In Section1.2 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=ma) implies that the velocity v of the mass m satisfies the equation
mdvdt=FG,
mdvdt=FG,
(1)
where FG=−mgFG=−mg is the (downward-directed) force of gravity, where the gravitational acceleration is g≈9.8m/s2g≈9.8m/s2 (in mks units; g≈32ft/s2g≈32ft/s2 in fps units).
Example1
No air resistance Suppose that a crossbow bolt is shot straight upward from the ground (y0=0y0=0) with initial velocity v0=49v0=49 (m/s). Then Eq.(1) with g=9.8g=9.8 gives
The bolt reaches its maximum height when v=−(9.8)t+49=0,v=−(9.8)t+49=0, hence when t=5t=5 (s). Thus its maximum height is
ymax=y(5)=−(4.9)(52)+(49)(5)=122.5(m).
ymax=y(5)=−(4.9)(52)+(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 Example1. The force FRFR exerted by air resistance on the moving mass m must be added in Eq.(1), so now
mdvdt=FG+FR.
mdvdt=FG+FR.
(2)
Newton showed in his Principia Mathematica that certain simple physical assumptions imply that FRFR is proportional to the square of the velocity: FR=kv2.FR=kv2. 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
FR=kvp,
FR=kvp,
where 1≦p≦21≦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=1p=1 for relatively low speeds and p=2p=2 for high speeds, whereas 1<p<21<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 FRFR is exactly of the form given here, with either p=1p=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 FGFG and a force FRFR of air resistance that is proportional to velocity (so that p=1p=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=0y=0 at ground level, then FG=−mgFG=−mg and
FR=−kv,
FR=−kv,
(3)
where k is a positive constant and v=dy/dtv=dy/dt is the velocity of the body. Note that the minus sign in Eq.(3) makes FRFR positive (an upward force) if the body is falling (v is negative) and makes FRFR 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=FR+FG=−kv−mg,
F=FR+FG=−kv−mg,
and Newton’s law of motion F=m(dv/dt)F=m(dv/dt) yields the equation
mdvdt=−kv−mg.
mdvdt=−kv−mg.
Thus
dvdt=−ρv−g,
dvdt=−ρv−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+gdv/dt=−ρv+g.
Equation(4) is a separable first-order differential equation, and its solution is
v(t)=(v0+gρ)e−ρt−gρ.
v(t)=(v0+gρ)e−ρt−gρ.
(5)
Here, v0=v(0)v0=v(0) is the initial velocity of the body. Note that
vτ=limt→∞v(t)=−gρ.
vτ=limt→∞v(t)=−gρ.
(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,
|vτ|=gρ=mgk.
|vτ|=gρ=mgk.
(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 substitute 0 for t and let y0=y(0)y0=y(0) denote the initial height of the body. Thus we find that C=y0+(v0−vτ)/ρ,C=y0+(v0−vτ)/ρ, and so
y(t)=y0+vτt+1ρ(v0−vτ)(1−e−ρt).
y(t)=y0+vτt+1ρ(v0−vτ)(1−e−ρt).
(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 y0y0 of the body, its initial velocity v0,v0, and the drag coefficientρ,ρ, the constant such that the acceleration due to air resistance is aR=−ρv.aR=−ρv. The two equations also involve the terminal velocity vτvτ defined in Eq.(6).
For a person descending with the aid of a parachute, a typical value of ρρ is 1.5, which corresponds to a terminal speed of |vτ|≈21.3|vτ|≈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 ρρ to perhaps as much as 0.5, which gives a terminal speed of |vτ|≈65|vτ|≈65 ft/s, about 44 mi/h. See Problems10 and 11 for some parachute-jump computations.
Example2
Velocity-proportional resistance We again consider a bolt shot straight upward with initial velocity v0=49m/sv0=49m/s from a crossbow at ground level. But now we take air resistance into account, with ρ=0.04ρ=0.04 in Eq.(4). We ask how the resulting maximum height and time aloft compare with the values found in Example1.
