs −-ρf)g vT = 9μ ,

we obtain

v ≈ 8.18 × 10− 4m ∕s. T

Example 6: force balance at terminal speed

A falling object has weight

W = 120N

and buoyant force

F = 8N. B

At terminal speed,

FD + FB = W,

FD = 120 − 8 = 112N.

12 Common mistakes

Drawing drag in the wrong direction. Drag opposes motion relative to the fluid, not necessarily motion relative to the ground.
Treating C_D as a universal constant independent of Reynolds number.
Assuming all drag is proportional to v².
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.

13 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 kg body has linear drag coefficient b = 0.20 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 0.632v_T after one time constant.
Derive the quadratic-drag terminal speed v_T = .
A skydiver doubles effective frontal area without changing C_D. 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 V .
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 kg falling body when buoyancy is negligible and v_T = 40 m∕s.
Explain why a single constant drag coefficient may fail over a very large range of speeds.
Starting from $m ˙v = mg − cv2,$
verify directly that
$v(t) = v tanh (gt∕v ) T T$
satisfies the equation for release from rest.

14 GRE-style speed checks

For linear drag F_D = bv, the SI units of b are (A) kg (B) kg∕s (C) N s²∕m (D) 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, $vT ∝$
(A) m (B) m² (C) (D) 1∕m. Answer: C.
If effective area is multiplied by four while all other quadratic-drag parameters remain unchanged, terminal speed is multiplied by (A) 4 (B) 2 (C) 1∕2 (D) 1∕4. 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.

15 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.

References

[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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