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Unformatted text preview: PHYS 183 - HW 1 Solutions Total points - 50 1 - 1.2
Brieﬂy describe the major levels of structure (such as planet, star, galaxy) in the universe. The largest scale is the universe itself, which is the sum total of all matter and energy. The largest-known organized structures are superclusters of galaxies (clusters of clusters!), then clusters of galaxies and groups of galaxies. Next come the individual galaxies (roughly 100 billion). Each galaxy contains billions of stars, and many or most stars may be orbited by planets. 7 points: -1 for each of the levels that is left out 2 - 1.4
What did Carl Sagan mean when he said that we are “star stuﬀ”? Most of the atoms in our bodies were made by stars well after the Big Bang. Atoms heavier than hydrogen and helium are formed in stars through nuclear fusion and then distributed throughout the universe when dying stars explode (supernova). So most of the stuﬀ in our bodies was once part of stars. 5 points 3 - 1.25
Which of the following correctly lists our “cosmic address” from small to large? 1 (a) Earth, solar system, Milky Way Galaxy, Local Group, Local Supercluster, universe. +one explanation sentence 3 points: 1 for correct answer, 2 for short explanation 4 - 1.26
When we say the universe is expanding, we mean that: (b) The average distance between galaxies is growing with time. +one explanation sentence 3 points: 1 for correct answer, 2 for short explanation 5 - 1.29
The star Betelgeuse is about 425 light-years away. If it explodes tonight: (c) We won’t know about it until 425 years from now. +one explanation sentence 3 points: 1 for correct answer, 2 for short explanation 6 - 1.33
An astronomical unit is: (b) Earth’s average distance from the Sun. +one explanation sentence 3 points: 1 for correct answer, 2 for short explanation 7 - 1.34
The fact that nearly all galaxies are moving away from us, with more distant ones moving faster, helped us to conclude that: 2 (a) The universe is expanding. +one explanation sentence 3 points: 1 for correct answer, 2 for short explanation 8 - 1.41
Raisin Cake Universe: Suppose that all the raisins in a cake are 1 cm apart before baking and 4 cm apart after baking: (a) Draw diagrams to represent the cake before and after baking. (b) Identify one raisin as the Local Raisin on your diagrams. Construct a table showing the distances and speeds of the other raisins as seen from the Local Raisin. Make sure the two diagrams are both to scale (relative to one another) and well labeled. The speeds will diﬀer between students depending on the baking time used and raisin conﬁguration. Example for a grid diagram and a 1 hour bake time (pretending that the unit spacing before baking is 1 cm!): Figure 1: Before Baking Table 1: Distance and Speed seen from the Local Raisin Raisin Before (cm) After (cm) Speed (cm/hr) 1 1 4 3 2 1.4 5.6 4.2 3 2 8 6 4 2.24 8.9 6.7 3 Figure 2: After baking. (c) Brieﬂy explain how your expanding cake is similar to the expansion of the universe. Explanation must makes sense and maybe talk about how they both expand uniformly or how the speed at which other raisins/structures are expanding is relative to which raisin/structure is designated as the local or starting point. 9 points - 3 for each part 4 9 - 1.46
We use radio waves, which travel at the speed of light, to communicate with robotic spacecraft. d d v= ⇒t= t v Speed of light − c = 2.998 × 108 m/s (a) How long does it take a message to travel from Earth to a spacecraft on Mars at its closest to Earth (about 56 million km)? (56 × 106 km)(103 m/km) 2.998 × 108 m/s = 187 seconds = 3.11 minutes t= (b) How long does it take a message to travel from Earth to a spacecraft on Mars at its farthest from Earth (about 400 million km)? (400 × 106 km)(103 m/km) 2.998 × 108 m/s = 1330 seconds = 22.2 minutes t= (c) How long does it take a message to travel from Earth to Pluto at its average distance from Earth (about 6 billion km)? (6 × 109 km)(103 m/km) 2.998 × 108 m/s = 2.00 × 104 seconds = 334 minutes = 5.56 hours t= 6 points - 2 for each part 10 - 1.53
Galactic Speed. We are located about 28000 light-years (ly) from the galactic center and we orbit the center once every 230 million years. How fast are we traveling 5 around the galaxy? Give your answer in km/hr. (Dont do mi/hr). Keep 3 signiﬁcant ﬁgures. Since everything is set out in the problem, all steps should be shown to get full points. (i.e. not just the numerical answers). (a) Assume that the orbit is a circle. The radius of this circle is then R = 28000 ly. Show that the circumference of this circle is S = 1.76 ×105 ly. The circumference of a circle is 2π times the radius of the circle: S = 2π R = 2π (28000 ly) = 1.76 × 105 ly (b) The orbital period is T = 2.3 ×108 year. The orbital speed is then given by v= Show that v = 7.65 × 10−4 c where c is the speed of light. Hint: 1 ly = c × year. v= = S T (1.76 × 105 ly)( c x yr ) ly (2) distance traveled S = time elapsed T (1) 2.3 × 108 yrs = 7.65 × 10−4 c (c) Use the speed of light c = 2.998 ×108 m/s = 2.998 ×105 km/s and calculate v in km/s. 2.998 × 105 km/s c v = (7.65 × 10 = 229 km/s −4 c) (d) Use the fact that 1 hr = 3600 s to convert your calculation to km/hr. 6 229 km 3600 s v= s hr = 824000 km/hr = 8.24 × 105 km/hr If more than 3 signiﬁcant ﬁgures kept throughout the problem the ﬁnal answer, to 3 sig ﬁgs, will be 8.26 × 105 km/hr. 8 points - 2 for each part BONUS
(a) Google Map tells me that the distance to Toronto is 539 km and it takes 5 hours 35 mins to get there. Calculate the speed in km/hr. Hint: Convert 5 hours 35 mins to minutes ﬁrst. How many signiﬁcant ﬁgures do you have? Then calculate the speed in km/min. Then convert it to km/hr. Remember that the relationship 1 hr = 60 min is exact. That is, the numbers 60 is an integer with an inﬁnite number of signiﬁcant digits. time = (5 hrs × 60 mins/hr) + 35 mins = 335 mins Keep all digits since 60 minutes to the hour is exact and the time is measured to the nearest minute. distance time 539 km = 1.61 km/min (3 digits since it is a division) 335 min 539 km 60 min 335 min hr 96.5 km/hr (with no rounding for the km/min calculation) 96.6 km/hr (if you used the 1.61 km/min value) v= = = = = 2 points (b) Alice is actually measuring the distance between her house in Montreal and a friends house in Toronto. This is how it went: From her house to the highway entrance, it was 2.2 km. From the highway entrance to the Yonge street exit in 7 Toronto, it was 5.31 ? 102 km. From the Yonge street exit to her friends house, it took 1.2 ? 10 km. How should Alice report the total distance? distance = = = 1 point 2.2 km + 5.31 × 102 km + 1.2 × 10 km 2.2 km + 531 km + 12 km 545.2 km 545 km (to the closest km since that is the smallest decimal place) 8 ...
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- Spring '10