sol8 - Massachusetts Institute of Technology Physics...

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Massachusetts Institute of Technology - Physics Department Physics-8.01 Fall 1999 Solutions for Assignment # 8 by Dru Renner [email protected] November 14, 1999 6:20 pm Problem 8.1 The subsequent motion will be elliptical with the CENTER of the planet at one focus. (Kepler’s First Law) Angular momentum of the satellite is conserved if we measure it relative to the center of the planet. (This is what is called “orbital angular momentum.”) Angular momentum is NOT conserved relative to ANY OTHER point! Initially the magnitude of angular momentum relative to the center of the planet is L = mRv 0 sin(20 ) where m is the mass of the satellite. At the point of maximum distance, the apogee, the magnitude of angular momentum is L = m 5 Rv sin(90 ) where v is the corresponding speed. (At the apogee and perigee, the angle between the velocity vector and the radius vector from either foci will always be 90 .) Conservation of angular momentum and hence conservation of the magnitude of angular momentum gives mRv 0 sin(20 )= m 5 Rv sin(90 v = sin(20 ) 5 · v 0 The gravitational force is conservative; therefore mechanical energy is conserved. Initially the mechanical energy is E = GMm R + 1 2 mv 2 0 At the point of maximum distance the mechanical energy is E = 5 R + 1 2 mv 2 = 5 R + 1 2 m à sin(20 ) 5 · v 0 ! 2 1
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Conservation of mechanical energy gives GMm R + 1 2 mv 2 0 = 5 R + 1 2 m à sin(20 ) 5 · v 0 ! 2 Solving the above for v 0 gives v 0 = v u u t 8 GM 5 R ³ 1 sin 2 (20 ) 25 ´ 1 . 27 · s GM R Notice that this satellite will slam into the planet before it has made one rotation about the planet (why?). Problem 8.2 (a) The distance apart is f 2 πR 1710 m. (b) n s T s = T ( n a f ) (Please see handout.); thus 2 T s = T (2 0 . 04). The period T s is smaller than T . The sandwich should be thrown backward. (c) Mary’s (and Peter’s) orbital period T =2 π ( R 3 /MG ) 0 . 5 = 5580 s 93 min. Mary must wait T ( n a f )=(2 0 . 04) × 93 min 3 . 04 hrs before she will make the catch. (d) n s T s = T ( n a f ), thus n s (4 π 2 a 3 /MG ) 0 . 5 =(4 π 2 R 3 /MG ) 0 . 5 ( n a f )wh ichleadsto a = R [( n a f ) /n s ] 2 / 3 ;thus a 6800(0 . 987) 6710 km. (e) (Need Picture) (f) The velocity of the spacecraft, v a ,is2 πR/T . v a 7 . 66 km / s. The velocity, v s ,o f the sandwich at the point X can be obtained from the relation ENERGY = GMm/ 2 a = 1 2 mv 2 s GMm/R R = 7000 km Since v a = q GM/R v s = v q 2 R/a v s 7 . 61 km / s The sandwich must be thrown relative to Peter’s motion with a velocity v s v a 7 . 61 7 . 66 ≈− 52 . 1m / s 116 . 5 mph. Too high for any human being! The “-” sign indicates that the sandwich must be thrown backwards. To reduce the speed with which the sandwich has to be thrown you could increase the number of orbits for both Mary and the sandwich before the catch is made. Using Dave Pooley’s program (as shown in lectures) we Fnd that for n s = n a = 3 the speed is about 77 . 1 mph (Roger Clemens could just do it!); for n s = n a = 4 the speed will be about 57.7 mph, and for n s = n a = 5 the speed is only about 46.0 mph (that is certainly doable!).
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This note was uploaded on 11/20/2011 for the course PHY 203 taught by Professor Staff during the Fall '09 term at Rutgers.

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sol8 - Massachusetts Institute of Technology Physics...

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