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/bò ^{w = v+mg/b} dw/w = -(m/b) ln[(v+mg/b) / (v_o+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; but 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 get:

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 v_o=0 to make the math easier)

Zero order:

F = -mg so -mg = m dv/dt which gives: dv = -g dt or v⁽⁰⁾(t) = v_o - gt = -gt .

First order:

F = -mg - bv⁽⁰⁾ so -mg -b[-gt] = m dv/dt which gives: dv = {-g -(b/m)[-gt]}dt

or: v⁽¹⁾(t) = - gt + (1/2)bgt²/m.

Second order:

F = -mg - bv⁽¹⁾ so -mg -b[-gt + (1/2)bgt²/m] = m dv/dt which gives:

dv = {-g -b/m[-gt + (1/2)bgt²/m]} dt

or: 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: F_AR = - bv² (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) _joò ^j (mg/b)^1/2cos(j)dj / [1 – sin²(j)] = (1/g) _joò ^j (mg/b)^1/2cos(j)dj / cos²(j) = (1/g) _joò ^j (mg/b)^1/2 dj / cos(j)

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

where j = sin^-1[(b/mg)^1/2v] and j_o = sin^-1[(b/mg)^1/2v_o] .

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 v_o, t should go to zero, and it does.

2) the terminal speed is when bv² = 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)^1/2v] 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 – 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.

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

Now 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 ò du/(1-u²) = ½ ò du/(1+u) + ½ ò du/(1-u) ; now let y = 1-u and z = 1+u so that
ò du/(1-u²) = 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 = 0, and y_o = 1-u_o = 1, and ln(1) = 0, we have
ò 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(j_o) + tan(j_o)}] }

where j = sin^-1[(b/mg)^1/2v] and j_o = sin^-1[(b/mg)^1/2v_o] .

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:

e^2at = {1+(b/mg)^1/2v} / {1-(b/mg)^1/2v} , or

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

(b/mg)^1/2v (e^2at + 1) = (e^2at - 1) , or

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

v(t) = (mg/b)^1/2 (e^at – e^-at) / (e^at + e^-at) , and since tanh(x) = (e^x – e^-x) / (e^x + e^-x), we have:

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

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