Solutions to all Odd Problems of

BASIC CALCULUS OF PLANETARY ORBITS AND INTERPLANETARY FLIGHT Chapter 1. Solutions of Odd Problems

Problem 1.1. In Figure 1.33, let V represent Mercury instead of Venus. Let α = ∠VES be the maximum angle that Copernicus observed after measuring it many times. As in the case of Venus, α is at its maximum when the line of sight EV is tangent to the orbit of Mercury (assumed to be a circle). This means that when α is at its maximum, the triangle EVS is a right triangle VS with right angle at V . Since Copernicus concluded from his observations that ES ≈ 0.38, we VS know that sin α = ES ≈ 0.38. Working backward (today we would push the inverse sine button 1 ◦ of a calculator) provides the conclusion that α ≈ 22 3 . So the maximum angle that Copernicus 1 ◦ observed was approximately 22 3 . From this he could conclude that EV ≈ 0.38. Problem 1.3. Since the lines of site BA and B0A of Figure 1.36a are parallel, Figure 1.36b tells us that (α + β) + (α0 + β0) = π as asserted. By a look at the triangle ∆BB0M, β + β0 + θ = π. So β + β0 = π − θ, and hence α0 + α + (π − θ) = π. So θ = α + α0.

Problem 1.5. Let R the radius of the circle of Figure 1.40b. Turn to Figure 1.40a, and notice that cos ϕ = R . The law of cosines applied to the triangle of Figure 1.40b tells us that (BB0)2 = rE R2 + R2 − 2R · R cos θ = 2R2(1 − cos θ). So BB0 = Rp2(1 − cos θ). Since the point B rotates on 1 the circle of Figure 1.40b at a constant rate, the ratio of 6 6 hours to 24 hours is the same as the 1 ◦ 6 6 ◦ ratio of θ expressed in degrees to 360 . So θ = 24 360 = 92.5 , and

0 p ◦ p ◦ BB = (rE cos ϕ) 2(1 − cos θ) = (6364.57 cos 52.92 ) 2(1 − cos 92.5 ) ≈ 5543.98 km.

a3 Problem 1.7. Newton’s computations of T 2 for Saturn’s five moons, were 1.953 2.53 3.53 83 243 45.312 ≈ 0.00361, 65.692 ≈ 0.00362, 108.422 ≈ 0.00365, 372.692 ≈ 0.00369, and 1903.82 ≈ 0.00381, where for each of the five moons, a is the radius of the circular orbit (and hence also the semimajor axis of the orbit) measured in outer ring radii, and T is the corresponding period of the orbit in hours. Given the difficulty of determining the radii and the periods of the orbits of these five moons with precision with the instruments that existed in Newton’s time, these results seem close enough to confirm Kepler’s third law for Saturn and these moons.

Figure Sol 1.1 shows a planet (comet or asteroid) at its aphelion position A and a short stretch of its orbit QA before its arrival at A. Its time of travel from Q to A is ∆t. It has already been

Q R

Δs

A S Sol 1.1 established that a planet (or comet or asteroid) attains its minimum speed vmin at aphelion.

1 Problem 1.9. i. Let arc QA be the length of the arc between Q and A. Since ∆t is the time of arc QA travel of the body, its average velocity vav during its trip from Q to A is equal to ∆t . For a very arc QA ∆s small ∆t, Q is close to A so that arc QA is close to ∆s, so that vav = ∆t ≈ ∆t . ii. For a small ∆s, the area ∆A of the elliptical sector SAQ is approximately equal to the area of 1 1 ∆A 2 (SA·∆s) the triangle ∆SAQ. Since the area of this triangle is 2 (SA · ∆s) it follows that κ = ∆t ≈ ∆t . 1 2 (a+c)·∆s abπ ∆s 2abπ From Figure 1.9, SA = a + c, so that κ ≈ ∆t . Since κ = T It follows that ∆t ≈ (a+c)T . ∆s iii. When ∆t is pushed to zero, Q gets pushed to A, so that the average speed vav ≈ ∆t , gets pushed to the speed vmin at aphelion. Simultaneously, the approximation of the area ∆A of the 1 1 elliptical sector SAQ by the area 2 (SA · ∆s) = 2 (a + c) · ∆s of ∆SAQ becomes tighter and tighter. 1 ∆A 2 (a+c)·∆s 1 abπ Therefore, the approximations κ = ∆t ≈ ∆t ≈ 2 (a + c)vmin snap to equalities. Since κ = T 2abπ this leads to the conclusion vmin = (a+c)T . √ √ 2 2 2 iv. Since c = εa and b = a − c , we get a + c = a(1 + ε) and b = a 1 − ε . So vmin = √ √ √ 2 2 2a( 1−ε)( 1+ε) q 2a 1−ε π √ π 2aπ 1−ε a(1+ε) T = ( 1+ε)2 T = T 1+ε .

Recall that 149,597,870.7 km 1 year 1 au/year = × ≈ 4.74 km/sec. 1 year 31,557,600 sec Problem 1.11. With Halley’s semimajor axis a = 17.83 au, eccentricity ε = 0.967, and period T = 75.32 years, we get

2aπ q 1+ε 2(17.83)π q 1+0.967 vmax = T 1−ε = 75.32 1−0.967 ≈ 11.483 au/year ≈ 54.43 km/sec and 2aπ q 1−ε 2(17.83)π q 1−0.967 vmin = T 1+ε = 75.32 1+0.967 ≈ 0.193 au/year ≈ 0.91 km/sec.

2 2 MME 4π M ME d 4π M ME d MME Problem 1.13. Using both F = G 2 and F = 2 , we get 2 = G 2 , hence d T ME +M T ME +M d 2 2 3 2 3 2 3 4π d G 4π d 4π d 4π d ME +M M 2 = 2 , and therefore 2 · = G. So · 2 = ME +M and · 2 = = 1+ . T ME +M d T ME +M G T GME T ME ME

Problem 1.15. i. Let F be the force of gravity of Earth on the Moon, ME the mass of Earth, and M that of the Moon. By Newton’s law of universal gravitation,

14 22 GME M (3.98600×10 )(7.3461×10 ) 20 F ≈ 2 = 8 2 = 1.98165 × 10 newtons. aM (3.844×10 )

ii. Let F be the force of gravity of the Sun on the Moon, MS the mass of the Sun, and M that of the Moon. By Newton’s law of universal gravitation,

20 22 GMS M (1.32712×10 )(7.3461×10 ) 20 F ≈ 2 = 11 2 = 4.35628 × 10 newtons. aE (1.49598×10 ) iii. So the gravitational force of the Sun on the Moon is more than twice than that of Earth on the Moon. So why is the Moon orbiting Earth? The fact is that both Earth and Moon are orbiting the Sun. As these orbiting motions occur, the Earth pulls the Moon into orbit around it.

−11 m3 Problem 1.17. With G = 6.67384 × 10 kg·sec2 in MKS, we get

−11 m3 −11 m3 (100 cm)3 1 kg 1 −8 cm3 G = 6.67384 × 10 kg·sec2 = 6.67384 × 10 kg·sec2 · 1 m3 · 1000 gr · sec = 6.67384 × 10 gr·sec2 .

2 Problem 1.19. We know from section 1G that 1 foot = 30.48 cm and 1 inch = 2.54 cm. With the 1 2.54 2.54 spheres of Figure 1.47 “distant but by 4 inch”, 2d = 4 so that d = 8 = 0.3175 cm. Since the 1 radius of the spheres is 2 foot, c = 15.24 + d = 15.24 + 0.3175 = 15.5575 cm. The volume of each 4 3 3 3 of the two spheres is 3 π(15.24) cm ≈ 14827 cm . Under the assumption that their densities are the same as that of the Earth, the mass of each sphere is (14827)(5.54) gr = 82,142 gr.

Problem 1.21. If can determine the time t it takes for the mutual force of attraction to move each of the spheres through a distance d, then the two spheres will touch after time t and the problem q 1 F 2 2 2mx(t) 2mx(t) is solved. The equality x(t) = 2 m t informs us that t = F and hence that t = F . With x(t) = d = 0.3175 cm and F equal to the magnitude of the maximal force of 0.485 dynes throughout the motion of the sphere, we get

q 2mx(t) q 2(82,142)(0.3175) t = F = 0.485 = 327.94 sec. q 2mx(t) Under the assumption that the minimal force acts throughout the motion, we get t = F = q 2(82,142)(0.3175) 0.465 = 334.92 sec and it takes a little longer for the two spheres to meet. Since the force of gravity between the two sphere varies from the smaller of these two values to the larger, 1 the time it takes for the two spheres to meet lies between the two calculated values, thus about 5 2 minutes. This conclusion is of course at odds with Newton’s assertion about the matter. Why? While Newton could compute GM with M the Earth’s mass (from his version of Kepler’s third law), he had no clue about G (as he had asserted himself).

Problem 1.23. i. Let a and ε be the semimajor axis and eccentricity of Sputnik 1’s elliptical orbit. The addition of the distance 942 km to Earth’s radius of 6371 km gives an aphelion distance of a + aε = 7313 km for the orbit, and adding 230 km to 6371 km gives a perihelion distance of a − aε = 6601 km. So 2a = (a + εa) + (a − εa) = 7313 + 6601 = 13,914 km and hence a = 6957 km. Since 2εa = (a + εa) − (a − εa) = 7313 − 6601 = 712 km, we get aε = 356 km, and hence 356 ε = 6957 ≈ 0.051. Sputnik 1’s period is T = 96.2 · 60 = 5772 in seconds. Using the speed formulas of Problems 1.8 and 1.9, we get

2aπ q 1+ε 2(6957)π q 1+0.051 vmax = T 1−ε = 5772 1−0.051 ≈ 7.97 km/sec and

2aπ q 1−ε 2(6957)π q 1−0.051 vmin = T 1+ε = 5772 1+0.051 ≈ 7.20 km/sec.

ii. Using Newton’s version of Kepler’s third law with M the Earth’s mass we get in MKS, GM = 4π2a3 4π2(6,957×103)3 14 3 2 T 2 = ((96.2)(60))2 ≈ 3.990 × 10 m /sec .

3 Chapter 2. Solutions of Odd Problems

Problem 2.1. With M the mass of the Moon, Newton’s version of Kepler’s third law tells us that 4π2a3 4π2(2413×103)3 12 m3 −11 m3 GM = T 2 = ((178.05)(60))2 ≈ 4.8601 × 10 sec2 . Taking G = 6.674 × 10 kg·sec2 from Chapter 1H, we find that M = 7.282 × 1022 kg.

Problem 2.3. Using the data from the initial orbit (with MKS), we get

4π2a3 4π2(13055×103)3 13 3 2 GM = T 2 = ((12.62)(3600))2 ≈ 4.2557 × 10 m /sec .

4π2a3 4π2(12631×103)3 13 3 2 For the post trim 1 orbit, GM = T 2 = ((11.97)(3600))2 ≈ 4.2843 × 10 m /sec , and for the post 4π2a3 4π2(12647×103)3 13 3 2 trim 2 orbit, GM = T 2 = ((11.99)(3600))2 ≈ 4.2863 × 10 m /sec . Problem 2.5 refers to videos about the Voyager missions.

Problem 2.7. Let M be the mass of Mars in kg. Using the orbital data of Table 2.2 for the moon 4π2a3 4π2(9378×103)3 13 m3 Phobos, we get GM = T 2 = ((0.31891)(24·60·60))2 ≈ 4.28871 × 10 sec2 . Using data for the orbit 4π2a3 4π2(23459×103)3 13 m3 of Deimos, GM = T 2 = ((1.26244)(24·60·60))2 ≈ 4.28390 × 10 sec2 . After dividing both results by −11 m3 23 23 G = 6.67384 × 10 kg·sec2 , we get M = 6.42615 × 10 kg and M = 6.41894 × 10 kg, respectively. (With regard to the “advertised” M = 6.42409 × 1023 kg and M = 6.39988 × 1023 kg, it is safe to say that their computation made use of a different value of G and different roundoff strategies.)

Problem 2.9 is another invitation to explore interesting videos related to the topics of this chapter.

a3 Problem 2.11. Kepler’s third law and Newton’s more explicit version of it tell us that T 2 is the same for any object in orbit around the Sun. Taking a = 1 au for Earth’s semimajor axis and T = 1 a3 year for its period, we get that in the units au and year, T 2 ≈ 1 for the Earth, and hence for any object in an elliptical orbit around the Sun. Therefore in these units T 2 ≈ a3. Recall the formula 2aπ q 1+ε vmax = T 1−ε for the maximum speed of any object in an elliptical orbit in terms of its orbital data. Turning to the units au and year and noting that ε < 1, we get √ 2aπ q 1+ε 2aπ q 1+ε 2π q 1+ε q 1+ε 2π· 2 vmax = ≈ 3 ≈ 1 ≈ 2π < √ , T 1−ε a 2 1−ε a 2 1−ε a(1−ε) a(1−ε) where vmax is expressed in au/year, and a and a(1 − ε) are the semimajor axis and perihelion distance in au, respectively.

Problem 2.13. Table 2.8 tells us that the orbit of Great Comet of 1680 had an eccentricity very close to 2 and a perihelion distance of 0.0062 au. It follows from Problem 2.11 that the maximum speed that the Great Comet of 1680 attained was close to √ √ q 1+ε 2π· 2 2π· 2 vmax ≈ 2π ≈ √ ≈ √ ≈ 112.85 au/year a(1−ε) a(1−ε) 0.0062 and hence (112.85)(4.74) ≈ 535 km/sec.

