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2.3 Acceleration–Velocity Models

In Section 1.2 we discussed vertical motion of a mass m near the surface of the earth under the influence of constant gravitational acceleration. If we neglect any effects of air resistance, then Newton’s second law (F=ma $(F = m a$ ) implies that the velocity v of the mass m satisfies the equation

m d v d t = F G,

$m \frac{d v}{d t} = F_{G},$ (1)

where FG=−mg $F_{G} = - m g$ is the (downward-directed) force of gravity, where the gravitational acceleration is g≈9.8 m/s2 $g \approx 9.8 {m/s}^{2}$ (in mks units; g≈32 ft/s2 $g \approx 32 {ft/s}^{2}$ in fps units).

Example 1

No air resistance Suppose that a crossbow bolt is shot straight upward from the ground (y0=0 $y_{0} = 0$ ) with initial velocity v0=49 $v_{0} = 49$ (m/s). Then Eq. (1) with g=9.8 $g = 9.8$ gives

d v d t = - 9.8, so v (t) = - (9.8) t + v 0 = - (9.8) t + 49.

$\frac{d v}{d t} = - 9.8, so v (t) = - (9.8) t + v_{0} = - (9.8) t + 49.$

Hence the bolt’s height function y(t) is given by

y (t) = \int [- (9.8) t + 49] d t = - (4.9) t 2 + 49 t + y 0 = - (4.9) t 2 + 49 t .

$y (t) = \int [- (9.8) t + 49] d t = - (4.9) t^{2} + 49 t + y_{0} = - (4.9) t^{2} + 49 t .$

The bolt reaches its maximum height when v=−(9.8)t+49=0, $v = - (9.8) t + 49 = 0,$ hence when t=5 $t = 5$ (s). Thus its maximum height is

y max = y (5) = - (4.9) (52) + (49) (5) = 122.5 (m) .

$y_{\max} = y (5) = - (4.9) (5^{2}) + (49) (5) = 122.5 (m) .$

The bolt returns to the ground when y=−(4.9)t(t−10)=0, $y = - (4.9) t (t - 10) = 0,$ and thus after 10 seconds aloft.

Now we want to take account of air resistance in a problem like Example 1. The force FR $F_{R}$ exerted by air resistance on the moving mass m must be added in Eq. (1), so now

m d v d t = F G + F R .

$m \frac{d v}{d t} = F_{G} + F_{R} .$ (2)

Newton showed in his Principia Mathematica that certain simple physical assumptions imply that FR $F_{R}$ is proportional to the square of the velocity: FR=kv2. $F_{R} = k v^{2} .$ But empirical investigations indicate that the actual dependence of air resistance on velocity can be quite complicated. For many purposes it suffices to assume that

F R = k v p,

$F_{R} = k v^{p},$

where 1≦p≦2 $1 ≦ p ≦ 2$ and the value of k depends on the size and shape of the body, as well as the density and viscosity of the air. Generally speaking, p=1 $p = 1$ for relatively low speeds and p=2 $p = 2$ for high speeds, whereas 1<p<2 $1 < p < 2$ for intermediate speeds. But how slow “low speed” and how fast “high speed” are depend on the same factors that determine the value of the coefficient k.

Thus air resistance is a complicated physical phenomenon. But the simplifying assumption that FR $F_{R}$ is exactly of the form given here, with either p=1 $p = 1$ or p=2, $p = 2,$ yields a tractable mathematical model that exhibits the most important qualitative features of motion with resistance.

Resistance Proportional to Velocity

Let us first consider the vertical motion of a body with mass m near the surface of the earth, subject to two forces: a downward gravitational force FG $F_{G}$ and a force FR $F_{R}$ of air resistance that is proportional to velocity (so that p=1 $p = 1$ ) and of course directed opposite the direction of motion of the body. If we set up a coordinate system with the positive y-direction upward and with y=0 $y = 0$ at ground level, then FG=−mg $F_{G} = - m g$ and

F R = - k v,

$F_{R} = - k v,$ (3)

where k is a positive constant and v=dy/dt $v = d y / d t$ is the velocity of the body. Note that the minus sign in Eq. (3) makes FR $F_{R}$ positive (an upward force) if the body is falling (v is negative) and makes FR $F_{R}$ negative (a downward force) if the body is rising (v is positive). As indicated in Fig. 2.3.1, the net force acting on the body is then

F = F R + F G = - k v - m g,

$F = F_{R} + F_{G} = - k v - m g,$

FIGURE 2.3.1.

Vertical motion with air resistance.

and Newton’s law of motion F=m(dv/dt) $F = m (d v / d t)$ yields the equation

m d v d t = - k v - m g .

$m \frac{d v}{d t} = - k v - m g .$

Thus

d v d t = - ρ v - g,

$\frac{d v}{d t} = - ρ v - g,$ (4)

where ρ=k/m>0. $ρ = k / m > 0 .$ You should verify for yourself that if the positive y-axis were directed downward, then Eq. (4) would take the form dv/dt=−ρv+g $d v / d t = - ρ v + g$ .

Equation (4) is a separable first-order differential equation, and its solution is

v (t) = (v 0 + g ρ) e - ρ t - g ρ .

$v (t) = (v_{0} + \frac{g}{ρ}) e^{- ρ t} - \frac{g}{ρ} .$ (5)

Here, v0=v(0) $v_{0} = v (0)$ is the initial velocity of the body. Note that

v τ = lim t \to \infty v (t) = - g ρ .

$v_{τ} = \lim_{t \to \infty} v (t) = - \frac{g}{ρ} .$ (6)

Thus the speed of a body falling with air resistance does not increase indefinitely; instead, it approaches a finite limiting speed, or terminal speed,

| v τ | = g ρ = m g k .

$| v_{τ} | = \frac{g}{ρ} = \frac{m g}{k} .$ (7)

This fact is what makes a parachute a practical invention; it even helps explain the occasional survival of people who fall without parachutes from high-flying airplanes.

We now rewrite Eq. (5) in the form

d y d t = (v 0 - v τ) e - ρ t + v τ .

$\frac{d y}{d t} = (v_{0} - v_{τ}) e^{- ρ t} + v_{τ} .$ (8)

Integration gives

y (t) = - 1 ρ (v 0 - v τ) e - ρ t + v τ t + C .

