Stokes Law Terminal Velocity Calculator
Find the Stokes' law terminal velocity v = 2r²(ρ_p−ρ_f)g/(9μ), the constant settling speed a small sphere reaches in a viscous fluid.
🔵 What is the Stokes Law Terminal Velocity Calculator?
This Stokes' law calculator finds v=2r²(ρ_p−ρ_f)g/(9μ), the constant terminal velocity a small sphere reaches settling (or rising) through a viscous fluid. Enter the particle radius, particle and fluid densities, and fluid viscosity, and it returns the terminal velocity and direction.
Stokes' law balances gravity (and buoyancy) against viscous drag for a small sphere at low Reynolds number, giving this simple closed-form terminal velocity.
Because velocity scales with radius squared, particle size has an outsized effect on settling speed, exactly why fine dust and smoke stay suspended in air far longer than sand or gravel.
This calculator is useful for fluid dynamics, sedimentation analysis, and materials science students studying particle settling behavior in liquids and gases.
📐 Formula
📖 How to Use This Calculator
Steps
💡 Example Calculations
Example 1 - Fine silt settling in water
Example 2 - Oil droplet rising in water
Example 3 - Fine dust settling in air
❓ Frequently Asked Questions
🔗 Related Calculators
What is Stokes' law?
Stokes' law describes the drag force on a small sphere moving through a viscous fluid at low Reynolds number, and by balancing this drag against gravity (minus buoyancy), it gives the constant terminal velocity the sphere eventually reaches, whether settling or rising.
What is the formula for Stokes' law terminal velocity?
v = 2r²(ρ_p−ρ_f)g/(9μ), where r is the particle's radius, ρ_p is the particle's density, ρ_f is the fluid's density, g is gravitational acceleration, and μ is the fluid's dynamic viscosity.
When is Stokes' law valid?
Stokes' law is only accurate for very low Reynolds number flow, generally Re well below 1 (some sources use Re<0.1 for high accuracy, others allow up to Re≈1). This calculator reports the resulting particle Reynolds number so you can check whether your specific case falls within this valid range.
Why does the formula give a negative velocity sometimes?
If the particle's density is less than the surrounding fluid's density (ρ_p < ρ_f), the term (ρ_p−ρ_f) becomes negative, correctly indicating the particle is buoyant and rises rather than sinks, exactly like an air bubble in water or a light oil droplet in a denser liquid.
Why does terminal velocity depend on radius squared?
Gravitational (and buoyant) force on a sphere scales with its volume, proportional to radius cubed, while Stokes drag scales only with radius. Balancing these two forces at terminal velocity leaves a net radius-squared dependence, which is why particle size has such an outsized effect on settling speed.
What is sedimentation analysis and how does this relate to Stokes' law?
Sedimentation analysis measures particle size distributions in a sample by timing how fast particles settle through a fluid column and applying Stokes' law in reverse (from measured settling velocity back to inferred particle radius), a classic technique in soil science, pharmaceuticals, and materials characterization.
Why do fine dust and smoke particles stay suspended so long in air?
Because Stokes' law terminal velocity scales with radius squared, very small particles (like fine dust, smoke, or fog droplets) settle extremely slowly, often slow enough that ordinary air currents keep them effectively suspended indefinitely, which is exactly why airborne dust and smog can persist for so long.
Does viscosity or density have a bigger effect on terminal velocity?
Terminal velocity is inversely proportional to viscosity (doubling viscosity halves velocity) and directly proportional to the density difference between particle and fluid, so both matter, but a highly viscous fluid can dramatically slow settling even for a fairly dense particle, which is why particles settle far more slowly in glycerin or honey than in water.
How is Stokes' law used in real centrifuge or sedimentation equipment?
Centrifuges effectively replace gravitational acceleration g with a much larger centrifugal acceleration, dramatically increasing the effective 'g' in the Stokes' law formula and speeding up settling by orders of magnitude, which is why centrifugation can separate particles that would take impractically long to settle under normal gravity alone.
What happens if Reynolds number is too high for this formula?
Beyond the low-Reynolds-number (Stokes) regime, drag no longer scales simply with velocity, and the terminal velocity formula requires more complex drag coefficient correlations (like those used in the general Drag Force and Drag Coefficient Calculator) instead of this simplified closed-form result.