Problem 2.15. This computation is identical to that of the Great Comet of 1843, so that for Lovejoy’s comet C/2011 W3,

1 √ √ q 1+ε 2π· 2 2π· 2 vmax ≈ 2π ≈ √ ≈ √ ≈ 119.82 au/year a(1−ε) a(1−ε) 0.0055 and hence (119.82)(4.74) ≈ 568 km/sec. The brief video

https://phys.org/news/2012-03-comet-lovejoy-survive-sun.html that depicts its hurried scramble around the Sun is worth a look.

2 Chapter 3. Solutions of Odd Problems

Problem 3.1. Using the equalities x = r cos θ and y = r sin θ, we get the Cartesian coordinates,

i. Since r = 0, this is the origin (0, 0).

π π ii. x = 5 cos 6 ≈ 4.33 and y = 5 sin 6 = 2.5.

5π 5π iii. x = 7 cos 4 ≈ −4.95 and y = 7 sin 4 ≈ −4.95.

9π 9π iv. x = −6 cos(− 4 ) ≈ −4.24 and y = −6 sin(− 4 ) ≈ 4.24. v. x = 7 cos 10 ≈ −5.87 and y = 7 sin 10 ≈ −3.81.

vi. x = −3 cos(−20) ≈ −1.22 and y = −3 sin(−20) ≈ −2.74.

Problem 3.3. The polar equations are obtained by substituting the equalities x = r cos θ and y = r sin θ into the Cartesian equations.

4 i. Since 2r cos θ + 3r sin θ = 4, we get r(2 cos θ + 3 sin θ) = 4, and hence r = 2 cos θ+3 sin θ . ii. From (r cos θ)2 + (r sin θ)2 = 4r sin θ, we get r2 = 4r sin θ, or either r = 0 or r = 4 sin θ. So the Cartesian equation is satisfied provided either of these two equations is satisfied. Since the polar origin (0, 0) is on the graph of r = 4 sin θ, the equation r = 4 sin θ is the relevant answer.

iii. Since (r cos θ)2+(r sin θ)2 = r cos θ((r cos θ)2−3(r sin θ)2), we get r2 = r3 cos θ(cos2 θ−3 sin2 θ). 1 So either r = 0 or r = cos θ(cos2 θ−3 sin2 θ) . Note that the polar point (0, 0) is not on the graph of this function.

Problem 3.5. The graphs turn out to be simple. Descriptions suffice.

i. Since the graph of r = −6 consists of all polar points (−6, θ) for any θ, this is a circle of radius 6.

8π 8π ii. The graph of θ = − 6 is the set of all polar points (r, − 6 ) for any r. So the graph is the 4π entire line determined by the ray θ = − 3 . A moment’s thought about the xy-plane tells us π ◦ that this is the line through the origin at an angle of 3 = 60 with the positive x-axis. iii. Since the polar origin lies on the graph of r = 4 sin θ, the graph of this equation is the same as the graph of r2 = r sin θ. In Cartesian coordinates this translates to x2 + y2 = 4y and in turn to x2 + y2 − 4y + 4 = 4. Hence the graph is that of the Cartesian equation x2 + (y − 2)2 = 4. This is a circle with center the Cartesian point (0, 2) and radius 2.

iv. The Cartesian version of r(sin θ + cos θ) = 1 is y + x = 1 or y = −x + 1. This is the line with slope −1 through the Cartesian point (0, 1).

1 Problem 3.7. With x = r cos θ and y = r sin θ, the equation ax + by + c = 0 transforms to −c r(a cos θ + b sin θ) = −c. If c 6= 0, then a cos θ + b sin θ 6= 0, and r = f(θ) = a cos θ+b sin θ is a polar function that has the line as its graph. (An ambitious reader might wish to determine a domain ? < θ < ? for which the entire line is traced out.) If c = 0, the line is not the graph a of a polar function. We can assume that b 6= 0 and consider the line y = − b x. Let θ0 with π π y a − 2 < θ0 < 2 be the angle the line makes with the polar axis. Note that tan θ0 = x = − b is the slope of the line. For any point (r, θ) on the line (other than the origin), θ must be one of the angles

θ0, θ0 ± π, θ0 ± 2π, . . . . Assume, if possible, that the line is the graph of a polar function r = f(θ). Since θ = θ0, θ0 ± π, θ0 ± 2π, . . . , are the only angles that arise, the graph of r = f(θ) consists of the points, (f(θ0), θ0), (f(θ0 +π), θ0 +π), (f(θ0 −π), θ0 −π), (f(θ0 +2π), θ0 +2π), (f(θ0 −2π), θ0 −2π),.... But there are not enough of these to fill the whole line.

d Problem 3.9. The solutions will rely on the discussion of section 3D and the equation r = 1+ε cos θ . i. By Figure 3.16, the directrix of the parabola is the vertical line x = d and the focus is the 10 origin. So d = 10 and ε = 1. Therefore the equation is r = 1+cos θ . ii. Since the semimajor and semiminor axes of the ellipse are a = 8 and b = 5 respectively, we 2 2 2 d √ d d d 1−ε 25 know that 2 = 8 and 2 = 5. Since 2 = 25, it follows that d = 2 · = . Since 1−ε 1−ε 1−ε √ 1−ε d 8 d 2 d 25 2 39 39 1−ε2 = 8, we get 1 − ε = 8 = 64 . So ε = 64 and ε = 8 . Therefore the equation of the 25 25 ellipse is r = √ 8 = √ . 39 8+ 39 cos θ 1+ 8 cos θ iii. The fact that the semimajor and semiminor axes of the hyperbola are a = 8 and b = 5 2 2 2 d √ d d d ε −1 25 respectively, tells us that 2 = 8 and 2 = 5. Since 2 = 25, we get d = 2 · = . ε −1 ε −1 ε −1 √ ε −1 d 8 d 2 d 25 2 89 89 Since ε2−1 = 8, ε − 1 = 8 = 64 . Therefore ε = 64 and ε = 8 . So the equation of the 25 25 hyperbola is r = √ 8 = √ . 89 8+ 89 cos θ 1+ 8 cos θ

d Problem 3.11. The ellipse discussed in section 3D part (ii) has semimajor axis a = 1−ε2 and semiminor axis b = √ d . Assuming that a = 7 and b = 4, we see that d = 7 and √ d = 4. 1−ε2 1−ε2 1−ε2 d2 d2 1−ε2 16 d 2 d Therefore 1−ε2 = 16 and it follows that d = 1−ε2 · d = 7 . Since 1−ε2 = 7, we see that 1−ε = 7 = √ 16 16 2 33 33 √ 7 49 . So ε = 49 and ε = 7 . This means that the ellipse (∗) given by r = 33 has semimajor 1+ 7 cos θ axis a = 7 and semiminor axis b = 4. The ellipse C and the ellipse (∗) can each be moved in the x2 y2 plane to coincide with the ellipse that has equation 72 + 42 = 1. Therefore the ellipses C and (∗) have the same shape.

d Problem 3.13. The ellipse with equation r = 1+ε cos θ and ε < 1 discussed in section 3D part (ii) 2 d √ d 2 d has semimajor axis a = 2 and semiminor axis b = 2 . Since b = 2 we see that d = 1−ε 1−ε 1−ε √ d2 1−ε2 b2 1−ε2 1 2 d b2 2 b2 a2−b2 a2−b2 1−ε2 · d = a . Since d = a , we get 1 − ε = a = a2 . Hence ε = 1 − a2 = a2 and ε = a . b2 It follows that the graph of the equation r = √ a is an ellipse with semimajor axis a and a2−b2 1+ a cos θ semiminor axis b. So the ellipse C and the ellipse (∗) can each be moved in the plane to coincide

2 x2 y2 with the ellipse that has equation a2 + b2 = 1. Therefore the ellipses C and (∗) have the same shape.

d Problem 3.15. The equation r = 1+ε sin θ , where d > 0 and ε ≥ 0, can be written as r+εr sin θ = d. The corresponding Cartesian equation is ±px2 + y2 + εy = d. Let’s begin with the case ε = 1. Since px2 + y2 ≥ y and d > 0, it follows that the minus alternative does not occur and hence that px2 + y2 + y = d. This is an equation of the parabola with focal point the origin O and directrix the horizontal line y = d. This can be seen as follows. Let P = (x, y) be any point on the parabola with focal point the origin and directrix the horizontal line y = d. Its distance from the origin is px2 + y2. Since d > 0 the directrix lies above the focal point. It follows that y < d and that the distance from P to the directrix is d − y. So px2 + y2 = d − y and hence px2 + y2 + y = d. Since this is identical to the earlier equation, it follows that the graph d of r = 1+sin θ is a parabola with focal point the origin O and directrix the line y = d. We’ll now suppose that ε 6= 1. Squaring both sides of ±px2 + y2 = d − εy, we get in successive steps (one of them a completion of a square) that x2 + y2 = d2 − 2εdy + ε2y2 x2 + (1 − ε2)y2 + 2εdy = d2 x2 2 2εd d2 1−ε2 + y + 1−ε2 y = 1−ε2 x2 2 2εd ε2d2 d2 ε2d2 1−ε2 + y + 1−ε2 y + (1−ε2)2 = 1−ε2 + (1−ε2)2 x2 εd 2 (1−ε2)d2+εd2 d 2 1−ε2 + (y + 1−ε2 ) = (1−ε2)2 = ( 1−ε2 ) , and (y+ εd )2 x2 1−ε2 d2 + d 2 = 1. ( 2 ) 1−ε2 1−ε Now proceed as in cases (ii) and (iii) of section 3D to show that the graph of this equation is an √ ellipse if ε < 1 and a hyperbola if ε > 1. In the case of the ellipse ε < 1, so 1 − ε2 < 1, and hence √ 1 − ε2 > 1 − ε2. So √ d < d . As in (ii) of section 3D, we’ll let a = d and b = √ d . The 1−ε2 1−ε2 1−ε2 1−ε2 x2 (y+aε)2 above equation becomes b2 + a2 = 1. Since a > b, the semimajor axis of the ellipse is a and its semiminor axis is b. However now, its focal axis is the y-axis and (see the discussion that concludes x2 y2 section 3C) it is obtained by shifting the ellipse b2 + a2 = 1 downward so that its upper focal point is at the origin. The case of the hyperbola ε > 1 is similar. Unlike the situation of (iii) of section 3D where the branches of the hyperbola open to the left and right, the branches of the hyperbola open up and down.

Problem 3.17. The graph of the polar function r = f(θ) = cos θ is sketched in Sol 3.4. A look at

O 1 ( 2 , 0)

Sol 3.4

3 π π 3π the table of Problem 3.6 tells us that over the intervals 0 ≤ θ ≤ 2 , 2 ≤ θ ≤ π, π ≤ θ ≤ 2 , 3π and 2 ≤ θ ≤ 2π, the function r = f(θ) = cos θ respectively, decreases, decreases, increases, and increases. This parallels the fact—see Figure 3.4—that f 0(θ) = − sin θ is negative over 0 < θ < π and positive over π < θ < 2π.

1 Problem 3.19. The graph of the function r = f(θ) = cos θ is the circle with radius 2 and center the 1 point C = ( 2 , 0) (in both polar and Cartesian coordinates). Since the radius CP is perpendicular to π the tangent at P , the angles γ = γ(θ) and ∠OPC add to 2 . See Sol 3.5b. Since the triangle ∆OCP π π 0 π is isosceles, ∠OPC = θ. So γ(θ) + θ = 2 and γ(θ) − 2 = −θ. To verify f (θ) = f(θ) · tan(γ(θ) − 2 ), sin(−θ) sin θ it remains to check that − sin θ = (cos θ)(tan(−θ)). But this is so since tan(−θ) = cos(−θ) = − cos θ . Problem 3.21. We’ll quickly recall the basic situation. Let P = (f(θ), θ) be any point on the graph of a polar function f(θ) with f(θ) 6= 0 (so that P is not the origin O). The point (f(θ+∆θ), θ+∆θ) is obtained by adding a small positive angle ∆θ to θ. The diagram of Sol 3.6a shows the circular arc (in red) with center O and radius f(θ) between the segments determined by θ and θ + ∆θ. Sol 3.6a sketches a situation (of current interest) where the graph of the function lies below the red circular

f (θ ) − f (θ + Δ θ ) B P tangent to the Δs P graph at P A tangent to the circle at P r = f (θ ) Δ θ Δ θ r = f (θ )

θ θ O O (a) (b) Sol 3.6 arc. As in the situation of Figure 3.22a, the curving triangle of the diagram is the beak at P . The ∆s 1 1 radian measure of ∆θ is ∆θ = f(θ) , where ∆s is the length of the circular arc. So ∆θ = ∆s · f(θ). After a substitution, f(θ + ∆θ) − f(θ) f(θ + ∆θ) − f(θ) = · f(θ). ∆θ ∆s Put in the tangent line to the graph of r = f(θ) at P , and let A be the point of intersection of the tangent with the ray determined by θ + ∆θ. Also put in the tangent line to the circle at P , and let B be the point of intersection of this tangent and the same ray. The two tangent lines and the ray form the triangle ∆AP B that we’ve called the triangle at P . In Sol 3.6b, the beak at P is drawn in red and the triangle at P in green. f(θ + ∆θ) − f(θ) We’ll now push ∆θ to 0 and investigate lim . As ∆θ is pushed to 0, the ∆θ→0 ∆s segment OAB rotates toward the segment OP . Both the beak at P and the triangle at P shrink

4 tangent to the B circle at P tangent to the A P graph at P

r = f (θ ) Δs

to the origin O

Sol 3.7 in the direction of their tips at P . The shrinking triangle approximates the shrinking beak better and better as the gap between OBA and OP closes. See the diagram of Sol 3.7. In the process, ∆s gets closer to BP and f(θ) − f(θ + ∆θ) = −(f(θ + ∆θ) − f(θ)) to AB. Therefore, as ∆θ is pushed to 0, −(f(θ + ∆θ) − f(θ)) AB closes in on the ratio . ∆s BP Because the tangent line to a circle at a point is perpendicular to its radius to the point, we know π that the angle at P between PO and PB is 2 . So as ∆θ shrinks to 0, the angle ∠PBA approaches π 2 , and the triangle ∆AP B approaches a right triangle with right angle at B. It follows that the AB π ratio BP closes in on the tangent of the angle ∠AP B. Because ∠AP O = γ and ∠BPO = 2 , the π angle ∠AP B = 2 − γ. By putting it all together, we have demonstrated that as ∆θ shrinks to 0 −(f(θ + ∆θ) − f(θ)) AB closes in on and this in turn on tan( π − γ). ∆s BP 2 π π Since tan( 2 − γ) = − tan(γ − 2 ) we have verified that f(θ + ∆θ) − f(θ) lim = tan(γ − π ). ∆θ→0 ∆s 2 0 π We have now arrived at the conclusion f (θ) = f(θ) · tan(γ(θ) − 2 ) in the case of Sol 3.6a.

