B lim f x 0 lim x x 3x x3 1 so no absolute

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Unformatted text preview: en R = R0 , then 2g sec α0 tan α0 R0 − 2v sec α0 R0 = 0, g tan α0 R0 − v = 0, tan α0 = v 2 /(gR0 ). (b) 2g sec2 α tan αR2 + 2g sec2 αR 2 (c) If α = α0 and R = R0 , then from part (a) g (sec2 α0 )R0 − 2v 2 (tan α0 )R0 − 2v 2 h = 0, but from 2 2 2 part (b) tan α0 = v /(gR0 ) so sec α0 = 1 + tan α0 = 1 + v 4 /(gR0 )2 thus 2 2 g [1 + v 4 /(gR0 )2 ]R0 − 2v 2 [v 2 /(gR0 )]R0 − 2v 2 h = 0, gR0 − v 4 /g − 2v 2 h = 0, 2 R0 = v 2 (v 2 + 2gh)/g 2 , R0 = (v/g ) tan α0 = v 2 /(v v 2 + 2gh and v 2 + 2gh) = v/ v 2 + 2gh, α0 = tan−1 (v/ v 2 + 2gh). 71. (a) v0 (cos α)(2.9) = 259 cos 23◦ so v0 cos α ≈ 82.21061, v0 (sin α)(2.9) − 16(2.9)2 = −259 sin 23◦ so v0 sin α ≈ 11.50367; divide v0 sin α by v0 cos α to get tan α ≈ 0.139929, thus α ≈ 8◦ and v0 ≈ 82.21061/ cos 8◦ ≈ 83 ft/s. (b) From part (a), x ≈ 82.21061t and y ≈ 11.50367t − 16t2 for 0 ≤ t ≤ 2.9; the distance traveled 2.9 (dx/dt)2 + (dy/dt)2 dt ≈ 268.76 ft. is 0 EXERCISE SET 14.7 1. The results follow from formulae (1) and (7) of Section 12.5. 2. (a) (rmax − rmin )/(rmax + rmin ) = 2ae/(2a) = e (b) rmax /rmin = (1 + e)/(1 − e), and the result follows. 3. (a) From (15) and (6), at t = 0, 2 2 C = v0 × b0 − GM u = v0 j × r0 v0 k − GM u = r0 v0 i − GM i = (r0 v0 − GM )i 2 (b) From (22), r0 v0 − GM = GM e, so from (7) and (17), v × b = GM (cos θi + sin θj) + GM ei, and the result follows. (c) From (10) it follows that b is perpendicular to v, and the result follows. (d) From Part (c) and (10), v × b = v v × b = GM b = vr0 v0 . From Part (b), √ (e + cos θ)2 + sin θ = GM e2 + 2e cos θ + 1. By (10) and Part (d), v × b = v 2 b = v (r0 v0 ) thus v = GM r0 v0 2 r0 v0 /(GM ) = 1 + e, GM/(r0 v0 ) = v0 /(1 + e) so v = e2 + 2e cos θ + 1. From (22), v0 1+e e2 + 2e cos θ + 1. 4. At the end of the minor axis, cos θ = −c/a = −e so v0 v0 1−e . e2 + 2e(−e) + 1 = 1 − e2 = v0 v= 1+e 1+e 1+e 5. vmax occurs when θ = 0 so vmax = v0 ; vmin occurs when θ = π so v0 1−e 1+e vmin = , thus vmax = vmin . e2 − 2e + 1 = vmax 1+e 1+e 1−e 6. If the orbit is a circle then e = 0 so from Part (e) of Exercise 3, v = v0 at all points on the orbit. Use (22) with e = 0 to get v0 = GM/r0 so v = GM/r0 . Chapter 14 Supplementary Exercises 520 3.99 × 105 /6640 ≈ 7.75 km/s. 7. r0 = 6440 + 200 = 6640 km so v = 8. From Example 1, the orbit is 22,352 mi above the Earth, thus v ≈ 1.24 × 1012 ≈ 6859.68 mi/h. 26,352 2(3.99) × 105 ≈ 10.88 km/s. 6740 9. From (23) with r0 = 6440 + 300 = 6740 km, vesc = 2π 3/2 4π 2 a3 a . But T = 1 yr = 365 · 24 · 3600 s, thus M = 10. From (29), T = √ ≈ 1.99 × 1030 kg. GT 2 GM 11. (a) At perigee, r = rmin = a(1 − e) = 238,900 (1 − 0.055) ≈ 225,760 mi; at apogee, r = rmax = a(1 + e) = 238,900(1 + 0.055) ≈ 252,040 mi. Subtract the sum of the radius of the moon and the radius of the Earth to get minimum distance = 225,760 − 5080 = 220,680 mi, and maximum distance = 252,040 − 5080 = 246,960 mi. a3 /(GM ) = 2π (b) T = 2π (238,900)3 /(1.24 × 1012 ) ≈ 659 hr ≈ 27.5 days. 12. (a) rmin = 6440 + 649 = 7,089 km, rmax = 6440 + 4,340 = 10,780 km so a = (rmin + rmax )/2 = 8934.5 km. (b) e = (10,780 − 7,089)/(10,780 + 7,089) ≈ 0.207. a3 /(GM ) = 2π (c) T = 2π (8934.5)3 /(3.99 × 105 ) ≈ 8400 s ≈ 140 min 1.24 × 1012 /4180 ≈ 17,224 mph 13. (a) r0 = 4000 + 180 = 4180 mi, v = 2 (4180)(17,824)2 r0 v0 −1 = −1 ≈ 0.07094. GM 1.24 × 1012 rmax = 4180(1 + 0.07094)/(1 − 0.07094) ≈ 4818 mi; the apogee altitude is 4818 − 4000 = 818 mi. (b) r0 = 4180 mi, v0 = 17,224+600 = 17,824 mi/h; e = 14. By equation (20), r = hence e ≥ 0. k , where k > 0. By assumption, r is minimal when θ = 0, 1 + e cos θ CHAPTER 14 SUPPLEMENTARY EXERCISES 2. (a) the line through the tips of r0 and r1 (b) the line segment connecting the tips of r0 and r1 (c) the line through the tip of r0 which is parallel to r (t0 ) 4. (a) speed (b) t 7. (a) r(t) = cos 0 dr dt distance traveled π u2 2 t du i + sin 0 2 = x (t)2 + y (t)2 = cos2 π t2 2 (c) π u2 2 distance of the particle from the origin du j; + sin2 π t2 2 = 1 and r(0) = 0 521 Chapter 14 π s2 i + sin 2 κ = r (s) = π |s| (b) r (s) = cos π s2 2 j, r (s) = −πs sin π s2 2 i + πs cos π s2 2 j, (c) κ(s) → +∞, so the spiral winds ever tighter. 8. (a) The tangent vector to the curve is always tangent to the sphere. (b) (c) 9. (a) v = const, so v · a = 0; the acceleration vector is always perpendicular to the velocity vector r(t) 2 = 1− 1 1 cos2 t (cos2 t + sin2 t) + cos2 t = 1 4 4 r(t) = 1, so by Theorem 14.2.7, r (t) is always perpendicular to the vector r(t). Then v(t) = Rω (− sin ωti + cos ωtj), v = v(t) = Rω (b) a = −Rω 2 (cos ωti + sin ωtj), a = a = Rω 2 , and a = −ω 2 r is directed toward the origin. (c) The smallest value of t for which r(t) = r(0) satisfies ωt = 2π , so T = t = 10. (a) F = F = m a = mRω 2 = mR v2 mv 2 = R2 R (b) R = 6440 + 3200 = 9600 km, 6.5 = v = Rω = 9600ω, ω = a = −a r 13 r = −Rω 2 = −Rω...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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