Solution
We substitute y0=0,v0=49,y0=0,v0=49, and vτ=−g/ρ=−245vτ=−g/ρ=−245 in Eqs.(5) and (9), and obtain
v(t)=294e−t/25−245,y(t)=7350−245t−7350e−t/25.
v(t)y(t)==294e−t/25−245,7350−245t−7350e−t/25.
To find the time required for the bolt to reach its maximum height (when v=0v=0), we solve the equation
v(t)=294e−t/25−245=0
v(t)=294e−t/25−245=0
for tm=25ln(294/245)≈4.558tm=25ln(294/245)≈4.558 (s). Its maximum height is then ymax=v(tm)≈108.280ymax=v(tm)≈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)=7350−245t−7350e−t/25=0.
y(t)=7350−245t−7350e−t/25=0.
Using Newton’s method, we can begin with the initial guess t0=10t0=10 and carry out the iteration tn+1=tn−y(tn)/y′(tn)tn+1=tn−y(tn)/y'(tn) 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≈9.411tf≈9.411 seconds (as opposed to 10 seconds without air resistance). It hits the ground with a reduced speed of |v(tf)|≈43.227m/s|v(tf)|≈43.227m/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≈4.853tf−tm≈4.853 s) than in ascent (tm≈4.558tm≈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:
FR=±kv2,
FR=±kv2,
(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<0FR<0 for upward motion (when v>0v>0) while FR>0FR>0 for downward motion (when v<0v<0). Thus the sign of FRFR is always opposite that of v, so we can rewrite Eq.(10) as
FR=−kv|v|.
FR=−kv|v|.
(10′)
Then Newton’s second law gives
mdvdt=FG+FR=−mg−kv|v|;
mdvdt=FG+FR=−mg−kv|v|;
that is,
dvdt=−g−ρv|v|,
dvdt=−g−ρv|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 y0y0 with initial velocity v0>0.v0>0. Then Eq.(11) with v>0v>0 gives the differential equation
dvdt=−g−ρv2=−g(1+ρgv2).
dvdt=−g−ρv2=−g(1+ρgv2).
(12)
In Problem13 we ask you to make the substitution u=v√ρ/gu=vρ/g−−−√ and apply the familiar integral
∫11+u2du=tan−1u+C
∫11+u2du=tan−1u+C
to derive the projectile’s velocity function
v(t)=√gρtan(C1−t√ρg)withC1=tan−1(v0√ρg).
v(t)=gρ−−√tan(C1−tρg−−√)withC1=tan−1(v0ρg−−√).
(13)
Because ∫tanudu=−ln|cosu|+C,∫tanudu=−ln|cosu|+C, a second integration (see Problem14) yields the position function
y(t)=y0+1ρln|cos(C1−t√ρg)cosC1|.
y(t)=y0+1ρln∣∣∣cos(C1−tρg−−√)cosC1∣∣∣.
(14)
Downward Motion: Suppose that a projectile is launched (or dropped) straight downward from the initial position y0y0 with initial velocity v0≦0.v0≦0. Then Eq.(11) with v<0v<0 gives the differential equation
dvdt=−g+ρv2=−g(1−ρgv2).
dvdt=−g+ρv2=−g(1−ρgv2).
(15)
In Problem15 we ask you to make the substitution u=v√ρ/gu=vρ/g−−−√ and apply the integral
∫11−u2du=tanh−1u+C
∫11−u2du=tanh−1u+C
to derive the projectile’s velocity function
v(t)=√gρtanh(C2−t√ρg)withC2=tanh−1(v0√ρg).
v(t)=gρ−−√tanh(C2−tρg−−√)withC2=tanh−1(v0ρg−−√).
(16)
Because ∫tanhudu=ln|coshu|+C,∫tanhudu=ln|coshu|+C, another integration (Problem16) yields the position function
it follows that in the case of downward motion the body approaches the terminal speed
|vτ|=√gρ
|vτ|=gρ−−√
(18)
(as compared with |vτ|=g/ρ|vτ|=g/ρ in the case of downward motion with linear resistance described by Eq.(4)).
Example3
Square-proportional resistance We consider once more a bolt shot straight upward with initial velocity v0=49m/sv0=49m/s from a crossbow at ground level, as in Example2. But now we assume air resistance proportional to the square of the velocity, with ρ=0.0011ρ=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)v2
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.
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=GMmr2,
F=GMmr2,
(19)
where G is a certain empirical constant (G≈6.6726×10−11 N·(m/kg)2G≈6.6726×10−11 N⋅(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 Example2 in Section1.2, but now we take account of lunar gravity.
Example4
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=4m/s2.T=4m/s2. At what height above the lunar surface should the retrorockets be activated to ensure a “soft touchdown” (v=0v=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/r2F/m=GM/r2 of Eq.(19), we get the (acceleration) differential equation
d2rdt2=T−GMr2,
d2rdt2=T−GMr2,
(20)
where M=7.35×1022M=7.35×1022 (kg) is the mass of the moon, which has a radius of R=1.74×106R=1.74×106 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=drdt,d2rdt2=dvdt=dvdr⋅drdt=vdvdr
v=drdt,d2rdt2=dvdt=dvdr⋅drdt=vdvdr
(as in Eq. (36) of Section1.6) and obtain the first-order equation
vdvdr=T−GMr2
vdvdr=T−GMr2
with the new independent variable r. Integration with respect to r now yields the equation
12v2=Tr+GMr+C
12v2=Tr+GMr+C
(21)
that we can apply both before ignition (T=0T=0) and after ignition (T=4T=4).
Before ignition: Substitution of T=0T=0 in (21) gives the equation
12v2=GMr+C1
12v2=GMr+C1
(21a)
where the constant is given by C1=v20/2−GM/r0C1=v20/2−GM/r0 with
v0=−1477kmh×1000mkm×1 h3600 s=−1477036ms
v0=−1477kmh×1000mkm×1 h3600 s=−1477036ms
and r0=(1.74×106)+53,000=1.793×106 mr0=(1.74×106)+53,000=1.793×106 m (from the initial velocity–position measurement).
After ignition: Substitution of T=4T=4 and v=0,r=Rv=0,r=R (at touchdown) into (21) gives
12v2=4r+GMr+C2
12v2=4r+GMr+C2
(21b)
where the constant C2=−4R−GM/RC2=−4R−GM/R is obtained by substituting the values v=0,r=Rv=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)=1.78187×106r=14(C1−C2)=1.78187×106 and finally h=r−R=41,870h=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=−450v=−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 v0v0 is necessary for the projectile to escape from the earth altogether. This will be so if its velocity v=dr/dtv=dr/dt 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
dvdt=d2rdt2=−GMr2,
dvdt=d2rdt2=−GMr2,
(22)
similar to Eq.(20), but with T=0T=0 (no thrust) and with M=5.975×1024M=5.975×1024 (kg) denoting the mass of the earth, which has an equatorial radius of R=6.378×106R=6.378×106 (m). Substitution of the chain rule expression dv/dt=v(dv/dr)dv/dt=v(dv/dr) as in Example4 gives
vdvdr=−GMr2.
vdvdr=−GMr2.
Then integration of both sides with respect to r yields
12v2=−GMr+C.
12v2=−GMr+C.
Now v=v0v=v0 and r=Rr=R when t=0,t=0, so C=12v20−GM/R,C=12v20−GM/R, and hence solution for v2v2 gives
v2=v20+2GM(1r−1R).
v2=v20+2GM(1r−1R).
