**Free Fall with air resistance near earth's surface**

Approximations: NEAR earth's surface means we will use F_{gravity} =
-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: F _{AR} = - 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/b}**

**e ^{-bt/m} = (v+mg/b) / (v_{o}+mg/b)**

**(v+mg/b) = (v _{o}+mg/b)
e^{-bt/m}**

**v(t) = v _{o}
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 v_{o} 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}**

v(small t) = - (mg/b)(1 - [1 - bt/m + (1/2)b^{2}t^{2}/m^{2}
- (1/6)(b^{3}t^{3}/m^{3}) + ...]

**v(small t) = -gt + (1/2)bgt ^{2}/m - (1/6)b^{2}gt^{3}/m^{2}
+ ...**

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 v_{o}=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) = v_{o}
- 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

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

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

**-------------------------------------**

**Case 2: F _{AR} = - bv^{2} **Note that since v is squared, we need to treat the sign for F

For now, let’s consider the case going downward, and choose down as the positive direction.

**SF = ma** leads to **mg – bv ^{2}
= m dv/dt** , or

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

**t = (1/g) **_{jo}**ò _{ }**

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

**t = (m/gb) ^{1/2} { ln [(1/cos(**

where j
= sin^{-1}[(b/mg)^{1/2}v] and j_{o} = sin^{-1}[(b/mg)^{1/2}v_{o}]
.

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/s^{2},
b has units of Nt/(m/s)^{2} = (kg*m/s^{2})/(m^{2}/s^{2})
= kg/m; therefore (m/gb)^{1/2}
has units of [kg/{(m/s^{2})*(kg/m)}]^{1/2}
= s.

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

__Limiting cases: __

1) as v goes to v_{o}, t should go to zero, and it does.

2) the terminal speed is when bv^{2} = mg which gives v_{t}
= (mg/b)^{1/2}. As v approaches
v_{t}, we would expect t to approach infinity. Since j = sin^{-1}[(b/mg)^{1/2}v]
and (b/mg)^{1/2} = 1/v_{t} , as v approaches v_{t}, 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 v_{t}
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 – bv ^{2} = m
dv/dt** , or

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

**t = (1/g) _{uo}**

To solve this integral, we note that 1/(1-u^{2}) =
A/(1-u) + B/(1+u) = {A(1+u) + B(1-u)}/(1-u^{2})

so that A+B = 1 and
A-B = 0 which has the solution of
A = ½ = B; so 1/(1-u^{2}) = ½ {1/(1+u)} + ½ {1/(1-u)}, so

**ò **du/(1-u^{2}) = ½ **ò **du/(1+u) + ½ **ò **du/(1-u) ;

now let y = 1-u and z = 1+u so that

**ò **du/(1-u^{2}) = ½ **ò **dz/z + ½ **ò -**dy/y = ½ {ln(z) – ln(z_{o})} - ½ {ln(y) –
ln(y_{o})}

but since z_{o} = 1+u_{o} = 1+0 = 1, and y_{o} = 1-u_{o}
= 1, and ln(1) = 0, we have

**ò **du/(1-u^{2}) = ½ {
ln(1+u) – ln(1-u) } = ½ ln{(1+u)/(1-u)} =

½ ln[ {1+(b/mg)^{1/2}v} / {1-(b/mg)^{1/2}v} ], so that

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

Although the two expressions for t,

**t = (m/gb) ^{1/2} { ln [{(1/cos(**

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

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)^{1/2}v} / {1-(b/mg)^{1/2}v} ]**

now let **a
=** **(gb/m) ^{1/2} **and we
get:

**e ^{2}**

**e ^{2}**

**(b/mg) ^{1/2}v (e^{2}**

**v(t) = (mg/b) ^{1/2} (e^{2}**

**v(t) = (mg/b) ^{1/2} (e**

**v(t) = (mg/b) ^{1/2} tanh[(gb/m)^{1/2}
t] ** =

__Limiting cases:__

For small t, which really means [(gb/m)^{1/2} t] = gt/v_{t
} is small, we need to look at the
behavior of tanh(x) = (e^{x} – e^{-x}) / (e^{x}
+ e^{-x}) by using the ^{x} and e^{-x} .

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

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

v(small t) = **(**mg/b)^{1/2}
tanh[(gb/m)^{1/2} t] = v_{t} tanh[gt/v_{t}]** ** ≈
v_{t} * [(gt/v_{t}]**
** = gt ,

which is the case for no air resistance.

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

__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 bv^{2} model for air resistance, and use b =
(1/2)C_{D}Aρ [see the earlier section on F(v)], and knowing the
density (ρ=m/V) of water (and ice) is close to 1 gm/cm^{3} = 1000
kg/m^{3} and air is close to 1.3 kg/m^{3}, and the formulas for
a sphere: cross sectional area (A) is
that of a circle = pr^{2}
and the volume (V) is (4/3)pr^{3}, and using C_{D} = ½ for a sphere,
we get:

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

A = pr^{2} = p (.01 m)^{2} = 3.1 x 10^{-4} m^{2};

b = (1/2)C_{D}Aρ_{air} = (1/2)(1/2)(3.1 x 10^{-4}m^{2})(1.3
kg/m^{3}) = 1.02 x 10^{-4} kg/m ;

v_{t} = [mg/b]^{1/2} = [(.0042 kg)(9.3 m/s^{2})/(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 v_{t} goes up by a factor of (10)^{1/2} =
3.16, so v_{t} goes up to 142 mph.