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Unformatted text preview: chaney (glc568) – Gravitational PE, Torque and Angular Momentum – murthy – (21118) 1 This print-out should have 37 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Energy is required to move a 520 kg mass from the Earth’s surface to an altitude h= 2 . 08 R E above the surface. The acceleration of gravity is 9 . 8 m / s 2 . The earth has the radius R E =6 . 37 × 10 6 m. What amount of energy is required to accom- plish this move? Hint: G and M E can be eliminated in favor of parameter of the gravitational acceleration at the surface of the earth; i.e. , g = 9 . 8 m / s 2 . The product GM E may be obtained from R 2 E , the gravitational acceleration of the earth’s surface. Correct answer: 2 . 19221 × 10 10 J. Explanation: Using the formulae U = − G mM E R and g = G M E R 2 E , where M E is the mass of the Earth and g is the gravitational acceleration at the Earth’s surface. We obtain that the change, Δ U , in the potential energy of the mass-Earth system is Δ U = ( U final − U initial ) = GmM E parenleftbigg 1 R initial − 1 R final parenrightbigg = GmM E parenleftbigg 1 R E − 1 R E + h parenrightbigg = mg R E parenleftbigg 1 − 1 1 + 2 . 08 parenrightbigg = 520 kg × 9 . 8 m / s 2 × 6 . 37 × 10 6 m × 2 . 08 1 + 2 . 08 = 2 . 19221 × 10 10 J . This is also the energy required to accomplish this move. 002 10.0 points Given: G = 6 . 67259 × 10 − 11 N m 2 / kg 2 A 1290 kg meteor comes in from outer space and impacts on the a moon’s surface (this moon is not necessarily the Earth’s moon). The mass of this moon is M = 5 × 10 22 kg, and the radius of this moon is R = 1 . 7 × 10 6 m. How much work is done by the Moon’s gravitational field? Correct answer: 2 . 53166 × 10 9 J. Explanation: Using the formula relating the work W done by the Moon’s gravitational field on the me- teor with the change Δ U in the potential energy of the meteor-Moon system, W = − Δ U = − ( U final − U initial ) = − parenleftbigg − GmM R − parenrightbigg , where M is the mass of the Moon, we obtain W = GmM R = (6 . 67259 × 10 − 11 N m 2 / kg 2 ) × (1290 kg) (5 × 10 22 kg) 1 . 7 × 10 6 m = 2 . 53166 × 10 9 J . 003 (part 1 of 2) 10.0 points G is the universal gravitational constant. g is the free fall acceleration at Earth’s surface. R E is the mean radius of the Earth. M E is the mass of the Earth. Very near the Earth’s surface, the gravita- tional attraction of the Earth on a body of mass m can be stated in terms of either g or G . Since potential energy is defined only up to an arbitrary constant, only changes in poten- tial energy are meaningful; therefore, when we talk about potential, we must give a reference value for the potential. Let U o be the gravi- tational potential energy set equal to zero at chaney (glc568) – Gravitational PE, Torque and Angular Momentum – murthy – (21118) 2 the surface of the Earth; let U ∞ be the grav-...
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This note was uploaded on 12/01/2008 for the course PHYSICS 160 taught by Professor Murthy during the Spring '08 term at Kentucky.

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solution_pdf4 - chaney(glc568 – Gravitational PE Torque...

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