(tan(γ− π ))θ Problem 3.23. Differentiating f(θ) = f(0)e 2 with the chain rule gives us

π d π π π π 0 tan(γ− 2 )θ tan(γ− 2 )θ f (θ) = f(0)e · dθ (tan(γ − 2 )θ) = f(0)e · tan(γ − 2 ) = f(θ) · tan(γ − 2 ).

Problem 3.25. We see from the graph of r = f(θ) = sin θ of Figure 3.14 and the area formula of Z π 1 2 1 section 3G that the integral 2 sin θ dθ is equal to the area of a circle of radius 2 . So its value is 0 1 2 π 2 1 π( 2 ) = 4 . Using the equality sin θ = 2 (1 − cos 2θ), we get Z π Z π 1 2 1 1 1 π π 2 sin θ dθ = 4 (1 − cos 2θ) dθ = 4 (θ − 2 sin 2θ) 0 = 4 . 0 0

5 √ Problem 3.27. The circle (x − 1)2 + (y − 1)2 = 2 has center (1, 1) and radius 2. Its graph is sketched in Sol 3.10. The Cartesian points (2, 0) and (0, 2) are on the circle and the segment joining them is on the line y = −x + 2. It follows that (1, 1) is on this line as well so that the segment is a

2

(1, 1)

0 2

Sol 3.10

diameter of the circle. To find the equivalent polar equation of the circle, we’ll let x = r cos θ and y = r sin θ to get

(x − 1)2 + (y − 1)2 = (r cos θ − 1)2 + (r sin θ − 1)2 = (r2 cos2 θ − 2r cos θ + 1) + (r2 sin2 θ − 2r sin θ + 1) = r2 − 2r(cos θ + sin θ) + 2. 2 So the polar equation of the circle is r − 2√r(cos θ√ + sin θ) = 0. Assuming that r 6= 0, we get 3π 2 2 r = 2(cos θ + sin θ). For θ = 4 , r = 2(− 2 + 2 ) = 0 so that the origin is on the graph of r = f(θ) = 2(cos θ + sin θ). Hence the graph of r = f(θ) = 2(cos θ + sin θ) is the entire circle. π Z 2 1 2 A look at Sol 3.10 tells us that 2 f(θ) dθ is the area consisting of half the circle plus the 0 area of them right triangle with base and height both equal to 2. So π Z 2 √ 1 2 1 2 1 2 f(θ) dθ = 2 π( 2) + 2 (2 · 2) = π + 2. 0 π Z 2 It also follows from the figure that pf(θ)2 + f 0(θ)2 dθ is equal one-half of the circumference of 0 √ √ 1 the circle. So the value of this integral is 2 (2π 2) = 2π. The two integrals can be solved directly. Since f(θ) = 2(cos θ + sin θ), f(θ)2 = 4(cos2 θ + 2 cos θ sin θ + sin2 θ) = 4(1 + 2 sin θ cos θ). π π Z 2 Z 2 π 1 2 2 2 Therefore 2 f(θ) dθ = 2(1 + 2 cos θ sin θ) dθ = (2θ + 2 sin θ) 0 = π + 2, as before. Since 0 0 f 0(θ) = 2(− sin θ + cos θ), we get f 0(θ)2 = 4(sin2 θ − 2 sin θ cos θ + cos2 θ) = 4(1 − 2 sin θ cos θ). π Z 2 √ π √ 2 0 2 p 2 0 2 2 So f(θ) + f (θ) = 8 and f(θ) + f (θ) dθ = 2 2θ 0 = 2π. 0

6 3 Problem 3.29. After writing r = f(θ) = sin θ+2 cos θ as r sin θ + 2r cos θ = 3, we see that the graph of this polar function is the line y = −2x + 3 with slope −2 and y-intercept 3 sketched in Sol 3.9c. π Z 2 1 2 The integral 2 f(θ) dθ is the area of the right triangle bounded by the graph of y = −2x + 3 0 and the x- and y-axes. It follows that π Z 2 1 2 1 3 9 2 f(θ) dθ = 2 ( 2 · 3) = 4 . 0 π Z 2 The integral pf(θ)2 + f 0(θ)2 dθ is the length of the hypotenuse of this triangle. So by the 0 Pythagorean theorem, π Z 2 q q √ p 2 0 2 3 2 2 45 3 f(θ) + f (θ) dθ = ( 2 ) + 3 = 4 = 2 5. 0

4 Problem 3.31. Section 3D tells us that the graph of r = f(θ) = 1 is an ellipse with 1+ 3 cos θ eccentricity ε = 1 and d = 4. By part (ii) of this section the semimajor and semiminor axes of this 3 q q √ ellipse are a = d = 4 = 4 · 9 = 9 and b = √ d = √ 4 = 4 · 9 = 4 · 9·2 = 3 2. 1−ε2 1−( 1 )2 8 2 1−ε2 1 2 8 8·2 3 1−( 3 ) Z π The area of an ellipse with semimajor and semiminor axes a and b is abπ. Since 1 4
2 dθ 2 1+ 1 cos θ 0 3 is one-half the area of the ellipse being considered, it follows that Z π Z π √ √ 8 dθ = 1 4
2 dθ = 1 ( 9 )(3 2)π = 27 2π. (1+ 1 cos θ)2 2 1+ 1 cos θ 2 2 4 0 3 0 3

θ θ 1 √ 1 √ 1 Problem 3.33. For the equiangular spiral f(θ) = e 3 , f 0(θ) = e 3 · √ , so that by the discussion 7 7 3 f 0(θ of equiangular spirals in section 3H, tan(γ(θ) − π ) = = √1 . Since tan π = √1 , it follows that 2 f(θ) 3 6 3 π π 2π γ(θ) − 2 = 6 and hence that γ(θ) = 3 . The graph of the spiral inSol 3.11 below was sketched by the polar graphing calculator

https://www.desmos.com/calculator/ms3eghkkgz

θ 1 √ Since f 0(θ) = √ e 3 , we see that the length of the spiral from θ = 0 to θ = 2π is given by 7 3 2π 2π 2π Z Z q 2 2 Z q 2 √ θ √ θ √ θ p 2 0 2 1 3 1 3 1 4 3 f(θ) + f (θ) dθ = 72 e + 3·72 e dθ = 7 3 e dθ 0 0 0 2π Z 1 √ 1 2π 2 √ θ 2 √ θ 2π
2 √
= √ e 3 dθ = √ 3e 3 = (e 3 − 1 ≈ 10.46. 7 3 7 3 0 7 0 The area that the spiral (along with the polar axis) encloses is

Z 2π Z 2π Z 2π √ √ √2 θ √2 θ √2 θ 2π √4π 1 2 1 1 3
1 3 1 3 3
3 3 2 f(θ) dθ = 2 72 e dθ = 2·72 e dθ = 2·72 2 e 0 = 22·72 (e − 1) ≈ 12.50. 0 0 0 The fact that the two shorter sides plus the bottom side of the green rectangle of the figure de- termined by the intervals [−1.05, 5.4] on the x-axis and [0, −2.55] on the y-axis (see the figure)

7 √θ 1 3 Sol 3.11. The graph of r = 7 e add to 2.55 + 2.55 + 6.45 = 11.55 and that its area is 2.55 · 6.45 ≈ 16.45 confirms the reasonableness of the two answers.

Problem 3.35. What about the argument that led to the conclusion that the length L of the polar graph of r = f(θ) between θ = a and θ = b is equal to Z b L = |f(θ)| dθ ? a This formula agrees with the correct version of section 3F in general only if f 0(θ) = 0, so only if f(θ) = c is a constant, and hence the graph of r = f(θ) is a circle with center the origin O. 1 π 1 π 1 tan(γ− 2 )θ (tan 4 )θ θ Let’s consider the equiangular spiral r = f(θ) = 4 e = 4 e = 4 e . See Sol 3.12a. 0 1 θ Because f (θ) = 4 e , Z 2π Z 2π √ Z 2π pf(θ)2 + f 0(θ)2 dθ = p2f(θ)2 dθ = 2 |f(θ)| dθ. 0 0 0 √ So the correct value differs from the incorrect value by a factor of 2. With regard to Figure 3.25 the simple fact is that for some graphs (or parts of graphs) the

length f(θi) dθ of the circular arc does not approximate the length of the graph between the rays determined by θi and θi+1 with sufficient accuracy. To give a quantitative sense of this, we’ll 1 θ experiment with f(θ) = sin θ rather than f(θ) = 4 e . The graph of f(θ) = sin θ is the circle sketched in Figure 3.14. Notice that as θ moves from 0 through small positive angles, the point (r, θ) on the graph recedes quickly from the origin. This part of the graph illustrates the problem π π π of “sufficient accuracy.” Consider the rays θ = 12 and θ = 9 . The ray θ = 12 cuts the graph at the π π point (sin 12 , 12 ) ≈ (0.26, 0.26). By the length formula for a circular arc, the circular arc centered π at O from this point to the ray θ = 9 has length

8 π π π π π π π π f( 12 )( 9 − 12 ) = (sin 12 )( 9 − 12 ) = (sin 12 )( 36 ) ≈ 0.0226. Let’s compare this against the length of the graph of the function between these two rays. This length is

π π Z 9 Z 9 √ π p 2 0 2 2 2 9 π π π f(θ) + f (θ) dθ = sin θ + cos θ dθ = θ π = ( 9 − 12 ) = 36 ≈ 0.0873, π π 12 12 12 or about 4 times the length of the circular arc. The diagram in Sol 3.13 illustrates the difference

π θ = 4

π θ = 9 π θ = 12

O θ = 0

Sol 3.13 between the lengths of this part of the graph of f(θ) = sin θ and that of the corresponding circu- lar arc.

9 Chapter 4. Solutions of Odd Problems

Problem 4.1. The figure below is derived from Figure 4.23. The angle at B is 180◦ − 25◦ − 30◦ = ◦ 125 . With F1 and F2 the magnitudes of the two forces and F = 115 pounds the magnitude of their C

30 o F F1

25 o

A F2 B Sol 4.1 resultant, we get by applying the law of sines to the triangle ABC that sin 25◦ sin 30◦ sin 125◦ = = . F1 F2 115 ◦ 115 0.42262 ◦ 115 0.5 So F1 = sin 25 sin 125◦ ≈ ( 0.81915 )115 ≈ 59.33 pounds and F2 = sin 30 sin 125◦ ≈ ( 0.81915 )115 ≈ 70.19 pounds.

Problem 4.3. Focus on Figure 4.4a. Given that the string has a length of PH = 1.6 meters and ◦ PS ◦ that θ = 60 , the radius PS of the circular path of the object P satisfies PH = cos 60 = 0.5, so that PS = 0.8. With the mass of P equal to 2 kg, its weight is W = 2g, where g is the 2 gravitational constant in MKS. Taking g = 9.8 m/sec , we get W = 2 · 9.8 = 19.6 newtons. This√ ◦ ◦ 3 is the magnitude of the vertical component of FH , so that FH sin 60 = 19.6. Since sin 60 = 2 , we get F = 19.6 = 19.6 √2 ≈ 22.62 newtons. The magnitude of the centripetal force on P is H sin 60◦ 3 ◦ F = FH cos 60 ≈ 11.36 newtons.

Problem 4.5. If the angle θ(t) of Figure 4.13 increases at a constant rate, then by the formula r(t)2θ0(t) = 2κ of section 4D, r(t)2 is constant and hence r(t) is constant. This means that the orbit  4κ2 d2r  4mκ2 is a circle. A look at the formula F (t) = m r(t)3 − dt2 = r(t)3 of section 4D informs us that F (t) is constant as well.

d Problem 4.7. The equation that defines the polar function r = f(θ) = a sin θ+b cos θ can be written as a(r sin θ) + b(r cos θ) = d, so that its Cartesian version is the equation of the line ay + bx = d. 1 1 Since d 6= 0, the origin is not on the line. Note that g(θ) = f((θ) = d (a sin θ + b cos θ). So 0 1 00 1 00 g (θ) = d (a cos θ − b sin θ) and g (θ) = d (−a sin θ − b cos θ). It follows that g (θ) = −g(θ). An application of the second form of the centripetal force equation of section 4D tells us that F (t) = 0.

Problem 4.9. The position of the point-mass P of mass m moving in an xy-plane is given by the equations x(t) = a cos ωt and y(t) = b sin ωt, where t ≥ 0 is time, and a, b, and ω are constants

1 y y

F (t) P y(t) P x ( ) Fy t F(t) x(t) O x O x

(a) (b)

Sol 4.6

√with a ≥ b > 0, and ω > 0. That the distance between P and O at any time t is r(t) = a2 cos2 ωt + b2 sin2 ωt follows directly from the Pythagorean theorem. A substitution shows x2 y2 quickly that the x- and y-coordinates of P satisfy the equation a2 + b2 = 1, so that P moves on this ellipse with semimajor axis a and semiminor axis b. Studying the x-and y-coordinates of P for increasing t ≥ 0 in combination with Figures 3.4 and 3.5, tells us that the point starts at (a, 0) and that it moves counterclockwise around the ellipse. By referring to Sol 4.6 and following what was done in Example 4.7, we can conclude that at any point (x(t), y(t)) the slopes of the segment PO and the vector determined by the resultant of 00 00 the forces Fx(t) = mx (t) and Fy(t) = my (t) are the same. It follows that this resultant force on P is a centripetal force√ in the direction of the origin O and that it has magnitude F (t) = p 2 2 2 2 2 2 2 2 Fx(t) + Fy(t) = mω a cos ωt + b sin ωt = mω r(t). The fact that P starts at time t = 0 and completes its first revolution when t satisfies ωt = 2π, 2π tells us that the period of the orbit is T = ω . Since the area of the ellipse is A = abπ, it follows that A ω abω Kepler constant of the orbit is κ = T = (abπ)( 2π ) = 2 . Since the force on P is centripetal, F (t)  4κ2 d2r  and r(t) satisfy the force equation F (t) = m r(t)3 − dt2 of section 4D. (This can also be verified directly for the given r(t) with a labor-intensive calculus exercise.) The case of a circular orbit was dealt with in Example 4.7, so we’ll now suppose that a > b. Assume, if possible, that F (t) satisfies m m 2 3 C an inverse square law F (t) = C r(t)2 . Then C r(t)2 = mω r(t) and hence r(t) = ω2 . The consequence that r(t) is constant contradicts the fact that the orbit is not a circle. So F (t) does not satisfy an inverse square law. Conclusion B of section 4G leaves us with the conclusion that the center of the centripetal force cannot be a focal point of the ellipse. But this is consistent with what we already know, namely that it is the center of the ellipse that is the center of the centripetal force of this example. In the circular case, the center of the ellipse is its one and only focal point and as we saw in Example 4.10, F (t) does satisfy an inverse square law.

2 The next few problems turn to the motion of a thrown object near Earth’s surface. The solutions of the problems require formulas developed in the paragraph Projectile Motion on Earth.

Problem 4.11. We’re assuming that the terrain is flat and horizontal and the x-axis lies along the g 2 ground, so that y(t) = 0 at the time t of impact. The expression y(t) = − 2 t + (v0 sin ϕ)t + y0 and the quadratic formula tell us that the time t at which impact occurs is √ √ −(v sin ϕ)± v2 sin2ϕ−4(− g )y (v sin ϕ)± v2 sin2ϕ+2gy t = 0 0 2 0 = 0 0 0 . −g g √ 2 2 (v0 sin ϕ)+ v0 sin ϕ+2gy0 Since t ≥ 0, the + option applies and the time of impact is timp = g . Since x(t) = (v0 cos ϕ)t, the projectile impacts

v0  p 2 2  x(timp) = g cos ϕ v0 sin ϕ + v0 sin ϕ + 2gy0 .

v0  p 2 2  downrange. The distance R = g cos ϕ v0 sin ϕ + v0 sin ϕ + 2gy0 is the range of the projectile.

Assume that y0 = 0. The trig formula sin 2ϕ = 2 sin ϕ cos ϕ tells us that

2 v0  p 2 2  v0 v0 R = g cos ϕ v0 sin ϕ + v0 sin ϕ = g cos ϕ(2v0 sin ϕ) = g sin 2ϕ.

2 π v0 So when y0 = 0, the maximal range is achieved for ϕ = 4 and is equal to Rmax = g . Problem 4.13. The speed of the projectile at any time t is

p 0 2 0 2 p 2 2 p 2 2 2 v(t) = x (t) + y (t) = (v0 cos ϕ) + (−gt + v0 sin ϕ) = v0 + g t − 2g(v0 sin ϕ)t.

0 y (t) −gt+v0 sin ϕ The slope of the trajectory at any time t is 0 = . If the terrain is flat and horizontal, x (t) v0 cos√ϕ 2 2 (v0 sin ϕ)+ v0 sin ϕ+2gy0 then by Problem 4.11, the time of impact is t = g . With t the time of impact, q 2 p 2 2 2 p 2 2 v(t) = v0 + [(v0 sin ϕ) + v0 sin ϕ + 2gy0 ] − 2(v0 sin ϕ)[(v0 sin ϕ) + v0 sin ϕ + 2gy0] q 2 p 2 2 2 2 2 2 p 2 2 = v0 + 2(v0 sin ϕ) v0 sin ϕ + 2gy0 + (v0 sin ϕ + 2gy0) − v0 sin ϕ − 2(v0 sin ϕ) v0 sin ϕ + 2gy0

p 2 = v0 + 2gy0, and √ √ 0 −(v sin ϕ)− v2 sin2ϕ+2gy +v sin ϕ − v2 sin2ϕ+2gy q 2 2 y (t) 0 0 0 0 0 0 v0 sin ϕ 2gy0 q 2 2gy0 0 = = = − 2 2 + 2 2 = − tan ϕ + 2 2 . x (t) v0 cos ϕ v0 cos ϕ v0 cos ϕ v0 cos ϕ v0 cos ϕ 0 y (t) q 2 2gy0 The formula 0 = − tan ϕ + 2 2 confirms what Figure 4.27 illustrates and this is that the x (t) v0 cos ϕ slope of the trajectory at the point of impact is negative. Refer to the solution of Problem 4.10 and v0 the fact that the projectile reaches its maximum height when t = g sin ϕ. The x-coordinate of the 2 2 v0 v0 v0 position of the projectile at this time is (v0 cos ϕ)( g sin ϕ) = g sin ϕ cos ϕ = 2g sin 2ϕ. A generalk property of the parabola tells us and Figure 4.27 illustrates that the trajectory of the projectile 2 v0 is symmetric about the vertical line x = 2g sin 2ϕ. It follows that the angle of the tangent to the path of the projectile at the elevation of y0 of its descent is −ϕ. Let φ be the angle of the path of q 2 2gy0 p 2 the projectile at impact. The fact that − tan ϕ + 2 2 ≤ − tan ϕ = − tan ϕ means that the v0 cos ϕ angle at impact is greater (in the sense of absolute value) than ϕ. Figure Sol 4.7 illustrates what

3 tangent at y elevation 0 tangent at ϕ impact y0

φ

R x Sol 4.7 has been observed. It is derived from Figure 4.27.

We take up issues that remain from the discussion in the paragraph Tossing the Hammer. v0  p 2 2  Putting the data provided in the paragraph into the formula R = g cos ϕ v0 sin ϕ+ v0 sin ϕ + 2gy0 for the range of a projectile, we get

30.7 ◦  ◦ p 2 2 ◦  R = 9.81 (cos 39.9 ) 30.7 sin 39.9 + (30.7 )(sin 39.9 ) + 2(9.81)(1.66) ≈ 96.50 meters for the distance that Yuriy Sedykh’s record setting throw would have attained in an air resistance free environment. (The discrepancy between the 96.50 meters and the 96.56 meters of the text is due to the differing round-off strategies that were used.) This tells us that air resistance and the retarding effect of the wire and the handle reduced this theoretical value of the record throw by 96.50 − 86.74 = 9.76 meters. The force with which the hammer pulled on Yuriy just before the moment of the hammer’s release was computed in the text to have been 3767 newtons. As was observed, this force was much greater than Yuriy’s weight of about 1079 newtons. So why didn’t Yuriy fly off with the ball just before he released it? This question has the same answer as the question as to why the Earth and Moon are not pushed to a collision by the gravitational force that attracts them toward each other. The answer is that the force of attraction between them is counterbalanced by the rapid motion of both around their common barycenter.

We turn next to the paragraph Supersized Reflecting Telescopes and the solution of Problem 4.15. Recall the fact that the parabolic boundary of the mirror of the GMT is given by the function 2π2τ 2 2 y = f(x) = g x . A look at the study of the parabola in Chapter 1C tells us that the focal point 2 1 2 of the parabola with equation is x = 4cy, or equivalently y = 4c x , where c > 0 is a constant, is 1 2π2τ 2 g the point (0, c). Solving 4c = g for c gives c = 8π2τ 2 , and tells us that the focal point of the g parabola of the mirror is a distance 8π2τ 2 meters above its lowest point.

2 1 Problem 4.15. The data g = 9.80 m/sec and τ = 12 revolutions per second, tells us that 2π2τ 2 2 g ≈ 0.01399 ≈ 0.014, and provides the approximate formula f(x) = 0.014x for the mirror’s parabola. That the vertical distance between the horizontal plane at the mirror’s rim and its central hole is f(4.2) − f(1.15) ≈ 0.014[(4.2)2 − (1.15)2] ≈ 0.23 m follows directly from Figure 4.32. The

4 g (9.80)(122)
focal point of the central mirror is the point (0, 8π2τ 2 ) ≈ 0, 8(9.87) ≈ (0, 17.87). So the focal point of the mirror lies about 18 meters above its central hole.

The next problem gives an example of a mass of uniform density that attracts a point-mass with a gravitational force of a magnitude that is distinctly different from the value provided by Newton’s law of universal gravitation.

M Problem 4.17. The uniform density of the cylinder—mass over volume—is ρ = πR2h . Let’s focus on one of the many typical thin discs of thickness dx that the cylinder has been sliced into. See Figure 4.34. The volume of the disc is πR2dx, so that its mass is πR2dx · ρ. The point-mass of mass m lies on the central axis of the cylinder at a distance c − x from the disc’s center. By the conclusion of Problem 4.16, the gravitational force with which the disc attracts the point-mass acts in the direction of the center of the disc with a magnitude of

2m(πR2dx·ρ) c−x
M c−x
2mM c−x
G 2 1 − √ = G · 2mπ( 2 )dx 1 − 1 = G 2 1 − 1 dx. R R2+(c−x)2 πR h (R2+(c−x)2) 2 R h (R2+(c−x)2) 2 Summing things up over all the discs from x = 0 to x = h, we find that the force with which the cylinder attracts the point mass is directed along the central axis of the cylinder with a magnitude of Z h Z h 2mM c−x
2mM x−c
F = G 2 1 − 1 dx = G 2 1 + 1 dx. R h 2 2 R h 2 2 0 (R +(c−x) ) 2 0 (R +(c−x) ) 2 Z h Z h Z h x−c
x−c x−c Since 1 + 1 dx = h + 1 dx, it remains to compute 1 dx. 2 2 2 2 2 2 0 (R +(c−x) ) 2 0 (R +(c−x) ) 2 0 (R +(c−x) ) 2 Z h Z h−c x−c u Start with the substitution u = x − c and du = dx, to get 1 dx = 1 du. 2 2 2 2 0 (R +(c−x) ) 2 −c (R +u ) 2 Then let v = R2 + u2 and dv = 2u du so that 2 2 Z h−c Z R +(h−c) R2+(h−c)2 u 1 − 1 1 2 2 1 2 2 1 1 du = v 2 dv = v 2 = (R + (h − c) ) 2 − (R + c ) 2 . 2 2 2 2 2 −c (R +u ) 2 R2+c2 R +c After putting things together, we have determined that the magnitude of the force of attraction of the cylinder of Figure 4.34 on the point-mass is

2mM 1 1  2 2 2 2 2 2  F = G R2h h − (R + c ) + (R + (c − h) ) . 1 The center of mass of the cylinder is the point 2 h on the horizontal axis, so that the value mM that Newton’s formula provides for the magnitude of this force is F = G 1 2 . When R and h (c− 2 h) are both very small, then the cylinder can be regarded as a point-mass and the two values are in 1 1 2 2 2 2 close agreement. To verify this, use L’Hospital’s rule to show that lim h−(R +c ) 2 +(R +(c−h) ) 2 = h→0 h c−h 1 − 1 and combine this result with the conclusion of Problem 4.16. (R+(c−h)2) 2

Problem 4.19. Using the estimate 7.35 × 1022 kg for the Moon’s mass and a conclusion of Prob- lem 4.18, we get 7.35 × 1022 ≈ 3.35 × 103 kg/m3 2.1968 × 1019 for the average density of the Moon. It was established in section 4H that Earth’s average density is 5.51 × 103 kg/m3.

5 Chapter 5. Solutions of Odd Problems

π Problem 5.1. Place Figure 5.16 on an xy-coordinate system as shown in Sol 5.1. The angle 3 1 π 2 25π determines the circular sector OAC of radius 5. The area of this circular sector is 2 ( 3 · 5 ) = 6 . The x-coordinate of the point B satisfies cos π = x , so that x = 5 cos π = 5 . Because x2 + y2 = 52 √ 3 5 3 2 is an equation of the circle and f(x) = 52 − x2 is a function that has the upper part of the circle y

A

3 π 3 5 O BC x Sol 5.1

q q √ 5 2 5 2 100−25 5 3 as its graph, AB = f( 2 ) = 5 − ( 2 ) = 4 = 2 . Therefore the area under the circle and √ √ 25π 1 5 5 3 π 3 over the segment BC is equal to 6 − 2 ( 2 2 ) = 25( 6 − 8 ). It follows that Z 5√ √ 2 2 π 3 5 − x dx = 25( 6 − 8 ). 5 2

2 √ x2 y 3 2 2 Since 52 + 32 = 1 is an equation of the ellipse, the function g(x) = 5 5 − x has the upper part of the ellipse as its graph. It follows that the shaded area of the figure is equal to Z 5 √ Z 5√ √ √ 3 2 2 3 2 2 3 π 3 π 3 5 5 − x dx = 5 5 − x dx = 5 (25( 6 − 8 )) = 15( 6 − 8 ) ≈ 4.61. 5 5 2 2

Problem 5.3. Sol 5.2 is a version of Figure 5.3 with P in the first or fourth quadrants, respectively.

It includes the surrounding circle with the angle β(t) and the point P0. The coordinate x = x(t) is now positive. As in the cases dealt with in section 5B, the distance OS between the center and the focus of the ellipse is equal to c = aε and a2 = b2 + c2. Consider the case where P is in the first quadrant. The Pythagorean theorem tells us that

2 2 2 2 2 2 2 2 b2 2 r(t) = (aε − x(t)) + y(t) = (aε) − 2aεx(t) + x(t) + y(t) = (aε) − 2aεx(t) + x(t) + a2 y0(t) . 2 2 2 Since the point P0 is on the circle of radius a, x(t) + y0(t) = a , so that

2 2 2 b2 2 2 2 2 2 b2 2 r(t) = (aε) − 2aεx(t) + x(t) + a2 (a − x(t) ) = (aε) + b − 2aεx(t) + x(t) − a2 x(t) 2 2 a2−(aε)2 2 2 2 2 (aε)2 2 = a − 2aεx(t) + x(t) − a2 x(t) = a − 2aεx(t) + x(t) − x(t) + a2 x(t) = a2 − 2aεx(t) + ε2x(t)2 = (a − εx(t))2.

1 Because a ≥ x(t) ≥ εx(t), a − εx(t) ≥ 0. Since r(t) ≥ 0 and r(t)2 = (a − εx(t))2, it follows that

y

P0 = (x, y 0 )

P = (x, y) at elapsed time t

r(t) α ( t) β ( t) x Q periapsis x t = 0 x apoapsis O c = aε S r(t)

P = (x, y) at elapsed time t

P0 = (x, y 0 )

Sol 5.2 r(t) = a − εx(t). The substitution x(t) = a cos β(t) provides the equality r(t) = a1 − ε cos β(t)
. In α(t) q 1+ε β(t) the case where P is in the first quadrant, the verification of Gauss’s equation tan 2 = 1−ε tan 2 is the same as before. Let’s turn to the situation in Sol 5.2 with P in the fourth quadrant. The Pythagorean theorem tells us in this case that

2 2 2 2 2 2 b2 2 r(t) = (x(t) − aε) + (−y(t)) = (x(t) − aε) + a2 (−y0(t)) .

2 2 2 Since P0 is on the circle of radius a, x(t) + (−y0(t)) = a , and hence

2 2 2 b2 2 2 2 2 2 b2 2 r(t) = (x(t) − aε) + a2 (a − (x(t) ) = (x(t) − aε) + b − a2 x(t) 2 2 2 2 b2 2 a2−b2 2 2 = x(t) − 2aεx(t) + a ε + b − a2 x(t) = a2 x(t) − 2aεx(t) + a = ε2x(t)2 − 2aεx(t) + a2 = (a − εx(t))2.

2 This implies that r(t) = a − εx(t), as before. And as before, the substitution x(t) = a cos β(t) provides the equality r(t) = a1 − ε cos β(t)
. In the case where P is in the fourth quadrant, the α(t) q 1+ε β(t) verification of Gauss’s equation tan 2 = 1−ε tan 2 needs to be slightly modified. As before, it relies on the link between α(t) and β(t) given by

x(t)−aε a cos β(t)−aε cos β(t)−ε cos α(t) = cos(−α(t)) = cos(2π − α(t)) = r(t) = a(1−ε cos β(t)) = 1−ε cos β(t) .

cos β(t)−ε Problem 5.5. From cos α(t) = 1−ε cos β(t) we get cos β(t) − ε = cos α(t)1 − ε cos β(t)
= cos α(t) − cos α(t)ε cos β(t)
.

ε+cos α(t) So cos β(t) + ε cos α(t) cos β(t) = ε + cos α(t), and hence cos β(t) = 1+ε cos α(t) . Insert this into the equation r(t) = a(1 − ε cos β(t)) to get

ε+cos α(t)
1+ε cos α(t)−ε2−ε cos α(t)
a(1−ε2) r(t) = a(1 − ε cos β(t)) = a 1 − ε 1+ε cos α(t) = a 1+ε cos α(t) = 1+ε cos α(t) .

a(1−ε2) By ignoring the fact that α is a function of t, we see that r is the function r(α) = 1+ε cos α of α. Problem 5.7. A look at Figure 5.4 tells us that the time t it takes the point-mass P to go from 2πt periapsis to apoapsis satisfies β(t) = π. It follows from Kepler’s equation β(t) − ε sin β(t) = T , 2πt T that π = T and that t = 2 . Since T is the period of the orbit, it takes the point-mass P another T 2 to return from apoapsis to periapsis. Problem 5.9. From Figure 5.1 we’ll take T = 0.6152 years, or (0.6152)(365.25) = 224.7018 days, for the period of Venus’s orbit and ε = 0.0068 for its eccentricity. Let t1, t2, and t3 be the number of days it takes for the angle α of Venus’s orbit to rotate from 0◦ to 60◦, from 0◦ to 120◦ and 0◦ ◦ π 2π to 180 , respectively. So α(t1) = 3 , α(t2) = 3 , and α(t3) = π. A look at Figure 5.18 tells us that T 224.7018 α(t) q 1+ε β(t) t3 = 2 . So t3 = 2 = 112.3509 days. Rewriting Gauss’s equation tan 2 = 1−ε tan 2 in β(t) q 1−ε α(t) the form tan 2 = 1+ε tan 2 and using it with t = t1 and t2, we get

β(t ) q q q tan 1 = 1−ε tan( 1 · π ) = 1−0.0068 tan π = 1−0.0068 √1 = 0.5734, and 2 1+ε 2 3 1+0.0068 6 1+0.0068 3 √ β(t2) q 1−ε 1 2π q 1−0.0068 π q 1−0.0068 tan 2 = 1+ε tan( 2 · 3 ) = 1+0.0068 tan 3 = 1+0.0068 3 = 1.7203.

β(t1) β(t2) By pushing the tan inverse button of a calculator (in degrees), we get 2 = 0.5206 and 2 = 1.0442. So β(t1) = 1.0412 and β(t2) = 2.0885. Inserting the term β(t1) into Kepler’s equation 2πt β(t) − ε sin β(t) = T , we get

2πt1 T = β(t1) − ε sin β(t1) = 1.0412 − (0.0068)(sin 1.0412) = 1.0353,

(1.0353)(224.7018) so that t1 = 2π = 37.0248 days. Doing so with β(t2), we get

2πt2 T = β(t2) − ε sin β(t2) = 2.0885 − (0.0068)(sin 2.0885) = 2.0826,

3 (2.0826)(224.7018) so that t2 = 2π = 74.4785 days. ◦ ◦ So Venus traces out the first 60 of its orbit in t1 = 37.0248 days, the second 60 in t2 − t1 = ◦ 74.4785−37.0248 = 37.4537 days, and the third 60 in t3 −t2 = 112.3509−74.4785 = 37.8724 days.

Problem 5.11. i. Table 5.1 informs us that for the eccentricity of ε of Mercury is ε = 0.20563593, i i so that ε < 0.20564. The i-th approximation βi of β(t) satisfies |β(t) − βi| < ε < (0.20564) . Repeated squaring tells us that ε2 < 0.04229, ε4 < 0.00179, and ε8 < 0.000004. So 8 steps should always suffice. The fact that ε6 < 0.00008 and ε7 < 0.00002 suggests that 6 steps will generally not be enough, but that 7 should be. The inequality |x + y| ≤ |x| + |y| holds for any real numbers x and y. After applying it in the 8 7 current situation, |β8 − β7| = |β8 − β(t) + β(t) − β7| ≤ |β8 − β(t)| + |β(t) − β7| < ε + ε < 0.000024. So |β8 − β7| rounds to zero. Hence |β7| and |β8| are equal when rounded to four decimal place accuracy. So β7 approximates β(t) for any t. ii. Our computations will carry six decimal places. So from Table 5.1, the eccentricity and period of Mercury’s orbit are respectively, ε = 0.205636 and T = 0.240849 years and hence (0.240849)(365.25) = 87.970097 days. Turn to section 5E and Kepler’s equation β(t) − ε sin β(t) = 2πt 2πt 2πt T . For t = 20 days, the successive approximation scheme β1 = T and βi+1 = T + ε sin βi = β1 + ε sin βi provides the solution

2πt 2π(20) β1 = T = 87.970097 = 1.428482,

β2 = β1 + ε sin β1 = 1.428482 + (0.205636) sin(1.428482) = 1.632039

β3 = β1 + ε sin β2 = 1.428482 + (0.205636) sin(1.632039) = 1.633732

β4 = β1 + ε sin β3 = 1.428482 + (0.205636) sin(1.633732) = 1.633711

β5 = β1 + ε sin β4 = 1.428482 + (0.205636) sin(1.633711) = 1.633711. of Kepler’s equation for β(t). So for t = 20 days, the approximation method tells us that after five iterations β(20) = 1.633711 radians (or about 93.60◦). The formula of Problem 5.8 tells us that Mercury travels the first quarter of its orbit in

T T ε 87.970097 (87.970097)(0.205636)) t = 4 − 2π = 4 − 2π ≈ 19.1134 days after perihelion. Therefore 20 days after perihelion, Mercury has traced out a little more that the first quarter of its orbit. iii. Taking β(20) = 1.633711 in the formula r(t) = a1−ε cos β(t)
of section 5B, we get r(20) = (57,909,227)1 − (0.205636) cos(1.633711)
= 58,657,934 km. To compute α(20) we’ll suppose that −1q 1+ε β(t)
Mercury is in its first post-perihelion orbit, so that the formula α(t) = 2 tan 1−ε tan 2 + 2πnt −1q 1+0.205636 1.633711
applies with nt = 0. It follows that α(20) = 2 tan 1−0.205636 tan 2 = 1.839085 radians, or about 105.37 degrees.

Problem 5.13. Suppose that you have a statement Si (mathematical or not) for each positive integer i ≥ 1. If you know that S1 is true and if you also know for any i ≥ 1 that the truth of Si implies the truth of Si+1, then the statement Si is true for all i ≥. That this is so is simple to see.

4 Since S1 is true and the truth of S1 implies the truth of S2, we know that S2 is true. This in turn implies the truth of S3, and in turn S4, and in turn S5, etc., etc. Since this can go on forever, all statements Si are true. This is a fact that is known as the principle of mathematical induction. 2πt Consider Kepler’s equation β(t)−ε sin β(t) = T , with t some given elapsed time from perihelion, for some body in an elliptical orbit of eccentricity ε and perihelion orbit T . Focus on the solution 2πt 2πt β(t) of the equation. Let β1 = T and for any i ≥ 1, let βi+1 = T + ε sin βi = β1 + ε sin βi. We i have defined βi for all i ≥ 1. The statement Si, for any i ≥ 1, is the assertion |β(t) − βi| ≤ ε . We’ll now verify that the conditions required for the principle of mathematical induction to hold are met

by the set of statements Si. 2πt 2πt 1 Since |β(t) − β1| = |( T + ε sin β(t)) − T | = |ε sin β(t))| ≤ ε, we know that |β(t) − β1| ≤ ε . So statement S1 is true. Next, we need to show that if Si is true, then Si+1 is true as well. So assume i that it is the case that |β(t) − βi| ≤ ε . Since

2πt i i+1 |β(t) − βi+1| = |( T + ε sin β(t)) − (β1 + ε sin βi)| = |ε sin β(t) − ε sin βi| ≤ ε|β(t) − βi| ≤ ε · ε = ε ,

(note the use of the inequality | sin x1 − sin x2| ≤ |x1 − x2| of section 5E), it follows that statement Si+1 is true. So for any i ≥ 1, the truth of Si implies the truth of Si+1. Since its conditions are met, the principle of mathematical induction tells us that statement Si is true for all i ≥ 1. Therefore, i |β(t) − βi| ≤ ε for all i ≥ 1. Since ε < 1, the sequence β1, β2, β3, β4,... converges to β(t).

2πt Problem 5.15. Earth’s eccentricity is ε = 0.016711 and in the situation in question β1 = T = 2π(170.000046) 365.259636 = 2.924336. The β(t) we need is the solution x of f(x) = x − ε sin x − β1 = 0. The β2 that Newton-Raphson gives us is

2πt f(β1) β1−ε sin β1− T 2.924336−(0.016711) sin(2.924336)−2.924336 β2 = β1 − 0 = β1 − = 2.924336 − = 2.9278799. f (β1) 1−ε cos β1 1−(0.016711) cos(2.924336)

Rounding this answer off to six decimal places gives β2 = 2.927880. Therefore the Newton-Raphson approximation provides the correct result after a single step.

Problem 5.17. If the orbit of the point-mass P is a circle, then a is its radius, r(t) = a and π π γ(t) = 2 throughout. So suppose that the orbit is not a circle. If γ(t1) = γ(t2) = 2 , then P is either at periapsis or at apoapsis at times t1 and t2. If P is at periapsis at both times or at apoapsis at both times, then r(t1) = r(t2). If P is at periapsis at one of these times and at apoapsis at the other, then r(t1) + r(t2) = a(1 − ε) + a(1 + ε) = 2a. In reference to the counter clockwise motion of P the times t1 and t2 refer to the elapsed times after some fixed perihelion. Beyond that there

are no assumptions on t1 and t2. In particular, t1 − t2 can be greater than the period T , so that

P can be in different orbits around S. √ a 1−ε2 Recall from section 5D that sin γ(t) = √ . Suppose that r(t1) = r(t2) or r(t1)+r(t2) = r(t)(2a−r(t)) 2a. In the first case, it is clear that sin γ(t1) = sin γ(t2). In the second case, r(t1) = 2a − r(t2) and 2a − r(t1) = r(t2). Since √ √ a 1−ε2 a 1−ε2 sin γ(t1) = √ = √ = sin γ(t2), r(t1)(2a−r(t1)) (2a−r(t2))r(t2)

sin γ(t1) = sin γ(t2) in this case also. Suppose conversely, that sin γ(t1) = sin γ(t2), it follows that 2 2 2 2 r(t1)(2a − r(t1)) = r(t2)(2a − r(t2)) and hence that a − 2ar(t1) + r(t1) = a − 2ar(t2) + r(t2) .

