Terminal Velocity Formula: Stokes' Law Derivation, Examples

The terminal velocity formula for a small sphere sinking through a liquid or gas is $v_t = \frac{2 g r^2 (\rho - \sigma)}{9 \eta}$ . It is the constant speed the sphere settles into once the upward forces on it exactly cancel its weight, so it stops speeding up.

This is the form taught in Class 11 under the mechanical properties of fluids, and it comes straight from Stokes' law for viscous drag. It holds for slow, smooth (laminar) flow around small spheres: oil drops, fine raindrops, ball bearings in glycerine. A falling skydiver also reaches a terminal speed, but there the drag grows with the square of speed, so a different formula applies. Both versions sit on the fluids formula sheet.

Quick facts

Formula
$v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta}$
SI unit
metre per second (m/s)
Dimensional formula
[M⁰ L¹ T⁻¹]
Based on
Stokes' law, $F = 6 \pi \eta r v$
Depends on
r², so doubling the radius gives four times the speed
Valid for
small spheres in slow laminar flow
If σ > ρ
the answer is negative: the sphere rises, like an air bubble in water

Symbols in the terminal velocity formula

What each letter in vₜ = 2r²(ρ − σ)g / 9η stands for
Symbol	Meaning	SI unit
$v_t$	Terminal velocity, the steady falling speed	m/s
$r$	Radius of the sphere (not its diameter)	m
$\rho$	Density of the sphere material	kg/m³
$\sigma$	Density of the fluid it falls through	kg/m³
$g$	Free fall acceleration, 9.8 m/s²	m/s²
$\eta$	Coefficient of viscosity of the fluid	Pa s

Some books write the fluid density as $\rho_0$ or $\rho_f$ instead of σ. The idea is the same: the bracket $(\rho - \sigma)$ is the density the sphere has in excess of its surroundings. A steel ball in water has a big excess and drops fast. A grain of sand in water has a smaller one, and fine dust in air has a tiny radius, so it hangs for a long time.

Derivation of the terminal velocity formula for Class 11

Three forces act on a sphere of radius r falling through a fluid:

Weight, downward: $W = \frac{4}{3}\pi r^3 \rho g$
Buoyant force (upthrust), upward: $U = \frac{4}{3}\pi r^3 \sigma g$
Viscous drag from Stokes' law, upward: $F = 6 \pi \eta r v$

At first the drag is small and the sphere accelerates. Drag grows with speed until the forces balance. At that point the net force is zero, so the speed stays fixed at $v_t$ :

$6 \pi \eta r v_t + \frac{4}{3}\pi r^3 \sigma g = \frac{4}{3}\pi r^3 \rho g$

$6 \pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho - \sigma) g \quad\Rightarrow\quad v_t = \frac{2 g r^2 (\rho - \sigma)}{9 \eta}$

The π cancels and $\frac{4}{3} \div 6 = \frac{2}{9}$ . That fraction is the one students most often get wrong, so rederive it once rather than memorising it. The drag term is covered on the viscous force formula page, and the upthrust term on the buoyant force formula page.

Worked examples

Example 1: a steel ball in glycerine

A steel ball of radius 1.0 mm (density 7800 kg/m³) is dropped into glycerine (density 1260 kg/m³, viscosity 1.5 Pa s). Find its terminal velocity.

Given: r = 1.0 × 10⁻³ m, ρ = 7800 kg/m³, σ = 1260 kg/m³, η = 1.5 Pa s, g = 9.8 m/s².

$v_t = \frac{2 (1.0\times10^{-3})^2 (7800 - 1260)(9.8)}{9 (1.5)} = \frac{0.1282}{13.5} = 9.5 \times 10^{-3}\ \text{m/s}$

That is about 9.5 mm/s, slow enough to time by eye in a tall jar.

Example 2: a fine raindrop, with a unit conversion

A cloud droplet has a radius of 0.010 mm. Take water density 1000 kg/m³, air density 1.2 kg/m³ and air viscosity 1.8 × 10⁻⁵ Pa s. Find its terminal velocity.

