An Object Falling through Air © 2004 by Karl Hahn

This is look-ahead material.

It addresses separable differential equations, which you will not need to know right now, but probably will have to know by the end of the semester. But you already possess the skills needed to solve the differential equations that occur in this problem. In addition, the development of this problem will put to work a number of different concepts you have learned up to this point, as well demonstrating how calculus provides the framework for solving problems like this.

So how do we find the formula for the velocity as a function of time of an object falling through air? Experiments in wind tunnels reveal that the force exerted by air flowing past an object (especially a spherical object) is proportional to the square of the velocity of the air flow. So we take that as a premise. It means that

   fa  =  kv2                                                     eq. b12-1.1

Even if you haven't taken a physics class, you have still probably heard of Newton's laws of motion. The second of these states that the force on any body is equal to its mass times the acceleration it is undergoing.

                  dv
   f  =  ma  =  m                                                 eq. b12.1-2
                  dt

   fg  =  gm                                                      eq. b12.1-3

The other force is f_a due to friction with the air. We already established a formula for that in equation b12.1-1. This force always acts in the opposite direction of the motion of the object. If we assume the object is falling downward, then f_a must pull upward.

So total force is a tug-of-war between gravity and air friction:

   f  =  fa - fg  =  kv2 - mg                                     eq. b12.1-4

f dv k = = m dt m

v2 - g = g

k mg

v2 - 1

= g

v2 - 1 c2

eq. b12.1-5a

where  c = √mg/k  is known (especially to sky divers) as the terminal velocity, which is a constant. Distilling eq. b12.1-5a down to its essence you have:

dv = g dt

v2 - 1 c2

eq. b12.1-5c

Observe that when the object begins falling, that is when its velocity is zero, equation b12.1-5c predicts that its acceleration will be -g. That is what we expect -- that when the object is at rest it is affected only by gravity, so its acceleration is the same as if it were dropped in a vacuum. Observe also that as the object's velocity approaches terminal velocity, the acceleration becomes closer and closer to zero -- or in other words the velocity changes less and less. This is also what we expect. Sky divers have long observed that once they reach terminal velocity their speed becomes steady.

So how do you solve equation b12.1-5c to find velocity, v, as a function of time, t? Equation b12.1-5c is what is known as a separable differential equation. The phrase, differential equation, means that it describes a relationship between a function, v(t), and its derivative, dv/dt. What makes this one separable is that it is susceptible to a method of solution called separation of variables. If you treat dv and dt as separate variables, then you can multiply this equation by dt, divide it by  v²/c² - 1,  and end up with

c2 dv = v2 - c2

g dt eq. b12.1-6a

On the left we have dv times a function purely of v (remember that c is constant), and on the right we have dt times a function purely of t (remember that g is also a constant, so what we have on the right is dt times a constant function of t).

Now that we have dv with a pure function of v and dt with a pure function of t, don't both sides of equation b12.1-6a look for all the world like a couple of integrands? Indeed that is how we solve this -- by treating each side as an integrand and then integrating both sides:

c2

dv = g v2 - c2

dt eq. b12.1-6b

The integral on the right is a piece of cake. The one on the left isn't very hard either. It's denominator is factored using the difference of squares, and then you can apply partial fractions to break it into easily integrated pieces. Upon integration (and the use of log identities) it becomes:

-

1 2

c ln

c + v c - v

= gt + K eq. b12.1-5c

Note that I only put the constant of integration, K, on one side of the resulting equation. That's because if you gave each side its own constant, you could subtract one of the constants from both sides and have only a single constant. So two constants are unnecessary. To find out what K is equal to, observe that when  t = 0,  we require that  v = 0  also. That is, at the moment you release the object (time zero), it is at rest (its velocity is zero). Putting in zero for both t and v, you find that on the left you have the log of one, which is zero, and on the right you have gt which is zero, consequently you must have  K = 0  to make everything balance in that first moment.

-

1 2

c ln

c + v c - v

= gt eq. b12.1-5d

To turn this into velocity, v, as a function of time, t, we just do some algebra. Here are the steps (which you should try on your own):

• Multiply both sides by -2
• Divide both sides by c
• Take the exponential of both sides.
• Multiply both sides by  c - v  and apply the distributive law to the right-hand side.
• Gather all the terms that have a v factor in them to the left of the equal and all those that don't to the right.
• Factor a v from the left side and a c from the right.
• Divide both sides by the expression that is multiplied by v. This puts v all alone on the left-hand side.

           e-2gt/c - 1
   v  =  c                                                        eq. b12.1-6a
           e-2gt/c + 1

So is this answer reasonable?? If you put in zero for t, the equation predicts that v is also zero, which jibes with releasing the object with no velocity at time zero. After that v is always negative, which indicates that the object is always falling down and never up, and that is what we expect it to do. As t goes to infinity, you can see that the quotient goes to -1 in the limit, so in the limit, v goes to minus the terminal velocity, -c, which is also what we expect. So this answer passes the smell-test.