5 2 2 Therefore (a − r(t1)) = (a − r(t2)) and hence a − r(t1) = ±(a − r(t2)). It follows that either r(t1) = r(t2) or r(t1) + r(t2) = 2a. In particular, if γ(t1) = γ(t2), then either r(t1) = r(t2) or r(t1) + r(t2) = 2a. We’ll now assume that sin γ(t1) = sin γ(t2) or, equivalently, that that either r(t1) = r(t2) or π r(t1) + r(t2) = 2a. If P is moving from apoapsis to periapsis at both t1 and t2, then 0 ≤ γ(t) ≤ 2 for both t = t1 and t = t2. Since sin γ(t1) = sin γ(t2), it follows from the graph of the sine in Figure 3.4, that γ(t1) = γ(t2). If P is moving from periapsis to apoapsis at both t1 and t2, then

γ ( t) r(t)

S apoapsis periapsis r(t 2 ) ( ) r(t1 ) γ t 2 γ ( t1 )

Sol 5.3

π 2 ≤ γ(t) ≤ π for both t = t1 and t = t2. Since sin γ(t1) = sin γ(t2), it again follows from the graph of the sine in Figure 3.4, that γ(t1) = γ(t2). The diagram in Sol 5.3 illustrates what is going on.

2 Problem 5.19. Solving x2 + y = 1 for y, we get y2 = b2(1 − x2 ) = b2 (a2 − x2), so that the a2 √b2 a2 a2 b b 1 x2 y2 2 2 2 2 2 graph of the function f(x) = a a − x = a (a − x ) is the upper half of the ellipse a2 + b2 = 1. The symmetry of this ellipse along with a standard integral formula for the length of the graph Z a of a function, tells us that the full length of the ellipse is equal to 4 p1 + f 0(x)2 dx. Since 0 0 1 b 2 2 − 1 b x 2 2 2 2 2 f (x) = (a − x ) 2 (−2x) = − 1 and a − b = c = a ε , we get 2 a a (a2−x2) 2 0 2 b2 x2 a2(a2−x2) b2x2 a4−(a2−b2)x2 a2−ε2x2 1 + f (x) = 1 + a2 a2−x2 = a2(a2−x2) + a2(a2−x2) = a2(a2−x2) = a2−x2 . √ With the trig substitution x = a sin θ with − π ≤ θ ≤ π and hence −a ≤ x ≤ a, we get a2 − x2 = √ 2 2 pa2(1 − sin2 θ) = a2 cos2 θ = a cos θ and dx = a cos θ dθ, and therefore

a a π π Z Z q Z 2 Z 2 p 0 2 a2−ε2x2 p 2 2 2 2 p 2 2 4 1 + f (x) dx = 4 a2−x2 dx = 4 a − ε a sin θ dθ = 4a 1 − ε sin θ dθ. 0 0 0 0

p(t) Problem 5.21. With t varying from t = 0 to t = √2π , the average value of is given by C s 2π 2π √ Z √ Z √ √ √ √2π C p(t) C 2π(q−s) C 2π(q−s) C dt = √2π q−s cos Ct dt = √  √1 sin Ct  = sin 2π − sin 0
= 0. 2π s C s Cs C Cs 0 0 0

We’ll conclude with a comment about the discussion in the paragraph The Perturbing Force of an Interior Planet. Consider the force with which a circular ring of mass M, radius R, and center S attracts a point-mass P of mass m that lies within the circle. The verification of the fact that the magnitudes of the vertical and horizontal components of this force on the point-mass P are equal to

6 Z φmax Z φmax GmMr sin φ cos φ GmMr cos2 φ 2 2 q 2 dφ and 2 2 q 2 dφ, πR(r −R ) 1− r sin2 φ πR(r −R ) 1− r sin2 φ −φmax R2 −φmax R2 respectively, relies on routine computations that are identical to those in part (i) of section 5I. The rest of the verification of the fact that this force on P acts in the direction of S with magnitude

GmM h 1 R 2 (1·3)2 1 R 4 (1·3·5)2 1 R 6 1·3·5·7
2 1 R 8 i G(r) = (r2−R2) 1 − 22 ( r ) − (2·4)2 3 ( r ) − (2·4·6)2 5 ( r ) − 2·4·6·8 7 ( r ) − · · · is carried out in the paragraph in detail.

7 Chapter 6. Solutions of Odd Problems

d1−q1 101−99 2 Problem 6.1. By a formula of section 6D, ε1 = = = = 0.01. By another d1+q1 101+99 200 formula from this section and the fact that GM for Eros is 4.4621 × 10−4 km3/sec2, the velocity of NEAR-Shoemaker at apoapsis of its post OCM-4 orbit was √ √ q GM √ 4.4621×10−4 2.112×10−2 v1 = 1 − ε1 = √ · 0.99 ≈ (0.9950) ≈ 0.00209 km/sec d1 101 10.050 or 2.09 m/sec. Since the OCM-5 burn tightened the periapsis distance of NEAR-Shoemaker’s orbit to q2 = 50 km while keeping the apoapsis distance at d2 = 101 km, we get that the eccentricity of d2−q2 101−50 51 this orbit was ε2 = = = = 0.3377. The velocity at apoapsis of this orbit was d2+q2 101+50 151 √ √ q GM √ 4.4621×10−4 2.112×10−2 v2 = 1 − ε2 = √ · 0.6623 ≈ (0.8138) ≈ 0.00171 km/sec d2 101 10.050 √ √ or 1.71 m/sec. Alternatively, by another formula of section 6D, v2 = √1−ε2 = √1−0.3377 = 0.8138 = v1 1−ε1 1−0.01 0.9950 0.8179. So as before, v2 = 0.8179v1 = (0.8179)(2.09) ≈ 1.71 m/sec.

Problem 6.3. The problem lists the periapsis velocities incorrectly as 3.25 m/sec and 3.07 m/sec, respectively. The correct values are 3.89 m/sec and 3.67 m/sec, respectively. The difference 3.89 − 3.67 = 0.22 m/sec (not 0.18 m/sec as incorrectly stated in the problem) is in close agreement

with the ∆v = 0.24 m/sec of the table. Using formulas of section 6D with q1 = 35 and d1 = 51, we d1−q1 know that the eccentricity of the orbit of NEAR-Shoemaker just prior to OCM-8 was ε1 = = d1+q1 51−35 16 51+35 = 86 = 0.1860 and that the speed at periapsis in this orbit was √ √ q GM √ 4.4621×10−4 2.112×10−2 v1 = 1 + ε1 = √ · 1.1860 ≈ (1.0890) ≈ 0.00389 km/sec q1 35 5.9161 or 3.89 m/sec. Since the OCM-8 burn tightened the apoapsis distance of NEAR-Shoemaker’s orbit

to d2 = 39 km while keeping the periapsis distance at q2 = 35 km, we get that the eccentricity of d2−q2 39−35 4 the post OCM-8 orbit was ε2 = = = = 0.0545 and that the spacecraft’s velocity at d2+q2 39+35 74 periapsis of this new orbit was √ √ q GM √ 4.4621×10−4 2.112×10−2 v2 = 1 + ε2 = √ · 1.0545 ≈ (1.0269) ≈ 0.00367 km/sec q2 35 5.9161 or = 3.67 m/sec.

Problem 6.5. To start, refer to Figure 6.26 and check the details of the discussion of the spacecraft’s

elliptical transfer orbit from the insertion point I to the rendezvous point P1. Now turn to the ◦ Hohmann transfer from I to P2. Let t2 be the time required for this transfer and let α(t2) = 100 and 8 r(t2) = 2.0700×10 km be the corresponding angle and distance for the rendezvous. Let a2 and ε2 be the semimajor axis and eccentricity of the transfer to P2. Note that the periapsis distance a2(1−ε2) 8 is the same 1.4710 × 10 km as before. After inserting the values cos α(t2) = cos(100) = −0.1736 8 and r(t2) = 2.0700 × 10 into the formula 2 a2(1−ε2) a2(1−ε2)(1+ε2) r(t2) = = , 1+ε2 cos α(t2) 1+ε2 cos α(t2) 8 8 (1.4710×10 )(1+ε2) we get 2.0700×10 = 1+ε (−0.1736) , and hence (2.0700)(1−0.1736ε2) = (1.4710)(1+ε2). Therefore
2 1.4710 + 0.1736(2.0700) ε2 = 2.0700 − 1.4710, so that

1 2.0700−1.4710 0.5990 ε2 = 1.4710+0.1736(2.0700) = 1.8304 = 0.3273.

8 8 This tells us that 0.6727a2 = a2(1 − ε2) = 1.4710 × 10 km and hence that a2 = 2.1867 × 10 km. Since the insertion point I of the craft into its solar transfer orbit is the perihelion of the orbit and since GM ≈ 1.3271 × 1020 m3/sec2 = 1.3271 × 1011 km3/sec2 for the Sun, Example 5.1 applies to tell us that the craft needs to be inserted into its transfer with a velocity of

q q 11 q 3 GM(1+ε2) 1.3271×10 )(1+0.3273) (1.3271)(1.3273)×10 v2 = ≈ 8 ≈ ≈ 34.604 km/sec. a2(1−ε2) 1.4710×10 1.4710

β(t2) The computation of t2 remains. By solving Gauss’s equation of Chapter 5B for tan , we get q 2 β(t2) 1−ε2 α(t2) π 10 tan = tan . Since α(t2) = 100( ) = π radians, it follows that 2 1+ε2 2 180 18

β(t2) q 1−0.3273 5 tan 2 ≈ 1+0.3273 tan( 18 π) ≈ 0.8484.

β(t2) −1 q GM So ≈ tan (0.8484) and β(t2) ≈ 1.4071. Solving Kepler’s equation β(t2)−ε2 sin β(t2) = 3 t2 2 a2 of Chapter 5C for t2 and inserting the values we have, we get

q 3 q 8 3 a2
(2.1867×10 )
t2 = GM β(t2) − ε2 sin β(t2) ≈ 1.3271×1011 1.4071 − 0.3273 sin(1.4071) ≈ 9,623,500 sec or about 111 days.

Problem 6.7. The diagram of Sol 6.1 below sketches the orbits of a spacecraft’s near Earth solar orbit, the orbit of Venus, as well as three elliptical solar transfer orbits (in shades of green)

P2 r(t 1 ) P1

α (t 1 ) P0 I transfer orbit S insertion

or orbit bit of Venus rth Ea ar- ne

Sol 6.1

2 from the insertion point I to the rendezvous points P0,P1, and P2, respectively.

Problem 6.9. We’ll verify the formulas for vin and vout in the situation of Figure 6.29. The various velocity vectors have the same meaning as in section 6J. The velocity of the planet P relative to S

is represented by the vector vP that has length the speed of P and direction given by the tangent to its orbital path. The figure depicts the vector vP and the angle θP between it and the vector v∞ of the craft’s entry into the planet’s gravitational sphere of influence (SOI). As before, we’ll assume

that the velocity vector vP of P relative to S is constant. Figure 6.29 also depicts the velocity vector v∞ at the point of the craft’s departure from the planet’s SOI. The respective resultants of vP and the two vectors v∞, each determined by the parallelogram law, are drawn into the figure as v in and v out in blue and green, respectively. The vector v in represents the velocity of the craft C relative to the Sun at the point of entry into the SOI and the vector v out the velocity of the craft relative to the Sun at the point of departure from the SOI. We know that the angle of deflection −1 1
δ is equal to δ = 2 sin ε . The magnitudes of v in and v out as well as the angle between them can be computed by making use of basic trigonometry. The computation of the magnitudes relies on Figure 6.30 (extracted from Figure 6.29). The law of cosines applied to the triangle that the

vectors vP and v in determine, tells us that (now also using v∞, vP , v in, and v out for the magnitudes 2 2 2 of these vectors) v in = v∞ + vP − 2v∞vP cos θP and hence that

p 2 2 v in = v∞ + vP − 2v∞vP cos θP .

2 By the law of cosines applied to the triangle formed by the vectors vP and v out, we get v out = 2 2 v∞ + vP − 2v∞vP cos(θP + δ). It follows that

p 2 2 v out = v∞ + vP − 2v∞vP cos(θP + δ).

