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Terminal velocity of a sphere
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Reaches constant speed when weight, buoyancy and drag balance.
v_t = 2r²(ρ−σ)g/9ηAt terminal speed the weight equals the buoyancy plus drag: v_t = 2r²(ρ − σ)g/(9η). The speed grows with the square of the radius, so large drops fall faster than small ones.
Fog droplets with r = 10 μm fall a few millimetres per second.
This formula needs Stokes' law to be valid.
In more detail
A sphere falling through a viscous fluid reaches terminal velocity when drag equals its weight minus buoyancy. For small spheres, this follows from Stokes' law and gives v = 2r²(ρ−ρf)g / 9η.
Real-world uses
- Sedimentation tests measure the size of particles in soil and water.
- Blood tests use settling rates to infer properties of blood.
- Raindrops and sedimentation tanks reach a steady fall speed when drag, weight and buoyancy balance.
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Frequently asked questions
Does a heavier sphere fall faster?
Yes, if it is denser, because the net force driving it down increases.
What is the role of buoyancy?
It reduces the net downward force, so it lowers the terminal velocity.
Why does a bigger sphere fall faster?
The terminal velocity depends on r squared, so a larger particle has a larger weight and falls faster.
Related topics in Newton's Laws & Forces
Pseudo forces · Drag force · Stokes' law · Spring force · Springs in series and parallel · Hooke's law limits