Here we will work on the **derivation of the Terminal Velocity equation or formula **using Stokes’ Law. We will consider a situation where a solid sphere moving slowly in a fluid to derive the Terminal Velocity equation. We will also solve a numerical problem using the terminal velocity equation.

## Derivation of Terminal Velocity Equation using Stokes’ law

When an object is falling through a fluid, in that case, if we want to analyze its motion (and find out its acceleration, if any) then we need to consider the weight of the object, the upthrust on the object applied by the displaced volume of the fluid, and the viscous drag force caused by the movement of the object in the fluid.

Usually, we consider the equilibrium situation, in which the weight of the object exactly balances the sum of upthrust and drag force. And in this equilibrium situation as the net force on the object is zero, hence the velocity of the object remains constant. This constant velocity is terminal velocity.

This is true for skydivers falling through the fluid air as well as for a ball bearing dropping through a column of oil.

To derive the Terminal Velocity equation we will consider simple situations, say for a solid sphere moving slowly in a fluid.

Now in equilibrium, i.e. when the solid sphere is moving with terminal velocity then:**weight of the sphere = upthrust on the sphere applied by the displaced fluid + Stokes’ force or viscous drag force**

**=>**weight of the sphere = weight of the displaced fluid + Stokes’ force or viscous drag force

**m _{s}g = weight of the displaced fluid + 6 πrȠV_{term} ……………….. (1)**

[ **Note:**

Here, m_{s} is the mass of the sphere. So, msg = weight of the sphere.

The Stokes’ Law formula for viscous drag force is represented in this way: F = 6 πrȠV

where r is the radius of the sphere, V is the velocity of the sphere and Ƞ is the coefficient of viscosity of the fluid.

Here in equilibrium condition in place of V, we will use **V _{term}** which is terminal velocity

]

Now, let’s expand equation (1) a bit more.**m _{s}** = mass of the sphere = volume of the sphere x density of the sphere material = (4/3) πr

^{3}ρ

_{s}

So in place of the weight of the sphere (

**m**) in equation 1 we can write, (4/3) πr

_{s}g^{3}ρ

_{s}g …….

**(a)**

Now, upthrust on the sphere = weight of the fluid displaced

= mass of the fluid displaced x g = volume of the fluid displaced x density of the fluid x g

= volume of the sphere x density of the fluid x g = (4/3) πr^{3} ρf g …. (b)

(Note that the volume of the fluid displaced is equal to the volume of the sphere.)

So from equations (1), (a) and (b) we get a new one,**(4/3) πr ^{3}ρ_{s}g = (4/3) πr^{3} ρ_{f} g + 6 πrȠV_{term} ……………….. (2)**

Now, let’s rearrange the equation to get **V _{term}**.

**V**= [

_{term}**(4/3) πr**(

^{3}**g****ρ**)] / [

_{s}–**ρ**_{f}**6 πrȠ**]

And after simplifying the equation we get the final **equation of terminal velocity** as follows

=> **V _{term} = [2 r^{2} g (ρ_{s} – ρ_{f} )] / [9Ƞ]**

## Numerical problem – solved using the terminal velocity formula

**Numerical Question**: A sphere is falling at terminal speed through a fluid.

The following data are available.

Diameter of sphere = 3.0 mm

Density of sphere = 2500 kgm^{-3}

Density of fluid = 875 kgm^{-3}

Terminal speed of sphere = 160 mms^{ -1}

*Determine the viscosity of the fluid.*

**Solution**

Using the terminal speed expression gives:

v_{t}=2r^{2}g ( **ρ _{s} – ρ_{f} ** ) /(9η )

=> η=2r^{2}g ( **ρ _{s} – ρ_{f} ** ) /(9v

_{t}) = [ 2x(1.5×10

^{-3})

^{2}x 9.81 x (2500-875)] / [9 x 0.16]

η= 50 mPas

## Summary

*Thus using Stokes’ law you can derive the terminal velocity equation* given below:

V

_{term}= [2 r^{2}g (ρ_{s}– ρ_{f})] / [9Ƞ]

We have also solved a sample numerical problem to find out **Ƞ** using the **equation of terminal velocity**.

Get more solved numerical on terminal velocity here.