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solutions_for_chapter_12

# solutions_for_chapter_12 - 11 Picture the Problem The three...

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11. Picture the Problem : The three masses are arranged at the vertices of an equilateral triangle as shown in the figure at right. Strategy: Each mass will be gravitationally attracted to the other two masses. The vector sum of the forces will be along a line toward the midpoint of the other two masses, due to symmetry. An examination of the geometry reveals that this force equals 2 cos30 F ° , where F is the gravitational force between any two masses. Use this relation together with equation 12-1 to find the force exerted on one of the masses. Solution: 1. (a) Find 2 cos30 F ° , where 2 2 F Gm r = from equation 12-1: ( 29 ( 29 2 2 11 2 2 2 9 6.75 kg N m 2 cos30 2 6.67 10 kg 1.25 m 3.37 10 N m F G r - - × = × ° = × ÷ = × 2. (b) If the side lengths are doubled the gravitational forces will be reduced by a factor of four. Insight: Doubling the masses will quadruple the gravitational force, because in this case the gravitational force depends upon the square of the mass of each object and the square of the separation distance. 20. Picture the Problem : Both the Earth and the Moon exert a force on other masses in their vicinity according to Newton’s Universal Law of Gravitation. Strategy: Use a ratio of gravitational forces to find the mass of the Moon in terms of the mass of the Earth. Note that if you solve Newton’s Universal Law of Gravitation (equation 12-1) for 2 m you find 2 2 1 . m F r Gm = Solution: Determine the ratio M E m m by solving equation 12-1 for 2 m : ( 29 ( 29 2 2 2 1 1 E E 6 4 M M M 1 M M M 2 2 2 E E E 1 E E E E 1 96 F r m F r Gm F r m m F r Gm F r F r = = = = = Insight: Using a ratio can be a powerful tool for solving a question like this one, where few hard data are given, only relationships between various quantities.

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