406_PartUniversity Physics Solution

406_PartUniversity Physics Solution - Gravitation 12-27 SET...

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Gravitation 12-27 SET UP: Use a coordinate system with the origin at the left-hand end of the rod and the -axis x along the rod, as shown in Figure 12.84. Divide the rod into small segments of length . dx (Use x for the coordinate so not to confuse with the distance x from the end of the rod to the particle.) Figure 12.84 EXECUTE: The mass of each segment is ( / ). dM dx M L = Each segment is a distance L xx −+ from mass m , so the force on the particle due to a segment is 22 . () Gm dM GMm dx dF L L Lxx == ′′ 00 0 2 1 L LL GMm dx GMm Fd F L Lx x L ⎛⎞ = ⎜⎟ ⎝⎠ ∫∫ 11 ( ) ()() GMm GMm L x x GMm F L xLx L x Lx +− =− = = ++ + EVALUATE: For x L >> this result become 2 /, FG M m x = the same as for a pair of point masses. 12.85. IDENTIFY: Compare E F to Hooke&s law. SET UP: The earth has mass 24 E 5.97 10 kg m and radius 6 E 6.38 10 m R . EXECUTE: For x Fk x , 2 1 2 Uk x = . The force here is in the same form, so by analogy 2 E 3 E . 2 Gm m Ur r R = This is also given by the integral of g from 0 to Fr with respect to distance. (b) From part (a), the initial gravitational potential energy is E E . 2 Gm m R Equating initial potential energy and final kinetic energy (initial kinetic energy and final potential energy are both zero) gives 23 E E , so 7.90 10 m/s. Gm vv R × EVALUATE: When 0 r = , ( ) 0 = , as specified in the problem. 12.86. IDENTIFY: In Eqs.(12.12) and (12.16) replace T by TT + Δ and r by rr + Δ . Use the expression in the hint to simplifying the resulting equations. SET UP: The earth has 24 E m and 6 R . E rhR = + , where h is the altitude above the surface of the earth. EXECUTE: (a) 32 E 2 r T GM π = therefore 1 /2 EE E E 2 3 3 1 1 2 ππ π π T GM GM GM GM Δ ΔΔ +Δ = = + + = + . Since 12 E E 3 ,. , GM r vT v G M r rv Δ = = and therefore ( ) 1 r v v GM r r GM r r −− Δ −Δ = = + and ( ) E 1. 2 2 E GM r vG M r v r r r Δ −= Δ Since E , . π r r Tv T GM Δ = (b) Starting with r T GM = (Eq.(12.12)), / Tr v = , and GM v r = (Eq.(12.10)), find the velocity and period of the initial orbit: 11 2 2 24 3 6 (6.673 10 N m /kg )(5.97 10 kg) 7.672 10 m/s, 6.776 10 m v ×⋅ × × × and 2 / 5549 s 92.5 v === min. We then can use the two derived equations to approximate and :

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12-28 Chapter 12 3 3 (100 m) 3 0.1228 s 7.672 10 m/s π r T v π Δ Δ= = = × and (100 m) 0.05662 m/s (5549 s) π π r v T Δ = = . Before the cable breaks, the shuttle will have traveled a distance d, 22 (125 m ) (100 m ) 75 m d =− = . (75 m) (0.05662 m/s) 1324.7 s 22 min t == = . It will take 22 minutes for the cable to break. (c) The ISS is moving faster than the space shuttle, so the total angle it covers in an orbit must be 2 radians more than the angle that the space shuttle covers before they are once again in line. Mathematically, () 2 vv t vt π rrr −Δ −= . Using the binomial theorem and neglecting terms of order ( ) 1 2 , 1 2 t vt v v r r vr t π rr r r r ΔΔ Δ + + = . Therefore, 2 rv T t π r v Tr r Δ ⎛⎞ + Δ+ ⎜⎟ ⎝⎠ . Since 2 a n d 3 vT T r π Δ = , 2 2 3 3 vT T t T tT ππ Δ + , as was to be shown.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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406_PartUniversity Physics Solution - Gravitation 12-27 SET...

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