(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,
v2>v20−2GMR,
v2>v20−2GMR,
so v will remain positive provided that v20≧2GM/R.v20≧2GM/R. Therefore, the escape velocity from the earth is given by
v0=√2GMR.
v0=2GMR−−−−−√.
(24)
In Problem27 we ask you to show that, if the projectile’s initial velocity exceeds √2GM/R2GM/R−−−−−−−√, then r(t)→∞r(t)→∞ as t→∞,t→∞, 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≈11,180v0≈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 Example4, we find that escape velocity from the lunar surface is v0≈2375m/s.v0≈2375m/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.dv/dt=−kv.(a) Show that its velocity and position at time t are given by
v(t)=v0e−kt
v(t)=v0e−kt
and
x(t)=x0+(v0k)(1−e−kt).
x(t)=x0+(v0k)(1−e−kt).
(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 Problem2, 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.dv/dt=−kv2. Show that
v(t)=v01+v0kt
v(t)=v01+v0kt
and that
x(t)=x0+1kln(1+v0kt).
x(t)=x0+1kln(1+v0kt).
Note that, in contrast with the result of Problem2, x(t)→+∞x(t)→+∞ as t→+∞.t→+∞. Which offers less resistance when the body is moving fairly slowly—the medium in this problem or the one in Problem2? Does your answer seem consistent with the observed behaviors of x(t) as t→∞t→∞?
Assuming resistance proportional to the square of the velocity (as in Problem4), how far does the motorboat of Problem3 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.dv/dt=−kv3/2. Show that
v(t)=4v0(kt√v0+2)2
v(t)=4v0(ktv0−−√+2)2
and that
x(t)=x0+2k√v0(1−2kt√v0+2).
x(t)=x0+2kv0−−√(1−2ktv0−−√+2).
Conclude that under a 3232-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 10ft/s2,10ft/s2, while air resistance provides 0.1ft/s20.1ft/s2 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 Problem7, with the sole difference that the deceleration due to air resistance now is (0.001)v2ft/s2(0.001)v2ft/s2 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
1000dvdt=5000−100v.
1000dvdt=5000−100v.
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 ρvft/s2,ρvft/s2, taking ρ=0.15ρ=0.15 without the parachute and ρ=1.5ρ=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 ρρ 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=640W=640 lb and volume 8 ft38 ft3—by dropping them into the ocean (v0=0v0=0). The force equation for a drum falling through water is
mdvdt=−W+B+FR,
mdvdt=−W+B+FR,
where the buoyant force B is equal to the weight (at 62.5 lb/ft362.5 lb/ft3) of the volume of water displaced by the drum (Archimedes’ principle) and FRFR 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=v√ρ/gu=vρ/g−−−√ to obtain the upward-motion velocity function given in Eq.(13) with initial condition v(0)=v0v(0)=v0.
Integrate the velocity function in Eq.(13) to obtain the upward-motion position function given in Eq.(14) with initial condition y(0)=y0y(0)=y0.
Separate variables in Eq.(15) and substitute u=v√ρ/gu=vρ/g−−−√ to obtain the downward-motion velocity function given in Eq.(16) with initial condition v(0)=v0v(0)=v0.
Integrate the velocity function in Eq.(16) to obtain the downward-motion position function given in Eq.(17) with initial condition y(0)=y0y(0)=y0.
Problems 17 and 18 apply Eqs.(12)–(17) to the motion of a crossbow bolt.
Consider the crossbow bolt of Example3, shot straight upward from the ground (y=0y=0) at time t=0t=0 with initial velocity v0=49m/s.v0=49m/s. Take g=9.8m/s2g=9.8m/s2 and ρ=0.0011ρ=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 Problem17, suppose that the bolt is now dropped (v0=0v0=0) from a height of y0=108.47y0=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=0v(0)=v0=0). Its motor provides a constant acceleration of 4ft/s2,4ft/s2, but water resistance causes a deceleration of v2/400ft/s2.v2/400ft/s2. Find v when t=10t=10 s, and also find the limiting velocity as t→+∞t→+∞ (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 v2/800v2/800 due to air resistance. How high in the air does it go?
