Unassigned%20Problems%201

# Unassigned%20Problems%201 - Chapter 1 Problems 1 Light...

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Chapter 1 Problems 1. Light travels at about 300,000 kilometers in a second. (a) 500 / 300,000 = 0.0017 s. (b) 10,000 / 300,000 = 0.033 s. (c) 400,000 / 300,000 = 1.3 s. (d) 0.3 × 150,000,000 / 300,000 = 150 s. (e) 3.1 × 10 13 / 300,000 = 10 8 s = 3.3 yr. So, the correct answer is (d) The Moon. 2. (a) 1000 = 1x10 3 ; 0.000001 = 1x10 -6 ; 1001 = 1.001 × 10 3 ; 1,000,000,000,000,000 = 1x10 15 ; 123,000 = 1.23 × 10 5 ; 0.000456 = 4.56 × 10 -4 . (b) 3.16 × 10 7 = 31,600,000; 2.998 × 10 5 = 299,800; 6.67 × 10 -11 = 0.0000000000667; 2 × 10 0 = 2. (c) 2000.01; 333,000; 9.47 × 10 12 . 4. The year 10,000 A.D. is about 8,000 years from now. Compared to the precession period of 26,000 years, this is 0.31 of a complete circle or 111°. With 12 constellations in the zodiac, their spacing is approximately 30° apart. The 111° degrees converts to 3.7 constellations. Using Figure 1.12, if the vernal equinox is now just entering Aquarius, it will be well into Scorpio by 10,000 A.D. 5. (a) In 7 days, the Moon would complete one sidereal month. But because the Earth has moved in its orbit, the Moon will have to revolve a little further in order to get back to its original orientation with the Sun and Earth. In 7 days, the Earth has moved approximately 7°; the Moon will have to move that same amount in its orbit. The amount of time that will take is simply 7/360 = t/7 or 0.14 days. So, the month would be 7.14 days long. (b) The easiest way to visualize this is to start with the Sun, Earth, and Moon lined up in that order. In one quarter of a year, the Earth will have moved 90°. But the Moon will also have moved 90°. Relative to the Earth, the Moon is not changing its position; the order is still Sun, Earth, and Moon. So there would be no month; the Moon would remain at a constant phase. 6. Hint: Because the Moon appears to orbit the Earth in 29.5 days, it appears to move about 12° per day. 7. The Moon is 384,000 km away from Earth. Calculate the circumference of its orbit and divide by its period, 27.3 days, in seconds. 2 × π × 384,000 km / 27.3 days × 24 hr/day × 3600 s/hr = 1.02 km/s 8. Simple trigonometry gives this distance to be (250)tan(30°) = 144 m. Students may need to do this graphically, however. 11. Again, from More Precisely 1-4 we find that parallax = baseline × 206,000" / distance, where 57.3° has been converted to arc seconds. The baseline = 12,800 km and the light year = 9,500,000,000,000 km (Review Question 3.) The result is 0.000065".

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12. Assuming a 2 cm thumb at a distance of 75 cm gives 2
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Unassigned%20Problems%201 - Chapter 1 Problems 1 Light...

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