Stokes' Law Calculator

Terminal velocity and drag for a sphere in viscous fluid (Re < 1)

Parameters

mⓘ
Pa·sⓘ
kg/m³ⓘ
kg/m³ⓘ
m/s²ⓘ
Show Trail

Controls

xⓘ

Calculated Values

Terminal Velocity:
14.82;m/s14.82;m/s
Drag Force at v_t:
0.00;N0.00;N

Examples

Steel ball in glycerin

r=1 mm, μ=1.5 Pa·s.

    Small pollen in air

    Low μ, low ρ_f.

      Visualization

      Stokes' Law — Viscous Drag on Spheres

      Stokes' law gives the drag force on a rigid sphere moving slowly through a viscous Newtonian fluid: F_d = 6πμrv, where μ is dynamic viscosity (Pa·s = N·s/m²), r is sphere radius (m), and v is speed relative to the fluid. The drag is linear in v and opposite to motion — characteristic of creeping flow (Reynolds number Re < 1).

      Reynolds number Re = ρvD/μ = ρv(2r)/μ compares inertial to viscous forces. Stokes law is valid for Re ≲ 1. For Re > 1, inertia dominates, drag scales roughly as v² (see drag coefficient C_d), and Stokes underestimates resistance.

      At terminal velocity in a gravitational field, forces balance: (ρ_s − ρ_f) × (4/3)πr³g = 6πμrv_t. Solving gives v_t = 2r²(ρ_s − ρ_f)g/(9μ). Larger, denser spheres fall faster; higher viscosity slows settling. Steel (ρ ≈ 7800 kg/m³) in water (ρ = 1000) falls much faster than in glycerin (μ ≈ 1.5 Pa·s).

      Buoyancy F_b = ρ_f V g acts upward on submerged volume V = (4/3)πr³. The net driving force is weight minus buoyancy, proportional to (ρ_s − ρ_f). In air, ρ_f is small so buoyancy is often neglected for dense particles.

      Applications in viscometry: drop a steel ball in an unknown liquid, measure v_t, and solve for μ. Conversely, known μ gives particle size from settling rate — used in sedimentation analysis and air quality monitoring.

      George Gabriel Stokes (1851) derived this from the Navier–Stokes equations by neglecting inertia. It underpins aerosol physics, blood sedimentation (ESR tests), fog and cloud microphysics, and microfluidics at low Re.

      Key Concepts

      • F_d = 6πμrv (sphere, Re < 1)
      • v_t = 2r²(ρ_s−ρ_f)g/(9μ)
      • Re = ρv(2r)/μ; creeping flow Re ≲ 1
      • Buoyancy: F_b = ρ_f × (4/3)πr³g
      • Drag linear in v (not v²)
      • μ in Pa·s; kinematic ν = μ/ρ in m²/s

      Real-World Applications

      • Laboratory viscometers (falling ball method)
      • Sedimentation tanks and centrifuges
      • Fog, mist, and fine dust settling in air
      • ESR and blood cell sedimentation in medicine
      • Class 11–12 terminal velocity problems
      • Colloid stability and Brownian motion context

      Explore Further

      More fluid mechanics tools

      Physics Equations

      Stokes Drag:
      Fd=6πμrvF_d = 6\pi\mu r v
      Terminal Speed:
      vt=2r2(ρs−ρf)g9μv_t = \frac{2r^2(\rho_s-\rho_f)g}{9\mu}

      Step-by-Step Solution

      See how the main results are calculated.

      1

      Step 1: Stokes Drag

      Equation:

      Fd=6πμrvF_d = 6\pi\mu r v

      Explanation:

      Laminar drag on sphere; valid for Re < 1.

      2

      Step 2: Terminal Velocity

      Equation:

      vt=2r2(ρs−ρf)g9μv_t = \frac{2r^2(\rho_s-\rho_f)g}{9\mu}

      Explanation:

      At terminal speed: weight = buoyancy + drag.

      3

      Step 3: Density Difference

      Calculation:

      ρs−ρf=7800−1000=6800.00 kg/m3\rho_s - \rho_f = 7800 - 1000 = 6800.00 \text{ kg/m}^3
      4

      Step 4: Compute v_t

      Calculation:

      vt=2×0.0012×6800.00×9.819×0.001=1.4824e+1 m/sv_t = \frac{2 \times 0.001^2 \times 6800.00 \times 9.81}{9 \times 0.001} = 1.4824e+1 \text{ m/s}

      Result:

      vt=1.4824e+1m/sv_t = 1.4824e+1 m/s
      5

      Step 5: Drag at Terminal Speed

      Calculation:

      Fd=6πμrvt=2.7943e−4 NF_d = 6\pi\mu r v_t = 2.7943e-4 \text{ N}

      Result:

      Fd=2.7943e−4NF_d = 2.7943e-4 N
      6

      Step 6: Validity

      Small spheres in viscous fluids.

      Explanation:

      Sedimentation, blood flow, fog droplets.

      Frequently Asked Questions (FAQ)

      Sphere vs cube?

      Stokes is for spheres; other shapes need different C_d.

      Rising bubble?

      Use (ρ_fluid − ρ_gas); buoyancy dominates.

      Units of μ?

      Pa·s = kg/(m·s); 1 Poise = 0.1 Pa·s.

      Brownian motion?

      Different regime — thermal fluctuations, not Stokes settling.

      Multiple spheres?

      Concentrated suspensions need corrections.

      Practice MCQs

      1. Stokes drag is proportional to:
      2. Larger radius r gives terminal speed:
      3. Stokes law fails when:
      4. F_d units:
      5. In honey vs water, sphere falls:
      6. Buoyancy opposes: