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Unformatted text preview: 4.19. CHAPTER 4, PROBLEM 19 223 4.19 Chapter 4, Problem 19 Problem: A vehicle is in a circular orbit of radius rA about the moon. To transfer to an orbit of larger radius rB , the vehicle is first placed on a Hohmann transfer orbit from Point A to Point B by firing its rocket momentarily at Point A and then again at Point B. Determine the changes in velocity, ∆vA and ∆vB , as functions of rA , rB , the mass of the moon, M , and the Universal Constant of Gravitation, G. Compute ∆vA and ∆vB for rA = 1200 mi, rB = 1800 mi, and M = 5.04 zettaslugs. .............. ................... ....... .... ..... .... .... ... .... ... . .. ... ... ... ... ................... . ... ... ........................ ... .. . ... ... .. ........ . . .................. ........ .................. .. .. . .. . .. . . . . .................. .. . ..... .. ................................... ...... ... . ......... .. ... .. .... .................................................. .. ............... . .. .. . ........................... . .... ... .......................................................... ... ..... . .. .. ................................................................... ... . .. .. . . .. .. .. .. . ................................................................... .. .. . . ................................................................................ ................................................................... .. . . . . . . . . ....... . . . . . . . . . . . . . . . . ... . . .. .. ................................................................................................................ .. . ............................................................ . . . . ... . . . ................................................................................. . . . . .................................................................................. . . .. . . . ................................................................................... . . . . .................................................................................... . .. . . . .............................................................................................................................. . . . . ................................................................................. . . . .................................................................................... . . . . . ..................................................................................... . . ... ......................................... . . .. . .................................................................................... . . . . . . .................................................................................... . . . . . ................................................................................. . . . . ................................................................................ . . . . .................................................................................. . . . ..... . . . . .............................................................................. . . . . . ............................................................................. . . . . ........................................................................... . . . .. .......................................................................... . . . .. ....................................................................... . . . ..... . . . .. .................................................................... ................................................................. .. . . .. ................................. . .. .. .... ... ....................................................... .. . ... .................................................... .. .. .. ... ........................................ ... . . ......... . ... ... ............................... ..... .. .. .............................. ... . ...... ......... ....... .. .. .. ........... .......... .. .. ... ... .. ... ... ... ... ... . ... ... .... ... ... .... ..... .... ........... ............ ................... . ...... . . .. B• Moon •A Solution: After the rocket has finished firing, denote the vehicle’s velocity at Point A by vA . Similarly, ˜ denote the vehicle’s velocity prior to firing the rocket by vB . Since angular momentum is conserved on the ˜ transfer orbit, we know that ˜ ˜ Ho = mrA vA = mrB vB Therefore, the velocities are related by vA = ˜ Because total energy is also conserved, we have 1 GM m 1 GM m = mvA − mvB − ˜2 ˜2 2 rB 2 rA Substituting for vA , this equation simplifies as follows. ˜ 1 GM m 1 =m mvB − ˜2 2 rB 2 rB rA 2 rB vB ˜ rA vB − ˜2 GM m rA =⇒ 1 r2 m 1− B 2 2 rA vB = ˜2 GM m rB 1− rB rA Dividing through by 1 m(1 − rB /rA ), we have 2 1+ rB rA vB = ˜2 2GM rB =⇒ vB = ˜2 2GM rA rB rB + rB Therefore, the velocities at Points A and B on the transfer orbit are vB = ˜ 2GM rA rB (rA + rB ) and vA = ˜ 2GM rB rA (rA + rB ) Since the vehicle is initially on a circular orbit of radius rA , its velocity at Point A prior to firing its rocket is GM vA = rA 224 CHAPTER 4. WORK AND ENERGY Similarly, after firing its rocket at Point B, its velocity on the circular orbit of radius rB is vB = GM rB Now, the changes in velocity at Points A and B are given by vA = vA + ∆vA ˜ From the results developed above, ∆vA is ∆vA = Similarly, ∆vB is ∆vB = GM − rB 2GM rA = rB (rA + rB ) GM 1− rB 2rB rA + rB 2GM rB − rA (rA + rB ) GM = rA GM rA 2rB −1 rA + rB and vB = vB + ∆vB ˜ We are given rA = 1200 mi, rB = 1800 mi, and M = 5.04 zettaslugs. Thus, noting that a zettaslug is 1021 slugs, ft3 ft3 5.04 · 1021 slug = 1.734 · 1014 GM = 3.44 · 10−8 slug · sec2 sec2 Therefore, the given data yield √ ft3/2 GM = 1.317 · 107 , sec Using these values, we have 2rB 6 = rA + rB 5 Therefore, ∆vA ∆vB = = 1.317 · 107 ft3/2 /sec √ 6.336 · 106 ft 1.317 · 107 ft3/2 /sec √ 9.504 · 106 ft 6 −1 5 1− 4 5 = 499 = 451 ft sec ft sec and 4 2rA = rA + rB 5 rA = 6.336 · 106 ft, rB = 9.504 · 106 ft ...
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