# Sol-Ch13 - Exercises 1. (a) a = Gm/r 2 = 6.67 × 10 –9...

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Unformatted text preview: Exercises 1. (a) a = Gm/r 2 = 6.67 × 10 –9 m/s 2 (b) E i = –Gm 2 /1, E f = mv 2 – Gm 2 /(0.5) 2 . E i = E f leads to v = 8.17 × 10 –5 m/s 2. F = GmM/r 2 . (a) 2.0 × 10 20 N; (b) 4.42 × 10 20 N 3. F = GmM/r 2 . (a) 2.33 × 10 –3 N; (b) 4.19 × 10 –1 N 4. Since F E = F M we find M E /x 2 = M M /(d – x) 2 which leads to 9(d – x) = +x, thus x = 0.9d. 5. F 31 = 50G, F 32 = 160G. F 3x = –(50 + 320/(5) 1/2 )G = –1.29 × 10 –8 N F 3y = 160G/(5) 1/2 = 4.77 × 10 –9 N 6. From 4/x 2 = 9/(1 – x) 2 , we find x = 0.4 m 7. (a) ∑ F x = –GM 2 /L 2 (2+2(2) 1/2 ) = –4.84 GM 2 /L 2 ∑ F y = GM 2 /L 2 (6 + 2(2) 1/2 ) = +8.82 GM 2 /L 2 (b) ∑ F x = –GM 2 /L 2 (12 + 1.5/(2) 1/2 ) = –13.1 GM 2 /L 2 ∑ F y = –GM 2 /L 2 (6 + 1.5/(2) 1/2 ) = –7.06 GM 2 /L 2 8. (a) ∑ F x : –(6 + 7.5)GM 2 /L 2 = –13.5 GM 2 /L 2 ; ∑ F y : 15sin 60° GM 2 /L 2 = 13.0 GM 2 /L 2 (b) ∑ F x : (15 – 10)cos 60° GM 2 /L 2 = 2.5 GM 2 /L 2 ∑ F y : –25sin 60° GM 2 /L 2 = –21.7 GM 2 /L 2 9. Δg = GM (1/R 1 2 – 1/R 2 2 ) = 6.48 cm/s 2 10. If M is mass of original sphere, M/8 is mass of removed sphere. F = GmM/d 2 – GmM/8 (d – R/2) 2 . 11. (a) g’ = g o (T’/T o ) 2 = (9.810) (1440/1441) 2 = 9.796 m/s 2 (b) (R + h)R 2 = g’/g o thus h = 4.55 km. 12. g o = GM/R 2 , g = GM/(R + h) 2 . The equation g = g o /N leads to h = [(N) 1/2 – 1]R CHAPTER 13 13. T 2 = T 1 (g 1 /g 2 ) 1/2 = 2(6) 1/2 = 4.92 s....
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## This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch13 - Exercises 1. (a) a = Gm/r 2 = 6.67 × 10 –9...

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