HW07-solutions - markowitz(am45362 – HW07 – distler...

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Unformatted text preview: markowitz (am45362) – HW07 – distler – (56295) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the escape speed for a rocket leaving the moon. The acceleration of gravity on the moon is 0 . 166 times that on earth and the moon’s radius is 0 . 273 R E . Correct answer: 2 . 38426 km / s. Explanation: Let : g m = 0 . 1666 g E and R m = 0 . 273 R E . The escape speed from the moon is v e.m = radicalbigg 2 GM m R m = radicalbig 2 g m R m (1) and the escape speed from Earth is v e.E = radicalBigg 2 GM E R E = radicalbig 2 g E R E . (2) Dividing, v e.m v e.E = radicalBigg g m R m g E R E v e.m = radicalBigg g m R m g E R E v e.E = radicalbig (0 . 166) (0 . 273)(11 . 2 km / s) = 2 . 38426 km / s . 002 10.0 points An object is projected upward from the sur- face of the earth with an initial speed of 5 km / s. The acceleration of gravity is 9 . 81 m / s 2 . Find the maximum height it reaches. Correct answer: 1592 . 83 km. Explanation: Let : R E = 6 . 37 × 10 6 m , v i = 5 km / s = 5000 m / s , and g = 9 . 81 m / s 2 . g = GM E R 2 E and the potential energy at a distance r from the surface of the earth is U ( r ) =- GM E m r Using conservation of energy to relate the initial potential energy of the system to its energy at its maximum height ( K f = 0), K f- K i + U f- U i = 0- K ( R E ) + U ( R E + h )- U ( R E ) = 0- 1 2 mv 2 i + GM E m R E + h- GM E m R E = 0 1 2 v 2 i = GM E h R E ( R E + h ) · R E R E v 2 i = 2 GM E R 2 E · hR E R E + h v 2 i = 2 g hR E R E + h ( R E + h ) v i 2 = 2 g hR E h =- v 2 i R E v 2 i- 2 g R E Since v 2 i- 2 g R E = (5000 m / s) 2- bracketleftBig 2 (9 . 81 m / s 2 ) × (6 . 37 × 10 6 m) bracketrightBig =- 9 . 99794 × 10 7 , h =- (5000 m / s) 2 (6 . 37 × 10 6 m)- 9 . 99794 × 10 7 · 1 km 1000 m = 1592 . 83 km . 003 (part 1 of 3) 10.0 points markowitz (am45362) – HW07 – distler – (56295) 2 A 2 g rock on a string is whirled fast enough (in a clockwise direction) to move in a vertical circle as shown. The rock has only enough velocity at the top of the loop to keep a very small (negligible) tension in the string. 2 g 2 m 7 3 ◦ ω center The angle 73 ◦ is measured from the vertical axis when the rock is at its lowest position ( i.e. , measured from the negative y axis in a counter-clockwise direction). What is the magnitude of the centripetal acceleration when θ = 73 ◦ ? Correct answer: 35 . 1305 m / s 2 . Explanation: Let : θ = 73 ◦ , R = 2 m , and m = 2 g . The mass m of the rock is not required and the length R of the string is not required. The free body diagram is T mg F net The tension along the string must point toward the center of the circle and the gravi- tational force is down, therefore we have summationdisplay F r : T- mg cos θ = ma r summationdisplay F θ : mg sin θ = ma θ ....
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This note was uploaded on 11/29/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas.

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HW07-solutions - markowitz(am45362 – HW07 – distler...

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