chp13_practice_solutions

# chp13_practice_solutions - at the Earth’s core where it...

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CHP13: Practice Sheet SOLUTIONS 1. T 2. F 3. F 4. F 5. x 6. D 7. B 8. A 9. D 10. A 11. D 12. C 13. B 14. g Jupiter = G (300*M Earth )/(11*R Earth ) 2 = 300/(11) 2 (G*M Earth )/( R Earth ) 2 = (300/121) g Earth = 2.48 g Earth = 2.48 * 9.8 m/s 2 = 24.3 m/s 2 = the acceleration due to gravity on Jupiter. The force due to this gravity on a 1 kg object (that is, its weight) is: F g = m g Jupiter = (1 kg) * (24.3 m/s 2 ) = 24.3 N 15. g (640 km up) = G * M Earth / (1.1 R Earth ) 2 = (1/1.1) 2 * (G*M Earth )/( R Earth ) 2 = (1/1.1) 2 g Earth = .83 * 9.8 m/s 2 = 8.1 m/s 2 16. The object would fall from the surface of the Earth where acceleration is greatest (9.8 m/s 2 ) gaining speed but undergoing less and less acceleration until it was traveling fastest
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Unformatted text preview: at the Earth’s core where it has zero acceleration. It would then overshoot the core, travelling slower and slower as it experiences larger and larger acceleration (directed back towards the Earth’s center) until it stops at the far surface of the Earth. There it has maximum acceleration again (9.8 m/s 2 ) and it begins its return trip. The object would oscillate between the two surfaces of the Earth. Summing up: acceleration is greatest at the two surfaces of the Earth and least (zero) at the center of the Earth. The velocity is least (zero) at the two surfaces, and greatest at the core....
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