Free Fall with air resistance near earth's surface

Approximations: NEAR earth's surface means we will use Fgravity = -mg where g = constant; also we will assume that the air density is constant and assume that the air resistance coefficient is indeed a constant.

Case 1:  FAR = - bv

SF = ma leads to -mg - bv  =  m dv/dt , or

dt  =  dv / [-g - (b/m)v]  =  -(m/b) dv / [v + mg/b]

Let w = v + mg/b so that dw = dv, and remember to adjust the limits of integration:

 t  =  -(m/b) wo=vo+mg/bς w = v+mg/b dw/w  =  -(m/b) { ln(v+mg/b) - ln(vo+mg/b)}  =  -(m/b) ln[(v+mg/b) / (vo+mg/b)]

e-bt/m  =  (v+mg/b) / (vo+mg/b)

(v+mg/b)  =  (vo+mg/b) e-bt/m

v(t)  =  vo e-bt/m - (mg/b)(1- e-bt/m)

Check units:

1) (mg/b) should have units of velocity.

2) (b/m) should have units of inverse time.

 Limiting cases:

1) as t goes to zero, v goes to vo as it should.

2) as t gets large, v goes to a terminal speed = -mg/b.

3) Behavior as m changes:  The bigger the m, the bigger the terminal speed and the longer it takes to reach that higher terminal speed.

4) Behavior as b changes:  The bigger the b, the smaller the terminal speed.  It is unclear what happens for small time with small b since b is in the denominator of the coefficient! We can find out by expanding e-bt/m in a power series.  To keep things as simple as possible, let’s assume that vo = 0 (we drop it):

v(small t)  =  - (mg/b)(1 - [1 - bt/m + (1/2)b2t2/m2 - (1/6)(b3t3/m3) + ...]

v(small t)  =  -gt + (1/2)bgt2/m - (1/6)b2gt3/m2 + ...

Thus, as b goes to zero, the speed becomes that of free fall without air resistance (v = -gt) like it should.

 

Method of Successive Approximation Technique

To put in effects that are smaller (such as air resistance compared to gravity), we can use a method called the method of successive approximations.
In this method, we first solve the problem without any of the smaller effects: (we will use vo=0 to make the math easier)

Zero order - assume for now that b=0, and solve for v(0)(t) :

F = -mg   so SF = ma leads to:   -mg = m dv/dt   which gives: dv = -g dt  , and integrating gives:   v(0)(t)  =  vo - gt  =  -gt .

 

First order – use zero order solution for v in the air resistance term of -bv, and solve for v(1)(t):

F  =  -mg - bv(0)   so SF = ma leads to:   -mg - b[-gt]  =  m dv/dt   which gives:   dv  =  {-g -(b/m)[-gt]}dt , and integrating gives:   v(1)(t)  =  - gt + (1/2)bgt2/m.

 

Second order – use first order solution for v in the air resistance term of –bv, and solve for v(2)(t):

F  =  -mg - bv(1)   so SF = ma leads to:   -mg - b[-gt + (1/2)bgt2/m]  =  m dv/dt which gives:  dv  =  {-g - b/m[-gt + (1/2)bgt2/m]} dt , and integrating gives:   v(2)(t)  =  -gt + (1/2)bgt2/m - (1/6)b2gt3/m2

 Note that this gives the same result as the series expansion of our analytical result!

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Case 2:  FAR = - bv2 
Note that since v is squared, we need to treat the sign for FAR differently going up versus going down. 
For now, let’s consider the case going downward, and choose down as the positive direction.

SF = ma leads to mg – bv2  =  m dv/dt , or   dt  =  dv / [g - (b/m)v2]  =  (1/g) dv / [1 – (b/mg)v2] .

Let (b/mg)1/2v = sin(j)  so that dv = (mg/b)1/2cos(j)dj , and remember to adjust the limits of integration: j = sin-1[(b/mg)1/2v]

 t  =  (1/g) joς j (mg/b)1/2cos(j)dj / [1 – sin2(j)]   =   (1/g) joς j (mg/b)1/2cos(j)dj / cos2(j)  =  (1/g) joς j (mg/b)1/2 dj / cos(j) .