$y (t) = - \frac{1}{ρ} (v_{0} - v_{τ}) e^{- ρ t} + v_{τ} t + C .$

We substitute 0 for t and let y0=y(0) $y_{0} = y (0)$ denote the initial height of the body. Thus we find that C=y0+(v0−vτ)/ρ, $C = y_{0} + (v_{0} - v_{τ}) / ρ,$ and so

y (t) = y 0 + v τ t + 1 ρ (v 0 - v τ) (1 - e - ρ t) .

$y (t) = y_{0} + v_{τ} t + \frac{1}{ρ} (v_{0} - v_{τ}) (1 - e^{- ρ t}) .$ (9)

Equations (8) and (9) give the velocity v and height y of a body moving vertically under the influence of gravity and air resistance. The formulas depend on the initial height y0 $y_{0}$ of the body, its initial velocity v0, $v_{0},$ and the drag coefficient ρ, $ρ,$ the constant such that the acceleration due to air resistance is aR=−ρv. $a_{R} = - ρ v .$ The two equations also involve the terminal velocity vτ $v_{τ}$ defined in Eq. (6).

For a person descending with the aid of a parachute, a typical value of ρ $ρ$ is 1.5, which corresponds to a terminal speed of |vτ|≈21.3 $| v_{τ} | \approx 21.3$ ft/s, or about 14.5 mi/h. With an unbuttoned overcoat flapping in the wind in place of a parachute, an unlucky skydiver might increase ρ $ρ$ to perhaps as much as 0.5, which gives a terminal speed of |vτ|≈65 $| v_{τ} | \approx 65$ ft/s, about 44 mi/h. See Problems 10 and 11 for some parachute-jump computations.

Example 2

Velocity-proportional resistance We again consider a bolt shot straight upward with initial velocity v0=49 m/s $v_{0} = 49 m/s$ from a crossbow at ground level. But now we take air resistance into account, with ρ=0.04 $ρ = 0.04$ in Eq. (4). We ask how the resulting maximum height and time aloft compare with the values found in Example 1.

Solution

We substitute y0=0, v0=49, $y_{0} = 0, v_{0} = 49,$ and vτ=−g/ρ=−245 $v_{τ} = - g / ρ = - 245$ in Eqs. (5) and (9), and obtain

v (t) y (t) = = 294 e - t / 25 - 245, 7350 - 245 t - 7350 e - t / 25 .

$\begin{aligned} v (t) & = & 294 e^{- t / 25} - 245, \\ y (t) & = & 7350 - 245 t - 7350 e^{- t / 25} . \end{aligned}$

To find the time required for the bolt to reach its maximum height (when v=0 $v = 0$ ), we solve the equation

v (t) = 294 e - t / 25 - 245 = 0

$v (t) = 294 e^{- t / 25} - 245 = 0$

for tm=25ln (294/245)≈4.558 $t_{m} = 25 \ln (294/245) \approx 4.558$ (s). Its maximum height is then ymax=v(tm)≈108.280 $y_{max} = v (t_{m}) \approx 108.280$ meters (as opposed to 122.5 meters without air resistance). To find when the bolt strikes the ground, we must solve the equation

y (t) = 7350 - 245 t - 7350 e - t / 25 = 0.

$y (t) = 7350 - 245 t - 7350 e^{- t / 25} = 0.$

Using Newton’s method, we can begin with the initial guess t0=10 $t_{0} = 10$ and carry out the iteration tn+1=tn−y(tn)/y'(tn) $t_{n + 1} = t_{n} - y (t_{n}) / y ′ (t_{n})$ to generate successive approximations to the root. Or we can simply use the Solve command on a calculator or computer. We find that the bolt is in the air for tf≈9.411 $t_{f} \approx 9.411$ seconds (as opposed to 10 seconds without air resistance). It hits the ground with a reduced speed of |v(tf)|≈43.227 m/s $| v (t_{f}) | \approx 43.227 m/s$ (as opposed to its initial velocity of 49 m/s).

Thus the effect of air resistance is to decrease the bolt’s maximum height, the total time spent aloft, and its final impact speed. Note also that the bolt now spends more time in descent (tf−tm≈4.853 $t_{f} - t_{m} \approx 4.853$ s) than in ascent (tm≈4.558 $t_{m} \approx 4.558$ s).

Resistance Proportional to Square of Velocity

Now we assume that the force of air resistance is proportional to the square of the velocity:

F R = \pm k v 2,

$F_{R} = \pm k v^{2},$ (10)

with k>0. $k > 0 .$ The choice of signs here depends on the direction of motion, which the force of resistance always opposes. Taking the positive y-direction as upward, FR<0 $F_{R} < 0$ for upward motion (when v>0 $v > 0$ ) while FR>0 $F_{R} > 0$ for downward motion (when v<0 $v < 0$ ). Thus the sign of FR $F_{R}$ is always opposite that of v, so we can rewrite Eq. (10) as

F R = - k v | v | .

$F_{R} = - k v | v | .$ (10′)

Then Newton’s second law gives

m d v d t = F G + F R = - m g - k v | v |;

$m \frac{d v}{d t} = F_{G} + F_{R} = - m g - k v | v |;$

that is,

d v d t = - g - ρ v | v |,

$\frac{d v}{d t} = - g - ρ v | v |,$ (11)

where ρ=k/m>0. $ρ = k / m > 0 .$ We must discuss the cases of upward and downward motion separately.

Upward Motion: Suppose that a projectile is launched straight upward from the initial position y0 $y_{0}$ with initial velocity v0>0. $v_{0} > 0 .$ Then Eq. (11) with v>0 $v > 0$ gives the differential equation

d v d t = - g - ρ v 2 = - g (1 + ρ g v 2) .

$\frac{d v}{d t} = - g - ρ v^{2} = - g (1 + \frac{ρ}{g} v^{2}) .$ (12)

In Problem 13 we ask you to make the substitution u=vρ/g−−−√ $u = v \sqrt{ρ / g}$ and apply the familiar integral

\int 1 1 + u 2 d u = tan - 1 u + C

$\int \frac{1}{1 + u^{2}} d u = \tan^{- 1} u + C$

to derive the projectile’s velocity function

v (t) = g ρ - - \sqrt tan (C 1 - t ρ g - - \sqrt) with C 1 = tan - 1 (v 0 ρ g - - \sqrt) .

