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/b 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, lets 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, lets 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) jo j (mg/b)1/2cos(j)dj / [1 sin2(j)] = (1/g) jo j (mg/b)1/2cos(j)dj / cos2(j) = (1/g) jo j (mg/b)1/2 dj / cos(j) .

From a table of integrals, we have 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) uo u (mg/b)1/2du / [1 u2] = (vt /g) uo 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 = = B; so 1/(1-u2) = {1/(1+u)} + {1/(1-u)}, so
du/(1-u2) = du/(1+u) + du/(1-u) ;
now let y = 1-u and z = 1+u so that
du/(1-u2) = dz/z + -dy/y = {ln(z) ln(zo)} - {ln(y) ln(yo)}
but since zo = 1+uo = 1+0 = 1, and yo = 1-uo = 1, and ln(1) = 0, we have
du/(1-u2) = { ln(1+u) ln(1-u) } = ln{(1+u)/(1-u)} =
ln[ {1+(b/mg)1/2v} / {1-(b/mg)1/2v} ], so that

t = (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 = (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 vs 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 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)CDAρ [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 = 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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