If a ball is projected upward from the ground with initial velocity v0v0 and resistance proportional to v2,v2, deduce from Eq.(14) that the maximum height it attains is
ymax=12ρln(1+ρv20g).
ymax=12ρln(1+ρv20g).
Suppose that ρ=0.075ρ=0.075 (in fps units, with g=32ft/s2g=32ft/s2) 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 Problem22 falls freely for 30 s with ρ=0.00075ρ=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 329,320329,320 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=3×108m/sc=3×108m/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 v0v0 less than escape velocity √2GM/R,2GM/R−−−−−−−√, then the maximum distance from the center of the earth attained by the projectile is
rmax=2GMR2GM−Rv20,
rmax=2GMR2GM−Rv20,
where M and R are the mass and radius of the earth, respectively. (b) With what initial velocity v0v0 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?
Suppose a projectile is launched vertically from the surface r=Rr=R of the earth with initial velocity v0=√2GM/R,v0=2GM/R−−−−−−−√, so v20=k2/Rv20=k2/R where k2=2GM.k2=2GM. Solve the differential equation dr/dt=k/√rdr/dt=k/r√ (from Eq.(23) in this section) explicitly to deduce that r(t)→∞r(t)→∞ as t→∞t→∞.
If the projectile is launched vertically with initial velocity v0>√2GM/Rv0>2GM/R−−−−−−−√, deduce that
drdt=√k2r+α>k√r.
drdt=k2r+α−−−−−−√>kr√.
Why does it again follow that r(t)→∞r(t)→∞ as t→∞t→∞?
(a) Suppose that a body is dropped (v0=0v0=0) from a distance r0>Rr0>R from the earth’s center, so its acceleration is dv/dt=−GM/r2.dv/dt=−GM/r2. Ignoring air resistance, show that it reaches the height r<r0r<r0 at time
t=√r02GM(√rr0−r2+r0cos−1√rr0).
t=r02GM−−−−−√(rr0−r2−−−−−−−√+r0cos−1rr0−−−√).
(Suggestion: Substitute r=r0cos2θr=r0cos2θ to evaluate ∫√r/(r0−r)dr∫r/(r0−r)−−−−−−−−√dr.) (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.v0<2GM/R−−−−−−−√. Then its height y(t) above the surface satisfies the initial value problem
d2ydt2=−GM(y+R)2;y(0)=0,y′(0)=v0.
d2ydt2=−GM(y+R)2;y(0)=0,y'(0)=v0.
Substitute dv/dt=v(dv/dy)dv/dt=v(dv/dy) and then integrate to obtain
v2=v20−2GMyR(R+y)
v2=v20−2GMyR(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
d2rdt2=−GMer2+GMm(S−r)2;r(0)=R,r′(0)=v0
d2rdt2=−GMer2+GMm(S−r)2;r(0)=R,r'(0)=v0
where MeMe and MmMm denote the masses of the earth and the moon, respectively; R is the radius of the earth and S=384,400S=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 v0v0 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/dtv=dy/dt 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
dPdt=F,
dPdt=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)≡0m'(t)≡0—when its rockets are turned off or burned out, for instance—then Eq.(1) gives
F=d(mv)dt=mdvdt+dmdtv=mdvdt,
F=d(mv)dt=mdvdt+dmdtv=mdvdt,
which (with dv/dt=adv/dt=a) is the more familiar form F=maF=ma of Newton’s second law.
But here m is not constant. Suppose m changes to m+Δmm+Δm and v to v+Δvv+Δv during the short time interval from t to t+Δt.t+Δt. Then the change in the momentum of the rocket itself is
ΔP≈(m+Δm)(v+Δv)−mv=mΔv+vΔm+ΔmΔv.