$v (t) = \sqrt{\frac{g}{ρ}} \tan (C_{1} - t \sqrt{ρ g}) with C_{1} = \tan^{- 1} (v_{0} \sqrt{\frac{ρ}{g}}) .$ (13)

Because ∫tanu du=−ln |cos u|+C, $\int \tan u d u = - \ln | \cos u | + C,$ a second integration (see Problem 14) yields the position function

y (t) = y 0 + 1 ρ ln ∣ ∣ ∣ cos ( C 1 - t ρ g - - \sqrt ) cos C 1 ∣ ∣ ∣ .

$y (t) = y_{0} + \frac{1}{ρ} \ln | \frac{\cos (C_{1} - t \sqrt{ρ g})}{\cos C_{1}} | .$ (14)

Downward Motion: Suppose that a projectile is launched (or dropped) straight downward from the initial position y0 $y_{0}$ with initial velocity v0≦0. $v_{0} ≦ 0 .$ Then Eq. (11) with v<0 $v < 0$ gives the differential equation

d v d t = - g + ρ v 2 = - g (1 - ρ g v 2) .

$\frac{d v}{d t} = - g + ρ v^{2} = - g (1 - \frac{ρ}{g} v^{2}) .$ (15)

In Problem 15 we ask you to make the substitution u=vρ/g−−−√ $u = v \sqrt{ρ / g}$ and apply the integral

\int 1 1 - u 2 d u = tanh - 1 u + C

$\int \frac{1}{1 - u^{2}} d u = \tanh^{- 1} u + C$

to derive the projectile’s velocity function

v (t) = g ρ - - \sqrt tanh (C 2 - t ρ g - - \sqrt) with C 2 = tanh - 1 (v 0 ρ g - - \sqrt) .

$v (t) = \sqrt{\frac{g}{ρ}} \tanh (C_{2} - t \sqrt{ρ g}) with C_{2} = \tanh^{- 1} (v_{0} \sqrt{\frac{ρ}{g}}) .$ (16)

Because ∫tanh u du=ln |coshu|+C, $\int \tanh u d u = \ln | \cosh u | + C,$ another integration (Problem 16) yields the position function

y (t) = y 0 - 1 ρ ln ∣ ∣ ∣ cosh ( C 2 - t ρ g - - \sqrt ) cosh C 2 ∣ ∣ ∣ .

$y (t) = y_{0} - \frac{1}{ρ} \ln | \frac{\cosh (C_{2} - t \sqrt{ρ g})}{\cosh C_{2}} | .$ (17)

(Note the analogy between Eqs. (16) and (17) and Eqs. (13) and (14) for upward motion.)

If v0=0, $v_{0} = 0,$ then C2=0, $C_{2} = 0,$ so v(t)=−g/ρ−−−√tanh(tρg−−√). $v (t) = - \sqrt{g / ρ} \tanh (t \sqrt{ρ g}) .$ Because

lim x \to \infty tanh x = lim x \to \infty sinh x cosh x = lim x \to \infty 1 2 ( e x - e - x ) 1 2 ( e x + e - x ) = 1,

$\lim_{x \to \infty} \tanh x = \lim_{x \to \infty} \frac{\sinh x}{\cosh x} = \lim_{x \to \infty} \frac{\frac{1}{2} (e^{x} - e^{- x})}{\frac{1}{2} (e^{x} + e^{- x})} = 1,$

it follows that in the case of downward motion the body approaches the terminal speed

| v τ | = g ρ - - \sqrt

$| v_{τ} | = \sqrt{\frac{g}{ρ}}$ (18)

(as compared with |vτ|=g/ρ $| v_{τ} | = g / ρ$ in the case of downward motion with linear resistance described by Eq. (4)).

Example 3

Square-proportional resistance We consider once more a bolt shot straight upward with initial velocity v0=49 m/s $v_{0} = 49 m/s$ from a crossbow at ground level, as in Example 2. But now we assume air resistance proportional to the square of the velocity, with ρ=0.0011 $ρ = 0.0011$ in Eqs. (12) and (15). In Problems 17 and 18 we ask you to verify the entries in the last line of the following table.

Air Resistance	Maximum Height (ft)	Time Aloft (s)	Ascent Time (s)	Descent Time (s)	Impact Speed (ft/s)
0.0	122.5	10	5	5	49
(0.04)`v`	108.28	9.41	4.56	4.85	43.23
(0.0011)v2 $(0.0011) v^{2}$	108.47	9.41	4.61	4.80	43.49

Comparison of the last two lines of data here indicates little difference—for the motion of our crossbow bolt—between linear air resistance and air resistance proportional to the square of the velocity. And in Fig. 2.3.2, where the corresponding height functions are graphed, the difference is hardly visible. However, the difference between linear and nonlinear resistance can be significant in more complex situations—such as, for instance, the atmospheric reentry and descent of a space vehicle.

FIGURE 2.3.2.

The height functions in Example 1 (without air resistance), Example 2 (with linear air resistance), and Example 3 (with air resistance proportional to the square of the velocity) are all plotted. The graphs of the latter two are visually indistinguishable.

Variable Gravitational Acceleration

Unless a projectile in vertical motion remains in the immediate vicinity of the earth’s surface, the gravitational acceleration acting on it is not constant. According to Newton’s law of gravitation, the gravitational force of attraction between two point masses M and m located at a distance r apart is given by

F = G M m r 2,

$F = \frac{G M m}{r^{2}},$ (19)

where G is a certain empirical constant (G≈6.6726×10−11 N⋅(m/kg)2 $G \approx 6.6726 \times 10^{- 11} N \cdot {(m / kg)}^{2}$ in mks units). The formula is also valid if either or both of the two masses are homogeneous spheres; in this case, the distance r is measured between the centers of the spheres.

The following example is similar to Example 2 in Section 1.2, but now we take account of lunar gravity.

Example 4

Lunar lander A lunar lander is free-falling toward the moon, and at an altitude of 53 kilometers above the lunar surface its downward velocity is measured at 1477 km/h. Its retrorockets, when fired in free space, provide a deceleration of T=4 m/s2. $T = 4 {m/s}^{2} .$ At what height above the lunar surface should the retrorockets be activated to ensure a “soft touchdown” (v=0 $v = 0$ at impact)?

