www.elsolucionario.net CHAPTER 1: Introduction, Measurement, Estimating Answers to Questions (a) Fundamental standards should be accessible, invariable, indestructible, and reproducible A particular person’s foot would not be very accessible, since the person could not be at more than one place at a time The standard would be somewhat invariable if the person were an adult, but even then, due to swelling or injury, the length of the standard foot could change The standard would not be indestructible – the foot would not last forever The standard could be reproducible – tracings or plaster casts could be made as secondary standards (b) If any person’s foot were to be used as a standard, “standard” would vary significantly depending on the person whose foot happened to be used most recently for a measurement The standard would be very accessible, because wherever a measurement was needed, it would be very easy to find someone with feet The standard would be extremely variable – perhaps by a factor of That also renders the standard as not reproducible, because there could be many reproductions that were quite different from each other The standard would be almost indestructible in that there is essentially a limitless supply of feet to be used There are various ways to alter the signs The number of meters could be expressed in one significant figure, as “900 m (3000 ft)” Or, the number of feet could be expressed with the same precision as the number of meters, as “914 m (2999 ft)” The signs could also be moved to different locations, where the number of meters was more exact For example, if a sign was placed where the elevation was really 1000 m to the nearest meter, then the sign could read “1000 m (3280 ft)” Including more digits in an answer does not necessarily increase its accuracy The accuracy of an answer is determined by the accuracy of the physical measurement on which the answer is based If you draw a circle, measure its diameter to be 168 mm and its circumference to be 527 mm, their quotient, representing , is 3.136904762 The last seven digits are meaningless – they imply a greater accuracy than is possible with the measurements The problem is that the precision of the two measurements are quite different It would be more appropriate to give the metric distance as 11 km, so that the numbers are given to about the same precision (nearest mile or nearest km) A measurement must be measured against a scale, and the units provide that scale Units must be specified or the answer is meaningless – the answer could mean a variety of quantities, and could be interpreted in a variety of ways Some units are understood, such as when you ask someone how old they are You assume their answer is in years But if you ask someone how long it will be until they are done with their task, and they answer “five”, does that mean five minutes or five hours or five days? If you are in an international airport, and you ask the price of some object, what does the answer “ten” mean? Ten dollars, or ten pounds, or ten marks, or ten euros? If the jar is rectangular, for example, you could count the number of marbles along each dimension, and then multiply those three numbers together for an estimate of the total number of marbles If the jar is cylindrical, you could count the marbles in one cross section, and then multiply by the number of layers of marbles Another approach would be to estimate the volume of one marble If we assume that the marbles are stacked such that their centers are all on vertical and horizontal lines, then each marble would require a cube of edge 2R, or a volume of 8R3, where R is the radius of a marble The number of marbles would then be the volume of the container divided by 8R3 © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher www.elsolucionario.net Chapter Introduction, Measurement, Estimating The result should be written as 8.32 cm The factor of used to convert radius to diameter is exact – it has no uncertainty, and so does not change the number of significant figures sin 30.0o Since the size of large eggs can vary by 10%, the random large egg used in a recipe has a size with an uncertainty