From a table of integrals, we have  ς dj / cos(j) = ln [(1/cos(j) + tan(j)]  so

t  =  (m/gb)1/2 { ln [(1/cos(j) + tan(j)]  - ln [(1/cos(jo) + tan(jo)]  }   =  (m/gb)1/2 { ln [{(1/cos(j) + tan(j)]} / {(1/cos(jo) + tan(jo)}] }

where j = sin-1[(b/mg)1/2v] and jo = sin-1[(b/mg)1/2vo] .

Note:  this is just t(v), it is not v(t).  Getting from t(v) to v(t) looks quite imposing.

Check units:

1) (m/gb)1/2 should have units of time:  m has units of kg; g has units of m/s2, b has units of Nt/(m/s)2 = (kg*m/s2)/(m2/s2) = kg/m;  therefore (m/gb)1/2 has units of  [kg/{(m/s2)*(kg/m)}]1/2 = s.

2) (b/mg)1/2 should have units of inverse velocity:  {(kg/m)/[kg*m/s2]}1/2 = s/m.

Limiting cases:

1) as v goes to vo, t should go to zero, and it does.

2) the terminal speed is when bv2 = mg which gives vt = (mg/b)1/2.  As v approaches vt, we would expect t to approach infinity.  Since j = sin-1[(b/mg)1/2v] and (b/mg)1/2 = 1/vt , as v approaches vt, j approaches p/2.  Now as j approaches p/2, 1/cos(j) approached infinity and tan(j) approaches infinity, and the ln[value approaches infinity] approaches infinity, the time approaches infinity as v approaches vt as we would expect.

 

Homework Problem:  Problem #10.  Use the method of successive approximation to find the first order expression for v(t) for the above Case 2.

 

Alternate solution for case 2:

SF = ma leads to mg – bv2  =  m dv/dt , or   dt  =  dv / [g - (b/m)v2]  =  (1/g) dv / [1 – (b/mg)v2] .

Let (b/mg)1/2v = u  so that dv = (mg/b)1/2du , and remember to adjust the limits of integration.  Note that (mg/b)1/2 = vt (see above), so  (b/mg)1/2 = 1/vt , so we can see that u = v/vt which is dimensionless as it should be if we are to subtract its square from 1.

 t  =  (1/g) uoς u (mg/b)1/2du / [1 – u2]   =   (vt /g) uoς u du / [1 – u2]  .

To solve this integral, we note that 1/(1-u2)  =  A/(1-u) + B/(1+u)  =  {A(1+u) + B(1-u)}/(1-u2)
so that  A+B = 1  and  A-B = 0  which has the solution of A = ½ = B; so 1/(1-u2) = ½ {1/(1+u)} + ½ {1/(1-u)}, so
 ς du/(1-u2)   =   ½ ς du/(1+u) + ½ ς du/(1-u) ; 
now let y = 1-u and z = 1+u so that
 ς du/(1-u2)   =   ½ ς dz/z + ½ ς -dy/y   =   ½ {ln(z) – ln(zo)} - ½ {ln(y) – ln(yo)}
but since zo = 1+uo = 1+0 = 1, and yo = 1-uo = 1, and ln(1) = 0, we have
 ς du/(1-u2)   =  ½ { ln(1+u) – ln(1-u) }  =   ½ ln{(1+u)/(1-u)}  =
½ ln[ {1+(b/mg)1/2v} / {1-(b/mg)1/2v} ], so that

t  =  ½ (m/gb)1/2 ln[ {1+(b/mg)1/2v} / {1-(b/mg)1/2v} ].

Although the two expressions for t,

t  =  (m/gb)1/2 { ln [{(1/cos(j) + tan(j)]} / {(1/cos(jo) + tan(jo)}] }  where j = sin-1[(b/mg)1/2v] and jo = sin-1[(b/mg)1/2vo] , and  

t  =  ½ (m/gb)1/2 ln[ {1+(b/mg)1/2v} / {1-(b/mg)1/2v} ]

look quite different, they are in fact equivalent.