Solution

Let r(t) denote the lander’s distance from the center of the moon at time t (Fig. 2.3.3). When we combine the (constant) thrust acceleration T and the (negative) lunar acceleration F/m=GM/r2 $F / m = G M / r^{2}$ of Eq. (19), we get the (acceleration) differential equation

d 2 r d t 2 = T - G M r 2,

$\frac{d^{2} r}{d t^{2}} = T - \frac{G M}{r^{2}},$ (20)

FIGURE 2.3.3.

The lunar lander descending to the surface of the moon.

where M=7.35×1022 $M = 7.35 \times 10^{22}$ (kg) is the mass of the moon, which has a radius of R=1.74×106 $R = 1.74 \times 10^{6}$ meters (or 1740 km, a little over a quarter of the earth’s radius). Noting that this second-order differential equation does not involve the independent variable t, we substitute

v = d r d t, d 2 r d t 2 = d v d t = d v d r \cdot d r d t = v d v d r

$v = \frac{d r}{d t}, \frac{d^{2} r}{d t^{2}} = \frac{d v}{d t} = \frac{d v}{d r} \cdot \frac{d r}{d t} = v \frac{d v}{d r}$

(as in Eq. (36) of Section 1.6) and obtain the first-order equation

v d v d r = T - G M r 2

$v \frac{d v}{d r} = T - \frac{G M}{r^{2}}$

with the new independent variable r. Integration with respect to r now yields the equation

1 2 v 2 = T r + G M r + C

$\frac{1}{2} v^{2} = T r + \frac{G M}{r} + C$ (21)

that we can apply both before ignition (T=0 $T = 0$ ) and after ignition (T=4 $T = 4$ ).

Before ignition: Substitution of T=0 $T = 0$ in (21) gives the equation

1 2 v 2 = G M r + C 1

$\frac{1}{2} v^{2} = \frac{G M}{r} + C_{1}$ (21a)

where the constant is given by C1=v20/2−GM/r0 $C_{1} = v_{0}^{2} / 2 - G M / r_{0}$ with

v 0 = - 1477 km h \times 1000 m km \times 1 h 3600 s = - 14770 36 m s

$v_{0} = - 1477 \frac{km}{h} \times 1000 \frac{m}{km} \times \frac{1 h}{3600 s} = - \frac{14770}{36} \frac{m}{s}$

and r0=(1.74×106)+53,000=1.793×106 m $r_{0} = (1.74 \times 10^{6}) + 53, 000 = 1.793 \times 10^{6} m$ (from the initial velocity–position measurement).

After ignition: Substitution of T=4 $T = 4$ and v=0, r=R $v = 0, r = R$ (at touchdown) into (21) gives

1 2 v 2 = 4 r + G M r + C 2

$\frac{1}{2} v^{2} = 4 r + \frac{G M}{r} + C_{2}$ (21b)

where the constant C2=−4R−GM/R $C_{2} = - 4 R - G M / R$ is obtained by substituting the values v=0, r=R $v = 0, r = R$ at touchdown.

At the instant of ignition the lunar lander’s position and velocity satisfy both (21a) and (21b). Therefore we can find its desired height h above the lunar surface at ignition by equating the right-hand sides in (21a) and (21b). This gives r=14(C1−C2)=1.78187×106 $r = \frac{1}{4} (C_{1} - C_{2}) = 1.78187 \times 10^{6}$ and finally h=r−R=41,870 $h = r - R = 41, 870$ meters (that is, 41.87 kilometers—just over 26 miles). Moreover, substitution of this value of r in (21a) gives the velocity v=−450 $v = - 450$ m/s at the instant of ignition.

Escape Velocity

In his novel From the Earth to the Moon (1865), Jules Verne raised the question of the initial velocity necessary for a projectile fired from the surface of the earth to reach the moon. Similarly, we can ask what initial velocity v0 $v_{0}$ is necessary for the projectile to escape from the earth altogether. This will be so if its velocity v=dr/dt $v = d r / d t$ remains positive for all t>0, $t > 0,$ so it continues forever to move away from the earth. With r(t) denoting the projectile’s distance from the earth’s center at time t (Fig. 2.3.4), we have the equation

d v d t = d 2 r d t 2 = - G M r 2,

$\frac{d v}{d t} = \frac{d^{2} r}{d t^{2}} = - \frac{G M}{r^{2}},$ (22)

FIGURE 2.3.4.

A mass `m` at a great distance from the earth.

similar to Eq. (20), but with T=0 $T = 0$ (no thrust) and with M=5.975×1024 $M = 5.975 \times 10^{24}$ (kg) denoting the mass of the earth, which has an equatorial radius of R=6.378×106 $R = 6.378 \times 10^{6}$ (m). Substitution of the chain rule expression dv/dt=v(dv/dr) $d v / d t = v (d v / d r)$ as in Example 4 gives

v d v d r = - G M r 2 .

$v \frac{d v}{d r} = - \frac{G M}{r^{2}} .$

Then integration of both sides with respect to r yields

1 2 v 2 = - G M r + C .

$\frac{1}{2} v^{2} = - \frac{G M}{r} + C .$

Now v=v0 $v = v_{0}$ and r=R $r = R$ when t=0, $t = 0,$ so C=12v20−GM/R, $C = \frac{1}{2} v_{0}^{2} - G M / R,$ and hence solution for v2 $v^{2}$ gives

v 2 = v 20 + 2 G M (1 r - 1 R) .

$v^{2} = v_{0}^{2} + 2 G M (\frac{1}{r} - \frac{1}{R}) .$ (23)

This implicit solution of Eq. (22) determines the projectile’s velocity v as a function of its distance r from the earth’s center. In particular,

v 2 > v 20 - 2 G M R,

$v^{2} > v_{0}^{2} - \frac{2 G M}{R},$

so v will remain positive provided that v20≧2GM/R. $v_{0}^{2} ≧ 2 G M / R .$ Therefore, the escape velocity from the earth is given by

v 0 = 2 G M R - - - - - \sqrt .