of about 5% Thus the amount of the other ingredients can also vary by about 5% and not adversely affect the recipe 0.500 10 In estimating the number of car mechanics, the assumptions and estimates needed are: the population of the city the number of cars per person in the city the number of cars that a mechanic can repair in a day the number of days that a mechanic works in a year the number of times that a car is taken to a mechanic, per year We estimate that there is car for every people, that a mechanic can repair cars per day, that a mechanic works 250 days a year, and that a car needs to be repaired twice per year (a) For San Francisco, we estimate the population at one million people The number of mechanics is found by the following calculation 106 people car people repairs yr year car 250 workdays mechanic repairs workday 1300 mechanics (b) For Upland, Indiana, the population is about 4000 The number of mechanics is found by a similar calculation, and would be mechanics There are actually two repair shops in Upland, employing a total of mechanics Solutions to Problems (a) 14 billion years (b) 1.4 1010 years 1.4 1010 y 3.156 107 s y (a) 214 significant figures (b) 81.60 significant figures (c) 7.03 (d) 0.03 4.4 1017 s significant figures significant figure (e) 0.0086 significant figures (f) 3236 significant figures (g) 8700 significant figures © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli (a) 1.156 1.156 100 (b) 21.8 2.18 101 (c) 0.0068 2.7635 101 (d) 27.635 (e) 0.219 (f) 444 2.19 10 (a) 8.69 10 (c) 8.8 10 86, 900 9,100 0.88 (d) 4.76 10 (e) 3.62 10 476 0.0000362 The uncertainty is taken to be 0.01 m 0.01 m % uncertainty 100% 1.57 m 0.25 m % uncertainty (a) % uncertainty 3.76 m (b) % uncertainty (c) 4.44 102 (b) 9.1 103 6.8 10 % uncertainty 100% 0.2 s 5s 0.2 s 50 s 0.2 s 300 s 1% 6.6% 100% 4% 100% 0.4% 100% 0.07% To add values with significant figures, adjust all values to be added so that their exponents are all the same 9.2 103 s 8.3 10 s 0.008 106 s 9.2 103 s 83 103 s 103 s 9.2 83 103 s 100 103 s 1.00 105 s When adding, keep the least accurate value, and so keep to the “ones” place in the parentheses 2.079 10 m 0.082 10 1.7 m When multiplying, the result should have as many digits as the number with the least number of significant digits used in the calculation 10 To find the approximate uncertainty in the area, calculate the area for the specified radius, the minimum radius, and the maximum radius Subtract the extreme areas The uncertainty in the area is then half this variation in area The uncertainty in the radius is assumed to be 0.1 104 cm © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher www.elsolucionario.net Chapter Introduction, Measurement, Estimating rspecified Aspecified 3.8 10 cm 4.5 109 cm Amin rmin 3.7 104 cm 4.30 109 cm Amax rmax 3.9 10 cm 4.78 109 cm A Amax Amin 4.78 109 cm Thus the area should be quoted as A 4.30 109 cm 4.5 0.2 0.24 109 cm 109 cm 11 To find the approximate uncertainty in the volume, calculate the volume for the specified radius, the minimum radius, and the maximum radius Subtract the extreme volumes The uncertainty in the volume is then half this variation in volume Vspecified rspecified Vmin 3 rmin Vmax 3 rmax V 4 2.77 m The percent uncertainty is 9.80 101 m 3 8.903 101 m 3 10.754 101 m 2.95 m Vmax Vmin 2.86 m 10.754 101 m 8.903 101 m 0.923 101 m V 9.80 101 m Vspecified 100 0.926 101 m 0.09444 9% 286.6 10 m 0.286 m 85 10 V 0.000 085 V 760 mg 760 10 kg 0.000 760 kg (if last zero is significant) (d) 60.0 ps 60.0 10 12 s 0.000 000 000 0600 s (e) 22.5 fm 22.5 10 15 m 0.000 000 000 000 022 m (f) 2.50 gigavolts 2.5 109 volts 2, 500, 000, 000 volts 12 (a) 286.6 mm (b) 85 V (c) 13 (a) 106 volts (b) 10 meters (c) 10 days megavolt Mvolt micrometers kilodays m kdays (d) 18 102 bucks 18 hectobucks (e) 10 pieces nanopieces 18 hbucks npieces 14 (a) Assuming a height of feet 10 inches, then '10" (b) Assuming a weight of 165 lbs, then 165 lbs 70 in m 39.37 in 0.456 kg lb 1.8 m 75.2 kg Technically, pounds and mass measure two separate properties To make this conversion, we have to assume that we are at a location where the acceleration due to gravity is 9.8 