The second form we can convert from t(v) into v(t):

 2t(gb/m)1/2  =  ln[ {1+(b/mg)1/2v} / {1-(b/mg)1/2v} ]

now let a = (gb/m)1/2  and we get:

e2at  =  {1+(b/mg)1/2v} / {1-(b/mg)1/2v} , or

e2at {1-(b/mg)1/2v}  =  {1+(b/mg)1/2v} , or putting all v’s on one side:

(b/mg)1/2v (e2at + 1)  =  (e2at - 1) ,  or

v(t)  =  (mg/b)1/2 (e2at - 1) / (e2at + 1) , or multiplying numerator and denominator by e-at

v(t)  =  (mg/b)1/2 (eat – e-at) / (eat + e-at) , and since tanh(x) = (ex – e-x) / (ex + e-x)  [note that tanh(x) is called the hyperbolic tangent function], we have:

v(t)  =  (mg/b)1/2 tanh[(gb/m)1/2 t]  =  vt tanh[(gb/m)1/2 t]  =  vt tanh[gt/vt]      since (mg/b)1/2 = vt .

Limiting cases:

For small t, which really means  [(gb/m)1/2 t]  =  gt/vt  is small, we need to look at the behavior of tanh(x) = (ex – e-x) / (ex + e-x) by using the Taylor series for each ex and e-x .

(ex – e-x) / (ex + e-x)      ({1+x)} – {1-x}) / ({1+x)} + {1-x})   =   2x / 2   =   x (to first order)

So for small t, (i.e., gt << vt), we have approximately:

v(small t)  =  (mg/b)1/2 tanh[(gb/m)1/2 t]  =  vt tanh[gt/vt]       vt * [(gt/vt]   =  gt ,

which is the case for no air resistance.

For small t to the next order in x using Taylor series for ex, e-x, and 1/(1+x): 
ex ≈ 1 + x + x2/2 + x3/6  and  e-x ≈ 1 - x + x2/2 - x3/6  so that
tanh(x) ≈ [(1 + x + x2/2 + x3/6 ) – (1 - x + x2/2 - x3/6)] / [(1 + x + x2/2 + x3/6 ) + (1 - x + x2/2 - x3/6)]
Now we use the approximation  1/(1+y) ≈ (1-y)  to get
tanh(x)    (2x + 2x3/6) / (2 + 2x2/2) ≈ (x + x3/6) * (1 - x2/2)    x + x3/6 – x3/2  = x – x3/3  so that (recalling that vt2 = mg/b)
v(t) = vt tanh(gt/vt) ≈ vt [gt/vt – g3t3/(3vt3)]  =  gt – bg2t3/(3m)  (where down is positive).
Compare this approximation to the one in Problem 10 above.

For large t,  tanh(x) = (ex – e-x) / (ex + e-x) goes to 1, and so v(large t) goes to (mg/b)1/2 = vt .

Estimation of a real case:

What is the terminal speed for an ice pellet (hail) if the pellet is a sphere of radius 1 cm?

If we work from the bv2 model for air resistance, and use b = (1/2)CD [see the earlier section on F(v)], and knowing the density (ρ=m/V) of water (and ice) is close to 1 gm/cm3 = 1000 kg/m3 and air is close to 1.3 kg/m3, and the formulas for a sphere:  cross sectional area (A) is that of a circle = pr2 and the volume (V) is (4/3)pr3, and using CD = ½ for a sphere, we get:

m = ρ*V = (1000 kg/m3)*(4/3)p(.01 m)3  = .0042 kg = 4.2 grams;

A = pr2  = p (.01 m)2 = 3.1 x 10-4 m2;

b =  (1/2)CDAρair = (1/2)(1/2)(3.1 x 10-4m2)(1.3 kg/m3) = 1.02 x 10-4 kg/m ;

vt = [mg/b]1/2 = [(.0042 kg)(9.3 m/s2)/(1.02 x 10-4 kg/m)]1/2 = 20.1 m/s = 45 mph.

If we increase the radius of the ice sphere by 10 (from 1 cm to 10 cm), m goes up by a factor of 1000, A goes up by a factor of 100, b goes up by a factor of 100, and vt goes up by a factor of (10)1/2 = 3.16, so vt goes up to 142 mph.

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