Derivation of Stoke’s Law

Have you wondered why the raindrops falling from a great height acting under the force of gravity do not kill humans? This can be explained by Stoke’s law. This law is an interesting example of the retarding force which is proportional to the velocity. In 1851, George Gabriel Stokes derived an equation for the frictional force, also known as the drag force. In this article, let us look at what is Stoke’s law and its derivation.

What is Stoke’s Law?

Stoke’s Law is a mathematical equation that expresses the settling velocities of the small spherical particles in a fluid medium. The law is derived considering the forces acting on a particular particle as it sinks through the loquid column under the influence of gravity. The force that retards a sphere moving through a viscous fluid is directly proportional to the velocity and the radius of the sphere, and the viscosity of the fluid. Sir George G. Stokes, an English scientist expressed clearly the viscous drag force F as:

\(F=6\pi \eta rv\)

Stokes’s law finds application in several areas such as:

  • Settling of sediment in freshwater
  • Measurement of the viscosity of fluids

In the next section, let us understand the derivation of Stoke’s Law.

Stoke’s Law Derivation

The viscous force acting on a sphere is directly proportional to the following parameters:

  • the radius of the sphere
  • coefficient of viscosity
  • the velocity of the object

Mathematically, this is represented as

\(F\propto\eta^ar^bv^c\)

Now let us evaluate the values of a, b and c.

Substituting the proportionality sign with an equality sign, we get

\(F=k\eta^ar^bv^c\) (1)

Here, k is the constant of proportionality which is a numerical value and has no dimensions.

Writing the dimensions of parameters on either side of equation (1), we get

[MLT–2] = [ML–1T–1]a [L]b [LT-1]c

Simplifying the above equation, we get

[MLT–2] = Ma ⋅ L–a+b+c ⋅ T–a–c (2)

According to classical mechanics, mass, length and time are independent entities.

Equating the superscripts of mass, length and time respectively from equation (2), we get

a = 1 (3)

–a + b + c = 1 (4)

–a –c = 2 or a + c = 2 (5)

Substituting (3) in (5), we get

1 + c = 2

c = 1 (6)

Substituting the value of (3) & (6) in (4), we get

–1 + b + 1 = 1

b = 1 (7)

Substituting the value of (3), (6) and (7) in (1), we get

\(F=k\eta rv\)

The value of k for a spherical body was experimentally obtained as \(6\pi\)

Therefore, the viscous force on a spherical body falling through a liquid is given by the equation

\(F=6\pi \eta rv\)

Terminal Velocity Formula

In the case of raindrops, initially, it is due to the gravity that it accelerates. As the velocity increases, the retarding force also increases. Finally, when viscous force and the buoyant force is equal to the force due to gravity, the net force becomes zero and so does the acceleration. The raindrop then falls with a constant velocity. Thus, in equilibrium, the terminal velocity vt is given by the equation

\(v_t=\frac{2a^2(\rho -\sigma )g}{9\eta }\)

where \(\rho\) and σ are mass densities of sphere and fluid respectively.

From the equation above, we can infer that the terminal velocity depends on the square of the radius of the sphere and inversely proportional to the viscosity of the medium.

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