m/s2 © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli 15 (a) 93 million miles 93 10 miles 1610 m mile 11 (b) 1.5 10 m 150 10 m 16 (a) ft (b) m ft m2 150 gigameters or 1.5 1011 m yd ft 1.5 1011 m 0.15 1012 m 0.15 terameters 0.111 yd 2 3.28 ft m 10.8 ft 17 Use the speed of the airplane to convert the travel distance into a time 1h 3600 s 1.00 km 3.8 s 950 km 1h 18 (a) 1.0 10 (b) 1.0 cm 10 m 1.0 10 1m 10 m 39.37 in m atom 100 cm 1.0 10 10 3.9 10 in 1.0 108 atoms m 19 To add values with significant figures, adjust all values to be added so that their units are all the same 1.80 m 142.5 cm 5.34 105 m 1.80 m 1.425 m 0.534 m 3.759 m 3.76 m When adding, the final result is to be no more accurate than the least accurate number used In this case, that is the first measurement, which is accurate to the hundredths place 20 (a) 1k h (b) 1m s (c) 1km h 0.621 mi 0.621mi h km 3.28 ft 1m 3.28 ft s 1000 m 1h km 3600 s 0.278 m s 21 One mile is 1.61 103 m It is 110 m longer than a 1500-m race The percentage difference is 110 m 100% 7.3% 1500 m 22 (a) 1.00 ly 2.998 108 m s 3.156 10 s 9.462 1015 m (b) 1.00 ly (c) 2.998 108 m s 9.46 1015 m AU 6.31 104 AU 11 1.00 ly 1.50 10 m AU 11 3600 s 1.50 10 m 7.20 AU h hr © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher www.elsolucionario.net Chapter Introduction, Measurement, Estimating r2 23 The surface area of a sphere is found by A (a) (b) DMoon AMoon DEarth DEarth AMoon DMoon DMoon 2.8 103 (b) 86.30 10 (c) AEarth 24 (a) 2800 0.0076 REarth d2 6.38 106 m 1.74 106 m 13.4 103 10 103 10 10 1.5 109 d 3.80 1013 m RMoon 103 8.630 103 7.6 10 (d) 15.0 108 3.48 106 m 10 109 104 109 25 The textbook is approximately 20 cm deep and cm wide With books on both sides of a shelf, with a little extra space, the shelf would need to be about 50 cm deep If the aisle is 1.5 meter wide, then about 1/4 of the floor space is covered by shelving The number of books on a single shelf level is then 3500 m book 0.25 m 8.75 10 books With shelves of books, the total number 0.04 m of books stored is as follows books 8.75 10 shelf level shelves 105 books 26 The distance across the United States is about 3000 miles 3000 mi km 0.621 mi hr 10 km 500 hr Of course, it would take more time on the clock for the runner to run across the U.S The runner could obviously not run for 500 hours non-stop If they could run for hours a day, then it would take about 100 days for them to cross the country 27 An NCAA-regulation football field is 360 feet long (including the end zones) and 160 feet wide, which is about 110 meters by 50 meters, or 5,500 m2 The mower has a cutting width of 0.5 meters Thus the distance to be walked is Area 5500 m d 11000 m 11 km width 0.5 m At a speed of km/hr, then it will take about 11 h to mow the field 28 A commonly accepted measure is that a person should drink eight 8-oz glasses of water each day That is about quarts, or liters of water per day Then approximate the lifetime as 70 years 70 y 365 d y L d 104 L 29 Consider the body to be a cylinder, about 170 cm tall, and about 12 cm in cross-sectional radius (a 30-inch waist) The volume of a cylinder is given by the area of the cross section times the height V r 2h 12 cm 170 cm 10 cm 10 cm © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli 30 Estimate one side of a house to be about 40 feet long, and about 10 feet high Then the wall area of that particular wall is 400 ft2 There would perhaps be windows in that wall, each about ft wide and feet tall, so 12 ft2 per window, or about 50 ft2 of window per wall Thus the percentage of wall 50 ft 100 12.5% Thus a rough estimate would be 10% 15% of area that is window area is 400 ft the house’s outside wall area 31 Assume that the tires last for years, and so there is a tread wearing of 0.2 cm/year Assume the average tire has a radius of 40 cm, and a width of 10 cm Thus the volume of rubber that is becoming pollution each year from one tire is the surface area of the tire, times the thickness per year that is wearing Also assume that there are 150,000,000 automobiles in the country – approximately one automobile for every two people So the mass wear per year is given by Mass Surface area