$v_{0} = \sqrt{\frac{2 G M}{R}} .$ (24)

In Problem 27 we ask you to show that, if the projectile’s initial velocity exceeds 2GM/R−−−−−−−√ $\sqrt{2 G M / R}$ , then r(t)→∞ $r (t) \to \infty$ as t→∞, $t \to \infty,$ so it does, indeed, “escape” from the earth. With the given values of G and the earth’s mass M and radius R, this gives v0≈11,180 $v_{0} \approx 11, 180$ (m/s) (about 36,680 ft/s, about 6.95 mi/s, about 25,000 mi/h).

Remark

Equation (24) gives the escape velocity for any other (spherical) planetary body when we use its mass and radius. For instance, when we use the mass M and radius R for the moon given in Example 4, we find that escape velocity from the lunar surface is v0≈2375 m/s. $v_{0} \approx 2375 m/s .$ This is just over one-fifth of the escape velocity from the earth’s surface, a fact that greatly facilitates the return trip (“From the Moon to the Earth”).

2.3 Problems

The acceleration of a Maserati is proportional to the difference between 250 km/h and the velocity of this sports car. If this machine can accelerate from rest to 100 km/h in 10 s, how long will it take for the car to accelerate from rest to 200 km/h?

Problems 2 through 8 explore the effects of resistance proportional to a power of the velocity.

Suppose that a body moves through a resisting medium with resistance proportional to its velocity v, so that dv/dt=−kv. $d v / d t = - k v .$ (a) Show that its velocity and position at time t are given by

$v (t) = v 0 e - k t$ $v (t) = v_{0} e^{- k t}$

and

$x (t) = x 0 + (v 0 k) (1 - e - k t) .$ $x (t) = x_{0} + (\frac{v_{0}}{k}) (1 - e^{- k t}) .$

(b) Conclude that the body travels only a finite distance, and find that distance.
Suppose that a motorboat is moving at 40 ft/s when its motor suddenly quits, and that 10 s later the boat has slowed to 20 ft/s. Assume, as in Problem 2, that the resistance it encounters while coasting is proportional to its velocity. How far will the boat coast in all?
Consider a body that moves horizontally through a medium whose resistance is proportional to the square of the velocity v, so that dv/dt=−kv2. $d v / d t = - k v^{2} .$ Show that

$v (t) = v 0 1 + v 0 k t$ $v (t) = \frac{v_{0}}{1 + v_{0} k t}$

and that

$x (t) = x 0 + 1 k ln (1 + v 0 k t) .$ $x (t) = x_{0} + \frac{1}{k} \ln (1 + v_{0} k t) .$

Note that, in contrast with the result of Problem 2, x(t)→+∞ $x (t) \to + \infty$ as t→+∞. $t \to + \infty .$ Which offers less resistance when the body is moving fairly slowly—the medium in this problem or the one in Problem 2? Does your answer seem consistent with the observed behaviors of x(t) as t→∞ $t \to \infty$ ?
Assuming resistance proportional to the square of the velocity (as in Problem 4), how far does the motorboat of Problem 3 coast in the first minute after its motor quits?
Assume that a body moving with velocity v encounters resistance of the form dv/dt=−kv3/2. $d v / d t = - k v^{3/2} .$ Show that

$v (t) = 4 v 0 ( k t v 0 - - \sqrt + 2 ) 2$ $v (t) = \frac{4 v_{0}}{{(k t \sqrt{v_{0}} + 2)}^{2}}$

and that

$x (t) = x 0 + 2 k v 0 - - \sqrt (1 - 2 k t v 0 - - \sqrt + 2) .$ $x (t) = x_{0} + \frac{2}{k} \sqrt{v_{0}} (1 - \frac{2}{k t \sqrt{v_{0}} + 2}) .$

Conclude that under a 32 $\frac{3}{2}$ -power resistance a body coasts only a finite distance before coming to a stop.
Suppose that a car starts from rest, its engine providing an acceleration of 10 ft/s2, $10 {ft/s}^{2},$ while air resistance provides 0.1 ft/s2 $0.1 {ft/s}^{2}$ of deceleration for each foot per second of the car’s velocity. (a) Find the car’s maximum possible (limiting) velocity. (b) Find how long it takes the car to attain 90% of its limiting velocity, and how far it travels while doing so.
Rework both parts of Problem 7, with the sole difference that the deceleration due to air resistance now is (0.001)v2 ft/s2 $(0.001) v^{2} {ft/s}^{2}$ when the car’s velocity is v feet per second.

Problems 9 through 12 illustrate resistance proportional to the velocity.

A motorboat weighs 32,000 lb and its motor provides a thrust of 5000 lb. Assume that the water resistance is 100 pounds for each foot per second of the speed v of the boat. Then

$1000 d v d t = 5000 - 100 v .$ $1000 \frac{d v}{d t} = 5000 - 100 v .$

If the boat starts from rest, what is the maximum velocity that it can attain?
Falling parachutist A woman bails out of an airplane at an altitude of 10,000 ft, falls freely for 20 s, then opens her parachute. How long will it take her to reach the ground? Assume linear air resistance ρv ft/s2, $ρ v {ft/s}^{2},$ taking ρ=0.15 $ρ = 0.15$ without the parachute and ρ=1.5 $ρ = 1.5$ with the parachute. (Suggestion: First determine her height above the ground and velocity when the parachute opens.)
Falling paratrooper According to a newspaper account, a paratrooper survived a training jump from 1200 ft when his parachute failed to open but provided some resistance by flapping unopened in the wind. Allegedly he hit the ground at 100 mi/h after falling for 8 s. Test the accuracy of this account. (Suggestion: Find ρ $ρ$ in Eq. (4) by assuming a terminal velocity of 100 mi/h. Then calculate the time required to fall 1200 ft.)
Nuclear waste disposal It is proposed to dispose of nuclear wastes—in drums with weight W=640 $W = 640$ lb and volume 8 ft3 ${8 ft}^{3}$ —by dropping them into the ocean (v0=0 $v_{0} = 0$ ). The force equation for a drum falling through water is

