problem12_85

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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12.85: a) , E 2 3 2 GM r T π = therefore ( 29 ( 29 . 1 1 ) ( E 2 1 E 2 3 E 2 3 E 3 2 3 2 2 3 2 2 3 2 GM r r π r r GM r π r r GM r π GM π T r r T T + = + + = + = + Since , . , 2 1 E 3 E - π = = = r GM v T v v r r GM therefore ( 29 ( 29 . 1 1 ) ( 2 3 2 2 2 1 E 2 1 2 1 2 1 r v r GM r GM r r GM v v r GM r r r r E E E - = - + = + = - - - - - Since . , 2 E 2 3 T r GM r π v T π = = b) Note: Because of the small change in r , several significant figures are needed to see the results. Starting with GM r T 2 3 2 π = (Eq.(12.14)), v r T 2 π = , and r GM v = (Eq.(12.12)) find the velocity and period of the initial orbit: s, m 10 672 . 7 m 10 776 . 6 ) kg 10 97 . 5 )( kg m N 10 673 . 6 ( 3 6 24 2 2 11 × = × × × = - v and 5 . 92 s 5549 2 = = = v r T π min. We then can use the two derived equations to approximate the s. m 05662 . and s, 1228 . 0 . and , and ) s 5549 ( ) m 100 ( s m 10 672 . 7 m) 100 ( 3 r 3 3 = = = = = = = × π π T r π T r π v π v T v T v T Before the cable breaks, the shuttle will have traveled a distance d, d 22 s 7 . 1324 ) s m 05662 (. ) m 75 ( So, m. 75 ) m 100 ( ) m 125 ( 2 2 = = = - = min. 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 π 2 radians more than the angle that the space shuttle covers before they are
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• International Space Station, GM E

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