Thickness wear density of rubber # of tires year tire year 0.4 m 0.1 m 0.002 m y 1200 kg m 600, 000, 000 tires 108 kg y At 32 For the equation v Bt , the units of At must be the same as the units of v So the units of A must be the same as the units of v t , which would be distance time Also, the units of Bt must be the same as the units of v So the units of B must be the same as the units of v t , which would be distance time 33 (a) The quantity vt has units of m s s for x The quantity 2at has units m s m s , which not match with the units of meters s m s , which also not match with the units of meters for x Thus this equation cannot be correct (b) The quantity v0 t has units of m s s m , and 2 at has units of m s s2 m Thus, s2 m Thus, since each term has units of meters, this equation can be correct (c) The quantity v0 t has units of m s s m , and 2at has units of m s since each term has units of meters, this equation can be correct 34 The percentage accuracy is 2m 100% 10 % The distance of 20,000,000 m needs to 10 m be distinguishable from 20,000,002 m, which means that significant figures are needed in the distance measurements 35 Multiply the number of chips per wafer times the number of wafers that can be made fro a cylinder 100 chips wafer 300 mm wafer 0.60 mm cylinder 50, 000 chips cylinder © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher www.elsolucionario.net Chapter Introduction, Measurement, Estimating 1.00 y 1.00 y (b) # of nanoseconds in 1.00 y: 1.00 y 1.00 y (c) # of years in 1.00 s: 1.00 s 36 (a) # of seconds in 1.00 y: 1.00 s 3.156 107 s 3.16 107 s 1y 3.156 107 s 109 ns 1y 1s 1y 3.16 1016 ns 3.17 10 y 3.156 10 s 37 Assume that the alveoli are spherical, and that the volume of a typical human lung is about liters, which is 002 m3 The diameter can be found from the volume of a sphere, 43 r r3 d d3 108 38 hectare 39 (a) (b) (c) (d) d3 10 m hectare 10 d 3.28 ft hectare 1m 10 15 kg proton or neutron bacterium 27 10 17 10 kg acre 104 ft proton or neutron DNA molecule 10 10 kg proton or neutron human 27 10 kg 41 10 kg proton or neutron galaxy 27 10 kg 2.69 acres 1010 protons or neutrons kg 2 10 m 1012 protons or neutrons kg 27 m3 108 104 m 1/ 3 10 29 protons or neutrons 1068 protons or neutrons 40 There are about 300,000,000 people in the United States Assume that half of them have cars, that they each drive 12,000 miles per year, and their cars get 20 miles per gallon of gasoline automobile 12, 000 mi gallon 108 people 1011 gallons y people 1y 20 mi 41 Approximate the gumball machine as a rectangular box with a square cross-sectional area In counting gumballs across the bottom, there are about 10 in a row Thus we estimate that one layer contains about 100 gumballs In counting vertically, we see that there are bout 15 rows Thus we estimate that there are about 1500 gumballs in the machine 42 The volume of water used by the people can be calculated as follows: 4 10 people 1200 L day 365 day 1000 cm km people 1y 1L 105 cm 4.4 10 km y © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli The depth of water is found by dividing the volume by the area V 4.4 10 km y 105 cm km d 8.76 10 8.76 cm y A 50 km y km 43 The volume of a sphere is given by V V 1ft 2000 lb 1T 1T 186 lb Then the radius is found by d 2r 3V 1/ cm y r For our 1-ton rock, we can calculate the volume to be 10.8 ft 10.8 ft 1/ 2.74 ft ft 44 To calculate the mass of water, we need to find the volume of water, and then convert the volume to mass 10 km 105 cm 1.0 cm km 10 kg cm ton 105 ton 10 kg To find the number of gallons, convert the volume to gallons 10 km 105 cm km 1.0 cm 1L 10 cm gal 3.78 L 108 gal 45 A pencil has a diameter of about 0.7 cm If held about 0.75 m from the eye, it can just block out the Moon The ratio of pencil diameter to arm length is the same as the ratio of Moon diameter to Moon distance From the diagram, we have the following ratios Pencil Moon Pencil Distance Moon Distance Pencil diameter Moon diameter Pencil distance Moon distance Moon diameter pencil diameter pencil distance Moon distance 10 m 0.75 m 3.8 105 km 3500 km 46 The person walks km h , 10 hours each day The radius of the Earth is about 6380 