$m d v d t = - W + B + F R,$ $m \frac{d v}{d t} = - W + B + F_{R},$

where the buoyant force B is equal to the weight (at 62.5 lb/ft3 ${62.5 lb/ft}^{3}$ ) of the volume of water displaced by the drum (Archimedes’ principle) and FR $F_{R}$ is the force of water resistance, found empirically to be 1 lb for each foot per second of the velocity of a drum. If the drums are likely to burst upon an impact of more than 75 ft/s, what is the maximum depth to which they can be dropped in the ocean without likelihood of bursting?
Separate variables in Eq. (12) and substitute u=vρ/g−−−√ $u = v \sqrt{ρ / g}$ to obtain the upward-motion velocity function given in Eq. (13) with initial condition v(0)=v0 $v (0) = v_{0}$ .
Integrate the velocity function in Eq. (13) to obtain the upward-motion position function given in Eq. (14) with initial condition y(0)=y0 $y (0) = y_{0}$ .
Separate variables in Eq. (15) and substitute u=vρ/g−−−√ $u = v \sqrt{ρ / g}$ to obtain the downward-motion velocity function given in Eq. (16) with initial condition v(0)=v0 $v (0) = v_{0}$ .
Integrate the velocity function in Eq. (16) to obtain the downward-motion position function given in Eq. (17) with initial condition y(0)=y0 $y (0) = y_{0}$ .

Problems 17 and 18 apply Eqs. (12)–(17) to the motion of a crossbow bolt.

Consider the crossbow bolt of Example 3, shot straight upward from the ground (y=0 $y = 0$ ) at time t=0 $t = 0$ with initial velocity v0=49 m/s. $v_{0} = 49 m/s .$ Take g=9.8 m/s2 $g = 9.8 {m/s}^{2}$ and ρ=0.0011 $ρ = 0.0011$ in Eq. (12). Then use Eqs. (13) and (14) to show that the bolt reaches its maximum height of about 108.47 m in about 4.61 s.
Continuing Problem 17, suppose that the bolt is now dropped (v0=0 $v_{0} = 0$ ) from a height of y0=108.47 $y_{0} = 108.47$ m. Then use Eqs. (16) and (17) to show that it hits the ground about 4.80 s later with an impact speed of about 43.49 m/s.

Problems 19 through 23 illustrate resistance proportional to the square of the velocity.

A motorboat starts from rest (initial velocity v(0)=v0=0 $v (0) = v_{0} = 0$ ). Its motor provides a constant acceleration of 4 ft/s2, $4 {ft/s}^{2},$ but water resistance causes a deceleration of v2/400 ft/s2. $v^{2} / 400 {ft/s}^{2} .$ Find v when t=10 $t = 10$ s, and also find the limiting velocity as t→+∞ $t \to + \infty$ (that is, the maximum possible speed of the boat).
An arrow is shot straight upward from the ground with an initial velocity of 160 ft/s. It experiences both the deceleration of gravity and deceleration v2/800 $v^{2} / 800$ due to air resistance. How high in the air does it go?
If a ball is projected upward from the ground with initial velocity v0 $v_{0}$ and resistance proportional to v2, $v^{2},$ deduce from Eq. (14) that the maximum height it attains is

$y max = 1 2 ρ ln (1 + ρ v 2 0 g) .$ $y_{max} = \frac{1}{2 ρ} \ln (1 + \frac{ρ v_{0}^{2}}{g}) .$
Suppose that ρ=0.075 $ρ = 0.075$ (in fps units, with g=32 ft/s2 $g = 32 {ft/s}^{2}$ ) in Eq. (15) for a paratrooper falling with parachute open. If he jumps from an altitude of 10,000 ft and opens his parachute immediately, what will be his terminal speed? How long will it take him to reach the ground?
Suppose that the paratrooper of Problem 22 falls freely for 30 s with ρ=0.00075 $ρ = 0.00075$ before opening his parachute. How long will it now take him to reach the ground?

Problems 24 through 30 explore gravitational acceleration and escape velocity.

The mass of the sun is 329,320 $329, 320$ times that of the earth and its radius is 109 times the radius of the earth. (a) To what radius (in meters) would the earth have to be compressed in order for it to become a black hole—the escape velocity from its surface equal to the velocity c=3×108 m/s $c = 3 \times 10^{8} m/s$ of light? (b) Repeat part (a) with the sun in place of the earth.
(a) Show that if a projectile is launched straight upward from the surface of the earth with initial velocity v0 $v_{0}$ less than escape velocity 2GM/R−−−−−−−√, $\sqrt{2 G M / R},$ then the maximum distance from the center of the earth attained by the projectile is

$r max = 2 G M R 2 G M - R v 2 0,$ $r_{max} = \frac{2 GMR}{2 G M - R v_{0}^{2}},$

where M and R are the mass and radius of the earth, respectively. (b) With what initial velocity v0 $v_{0}$ must such a projectile be launched to yield a maximum altitude of 100 kilometers above the surface of the earth? (c) Find the maximum distance from the center of the earth, expressed in terms of earth radii, attained by a projectile launched from the surface of the earth with 90% of escape velocity.
Suppose that you are stranded—your rocket engine has failed—on an asteroid of diameter 3 miles, with density equal to that of the earth with radius 3960 miles. If you have enough spring in your legs to jump 4 feet straight up on earth while wearing your space suit, can you blast off from this asteroid using leg power alone?
1. Suppose a projectile is launched vertically from the surface r=R $r = R$ of the earth with initial velocity v0=2GM/R−−−−−−−√, $v_{0} = \sqrt{2 G M / R},$ so v20=k2/R $v_{0}^{2} = k^{2} / R$ where k2=2GM. $k^{2} = 2 G M .$ Solve the differential equation dr/dt=k/r√ $d r / d t = k / \sqrt{r}$ (from Eq. (23) in this section) explicitly to deduce that r(t)→∞ $r (t) \to \infty$ as t→∞ $t \to \infty$ .
2. If the projectile is launched vertically with initial velocity v0>2GM/R−−−−−−−√ $v_{0} > \sqrt{2 G M / R}$ , deduce that
  
  $d r d t = k 2 r + α - - - - - - \sqrt > k r \sqrt .$ $\frac{d r}{d t} = \sqrt{\frac{k^{2}}{r} + α} > \frac{k}{\sqrt{r}} .$
  
