FREE FALL WITH AIR RESISTANCE - A FEATHER, A COIN AND A HUMAN

In my previous article, while dealing with the perfect free fall motion of a body, I have only given a rough idea regarding the effects of air resistance. This time I go a little further. 

 Let us, as before, consider a body of mass m, situated at a height h above the Earth's surface, where h << Rₑ, the latter being the Earth's radius. We use this constraint to neglect the variation of g with increasing height from the surface and take it to be around 9.8 m/s² for simplicity. Initially, the body is held by some support so that at t = 0, when the body is released, it starts falling with an initial velocity which is obviously 0. Once again, we can set up a suitable choice of coordinates (as shown below) where the z-axis denotes vertical motion taking the downward direction positive, and the horizontal x-axis represents the Earth's surface. The velocity and displacement of the body at t seconds are ż(t) and z(t), respectively. 

As the body falls through the air, i.e., essentially through the atmosphere, it experiences two kinds of forces. The first one is the normal acceleration due to gravity pulling the body towards the surface and necessarily towards the Earth's centre, i.e., towards the centre of the gravitational field. The second type of force arises from the viscous drag of air. The viscous force, also known as the drag force, is an intrinsic property of all fluids (liquid or gas) by virtue of which they offer resistance to the free motion of a typical solid object as it moves through a column of fluid. When a body freely falls in a vacuum environment, it experiences no other force except uniform gravitational acceleration. But the same body, when falling through the air or any fluid, encounters this viscous drag, experiencing an impeding resistance to otherwise perfect free fall. Being a frictional force arising from the collision of air molecules with the falling body, it acts in the opposite direction. The drag force depends on factors such as the shape, size, density and surface characteristics of the falling body, density of air, pressure gradient and so on. I will come to the details of fluid dynamics and air resistance in a fleeting moment. But for now, it suffices to say that in the simplest case of a body falling from some height above the Earth's surface, we can assume this viscous force to be directly proportional to the instantaneous velocity of the falling body and linked by a constant factor k. By this assumption that the force of air resistance is a linear function of instantaneous velocity, we can write the equation of motion of the falling body as follows: 

Equation 1.0

Upon performing the first integration and applying the suitable boundary conditions, i.e., at t = 0, initial velocity ż(t) = 0, we obtain the velocity of the falling body to be 

Equation 1.1

On performing the second successive integration we get the displacement of the particle from its starting position, and applying the suitable boundary condition, i.e., at t = 0, z(t) = 0, we arrive at 

Equation 1.2

We have already seen that in a vacuum, the velocity of the freely falling body bears a linear relation to time, increasing progressively with each passing second. So does its displacement, which turns out to be a quadratic function of time, increasing as the sum of squares. However, when there is linear air resistance, we see from equations 1.1 and 1.2 that velocity and displacement take on an exponentially decreasing function of time, joined by some constant terms. In the limiting case, when

or when t is sufficiently large, the exponential factor becomes negligible and the expression for velocity as derived from equation 1.1 simply becomes 

Equation 1.1.1

This term, vₜ, represents what we refer to as the terminal velocity of the body. The terminal velocity is the maximum velocity with which a body falls through a fluid. I find it tiresome to explicitly derive the limiting case of equation 1.2 or re-derive both equations for a body falling with an initial velocity and their limiting cases, for I suppose my readers will do it themselves. 

So far, I have followed the standard textbook method of characterising free fall with air resistance. I have not explained why I have assumed air resistance to be simply proportional to the instantaneous velocity of the falling body and why not v², v³, v⁴⁸⁷ or some arbitrary vⁿ. Other than that, I have not specified this k. What is k? If you were careful enough, you would have already noticed that equation 1.1.1 does not seem to be dimensionally correct. Therefore this k must be a constant of force and has to be related to the body's mass plus some other quantity. So, to tackle this problem effectively, we need the theoretics of fluid dynamics. 

Air is essentially a fluid. From Archimedes' principle, we know that a body when immersed completely in a fluid, experiences forces on all sides. But the forces orthogonal to the downward pull of gravity cancel out, and those that remain are the downward pull of gravity and the normal reaction force oppositely directed. This upward-directed normal force is essentially the buoyant force. The origin of the buoyant force is simple to understand. If a solid object is to fall through the air (or any fluid), it must push away the air (or whatever fluid) molecules out of the way. Otherwise, how can it move? As the object presses on the fluid, it exerts a force on the latter, and by Newton's third law, following action and opposite reaction, the body experiences the same force with which it pushes the fluid. The buoyant force is this upward force exerted by a fluid that opposes the weight of the immersed body. If the buoyant force equals the weight of the body, the body floats. If the former is greater or less, the body rises and sinks, respectively. The buoyant force arises as a result of the difference in pressure between the upper and bottom surfaces of a submerged body. Since the pressure of a fluid increases with depth, the bottom surface of the body experiences more pressure due to the total mass of fluid on top of it than the upper surface. Thus there arises a net pressure difference and hence, the force of buoyancy. As pointed out by Archimedes, the buoyant force depends on the density of the fluid, the density of the body, the volume of liquid displaced by the body, and on the acceleration due to gravity.  

However, in the case of a body falling through the air, the force of buoyancy exists as long as the size of the body is significantly smaller, for example, a water droplet. But for heavier bodies in the air, the effect due to buoyancy turns out to be negligible. This is because air is the least dense compared to everyday objects. A feather, coin, human or even a T. Rex will readily fall through a column of air, for they are far heavier and denser than air. If our atmosphere were as dense as water, the feather would float, as the expression goes, ''in thin air''. I should say, ''in dense air''. Similarly, if we could compress air 13,600 times to reach the density of mercury, then pennies and pounds would float just on top of the upper atmosphere. Interesting, eh? 