Now let  τ = c/(2g), and the equation becomes

e-t/τ - 1 v = c eq. b12.1-6b e-t/τ + 1

Note that τ has units of time. For a sky diver falling with arms and legs outstretched, terminal velocity, c, is typically around 60 meters/second. Dividing that by twice g, we find that τ for the sky diver is about 3 seconds.

Table b12-1: Normalized Falling Velocity

Elapsed Time	% of terminal velocity
0	0
τ	46.2
2τ	76.2
3τ	90.5
4τ	96.4
5τ	98.7
6τ	99.5

The table on the right shows some sample values from equation b12.1-6b. We say that the table is normalized because it holds regardless of the value of g or c. With the table constructed this way, it is valid for any kind of object falling through air. For the falling sky diver we see that in just 18 seconds after jumping, he or she is already at 99.5% of terminal velocity.

To find distance fallen, x(T), (which will be a negative number because down is the negative direction) , we integrate equation b12.1-6b dt from  t = 0  to  t = T,  where T is elapsed time after dropping the object. At first this looks like difficult integral, but once you multiply top and bottom by e^t/(2τ), this one becomes susceptible to simple substitution.

   x(T)  =  c

 T
 
 0

 e-t/(2τ) - et/(2τ)
                   dt                             eq b12.1-7a
 e-t/(2τ) + et/(2τ)

Now substitute  u = e^-t/(2τ) + e^t/(2τ),  and  du = (-1/(2τ)) (e^-t/(2τ) - e^t/(2τ)) dt.  With what you've learned so far about taking integrals, you should be able to get as far as

   x(T)   =   -2τc

ln

e-t/(2τ) + et/(2τ)

T

0

                     eq. b12.1-7b

When  t = 0,  the thing inside the log is equal to 2. So evaluating the definite integral gives:

   x(T)  =  -2τc

ln

e-T/(2τ) + eT/(2τ)

  -  ln(2)

               eq. b12.1-7c

Of course we have to see if this equation passes the smell test. When  T >> τ,  we know that the object has been falling with very nearly constant velocity of c for a while. So we expect that x(T) should be near to  -cT.  You can see that when T is a large multiple of τ, you have  e^-T/(2τ) ≈ 0.  So dropping the  e^-T/(2τ)  from equation b12.1-7c and simplifying (observing that ln and the exponential are inverses of each other), we find

   x(T)   ≈   -cT  +  2cτ ln(2)                                  eq. b12.1-8

What about the case where T is a only very small fraction of τ? In this case we expect that air resistance hasn't had much effect yet, and the object's falling behavior should be similar to its falling in a vacuum. That is, we expect that  x(T)  should be very near to  -gT²/2.  Demonstrating this is more difficult than showing what happens when T is large. To do it we need to use Maclaurin and Tayor series. But first, to make it easier we rewrite equation b12.1-7c substituting  s = T/(2τ)  and combining the logs using the log identity:

   x(T)  =  -2τc ln

e-s + es
        
    2

                                   eq. b12.1-9a

Recall the Maclaurin series for  e^s  is

   es  =  1 + s +

 s2
    +
 2!

 s3
    +
 3!

 s4
    +  ...
 4!

When you apply that formula to both  e^s  and  e^-s,  add the two series term-by-term, then divide by two, you find that

   es + e-s
             =  1 +
       2

 s2
    +
 2!

 s4
    +
 4!

 s6
    +
 6!

 s8
    +  ...
 8!

Remember that  s = T/(2τ),  and that T is a small fraction of τ. This means that not only is  0 < s < 1,  but s is much closer to zero than to one. So the higher power terms of the above series head toward zero very fast. Making the approximation for small T, then, we drop all the terms except the first one containing an s, and we have for s close to zero:

   x(T)  ≈  -2τc ln

1 +

 s2
   
 2!

                                     eq. b12.1-9b

Now recall that the Taylor series for  ln(1 + x)  is

   ln(1 + x)  =  x -

 x2
    +
  2

 x3
    -
  3

 x4
    +
  4

 x5
    -  ...
  5

for  -1 < x < 1. Clearly  s²/2  falls well within this range. We put  s²/2  in for x into the above Taylor series. And again, because s is assumed to be close to zero, we drop the higher powered terms and we're left with

   x(T)  ≈  -2τc

 s2
                                                eq. b12.1-10a
 2!

and  2! = 2.  Now all that's left to do is lots of back-substitution and cancelling. Recall that  s = T/(2τ).  The above becomes:

x(T) ≈ -2τc

s2 = -τcs2 = -τc 2

T2 = -c 4τ2

T2 eq. b12.1-10b 4τ

Recall also that  τ = c/(2g).  So for T much smaller than τ:

   x(T)  ≈  -c

 T2
     =  -c
 4τ

 2gT2
       =  -
  4c

 gT2
                             eq. b12.1-10c
  2

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