  Why does it again follow that r(t)→∞ $r (t) \to \infty$ as t→∞ $t \to \infty$ ?
(a) Suppose that a body is dropped (v0=0 $v_{0} = 0$ ) from a distance r0>R $r_{0} > R$ from the earth’s center, so its acceleration is dv/dt=−GM/r2. $d v / d t = - G M / r^{2} .$ Ignoring air resistance, show that it reaches the height r<r0 $r < r_{0}$ at time

$t = r 0 2 G M - - - - - \sqrt (r r 0 - r 2 - - - - - - - \sqrt + r 0 cos - 1 r r 0 - - - \sqrt) .$ $t = \sqrt{\frac{r_{0}}{2 G M}} (\sqrt{r r_{0} - r^{2}} + r_{0} \cos^{- 1} \sqrt{\frac{r}{r_{0}}}) .$

(Suggestion: Substitute r=r0cos2 θ $r = r_{0} \cos^{2} θ$ to evaluate ∫r/(r0−r)−−−−−−−−√dr $\int \sqrt{r / (r_{0} - r)} d r$ .) (b) If a body is dropped from a height of 1000 km above the earth’s surface and air resistance is neglected, how long does it take to fall and with what speed will it strike the earth’s surface?
Suppose that a projectile is fired straight upward from the surface of the earth with initial velocity v0<2GM/R−−−−−−−√. $v_{0} < \sqrt{2 G M / R} .$ Then its height y(t) above the surface satisfies the initial value problem

$d 2 y d t 2 = - G M ( y + R ) 2; y (0) = 0, y' (0) = v 0 .$ $\frac{d^{2} y}{d t^{2}} = - \frac{G M}{{(y + R)}^{2}}; y (0) = 0, y ′ (0) = v_{0} .$

Substitute dv/dt=v(dv/dy) $d v / d t = v (d v / d y)$ and then integrate to obtain

$v 2 = v 20 - 2 G M y R ( R + y )$ $v^{2} = v_{0}^{2} - \frac{2 G M y}{R (R + y)}$

for the velocity v of the projectile at height y. What maximum altitude does it reach if its initial velocity is 1 km/s?
In Jules Verne’s original problem, the projectile launched from the surface of the earth is attracted by both the earth and the moon, so its distance r(t) from the center of the earth satisfies the initial value problem

$d 2 r d t 2 = - G M e r 2 + G M m ( S - r ) 2; r (0) = R, r' (0) = v 0$ $\frac{d^{2} r}{d t^{2}} = - \frac{{G M}_{e}}{r^{2}} + \frac{{G M}_{m}}{{(S - r)}^{2}}; r (0) = R, r ′ (0) = v_{0}$

where Me $M_{e}$ and Mm $M_{m}$ denote the masses of the earth and the moon, respectively; R is the radius of the earth and S=384,400 $S = 384, 400$ km is the distance between the centers of the earth and the moon. To reach the moon, the projectile must only just pass the point between the moon and earth where its net acceleration vanishes. Thereafter it is “under the control” of the moon, and falls from there to the lunar surface. Find the minimal launch velocity v0 $v_{0}$ that suffices for the projectile to make it “From the Earth to the Moon.”

2.3 Application Rocket Propulsion

Suppose that the rocket of Fig. 2.3.5 blasts off straight upward from the surface of the earth at time t=0. $t = 0 .$ We want to calculate its height y and velocity v=dy/dt $v = d y / d t$ at time t. The rocket is propelled by exhaust gases that exit (rearward) with constant speed c (relative to the rocket). Because of the combustion of its fuel, the mass m=m(t) $m = m (t)$ of the rocket is variable.

To derive the equation of motion of the rocket, we use Newton’s second law in the form

d P d t = F,

$\frac{d P}{d t} = F,$ (1)

where P is momentum (the product of mass and velocity) and F denotes net external force (gravity, air resistance, etc.). If the mass m of the rocket is constant so m'(t)≡0 $m ′ (t) \equiv 0$ —when its rockets are turned off or burned out, for instance—then Eq. (1) gives

F = d ( m v ) d t = m d v d t + d m d t v = m d v d t,

$F = \frac{d (m v)}{d t} = m \frac{d v}{d t} + \frac{d m}{d t} v = m \frac{d v}{d t},$

which (with dv/dt=a $d v / d t = a$ ) is the more familiar form F=ma $F = m a$ of Newton’s second law.

But here m is not constant. Suppose m changes to m+Δm $m + Δ m$ and v to v+Δv $v + Δ v$ during the short time interval from t to t+Δt. $t + Δ t .$ Then the change in the momentum of the rocket itself is

Δ P \approx (m + Δ m) (v + Δ v) - m v = m Δ v + v Δ m + Δ m Δ v .

$Δ P \approx (m + Δ m) (v + Δ v) - m v = m Δ v + v Δ m + Δ m Δ v .$

But the system also includes the exhaust gases expelled during this time interval, with mass −Δm $- Δ m$ and approximate velocity v−c. $v - c .$ Hence the total change in momentum during the time interval Δt $Δ t$ is

Δ P \approx = (m Δ v + v Δ m + Δ m Δ v) + (- Δ m) (v - c) m Δ v + c Δ m + Δ m Δ v .

$\begin{aligned} Δ P & \approx & (m Δ v + v Δ m + Δ m Δ v) + (- Δ m) (v - c) \\ = & m Δ v + c Δ m + Δ m Δ v . \end{aligned}$

Now we divide by Δt $Δ t$ and take the limit as Δt→0, $Δ t \to 0,$ so Δm→0, $Δ m \to 0,$ assuming continuity of m(t). The substitution of the resulting expression for dP/dt in (1) yields the rocket propulsion equation

m d v d t + c d m d t = F .

$m \frac{d v}{d t} + c \frac{d m}{d t} = F .$ (2)

If F=FG+FR, $F = F_{G} + F_{R},$ where FG=−mg $F_{G} = - m g$ is a constant force of gravity and FR=−kv $F_{R} = - k v$ is a force of air resistance proportional to velocity, then Eq. (2) finally gives

m d v d t + c d m d t = - m g - k v .

