# CH13 - E x e r c ises 8(a ∑ F x ∑F y(b ∑ F x ∑F y...

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Unformatted text preview: E x e r c ises 8. (a) ∑ F x ∑F y : (b) ∑ F x ∑F y : – (6 + 7.5)GM 2 /L 2 = – 13.5 GM 2 /L 2 ; 1 5sin 60 ° G M 2 /L 2 = 1 3.0 GM 2 /L 2 : ( 15 – 10)cos 60 ° G M 2 /L 2 = 2 .5 GM 2 /L 2 : – 25sin 60 ° G M 2 /L 2 = – 21.7 GM 2 /L 2 10. If M is mass of original sphere, M/8 is mass of removed s phere. F = GmM/d 2 – G mM/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 t hus h = 4.55 km. 14. F = GmM[1/(r – a) 2 – 1 (r + a) 2 ] = 4GmMra/(r 2 – a 2 ) If r >>a, then F ≈ 4 GmMa/r 3 2 26. (a) E = –GmM/2a; where 2a = r A + r P . Thus e = –3.38 × 10 8 J (b) T 2 = κ a 3 ; thus T = 125 min (c) From Eq. 13.73 : v P 2 = ( GM/1)(r A /r P ), thus v P = 8 .49 km/s 27. since L = (3) 1 /2 r; F = (2Gm 2 /L 2 ) ((3) 1 /2 /2) = Gm 2 /r 2 ( 3) 1 /2 ; Also F = m ω 2 r ; thus ω 2 = G m/r 3 ( 3) 1 /2 Problems 4. (a) F r = – GmMr/R 3 (b) U(r) – U(R) = (GmM/R 3 ) ∫ r dr – (GmM/2R)(r 2 /R 2 – 1 ) But U (R) = –GmM/R, thus 1 2. (a) E = 1/2 m(v r 2 + v θ 2 ) – GmM/r L = m v θ r, so v θ 2 = ( L/mr) 2 (b) v r = 0 at these points 13. dM = ρ dv = [ ρ o ( 1 – r/2R)](4 π r 2 d r) Thus M(r) = ∫ d M = 4 π ρ o r 3 ( 1/3 – 4/8R) g(r) = GM(r) /r 2 = 4 π ρ o G (r/3 – r 2 /8R) 15. m 1 ω 2 r 1 = m 2 ω 2 r 2 = G m 1 m 2 /(r 1 + r 2 ) 2 t hus GM : r 2 = m 1 (r 1 + r 2 )/(m 1 + m 2 ) and r 1 = m 2 (r 1 + r 2 )/(m 1 + m 2 ) ω 2 = G m 1 /r 2 ( r 1 + r 2 )/G(m 1 + m 2 )/(r 1 + r 2 ) 3 T 2 = 4 π 2 ( r 1 + r 2 ) 3 /G(m 1 + m 2 ) 1 7. From the law of sines : Sin θ /m ω 2 r = sin( λ + θ )/mg r = R cos λ, and if λ >> θ , then sin( λ + θ ) = sin λ Thus, sin θ = ω 2 R sin2 λ /2g ...
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CH13 - E x e r c ises 8(a ∑ F x ∑F y(b ∑ F x ∑F y...

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