km, and the distance around the world at the equator is the circumference, REarth We assume that the person can “walk on water”, and so ignore the existence of the oceans 1h 1d 6380 km 103 d km 10 h © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher www.elsolucionario.net Elementary Particles Chapter 32 2KE th mp c = 4KE th mp c + 4mp2 c − 16mp2 c → 2KE th mp c = 12mp2 c → KE th = 6mp c = Q 59 To find the length in the lab, we need to know the speed of the particle which is moving relativistically Start with Eq 26-6 ⎛ ⎞ KE = m0c ⎜ − 1⎟ → v = c − ⎜ − v2 c2 ⎝ ∆tlab = ∆t0 1− v c 2 = ⎟ ⎠ 2.91×10−13 s − ( 0.603) ( ⎛ KE ⎞ ⎜ m c + 1⎟ ⎝ ⎠ = c 1− ⎛ 450 ×106 eV ⎞ ⎜ 1777 ×106 eV + 1⎟ ⎝ ⎠ = 0.603c = 3.65 ×10−13 s )( ) ∆xlab = v∆tlab = ( 0.603) 3.00 ×108 m s 3.65 ×10−13 s = 6.60 ×10−5 m 60 (a) To conserve charge, the missing particle must be neutral To conserve baryon number, the missing particle must be a meson To conserve strangeness, charm, topness, and bottomness, the missing particle must be made of up and down quarks and antiquarks only With all this information, the missing particle is π = u u + d d (b) This is a weak interaction since one product is a lepton To conserve charge, the missing particle must be neutral To conserve the muon lepton number, the missing particle must be an antiparticle in the muon family With this information, the missing particle is ν µ 61 One u quark from the antiproton joins with a d quark from the neutron to make the π − The other four quarks form the π 62 A relationship between total energy and speed is given by Eq 26-8 E= m0 c → − v2 c2 1− v c = 2 ⎛ m c2 ⎞ → 1− = ⎜ ⎟ c ⎝ E ⎠ m0 c v2 E → ⎛ m0c2 ⎞ ⎛ 9.38 ×108 eV ⎞ = 1− ⎜ ⎟ = 1− ⎜ ⎟ = ( to digits ) → v = c 12 c ⎝ 7.0 ×10 eV ⎠ ⎝ E ⎠ v Use the binomial expansion to express the answer differently 2 ⎛ 9.38 ×108 eV ⎞ ⎛ 9.38 ×108 eV ⎞ = 1− ⎜ ≅ 1− ⎜ = − 9.0 ×109 → v = c (1 − 9.0 ×109 ) ⎟ ⎟ 12 12 c ⎝ 7.0 ×10 eV ⎠ ⎝ 7.0 ×10 eV ⎠ v © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 394 www.elsolucionario.net CHAPTER 33: Astrophysics and Cosmology Answers to Questions Long ago, without telescopes, it was difficult to see the individual stars in the Milky Way The stars in this region of the sky were so numerous and so close together and so tiny that they all blended together to form a cloudy or milky stripe across the night sky Now using more powerful telescopes, we can see the individual stars that make up the Milky Way galaxy When a star generates more energy than it radiates, its temperature increases, which produces a greater outward pressure that overcomes the inward gravitational forces and the star increases in size When a star generates less energy than it radiates, its temperature decreases, which produces a smaller outward pressure and the star decreases in size as the inward gravitational forces start to overcome the outward radiation pressure A red giant star is extremely large in size, it has a very high luminosity, but it has a relatively cool surface temperature that causes it to be red in color See the H-R diagram: (1) Star is getting cooler and redder, it is increasing in size and mass (higher luminosity) (2) Star is getting hotter and whiter, it is decreasing in size and mass (lower luminosity) (3) Star is getting cooler and redder, it is decreasing in size and mass (lower luminosity) (4) Star is getting hotter and whiter, it is increasing in size and mass (higher luminosity) (5) Star is getting cooler and redder, it is increasing in size, but has same mass and luminosity (6) Star is getting hotter and whiter, it is decreasing in size, but has same mass and luminosity (7) Star has same temperature and color, but it is increasing in size and mass (higher luminosity) (8) Star has same temperature and color, but it is decreasing in size and mass (lower luminosity) Although the H-R diagram only directly relates the surface temperature of a star to its absolute luminosity (and thus doesn’t directly reveal anything about the core), the H-R diagram does provide clues regarding what is happening at the core of a star Using the current model of stellar evolution and the H-R diagram, we can infer that the stars on the main