Chapter 12 Hmwk 8 Solutions

Chapter 12 Hmwk 8 Solutions - Solutions to Homework #8...

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Unformatted text preview: Solutions to Homework #8 Chapter 12 8. Picture the Problem : Both the Earth and the Sun exert attractive gravitational forces on the Moon. The three objects are arranged as shown. Strategy: Use the Universal Law of Gravity (equation 12- 1) to find the components of the force acting on the Moon. The Earth exerts a downward force on the Moon and the Sun exerts a force toward the left. The net force is therefore at an angle below the line that connects the Moon to the Sun. Solution: 1. Use equation 12- 1 to find S F : ( 29 ( 29 ( 29 ( 29 22 30 11 2 2 20 M S S 2 2 11 M-S 7.35 10 kg 2.00 10 kg 6.67 10 N m /kg 4.36 10 N 1.50 10 m M M F G r- = = = 2. Now find E F : ( 29 ( 29 ( 29 ( 29 22 24 11 2 2 20 M E E 2 2 11 M-E 7.35 10 kg 5.97 10 kg 6.67 10 N m /kg 1.98 10 N 3.84 10 m M M F G r- = = = 3. Add the components: ( 29 ( 29 2 2 2 2 20 20 22 4.36 10 N 1.98 10 N 4.79 10 N x y F F F = + = + = 4. Find the direction : 20 1 1 20 1.98 10 N tan tan 4.36 10 N 24.4 toward the Earth off the line from the Moon to the Sun y x F F -- = = = Insight: Note that the Sun exerts a force on the Moon (4.3610 20 N) that is 2.2 times larger than the force the Earth exerts on the Moon (1.9810 20 N). 20. Picture the Problem : The volcano on Io spews material at high speed straight upward. The mass slows down, rising to a height of 5.00 km before coming to rest momentarily under the influence of Ios gravitation. Strategy: Use conservation of energy to relate the initial kinetic energy of the ejected material to its potential energy at the maximum altitude. This relation will allow the calculation of the surface gravity of Io. Then use equation 12-4 and the given radius of Io to find the mass of Io. Solution: 1. (a) 2 2 1 i f 2 Use to find , and use to find . mv mgh g g GM R M = = 2. (b) Set i f E E = and solve for g : ( 29 ( 29 2 1 i f 2 2 2 2 i 3 f 134 m/s 1.80 m/s 2 2 5.00 10 m mv mgh v g h = = = = 3. (c) Solve equation 12-4 for M : ( 29 ( 29 2 2 6 2 22 11 2 2 1.80 m/s 1.82 10 m 8.94 10 kg 6.67 10 N m /kg gR M G- = = = Insight: The use of conservation of energy to find g is equivalent to solve equations 4-6, 2 2 2 v v g y =- , for g . This approach assumes that g is constant over the 5.00 km that the ejected material rises. This is a pretty good assumption, because you will get the same answer to 3 significant figures if you apply the more exact method of section 12-5. The mass of Io according to NASAs Solar System Exploration web site is 8.931610 22 kg. 28. Picture the Problem : The tiny moon Dactyl travels around 243 Ida in an approximately circular orbit. Strategy: Solve Keplers Third Law (equation 12-7) for the mass of 243 Ida, using the orbit distance and period given in the problem....
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Chapter 12 Hmwk 8 Solutions - Solutions to Homework #8...

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