Stokes Law: Formula, Derivation, Terminal, Examples

Stokes Rule: According to Newton’s law of gravitation, the earth exerts a downward force on all objects on our globe. So how come we don’t see any clouds falling? The Stokes Law is the reason behind this. Sir George G. Stokes, a British scientist, developed Stokes Law formula in 1851. He considered the forces operating on a specific particle that sinks through a liquid column under gravity’s influence.

Stokes Law formula is a mathematical expression for the drag force that prevents tiny spherical particles from falling through a fluid medium. The amount of water in clouds is enormous. Clouds contain microscopic water droplets that descend slowly. The drag force of the air, on the other hand, outweighs the gravitational force for microscopic particles. The drag force increases as the size of an object grow. Read the complete article to obtain a better grasp to state Stokes Law, derivation and definition.

What is Stokes Law in Physics?

To understand what is Stokes law, it describes the viscous forces acting on a body as it is dropped into a fluid. Due to the viscous forces, a body falling through a liquid falls slowly or, in simple words, its velocity does not achieve a higher value. As a spherical body falls under the effect of its weight, the body deaccelerates, the velocity of the body increases but it cannot increase beyond a maximum value known as the terminal velocity.

The Stokes law states when a body falls through a fluid, it drags the layer of the fluid in contact with it. A relative motion is developed between the different layers of the fluid, and as a result, the body experiences a retarding force. This retarding force acting on a body as it passes through a viscous fluid is directly proportional to the sphere’s velocity, radius, and fluid viscosity. 
A few common examples of such a motion include a person falling with a parachute, pouring raindrops, and swinging a pendulum bob. It is seen that the viscous force is proportional to the velocity of the object and is opposite to the direction of motion.

Stokes Law Formula

To derive the Stokes Law Formula, he performed several experiments to understand the motion of small spherical bodies in different fluids, and he concluded that the viscous force \(F\) acting on a spherical body of radius \(r\) depends directly on the following quantities:
i) The radius \((r)\) of the sphere falling through the liquid.
ii) The velocity \((v)\) of the sphere falling through the liquid.
iii) The coefficient of viscosity \((η)\) of the liquid.

Thus, The stokes Law formula states that the drag force \(F\) acting upward in resistance to the fall can be given as:
\(F = 6πrηv\)

Stokes Law Derivation

When a sphere or a body moves through a fluid, a friction force must be overcome. The Stokes Law, named after the scientist George Gabriel Stokes, quantitatively gives the relationship between the frictional force that acts between a liquid and a spherical body that falls through it based on quantities such as its radius and velocity. Mathematically, this can be written as:

\(F ∝ η^a r^b v^c\)
Where \((r)\) represents radius, \((v)\) represents the velocity of the sphere falling through the liquid, and \((η)\) represents the coefficient of viscosity of the liquid and the powers \(a,\,b\) and \(c\) are some unknowns.

Here \(F\) is the drag force, but we do not know how the quantities are related to it. To determine the relationship between them, we use dimensional analysis. To calculate the values of \(a,\,b\) and \(c\) substitute the proportionality sign with an equality sign as:
\(F = kη^a r^b v^c\) —–(i)
Here \(k\) represents the constant of proportionality.
Write the dimensions of the quantity on both the sides of equation (i), we get:
\(\left[ {{M^1}{L^1}{T^{ – 2}}} \right] = k{\left[ {{M^1}{L^1}{T^{ – 1}}} \right]^a}{\left[ {{M^0}{L^{ – 1}}{T^0}} \right]^b}{\left[ {{M^0}{L^1}{T^{ – 1}}} \right]^c}\)

On simplifying the equation, we get:
\(\left[ {{M^1}{L^1}{T^{ – 2}}} \right] = \left[ {{M^a}{L^{ – a + b + c}}{T^{ – a – c}}} \right]\) —–(ii)
The independent entities of mass, length and time can be equated across the left and right-hand side of the equation as per classical mechanics; equating the powers of quantities, we get:
\( \Rightarrow a = 1\) —–(iii)
\(- a + b + c = 1\) ——(iv)
Or \(- a – c = – 2\)
\(a + c = 2\) ——-(v)
Substitute the value from equation (iii) into equation (v), we get:
\(1 + c = 2\)
\(c = 1\)
Using the values obtained above in the equation (iv), we get:
\(- 1 + b + 1 = 1\)
\(b = 1\)
Substituting these values in the equation (i), we get:
\(F = kηrv\)
Various experiments were conducted, and the value of \(k\) was found to be equal to \(6π\). Thus, the drag force acting on a spherical body falling through a liquid becomes:
\(F = 6πηrv\)
This is the equation of Stokes Law derivation.

Requirements for Stokes Law

1. The body should fall through a liquid that has infinite extension.
2. The spherical body falling through the liquid must be entirely smooth and rigid.
3. No slipping between fluid and rigid body falling through it.
4. The motion of the body should be streamlined and not turbulent.
5. The size of the ball, although relatively small, must be larger than the distance between the molecules of the liquid.
6. The liquid medium must be homogeneous and continuous.

