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mat-sln-asn-pt7-tst4-sum05

mat-sln-asn-pt7-tst4-sum05 - Solution for Practice Test 7...

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Solution for Practice Test 7 for Test 4 Solution to Practice Test Problem 7.1() Problem: (a) NSF exam # NSF point total (b) Grading Key: 45 Points Grading Key: 25 Points Total Points for Problem: 45 Points Solution to Practice Test Problem 7.2() Problem: You are travelling across the solar system in your spaceship. As you are nearing a planet, the ship’s engine sputters and dies, and you end up in circular orbit around the planet. You try several times to restart the engine, each time altering your orbit, until finally you decide to try something else. You start reading the ship’s log looking for ideas, and you find the following data for each orbit: Height Speed 1 × 10 7 m 6 × 10 4 m / s 1 . 5 × 10 7 m 5 × 10 4 m / s 2 × 10 7 m 4 × 10 4 m / s 2 . 5 × 10 7 m 3 . 5 × 10 4 m / s 3 × 10 7 m 3 × 10 4 m / s Assume that “Height” measures the distance from the center of the planet, and that each orbit is a stable, uniform circular orbit. (a)Independantly of the data, derive an equation relating v 2 to r (that is, relating the square of the orbital speed to the radial distance from the planet’s center). (b)Make a table of v 2 and 1 /r values. v 2 1 r
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(c)Graph v 2 vs. 1 /r . Label your axes with units. Draw a best-fit line. (d)Calculate the slope of your best-fit line. Unclude units. (e)Comparing your equation from part (a) to your graph in (c), give one reason related to the ship’s instru- ments why the data doesn’t go through the point (0,0). What type of error does this suggest? (f)Using the graph and techniques practiced in lab, decide which planet you are orbiting. Planet Mass Jupiter 1 . 9 × 10 27 kg Saturn 5 . 69 × 10 26 kg Uranus 8 . 68 × 10 25 kg (g)Are you confident with your choice? Do a percent deviation analysis. Solution to Part (a) The only force on the spacecraft ( m ) is gravity due to the planet ( M ), so Newton II for the spacecraft reads | Σ vector F | = ma = GMm r 2 Assuming a circular orbit, the acceleration is centripetal, so a = v 2 /r , m parenleftbigg v 2 r parenrightbigg = GMm r 2 v 2 = GM r Grading Key: Part (a) 2 Points 1 point(s) : Correct answer. 1 point(s) : Reasonable derivation. Solution to Part (b)
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v 2 1 r 3 . 6 × 10 9 m 2 / s 2 10 . 0 × 10 - 8 m - 1 2 . 5 × 10 9 m 2 / s 2 6 . 7 × 10 - 8 m - 1 1 . 6 × 10 9 m 2 / s 2 5 . 0 × 10 - 8 m - 1 1 . 2 × 10 9 m 2 / s 2 4 . 0 × 10 - 8 m - 1 0 . 9 × 10 9 m 2 / s 2 3 . 3 × 10 - 8 m - 1 Grading Key: Part (b) 2 Points 1 point(s) (2 times) : Two middle values correct.
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