Problem 6.11. We turn to Voyager 2’s flyby of Saturn. Let P designate the planet Saturn and turn 3 2 to Tables 6.5 and 6.6 for the needed data. For MP the mass of Saturn, GMP = 37,940,585 km /sec . The Sun-relative speed of Saturn at the time of the flyby was vP = 9.59 km/sec and the angle between Saturn’s Sun-relative velocity and the direction of Voyager 2’s approach to Saturn (as ◦ described in Figure 6.17) was θP = 98.2 . Using data in Table 6.5, we get that q q GMP 37,940,585 v∞ = a = 332,965 ≈ 10.67 km/sec for Voyager 2’s flyby and a(ε−1) = 332,965(1.482601−1) = 160,689 km for the distance of the craft −1 1 from Saturn’s center of mass at periapsis of the flyby. Since ε = 1.482601, we get δ = 2 sin ( ε ) = −1 1 ◦ 2 sin ( 1.482601 ) = 84.83 for the angle of deflection. It follows that

p 2 2 p 2 2 v in = v∞ + vP − 2v∞vP cos θP ≈ 10.67 + 9.59 − 2(10.67)(9.59) cos 98.2 ≈ 15.33 km/sec, and

p 2 2 p 2 2 v out = v∞ + vP − 2v∞vP cos(θP + δ) ≈ 10.67 + 9.59 − 2(10.67)(9.59) cos(98.2 + 84.83) ≈ 20.25 km/sec. Voyager 2’s flyby of Saturn resulted in a change of direction of

3 −1 v∞
−1 v∞
ϕ = ϕ1 + ϕ2 = sin sin θP − sin sin(θP + δ) v in v out −1 10.67
−1 10.67
≈ sin 15.33 sin 98.2 − sin 20.25 sin(98.2 + 84.83) ≈ 43.54◦ − (−1.60◦) = 45.14◦. We’ll consider Voyager 2’s flyby of Uranus next. Let P designate Uranus and turn to Tables 6.5 3 2 and 6.6 for the needed data. For MP the mass of Uranus, GMP = 5,794,585 km /sec . The Sun- relative speed of Uranus at the time of the flyby was vP = 6.71 km/sec and the angle between the ◦ Sun-relative velocity of Uranus and the direction of Voyager 2’s approach to Uranus was θP = 106.0 . Using data in Table 6.5, we get that q q GMP 5,794,585 v∞ = a = 26,694 ≈ 14.73 km/sec for Voyager 2’s flyby and a(ε − 1) = 26,694(5.014153 − 1) = 107,154 km for the distance of the craft from the center of mass of Uranus at periapsis of the flyby. Since ε = 5.014153, we get −1 1 −1 1 ◦ δ = 2 sin ( ε ) = 2 sin ( 5.014153 ) = 23.01 for the angle of deflection. It follows that

p 2 2 p 2 2 v in = v∞ + vP − 2v∞vP cos θP ≈ 14.73 + 6.71 − 2(14.73)(6.71) cos 106.0 ≈ 17.79 km/sec, and

p 2 2 p 2 2 v out = v∞ + vP − 2v∞vP cos(θP + δ) ≈ 14.73 + 6.71 − 2(14.73)(6.71) cos(106.0 + 23.01) ≈ 19.66 km/sec. Voyager 2’s flyby of Uranus resulted in a change of direction of

−1 v∞
−1 v∞
ϕ = ϕ1 + ϕ2 = sin sin θP − sin sin(θP + δ) v in v out −1 14.73
−1 14.73
≈ sin 17.79 sin 106.0 − sin 19.66 sin(106.0 + 23.01) ≈ 52.74◦ − 35.60◦ = 17.14◦ (Notice that the direction changes 45.14◦ for the Saturn flyby and 17.14◦ for the Uranus flyby derived above differ from the 45.28◦ and 17.31◦ announced in the statement of Problem 6.11. The reason? The computations of the announced changes were rounded off differently.)

2 2 Problem 6.13. Solving x2 − y = 1 for y we get, y = x2 −1 = 1 (x2 −a2), so that y2 = b2 (x2 −a2) √ a2 b2 √b2 a2 a2 a2 b 2 2 b 2 2 and y = ± a x − a . It follows that f(x) = a x − a is a function that has the upper half of b the hyperbola as graph. (Note that the coefficient a was erroneously omitted in the first line of the statement of the problem.) Consider Figure 6.12 and 6.9. Revolving the diagram of Figure 6.12 around the y-axis moves (−a, 0) to (a, 0), (−a cosh β(t), 0) to (a cosh β(t), 0), and the left branch of √ b 2 2 the graph of f(x) = a x − a to the right branch. The formula Z a cosh β(t) √ 1 b 2 2 B(t) = − 2 x(t)y(t) − a x − a dx. a √ is a direct consequence (note that x(t) < 0). The formula cosh−1u = ln(u + u2 − 1), where u ≥ 1 can be found in standard calculus texts, for example in reference 2 for Chapter 3. (It is not hard −1 ev+e−v v to derive. Let v = cosh u and note that u = cosh v = 2 , so that e − 2u + e−v = 0, and

4 hence (ev)2 − 2uev + 1 = 0. Now use the quadratic formula to solve for ev and apply the natural x log function ln.) Substituting u = a with x ≥ a into the formula, we get q √ √ √ −1 x x x2 x 1 2 2 1 2 2
1 2 2
cosh a = ln( a + a2 − 1) = ln( a + a x − a ) = ln a x + x − a = ln a + ln x + x − a . Notice that with x = a, we see that cosh−1(1) = ln(1 + 0) = ln 1 = 0. Z √ −1 x 2 2 Inserting this expression for cosh a into the formula for x − a dx tells us that Z √ √ √ √ 2 2 x 2 2 a2 2 2 x 2 2 a2 −1 x 1 x − a dx = 2 x − a − 2 ln(x + x − a ) + C = 2 x − a − 2 (cosh a − ln a ) + C.

1 After absorbing ln a into the constant C, Z √ √ 2 2 x 2 2 a2 −1 x x − a dx = 2 x − a − 2 cosh a + C.

Since cosh−1(1) = 0, it follows that Z a cosh β(t) √ q b 2 2 b 1 2 2 2 a2 −1
a x − a dx = a 2 (a cosh β(t)) a cosh β(t) − a − 2 cosh (cosh β(t)) a 1 1 1 1 = 2 (a cosh β(t))(b sinh β(t)) − 2 abβ(t) = 2 x(t)y(t) − 2 abβ(t), 1 Therefore B(t) = 2 abβ(t). Problem 6.15. Superimpose a polar coordinate system over the Cartesian system of Figure 6.31 as described in Chapter 3C. It follows from the connection between the polar and Cartesian systems developed there, in particular by the discussion of Figure 3.13, that P = (a cos β, a sin β) for some positive real number β. The polar equation of the circle is r = f(θ) = a, so that by the polar area Z β 1 2 1 2 β 1 2 integral of Chapter 3G, the highlighted sector has area B = 2 a dθ = 2 a θ 0 = 2 a β. 0 Problem 6.17. We are given that y = f(x) is a differentiable function that satisfies f(0) = 0, f 0(x) > 0 for x 6= 0, and that the graph of y = f(x) is concave up for x ≥ 0 and concave down

for x ≤ 0. Consider y = f(x) − c with c ≥ 0 a constant and suppose that f(x0) − c = 0. Since f(0) − c = −c < 0, and f(x0) − c = 0, it follows that x0 ≥ 0. Consider the tangent line to the graph of y = f(x) − c at the point (x1, f(x1) − c). The 0 point-slope form of the equation of this line is y − (f(x1) − c) = f (x1)(x − x1). Let x2 be the x-coordinate of the point of intersection of the line with the x-axis. Since y = 0 at that point, 0 f(x1)−c −(f(x1) − c) = f (x1)(x2 − x1). It follows that x2 = x1 − 0 . The transition from x1 to x2 f (x1) 1 is illustrated in Sol 6.3 in the special case of f(x) = 1.2 sinh x − x and c = 3 . Repeating the procedure that led from x1 to x2, with x2 gives rise to x3, then to x4, and so on. Sol 6.3 shows f(xi)−c how and why the sequence x1, x2, x3, . . . , xi,... , where for each xi, xi+1 = xi − 0 , converges to f (xi) the x0 that satisfies f(x0) − c = 0. If the initial stab x1 at a solution is to the left of x0, again see Sol 6.3, then the succeeding x2 falls to the right and we’re back in the earlier situation. The graph of Sol 6.3 illustrates the recipe: from any approximation—this is a point on the x-axis—starting with x1, go vertically to the graph and slide down (or up) the tangent back to x-axis to get the next

5 x1 x0 x3 x2 x1 − c

( ) 1 f x − c = 1.2 sinh x − x − −3

1 Sol 6.3. The graph of f(x) = 1.2 sinh x − x − 3 as drawn with https://www.desmos.com/calculator. approximation. (If f 0(0) = 0, then x = 0 is problematic in this regard, because the tangent line at (0, −c) is parallel to the x-axis and does not intersect it.) q GM Consider the hyperbolic Kepler equation ε sinh x − x = a3 t, where t ≥ 0 is a constant. q GM What was just set out applies with y = f(x) = ε sinh x − x, c = a3 t, and x0 = β(t). The facts developed in section 6E and a look at the graphs of Figure 6.10 confirm that f(x) satisfies all the requirements set out in the problem: f(0) = 0, f 0(x) = ε cosh x − 1 ≥ ε − 1 > 0, and since f 00(x) = ε sinh x, f(0) = 0, f 0(x) > 0 for x 6= 0, the graph of f(x) is concave up for −1 1 q GM
x ≥ 0 and concave down for x ≤ 0. With β1 = sinh ε a3 t as a good first stab at β(t) and q (ε sinh β −β )− GM t i i a3 βi+1 = βi − , the sequence β1, β2,..., converges to x0 = β(t). Sol 6.3 considered ε cosh βi−1 1 the representative case y = f(x) = 1.2 sinh x − x with c = 3 . Problem 6.19. The diagram of Sol 6.4 shows how the graph of the function f(x) = sinh−1 x (in green) is obtained by reflecting the the graph of y = sinh x (in blue) across the line y = x. Turn to −1 −1 Figure 6.33. Let P1 = (x1, y1) = (x1, sinh x1) and let L be the tangent to the graph of y = sinh x at P . Because f 0(x) = √ 1 (by Example 6.10), the slope of L is √ 1 . Therefore the point-slope 1 x2+1 2 x1+1

6 y = sinh x y = x

1 y = sinh − x

Sol 6.4

1 −1 0 2 − 1 √ 2 form of the equation of the tangent is y = 2 (x − x1) + sinh x1. Since f (x) = (x + 1) , x1+1

00 1 2 − 3 x f (x) = − (x + 1) 2 (2x) = − 3 . 2 (x2+1) 2

00 −1 Since f (x) < 0 for x ≥ 0, the graph of f(x) = sinh x is concave down for x ≥ 0. Let x2 √ 1 −1 satisfy x2 > x1 and let y2 = 2 (x2 − x1) + sinh x1 be the y-coordinate of the point on L with x1+1 x-coordinate x2. A look at Figure 6.33 tells us that −1 −1 √ 1 −1 −1 √ 1 | sinh x2 − sinh x1| < |y2 − y1| = 2 (x2 − x1) + sinh x1 − sinh x1| = 2 |x2 − x1|. x1+1 x1+1 Refer to the solution of the hyperbolic Kepler equation of section 6H and note that the inequality −1 −1 above is stronger than the inequality | sinh x2 − sinh x1| < |x2 − x1| (when x1 6= 0). This means that the sequence β1, β2, β3,... should in general converge to the solution β(t) of Kepler’s equation 1 more quickly than the inequality |β(t) − βi| ≤ εi |β(t)| suggests. We’ll look at Problems 6.20, 6.21, and 6.22 together. To understand the printouts of the HORIZONS ephemerides system for the changing parameters of Cassini’s trajectory, one needs to know the following abbreviations as well as the concepts to which they refer. Several of them, e.g., eccentricity, periapsis distance, semimajor axis, ... have been used throughout the text, others are illustrated in Figure 6.25. EC = eccentricity of the trajectory, denoted by ε in this text. QR = periapsis distance, often denoted by q in this text.

7 IN = angle of inclination of the orbit relative to Earth’s orbital plane as reference plane, denoted by i in Figure 6.25. OM = longitude of the ascending node, the angle denoted by Ω in Figure 6.25. W = argument of periapsis, the angle denoted by ω in Figure 6.25. Referred to as “argument of perifocus” in HORIZONS. Tp = time of periapsis (in Julian day number). N = mean or average motion in degrees/day. q GM MA = mean anomaly, the angle a3 t in radians, where GM is the gravitational constant of the attracting body, a the semimajor axis of the orbit, and t the elapsed time from periapsis. TA = true anomaly, the angle denoted by α = α(t) in this text, where t is the elapsed time from some periapsis. Often denoted by ν in the literature. A = semimajor axis. Listed with negative numbers in HORIZONS in hyperbolic situations. AD = apoapsis distance. Not defined, or infinite, in hyperbolic situations. When users found this problematic in some computer applications, HORIZONS set it equal to the huge, artificial number 9.9999999 × 1099 km or its equivalent 6.845864538097 × 1091 au. (Given the estimate 8.8 × 1023 km for the diameter of the universe, this is “essentially infinite.”) PR = sidereal period. The notation E−05, E+00, E+6, ... , E+91, E+99 in HORIZONS following a number tells us that the number is multiplied by 10−5, 100 = 1, 106,... or (see the discussion of AD) 1091 or 1099. Go to the website https://ssd.jpl.nasa.gov/horizons.cgi and consider Current Settings. Fol- low the instructions: Under Ephemeris Type [change], click on change, Select Orbital Elements, and click on Use Selection Above. Under Target Body [change], click on change and type Cassini into the box Lookup the specified body, then Search, and Select MB: Cassini (spacecraft), and click on Select Indicated Body. Under Center [change] click on change and type @Saturn into the box Specify Center, then Search, Select Saturn (body center), and click Use Selected Location. Relevant Time Span settings follow in the problems below. For Table Settings and Display/Output use the defaults. For much more information about HORIZONS the reader is directed to https://ssd.jpl.nasa.gov/?horizons doc. Continue with the instructions under Current Settings. For Time Span [change], click on change. Under Start Time insert 2004-July-01 01:12, under Stop Time insert 2004-July-01 02:48 for the SOI (or the time spans for OCM-2 or the flyby of Titan), and under Step Size insert 1 and minutes. Then click Use Specified Time. Finally click Generate Ephemeris. The table that HORIZONS generates provides the data for a minute by minute picture of Cassini’s changing trajectory during the SOI (or OCM-2 or the flyby of Titan). The tables Horizons 6.20, Horizons 6.21, and Horizons 6.22 are copies of the output of the data for the Saturn-specific trajectory of Cassini that HORIZONS generates for each of the three maneuvers. To limit the output to a single page, the data is presented in 5 or 10 minute intervals rather than in 1 minute intervals that the problems prescribe. Let’s start with the SOI and Table 6.20. The flow of numbers that describe Cassini’s evolving hyperbolic approach to Saturn tell us that the eccentricities EC decreased, so that the hyperbolas

8 tightened (refer to the conclusions of Problem 6.8), that its periapsis distances QR decreased slightly, while the semimajor axes A increased. Finally, during the last part of the maneuver, the trajectory flipped from tightly hyperbolic (EC slightly greater than 1) to tightly elliptical (EC slightly less than 1). A comparison of the last set of data in Table 6.20 with the first set of Table 6.21 tells us

Table 6.20. Orbital ephemerides for the insertion of Cassini into orbit around Saturn (Saturn Orbit Insertion or SOI) on July 1st in 10 minute intervals.