Convert: r = 0.010 mm = 1.0 × 10⁻⁵ m. The density difference is 1000 − 1.2 = 998.8 kg/m³.

$v_t = \frac{2 (1.0\times10^{-5})^2 (998.8)(9.8)}{9 (1.8\times10^{-5})} = \frac{1.958\times10^{-6}}{1.62\times10^{-4}} = 0.012\ \text{m/s}$

About 1.2 cm/s. This is why tiny droplets seem to float in a cloud.

Example 3: finding the viscosity

A steel ball of radius 2.0 mm (7800 kg/m³) reaches a steady 2.5 cm/s in a thick oil of density 900 kg/m³. Find the oil's viscosity.

Rearrange for viscosity: $\eta = \frac{2}{9}\,\frac{r^2 (\rho - \sigma) g}{v_t} = \frac{2 (2.0\times10^{-3})^2 (6900)(9.8)}{9 (0.025)}$

$\eta = \frac{0.5410}{0.225} = 2.4\ \text{Pa s}$

Convert the radius to metres before squaring. A radius left in millimetres makes the answer a million times too big.

Formula for terminal velocity at high speed

Stokes' law fails once the flow behind the object turns turbulent. For a skydiver, a parachute or a falling hailstone, the drag is close to $\frac{1}{2} \rho_{air} C_d A v^2$ , with $C_d$ the drag coefficient and A the frontal area. Setting this equal to the weight mg gives

$v_t = \sqrt{\frac{2 m g}{\rho_{air} C_d A}}$

Here the speed grows with the square root of the mass, not with r². Spreading the arms raises A and lowers $v_t$ , which is how a skydiver controls the fall. US courses meet this version through the drag force formula; the Class 11 syllabus sticks to the Stokes form.

Graph and special cases

A velocity time graph for a sphere released from rest starts with slope close to g reduced by buoyancy, then curves over and flattens at $v_t$ . The acceleration falls to zero as the curve levels out. Displacement keeps rising in a straight line after that.

Fluid denser than the sphere (σ > ρ): the formula gives a negative value, meaning a steady upward speed. Air bubbles in a fizzy drink rise this way.
Equal densities: $v_t = 0$ . The sphere stays wherever you put it.
Hot fluid: liquids get less viscous when heated, so η drops and $v_t$ rises. Gases behave the other way.

The falling ball method turns this round to measure η, as described on the coefficient of viscosity formula page.

Common mistakes

Using the diameter for r. Halve a given diameter first.
Writing 9/2 instead of 2/9. Check by dimensions: the answer must come out in m/s.
Leaving out σ. Dropping buoyancy is only fair for a dense object in air.
Applying it to large objects. Footballs and people are in turbulent flow, so use the square root form.
Calling terminal velocity a maximum acceleration. At terminal velocity the acceleration is zero.

Frequently asked questions

What is the formula of terminal velocity in Class 11?

For a sphere in a viscous fluid, vₜ = 2r²(ρ − σ)g / 9η, where r is the radius, ρ and σ are the densities of the sphere and fluid, and η is the viscosity. It comes from balancing weight against upthrust plus Stokes drag.

What is the SI unit of terminal velocity?

It is a speed, so its unit is the metre per second, and its bracket dimensions are [M⁰ L¹ T⁻¹] like any velocity.

How does terminal velocity depend on the radius?

In the Stokes form it is proportional to r². A drop with twice the radius falls four times as fast, which is why big raindrops reach the ground quickly.

What is the net force at terminal velocity?

Zero. Weight is exactly balanced by the buoyant force and the viscous drag, so the object moves at constant velocity with no acceleration.

Is there a formula for terminal velocity of a skydiver?

Yes. With drag proportional to v², vₜ = √(2mg / ρC_dA), where ρ is the air density, C_d the drag coefficient and A the frontal area.