ΔP≈(m+Δm)(v+Δv)−mv=mΔv+vΔm+ΔmΔv.
But the system also includes the exhaust gases expelled during this time interval, with mass −Δm−Δm and approximate velocity v−c.v−c. Hence the total change in momentum during the time interval ΔtΔt is
ΔP≈(mΔv+vΔm+ΔmΔv)+(−Δm)(v−c)=mΔv+cΔm+ΔmΔv.
ΔP≈=(mΔv+vΔm+ΔmΔv)+(−Δm)(v−c)mΔv+cΔm+ΔmΔv.
Now we divide by ΔtΔt and take the limit as Δt→0,Δt→0, so Δm→0,Δm→0, assuming continuity of m(t). The substitution of the resulting expression for dP/dt in (1) yields the rocket propulsion equation
mdvdt+cdmdt=F.
mdvdt+cdmdt=F.
(2)
If F=FG+FR,F=FG+FR, where FG=−mgFG=−mg is a constant force of gravity and FR=−kvFR=−kv is a force of air resistance proportional to velocity, then Eq.(2) finally gives
mdvdt+cdmdt=−mg−kv.
mdvdt+cdmdt=−mg−kv.
(3)
Constant Thrust
Now suppose that the rocket fuel is consumed at the constant “burn rate” ββ during the time interval [0,t1],[0,t1], during which time the mass of the rocket decreases from m0m0 to m1.m1. Thus
m(0)=m0,m(t1)=m1,m(t)=m0−βt,dmdt=−βfor t≤t1,
m(0)m(t)==m0,m0−βt,m(t1)dmdt==m1,−βfor t≤t1,
(4)
with burnout occurring at time t=t1t=t1.
Problem1 Substitute the expressions in (4) into Eq.(3) to obtain the differential equation
(m0−βt)dvdt+kv=βc−(m0−βt)g.
(m0−βt)dvdt+kv=βc−(m0−βt)g.
(5)
Solve this linear equation for
v(t)=v0Mk/β−gβtβ−k+(βck+gm0β−k)(1−Mk/β),
v(t)=v0Mk/β−gβtβ−k+(βck+gm0β−k)(1−Mk/β),
(6)
where v0=v(0)v0=v(0) and
M=m(t)m0=m0−βtm0
M=m(t)m0=m0−βtm0
denotes the rocket’s fractional mass at time t.
No Resistance
Problem2 For the case of no air resistance, set k=0k=0 in Eq.(5) and integrate to obtain
v(t)=v0−gt+clnm0m0−βt.
v(t)=v0−gt+clnm0m0−βt.
(7)
Because m0−βt1=m1,m0−βt1=m1, it follows that the velocity of the rocket at burnout (t=t1t=t1) is
It follows that the rocket’s altitude at burnout is
y1=y(t1)=(v0+c)t1−12gt21−cm1βlnm0m1.
y1=y(t1)=(v0+c)t1−12gt21−cm1βlnm0m1.
(10)
Problem4 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.
Problem5 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=0v=0 in (3), then the resulting initial acceleration
dvdt=cmdmdt−g
dvdt=cmdmdt−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,m0g/β, and that the time tBtB which then elapses before actual blastoff is given by
tB=m0g−βcβg.
tB=m0g−βcβg.
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=0g=0 in Eq.(8) we see that, as the mass of the rocket decreases from m0m0 to m1,m1, its increase in velocity is
Δv=v1−v0=clnm0m1.
Δv=v1−v0=clnm0m1.
(11)
Note that ΔvΔv depends only on the exhaust gas speed c and the initial-to-final mass ratio m0/m1,m0/m1, but does not depend on the burn rate β.β. For example, if the rocket blasts off from rest (v0=0v0=0) and c=5c=5 km/s and m0/m1=20,m0/m1=20, then its velocity at burnout is v1=5ln20≈15v1=5ln20≈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.