Chapter 11 Gravity

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242 Chapter 11 Because MN = 17.1ME and MS = 333,000ME: F F g, N-U g, U-S = 5 3. 33× 10 M 17. 1M E E ⎛ ⎜⎛ T ⎜ ⎜ ⎝⎝ T N U ⎞ ⎟ ⎠ 2 3 Substitute numerical values and evaluate F g, N-U F g, U-S ⎞ −1⎟ ⎟ ⎠ F 2 g, N-U F g, U-S 17. 1 = ⎛ 5 N 3. 33 10 ⎜⎛ T ⎞ × ⎜ ⎜ ⎟ U ⎝⎝ T ⎠ 17. 1 −4 = ≈ 2× 10 2 3 2 ⎛ 5 164. 8 y ⎞ 3. 33 10 ⎜⎛ ⎞ × ⎜ ⎟ −1⎟ ⎜ 84. 0 y ⎟ ⎝⎝ ⎠ ⎠ : 2 3 ⎞ −1⎟ ⎟ ⎠ Because this ratio is so small, during the time at which Neptune is closest to Uranus, the force exerted on Uranus by Neptune is much less than the force exerted on Uranus by the Sun. 97 •• [SSM] Four identical planets are arranged in a square as shown in Figure 11-29. If the mass of each planet is M and the edge length of the square is a, what must be their speed if they are to orbit their common center under the influence of their mutual attraction? Picture the Problem Note that, due to the symmetrical arrangement of the planets, each experiences the same centripetal force. We can apply Newton’s second law to any of the four planets to relate its orbital speed to this net (centripetal) force acting on it. M a M ∑ r F F r = ma 1 M Applying radial radial the planets gives: θ θ F 2 F 1 to one of c 1 a M F = 2F cosθ + F 2 2
Substituting for F1, F2, and θ and 2 2GM GM F = cos45° + simplifying yields: c 2 a ( 2a) Because 2GM = 2 a GM = 2 a 2 2 ⎛ ⎜ ⎝ 1 GM + 2 2a 1 ⎞ 2 + ⎟ 2 ⎠ 2 2 Mv 2Mv Fc a 2 a = = 2 2 2Mv GM ⎛ 1 ⎞ = ⎜ 2 + 2 ⎟ a a ⎝ 2 ⎠ Solve for v to obtain: GM ⎛ 1 ⎞ v = ⎜1 + ⎟ = 1. 16 a ⎝ 2 2 ⎠ 2 2 Gravity 2 2 GM a 103 ••• In this problem you are to find the gravitational potential energy of the thin rod in Example 11-8 and a point particle of mass m0 that is on the x axis at x = x0. (a) Show that the potential energy shared by an element of the rod of mass 1 dm (shown in Figure 11-14) and the point particle of mass m0 located at x0 ≥ 2 L is given by Gm0dm GMm0 dU = − = dxs x0 − xs L( x0 − xs ) where U = 0 at x0 = ∞. (b) Integrate your result for Part (a) over the length of the rod to find the total potential energy for the system. Generalize your function U(x0) to any place on the x axis in the region x > L/2 by replacing x0 by a general coordinate x and write it as U(x). (c) Compute the force on m0 at a general point x using Fx = –dU/dx and compare your result with m0g, where g is the field at x0 calculated in Example 11-8. Picture the Problem Let U = 0 at x = ∞. The potential energy of an element of the stick dm and the point mass m0 is given by the definition of gravitational potential energy: = −Gm dm r where r is the separation of dm and m0. dU 0 (a) Express the potential energy of Gm0dm dU = − x − x the masses m0 and dm: 0 s The mass dm is proportional to the size of the element dxs : Substitute for dm and λ to express dm = λ dxs M where λ = . L Gm0λ dx dU = − dU in terms of xs: x − x L( x − x ) 0 s s = GMm − 0 0 dx s s 243
Page 1 and 2: Chapter 11 Gravity Conceptual Probl
Page 3 and 4: Squaring both sides of the equation
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Page 7 and 8: Gravity 16.7 d. Show that these dat
Page 9 and 10: Gravity Picture the Problem We can
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Page 19 and 20: Substitute for cosθ and simplify t
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Chapter 11 Gravity

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