Continuing as before, for most objects, apart from the buoyant force, there lies a more dominant force. As the body falls, it displaces the surrounding air molecules giving rise to material contact forces or what we simply know as friction. This is what we call air resistance. More precisely, we can say that the body encounters the viscous drag of air. As the name suggests, the viscous drag arises from the viscosity of air. Viscosity, as we all know, in the case of any fluid, be it air, water, or some other liquid/gas, is its intrinsic resistance to flow. In the standard textbook language, viscosity is the internal friction arising between the layers of fluid via which it opposes the relative motion between its different layers. In light of our current discussion, viscosity can be taken as the internal resistance manifest between the air layers by which they oppose the free fall motion of a body falling through the air.

A sphere falling through a liquid. Fg is the downward force of gravity and Fd is the drag force directed opposite. The air layers streamline across the spherical body. 
Image Credits: Public Domain, via Wikimedia Commons  

Air resistance is complicated. But we can simplify it for easy understanding. Suppose we have a column of air as shown above. As the body falls through the air, by virtue of air friction, it tries to speed up the air molecules in its immediate vicinity. But the layers a little farther off are not being directly moved by the falling body, which in turn tries to slow down the air layer speeded up by the body's motion. This speeding up and slowing down of the air molecules and the relative friction between the air layers gives rise to the viscous drag of air. In 1851, George Gabriel Stokes derived an expression for the drag force experienced by a small spherical body falling through a viscous fluid with a very small Reynolds number. The Reynolds number, as we know, is a dimensionless number that determines whether a fluid will flow in a streamlined pattern or be turbulent. For the small relative velocity of the falling object v₁, Stokes derived the drag force as follows: 

Equation 1.3

Here Fd is the frictional force, also known as Stokes' Drag, acting on the interface between the fluid layer and the moving body, η is the coefficient of viscosity, and R is the radius of the spherical object. 

Unfortunately, Stokes' law is applicable exclusively for smaller bodies falling with low velocities through a viscous fluid, where we need to consider the effect of buoyancy. One such scenario where we can use Stokes' law is in Millikan's Oil Drop Experiment for the determination of the electronic charge. Further details about the experiment can be found in the following article, that I wrote a while ago. At first, Millikan's experimental set-up allowed a single oil drop to fall through the air medium, being pulled downwards via gravity with a terminal velocity (v1) arising due to air resistance. The equation of motion can be expressed as follows:  

Equation 1.3.1

Adjusting for the buoyancy of air, the above equation reads

Equation 1.3.2

where σ is the density of oil, 𝜌 is the density of air, 𝜼 is the viscosity of air, m is the mass of the oil drop, R is its radius, v1 is the terminal velocity and g is the acceleration due to gravity.  

Stokes' law is more frequently referred to as the linear law of air resistance for low Reynolds number where the fluid flow is laminar. However, in the case of larger bodies falling with greater velocities, we can not use this simple linear relationship because the fluid may not streamline across the falling object. For that, we need to use the Drag Equation, as was discovered by Lord Rayleigh. The drag equation given below follows a quadratic relationship. 

Equation 1.4

Here Fd  is the drag force, 𝝆 is the density of the fluid, u is the velocity of the falling body (relative to the fluid), A is the body's net surface area and Cd  is the drag coefficient which is again a dimensionless constant that depends primarily upon the geometry of the object. Let us have it this far.

At this length of the article, you have the right to ask me where these expressions come from or how we know which one to use. Well, in that case, I must say that these equations are obtained from dimensional analysis by which we can relate different physical quantities. In dimensional analysis, we can assume air resistance to be an nth-degree polynomial function of velocity linked via constants that are to be experimentally determined. Air resistance is a complicated thing. There is no means to assume that air resistance is either linear or quadratic. It can depend on any nth power of velocity for greater speeds and in more dynamic scenarios, as in the case of a bullet. While writing this article, I came across an old book on air resistance written by Richard de Villamil in which he gives a beautiful account of the velocity dependence of air resistance. In fact, if you try to think a little bit, you will intuitively understand why air resistance increases with increasing velocity. 

For an object like a human being, a coin or a T. Rex falling from some height above the Earth's surface, we can consider the quadratic law to dominate. We can neglect the effect of buoyancy for obvious reasons because it has too small a contribution. This is due to the fact that air is just not dense enough to support the weight of a human body, a T.Rex or most everyday objects. Indeed, as we walk or try to fall from a great height, we will be buoyed up. But we are far heavier than air to feel the buoyant force. If only we could compress air 800 times to give it the density of water, we would be able to float, as the expression goes, ''in thin air''. Likewise, if we compress air 13,600 times to give it the density of mercury, all our pennies and pounds would effortlessly float above the outer boundary of the atmosphere. Interesting, eh? On this head, I challenge you to point out one uninteresting thing in the whole universe. 

Thus when an object falls through the air, the viscous drag of the latter starts to slow down the object's fall. When the object reaches the terminal velocity, the gravitational force and the drag force balance each other and it stops accelerating only to fall with a constant terminal velocity. Using the quadratic equation and neglecting buoyancy effects we have the equation of motion for a body of falling through the atmosphere as follows: 

Equation 1.4.1

The symbols have their standard meaning. 

At equilibrium no force acts on the particle from where we get 

Equation 1.4.2

which spits out the expression for terminal velocity as 

Equation 1.4.3

The above expression tells us why feathers fall slower than pennies unless otherwise. But they accelerate equally. And that will be our next topic - why all objects accelerate equally under a constant gravitational field. 

                                                                                                                     To be continued...