Terminal Velocity

The maximum velocity that can be attained by a body falling under the viscous drag of the fluid is called terminal velocity. A relative motion occurs between the layers of the medium as the body falls through a liquid. Due to its motion, a viscous drag force acts on the body that would retard the body’s motion. The viscous force acting on the body, according to Stokes law, can be given as \(F = 6πηrv\); as the velocity increases, so does the force acting on the body.

When the body’s weight gets balanced by the sum of the upthrust and viscous force, no net force acts on the body. Hence its acceleration becomes zero. The body thus moves with a constant velocity. This is terminal velocity. The formula for terminal velocity \((v_t)\) is,
\({v_t} = 2{r^2}\frac{{\left( {\rho – \sigma } \right)g}}{{9\eta }}\)
Where \(ρ\) and \(σ\) are mass densities of sphere and fluid, respectively.

Real-life Examples Based on Stokes’ Law

a) Working of a Parachute

We have often seen videos of people going for skydives. An aeroplane is taken to a reasonable height. From a flying aeroplane, people jump one after the after with a parachute attached to their back. As a person falls, they go down towards the surface with acceleration due to gravity \(g\).

Falling downwards, the acceleration of each person goes on decreasing due to the viscous forces acting on him due to air and ultimately achieving terminal velocity. Thus, as each individual attains a constant velocity, they open their parachute while they are at a safe distance from the earth’s surface. This precalculated opening of the parachute after achieving the terminal velocity helps them land safely and within a safe distance.

b) Raindrops

Ever wondered how the drops of water falling at acceleration due to gravity do not hit us like mini bullets?  Under the effect of the earth’s gravitational force, raindrops fall towards the earth and encounter the viscous drag force in their path. These viscous forces, when they become equal to the weight of the droplets, the net force acting on the raindrops become zero, and hence, these drops achieve a terminal velocity. This ensures that these raindrops do not have high kinetic energy as they reach the earth’s surface and do not harm people.

c) Beer Bubbles

The rising bubbles in a glass full of beer exhibit the inverse effect to the raindrops. Bubbles being lighter than the surrounding liquid, feel an upward push due to gravity. According to Stokes law, they rise with a constant velocity. If a glass of beer is noticed carefully, it can be seen that sometimes the bubbles move down due to the circulation of liquid in the beer glass. The rising bubbles in the centre of the glass drag some liquid along with them, reaching the top; some small bubbles are dragged downwards against the force of gravity. Similarly, small raindrops are pulled up by air currents against the force of gravity.

Shortcomings of Stokes Law

1. Stokes equation cannot be applied when the density difference between the liquid and the body in the equation is negative. When the particles of the sinking body are less dense than the dispersing medium, floatation or creaming takes place.
2. When the solid content is high, Stokes equation may not show the real sedimentation rate. High solid content imparts additional viscosity to the system, which must be considered if the correct settling rate is determined. The equation contains only the viscosity of the medium.
3. The dielectric constant is a significant parameter that is ignored in Stokes equation. As we know, the electrical potential between two charges is inversely proportional to the medium’s dielectric constants. In the case of automobile liquids, the absence of a dielectric factor leads to significantly incorrect results.
4. The collisions between molecules due to the zig-zag motion, known as Brownian motion, significantly alter the result obtained from Stokes law.

Summary

Stokes law describes the viscous forces acting on a body as it is dropped into a fluid. As a body falls through a fluid, it drags the layer of the fluid in contact with it. A relative motion is developed between the different layers of the fluid, and as a result, the body experiences a retarding force. This retarding force acting on a body as it passes through a viscous fluid is directly proportional to the sphere’s velocity, radius, and fluid viscosity. This is Stokes law. This equation of Stokes Law can be given as, \(F = 6 πηrv\). Also, students should memorise the Stokes Law definition and write it verbatim in the exam.

A few examples of stokes law include the motion of a person falling with a parachute, pouring raindrops, and swinging a pendulum bob.

FAQs

Q.1. What is Stokes Law?
Ans: Stokes law is the mathematical equation that expresses the drag force resisting the fall of small spherical particles through a fluid medium.

Q.2. What is terminal velocity?
Ans: The maximum velocity with which a spherical body falls through a viscous fluid is called the terminal velocity.

Q.3. Write the expression for the viscous force acting on a spherical body.
Ans: The viscous force acting on a body can be given as \(F = 6 πηrv.\)

Q.4. Write a few examples of Stokes law.
Ans: Falling of raindrops, formation of oil drops, and Millikan’s oil drop experiment.

Q.5. What are the factors that affect the force acting on the spherical body falling through a viscous fluid?
Ans: The factors affecting the viscous force acting on a spherical body are the radius, velocity of the spherical body, and coefficient of viscosity of the medium.

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