9 Table 6.21. Orbital ephemerides in 5 minute intervals for the orbit of Cassini around Saturn 2004 during OTM-2 on August 23, .

10 that over the seven weeks after the SOI, Cassini’s elliptical orbit became a little rounder, while the increasing periapsis distance and the fact that the apoapsis distance and semimajor axes more than doubled tell us that the orbit expanded. Seven and a half weeks after Cassini’s SOI, the space- craft underwent OCM-2. The decreasing eccentricity data EC of Table 6.21 shows that Cassini’s

Table 6.22. Orbital ephemerides in 10 minute intervals for the orbit of Cassini during the Titan flyby of October 26, 2004.

11 trajectory became steadily rounder. The increasing periapsis distances QR along with the stable apoapsis distances AD confirm that its orbit continued to expand. The flyby of Saturn’s moon Titan had an appreciable effect on Cassini’s elliptical orbit. The data of Table 6.22 inform us that the eccentricities EC decreased, rounding the ellipse further. It also tightened the orbit by decreasing the periapsis distance QR by nearly one-third and the apoapsis distance AD by close to half. During the entire time from Cassini’s approach to Saturn on July 1st, 2004 until its departure from the gravitational influence of Titan on October 26th, 2004, the inclination IN of Cassini’s orbital plane relative to that of Earth increased slowly, only to decrease again towards the end of the passage of Titan. The angles measuring the orientation OM of the nodal line and the orientation W of the focal axis of the trajectory were very stable throughout. But toward the end of Cassini’s swing around Titan, OM decreased and W increased. All in all, the data in the tables that describe Cassini’s three important maneuvers are aligned with the changing orbital geometry that diagram of Figure 6.24 depicts. Notice that these three maneuvers and the evolving modifications in the craft’s trajectory that they effected all occurred during the craft’s critical first loop around Saturn.

Problem 6.23. Consider the parabola of Figure 6.34 with its focal point (d, 0) and its directrix the line x = −d. By definition, a point P = (x, y) is on the parabola, if the distance from (x, y) to (d, 0) is equal to the distance from (x, y) to the line vertical line through x = −d. So P = (x, y) is on the parabola precisely if p(x − d)2 + y2 = x + d. After squaring both sides, x2 − 2dx − d2 + y2 = x2 + 2dx + d2. So P = (x, y) is on the parabola, precisely if y2 = 4dx. √ 1 The equation of the upper half of the parabola is y = 2 dx = 2(dx) 2 . A look at the figure tells us that the area B is the difference between a parabolic section and a triangle. In particular, Z x 1 1 2 B = 2(dx) dx − 2 xy. 0 There is a lesson to be drawn from this conclusion. What was just derived is problematic for two reasons. The first is that x serves simultaneously in two different roles. In addition to being the x coordinate of P , it is also the variable of integration. The second problem is the choice of the constant d in the definition of the parabola, and the consequence that the two dx in the integrand have two totally different meanings. In response, let’s write the constant d as c and the point P as

(x1, y1). We now get

Z x1 1 3 3 3 1 1 1 √ 2 3 x1 1 √ 4 √ √ 1 √ 2 2 2 2 2 2 2 B = 2c x dx − 2 x1y1 = 2 c 3 x 0 ) − 2 x1(2 cx1 ) = 3 cx1 − cx1 = 3 cx1 . 0 √ 1 3 2 We can now translate back to get that the area B of Figure 6.34 is equal to 3 dx . Turn to the paragraph Parabolic Trajectories. The discussion in the paragraph is complete, but we’ll fill in a few details. Focus on the parabolic trajectory of C with focal point S and its motion along it as illustrated in Figure 6.35. Review the meaning of the time variable t, angle α(t), and the function A(t). Recall that α(t) ≥ 0 when t ≥ 0, and α(t) < 0 when t < 0. For t ≥ 0, A(t) is the area swept out by the segment SC from t = 0 to t. For a negative t, A(t) is minus the area swept out from t to t = 0. With q the periapsis distance of the trajectory of C, the focal point is positioned at (0, −q). The directrix is the line x = q and (by applying the first part of Problem 6.23) the equation of the parabola is

12 y2 = −4qx.

2 √ 1 √ 1 Since x ≤ 0, the two solutions of y = −4qx for y are y = +2 q(−x) 2 and y = −2 q(−x) 2 . The first is the equation for the upper half of the parabola and the second is the equation for the lower √ 1 half. Another look at Figure 6.35 informs us that with x = −q, the length 2y = 4 q(−q) 2 = 4q is the latus rectum of the parabola. Recall that the coordinates of the position C of the craft at time t are x(t) and y(t), so that π C = (x(t), y(t)), and that r(t) is the distance from S to C. Suppose 0 ≤ α(t) ≤ 2 and refer to q+x(t) π diagram (a) of Sol 6.5. Since x(t) ≤ 0, cos α(t) = r(t) . If 2 ≤ α(t) ≤ π, then by diagram (b), −x(t)−q q+x(t) cos(π − α(t)) = r(t) . Since cos(π − α(t)) = − cos α(t), we get cos α(t) = r(t) in this case also.

r(t) r(t) ( ) α ( t) α t − q x(t) 0 x(t) − q 0

(a) (b) Sol 6.5

Since cos α(t) = cos(−α(t)), this equality holds for negative α(t) as well. Recall that β(t) is defined 1 α(t) (−x(t)) 2 √ by β(t) = tan 2 and that β(t) = ± q with the + in place when t ≥ 0, and the − when t ≤ 0. The paragraph goes on to establish the parabolic version

1 3 κt q GM β(t) + 3 β(t) = q2 = 2q3 t of Kepler’s equation, where κ is the Kepler constant of the trajectory and GM is the gravitational constant of the attracting body. The solution of Kepler’s equation for β(t) in terms of t is compli- cated, but it can be cast in an explicit form (unlike those of the elliptical situation of Chapter 5E and the hyperbolic situation in section 6H).

2 9 2 Problem 6.25. For f(x) = x(x + 3q) − 2 GMt and t a constant, we get f 0(x) = (x + 3q)2 + 2x(x + 3q) = (x + 3q)(x + 3q + 2x) = 3(x + 3q)(x + q) and f 00(x) = 3(x + q) + 3(x + 3q) = 6(x + 2q) by applying both the chain and product rules. The second derivative test tells us the following: Since f 0(−3q) = 0 and f 00(−3q) = −6q < 0, y = f(x) has a local maximum at x = −3q; since f 0(−q) = 0 and f 00(−q) = 6q > 0, y = f(x) has a local minimum at x = −q; and since f 00(−2q) = 0 with f 00(x) < 0 for x < −2q and f 00(x) > 0 for x > −2q, y = f(x) has a point of inflection at

13 9 2 3 9 2 0 x = −2q. Notice that f(−3q) = − 2 GMt and f(−q) = −4q − 2 GMt . For x > −q, f (x) = 3(x + 3q)(x + q) > 0 and f 00(x) = 6(x + 2q) > 6q > 0. Therefore y = f(x) is increasing and 9 2 concave up for x > −q, and hence for x ≥ 0. Because f(0) = − 2 GMt , it follows that y = f(x) has a unique positive real root.

To see that qβ(t)2 is the positive real root of f(x) = x(x+3q)2 − 9 GMt2 requires details. Return p √ 2 to Kepler’s equation and rewrite it as q3β(t)(1+ 1 β(t)2) = √1 GM ·t. After multiplying through 3 2 p √ by 3, we get q3β(t)(3+β(t)2) = √3 GM ·t. By squaring both sides, q3b(t)2(3+β(t)2)2 = 9 GMt2. 2 2 Finally, after rewriting things once more,

2 2
2 9 2 qβ(t) 3q + qβ(t) = 2 GMt ,

2 2 9 2 So qβ(t) is the unique positive real root of f(x) = x(x + 3q) − 2 GMt . The rest of the discussion of the Parabolic Trajectories in text is complete. This includes the formula

q 1 1 1 9 2 3
3 1 9 2 3
3 β(t) = ± q 4 GMt + z(t) + q + q 4 GMt − z(t) + q − 2 q 9 2 3 2 6 for β(t), with z(t) = ( 4 GMt + q ) − q , where the + applies if t ≥ 0, and the − if t < 0. It also includes the parabolic velocity formulas.

We will now apply these results to the study of Lovejoy’s comet C/2011 W3. In reference to its trajectory, the perihelion distance was 0.00555381 au. With 1 au = 149597870.7 km, we’ll take q = 8.3083815×105 km. Since its eccentricity of ε = 0.99992942 was very close to 1, we will assume that the comet’s trajectory was parabolic.

Problem 6.27. Since seven days is equal to 7(86,000) = 6.04800 × 105 seconds, the comet was t = 6.04800 × 105 sec past perihelion. The assumption that the comet’s orbit is parabolic along with the data for the orbit that was supplied, calls for the solution of the parabolic Kepler equation

1 3 q GM q 1.3271244×1011 5 β(t) + 3 β(t) = 2q3 t = 2(8.3083815×105)3 (6.04800 × 10 ) = 205.7206826

for β(t). Since t ≥ 0,

q 1 1 1 9 2 3
3 1 9 2 3
3 β(t) = q 4 GMt + z(t) + q + q 4 GMt − z(t) + q − 2 , q 9 2 3 2 6 where z(t) = ( 4 GMt + q ) − q . Computing z(t), we get q  9 11 5 2 5 32 5 6 z(t) = 4 (1.3271244 × 10 )(6.04800 × 10 ) + (8.3083815 × 10 ) − (8.3083815 × 10 ) = 10.92244830 × 1022. It follows that 1 9 2 3
3 4 GMt + z(t) + q 1 9 11 5 2 22 5 3
3 = 4 (1.3271244 × 10 )(6.04800 × 10 ) + (10.92244830 × 10 ) + (8.3083815 × 10 )

14 = 6.02259045 × 107, and 1 9 2 3
3 4 GMt − z(t) + q 1 9 11 5 2 22 5 3
3 = 4 (1.3271244 × 10 )(6.04800 × 10 ) − (10.92244830 × 10 ) + (8.3083815 × 10 ) = −0.00352381 × 107. By substituting into the formula for β(t), we get

q 1 7 1 7 β(t) = (8.3083815×105) (6.02259045 × 10 ) + (8.3083815×105) (−0.00352381 × 10 ) − 2 = 8.39319475. Plugging the value β(t) = 8.39319475 into the formula r(t) = q(β(t)2 + 1) and the result into √ √ q 2 −1 GMq(1+ε) v(t) = GM r(t) and γ(t) = π − sin r(t)v(t) , we get the following information about the trajectory of Lovejoy’s comet at time t, r(t) = q(β(t)2 + 1) = (8.3083815 × 105)(8.393194752 + 1) ≈ 59,359,828 km, √ √ q 2 11q 2 v(t) = GM r(t) = 1.3271244 × 10 59,359,828 = 66.86894909 ≈ 66.87 km/sec, and √ √ 11 5 −1 2GMq −1 2(1.3271244×10 )(8.3083815×10 ) γ(t) = π − sin r(t)v(t) = π − sin (5.93598283×107)(66.86894909) = π − 0.11858513 radians ≈ 180◦ − 6.7944◦ ≈ 173.21◦.

v(t ) √ Problem 6.29. Since √ 0 is a positive constant and the function f(x) = x, 0 ≤ x < ∞, is GM √ v(t ) a continuous function, there is a positive number x such that x = √ 0 and hence v(t ) = 0 0 GM 0 √ √ v(t ) GM x . If this were not so, the horizontal line y = √ 0 would not intersect the graph of √0 GM f(x) = x. This would mean that the graph of the function, see Figure 6.36, would have at least two disjoint components, thus violating its continuity. √ √ √ q 2 √ q 2 2 If v(t0) < GM , then GM x0 < GM . Since g(x) = x is an increasing r(t0) r(t0) √ 2 2 √ function, x0 < . Set − x0 = D. Note that D is positive and that v(t0) = GM x0 = √ r(t0 r(t0 q 2 GM r(t ) − D. 0 √ √ √ q 2 √ q 2 2 If v(t0) > GM r(t ) , then GM x0 > GM r(t ) , so that x0 > r(t . In this case, set 0 √0 √ 0 2 √ q 2 D = x0 − . Again D is positive, and this time v(t0) = GM x0 = GM + D. r(t0 r(t0)

15 http://www.springer.com/978-3-030-24867-3