$m \frac{d v}{d t} + c \frac{d m}{d t} = - m g - k v .$ (3)

Constant Thrust

Now suppose that the rocket fuel is consumed at the constant “burn rate” β $β$ during the time interval [0,t1], $[0, t_{1}],$ during which time the mass of the rocket decreases from m0 $m_{0}$ to m1. $m_{1} .$ Thus

m (0) m (t) = = m 0, m 0 - β t, m (t 1) d m d t = = m 1, - β for t \leq t 1,

$\begin{aligned} m (0) & = & m_{0}, & m (t_{1}) & = & m_{1}, \\ m (t) & = & m_{0} - β t, & \frac{d m}{d t} & = & - β for t \leq t_{1}, \end{aligned}$ (4)

with burnout occurring at time t=t1 $t = t_{1}$ .

Problem 1 Substitute the expressions in (4) into Eq. (3) to obtain the differential equation

(m 0 - β t) d v d t + k v = β c - (m 0 - β t) g .

$(m_{0} - β t) \frac{d v}{d t} + k v = β c - (m_{0} - β t) g .$ (5)

Solve this linear equation for

v (t) = v 0 M k / β - g β t β - k + (β c k + g m 0 β - k) (1 - M k / β),

$v (t) = v_{0} M^{k / β} - \frac{g β t}{β - k} + (\frac{β c}{k} + \frac{g m_{0}}{β - k}) (1 - M^{k / β}),$ (6)

where v0=v(0) $v_{0} = v (0)$ and

M = m ( t ) m 0 = m 0 - β t m 0

$M = \frac{m (t)}{m_{0}} = \frac{m_{0} - β t}{m_{0}}$

denotes the rocket’s fractional mass at time t.

No Resistance

Problem 2 For the case of no air resistance, set k=0 $k = 0$ in Eq. (5) and integrate to obtain

v (t) = v 0 - g t + c ln m 0 m 0 - β t .

$v (t) = v_{0} - g t + c \ln \frac{m_{0}}{m_{0} - β t} .$ (7)

Because m0−βt1=m1, $m_{0} - β t_{1} = m_{1},$ it follows that the velocity of the rocket at burnout (t=t1 $t = t_{1}$ ) is

v 1 = v (t 1) = v 0 - g t 1 + c ln m 0 m 1 .

$v_{1} = v (t_{1}) = v_{0} - g t_{1} + c \ln \frac{m_{0}}{m_{1}} .$ (8)

Problem 3 Start with Eq. (7) and integrate to obtain

y (t) = (v 0 + c) t - 1 2 g t 2 - c β (m 0 - β t) ln m 0 m 0 - β t .

$y (t) = (v_{0} + c) t - \frac{1}{2} g t^{2} - \frac{c}{β} (m_{0} - β t) \ln \frac{m_{0}}{m_{0} - β t} .$ (9)

It follows that the rocket’s altitude at burnout is

y 1 = y (t 1) = (v 0 + c) t 1 - 1 2 g t 21 - c m 1 β ln m 0 m 1 .

$y_{1} = y (t_{1}) = (v_{0} + c) t_{1} - \frac{1}{2} g t_{1}^{2} - \frac{c m_{1}}{β} \ln \frac{m_{0}}{m_{1}} .$ (10)

Problem 4 The V-2 rocket that was used to attack London in World War II had an initial mass of 12,850 kg, of which 68.5% was fuel. This fuel burned uniformly for 70 seconds with an exhaust velocity of 2 km/s. Assume it encounters air resistance of 1.45 N per m/s of velocity. Then find the velocity and altitude of the V-2 at burnout under the assumption that it is launched vertically upward from rest on the ground.

Problem 5 Actually, our basic differential equation in (3) applies without qualification only when the rocket is already in motion. However, when a rocket is sitting on its launch pad stand and its engines are turned on initially, it is observed that a certain time interval passes before the rocket actually “blasts off” and begins to ascend. The reason is that if v=0 $v = 0$ in (3), then the resulting initial acceleration

d v d t = c m d m d t - g

$\frac{d v}{d t} = \frac{c}{m} \frac{d m}{d t} - g$

of the rocket may be negative. But the rocket does not descend into the ground; it just “sits there” while (because m is decreasing) this calculated acceleration increases until it reaches 0 and (thereafter) positive values so the rocket can begin to ascend. With the notation introduced to describe the constant-thrust case, show that the rocket initially just “sits there” if the exhaust velocity c is less than m0g/β, $m_{0} g / β,$ and that the time tB $t_{B}$ which then elapses before actual blastoff is given by

t B = m 0 g - β c β g .

$t_{B} = \frac{m_{0} g - β c}{β g} .$

Free Space

Suppose finally that the rocket is accelerating in free space, where there is neither gravity nor resistance, so g=k=0. $g = k = 0 .$ With g=0 $g = 0$ in Eq. (8) we see that, as the mass of the rocket decreases from m0 $m_{0}$ to m1, $m_{1},$ its increase in velocity is

Δ v = v 1 - v 0 = c ln m 0 m 1 .

$Δ v = v_{1} - v_{0} = c \ln \frac{m_{0}}{m_{1}} .$ (11)

Note that Δv $Δ v$ depends only on the exhaust gas speed c and the initial-to-final mass ratio m0/m1, $m_{0} / m_{1},$ but does not depend on the burn rate β. $β .$ For example, if the rocket blasts off from rest (v0=0 $v_{0} = 0$ ) and c=5 $c = 5$ km/s and m0/m1=20, $m_{0} / m_{1} = 20,$ then its velocity at burnout is v1=5ln 20≈15 $v_{1} = 5 \ln 20 \approx 15$ km/s. Thus if a rocket initially consists predominantly of fuel, then it can attain velocities significantly greater than the (relative) velocity of its exhaust gases.

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Table of Contents for
2.3 Acceleration–Velocity Models Acceleration–Velocity Models

2.3 Acceleration–Velocity Models

Example 1

Resistance Proportional to Velocity

FIGURE 2.3.1.

Example 2

Solution

Resistance Proportional to Square of Velocity

Example 3

FIGURE 2.3.2.

Variable Gravitational Acceleration

Example 4

Solution

FIGURE 2.3.3.

Escape Velocity

FIGURE 2.3.4.

Remark

2.3 Problems

2.3 Application Rocket Propulsion

FIGURE 2.3.5.

Constant Thrust

No Resistance

Free Space

Table of Contents for 2.3 Acceleration&#8211;Velocity Models Acceleration–Velocity Models

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Table of Contents for
2.3 Acceleration–Velocity Models Acceleration–Velocity Models