sequence are fusing hydrogen nuclei to helium nuclei at the core and that stars in the red giant region are fusing helium and beryllium to make heavier nuclei such as carbon and that this red giant process will continue until fusion can no longer occur and the star will collapse The initial mass of a star determines its final destiny If, after the red giant stage of a star’s life, its mass is less than 1.4 solar masses, then the star cools as it shrinks and it becomes a white dwarf If its mass is between 1.4 and 2-3 solar masses, then the star will condense down to a neutron star, which will eventually explode as a supernova and become a white dwarf If its mass is greater than © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 395 www.elsolucionario.net Chapter 33 Astrophysics and Cosmology 2-3 solar masses, then the star will collapse even more than the neutron star and it will form a black hole Yes Since the population of hotter stars on the H-R diagram is relatively low compared to other regions of the diagram, it seems to imply that these hotter stars stay on the main sequence for a shorter period of time and, thus, have shorter lives (This also makes sense from a fuel standpoint, since these hotter stars will “burn” up their fuel more quickly.) When measuring star parallaxes from the Moon, there are two different cases: (1) If you did the measurements two weeks apart (one at full moon and one at new moon), you would need to assume that the Earth did not move around the Sun very far, and then the d shown in Figure 33-12 would be the Earth-Moon distance instead of the Sun-Earth distance (2) If you did the measurements six months apart and at full moon, then the d shown in Figure 33-12 would be the Sun-Earth distance plus the Earth-Moon distance instead of just the Sun-Earth distance From Mars, then the d shown in Figure 33-12 would be the Sun-Mars distance instead of the Sun-Earth distance You would also need to know the length of a Mars “year” so you could take your two measurements at the correct times Measure the period of the changing luminosity of a Cepheid variable, and then use the definite relationship between the period and absolute luminosity to determine the absolute luminosity (L) of the star Then use ( ) = L 4π d to find the relationship between the apparent brightness ( ) and the distance to the star ( d ) Once the apparent brightness is experimentally determined, we can then find the distance to the star 10 A geodesic is the shortest distance between two points on a surface Its role in General Relativity is to help us explain curved space-time Light always travels by the shortest and most direct path between two points (a geodesic) and if this path is curved it means that space itself is curved Seeing gravity as the cause of this curvature is a fundamental idea of General Relativity 11 If the redshift was due to something besides expansion, then the Big Bang theory and the view of an expanding universe would be called into question, since the redshift is a major piece of evidence for the expanding universe Also, the cosmic microwave background data would be conflicting evidence to this new data Cosmic microwave background is seen as a confirmation of an expanding universe Thus if there were some other explanation for the redshift, there might be two distinct schools of thought based on these two conflicting pieces of evidence 12 No, just because everything appears to be moving away from us does not mean we are at the center of the universe Here is an analogy If we were sitting on the surface of a balloon and then more air was put into the balloon causing it to expand, notice that every other point on the balloon is now farther away from you The points close to you are farther away because of the expansion of the rubber and the points on the other side of the balloon are farther away from you because the radius of the balloon is now larger 13 If you were located in a galaxy near the boundary of our observable universe, galaxies in the direction of the Milky Way would be receding from you The outer “edges” of the observable universe are expanding at a faster rate than the points more “interior”, thus, due to a relative velocity argument, the slower galaxies in the direction of the Milky Way would look like they are receding from your faster galaxy near the outer boundary Also see Figure 33-20 © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 396 www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli 14 An explosion on Earth blows pieces out into the space around it, but the Big Bang was the start of the expansion of space itself In an explosion on Earth, the pieces that are blown outward will slow down due to air resistance and the farther away they are the slower they will be moving and then they will eventually come to rest, but with the Big Bang, the farther away galaxies are from each other the faster they are moving away from each other In an explosion on Earth, the pieces with the higher initial speeds end up farther away from the explosion before coming to rest, but the Big Bang appears to be relatively uniform where the farthest galaxies are moving fastest and the nearest galaxies are moving the slowest An explosion on Earth would correspond to a closed universe, since the pieces would eventually stop, but we would not see a “big crunch” due to gravity as we would with an actual closed universe 15 To “see” a black hole in space we need indirect evidence If a large visible star or galaxy was rotating quickly around a non-visible gravitational companion, the non-visible companion could be a massive black hole Also, as matter begins to accelerate toward a black hole, it will emit characteristic X-rays, which we could detect on Earth Another way we could “see” a black hole is if it caused gravitational lensing of objects behind it Then we would see stars and galaxies in the “wrong” place as their light is bent as it passes past the black hole on its way to Earth 16 The Schwarzschild radius of a hydrogen atom in its ground state would be caused by a mass of: 6.67 × 10 −11 N i m kg M → M = 3.57 × 1016 kg R = 5.29 × 10−11 m = × 10 m s ( ( ) ) 17 If we look closely at the top of the right-hand side of Figure 33-25: The Earth formed billion years ago and people evolved million years ago These events are both very close to the very end of the timeline shown in Figure 33-25 18 The material contained in the universe is at some average temperature and, thus, it radiates a blackbody spectrum that is characteristic of this temperature This temperature is now so low because of the expansion of the universe When the universe was extremely small and the sum total energy of the universe was trapped in that extremely small volume, the average temperature of the universe was amazingly high As this energy has been spread out over our expanding universe, the average temperature has decreased to its current low level 19 Atoms were unable to exist until hundreds of thousands of years after the Big Bang because the temperature of the universe was still too high until that time At those incredibly high temperatures, the free electrons and nuclei were moving so fast and had so much kinetic energy, and they had so many high energy photons colliding with them, that they could never combine together to form atoms Once the universe cooled below 3000 K, this coupling could take place, and atoms were formed © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 397 www.elsolucionario.net Chapter 33 Astrophysics and Cosmology 20 The universe would eventually collapse in on itself if it had a positive curvature This would mean that there is so much mass in the universe that the gravitational attraction would pull all of the mass of the universe back into a “big crunch.” In this situation, the average density of the universe would be greater than the critical density Solutions to Problems See Figure 13-11 in the textbook The parsec is the distance D when the angle φ is second of arc φ = 1′′ × tan φ = 1o 3600′′ d D × π rad 180o → D= = 4.848 × 10 −6 rad d tan φ = 1.496 × 1011 m ( −6 tan 4.848 × 10 rad ) = 3.086 × 1016 m 1ly ⎛ ⎞ ⎟ = 3.26 ly 15 ⎝ 9.461 × 10 m ⎠ D = 3.086 × 1016 m ⎜ Use the angle to calculate the distance in parsecs, and then convert to other units ⎛ 3.26 ly ⎞ 1 d ( pc ) = = = 2.63 pc ⎜ ⎟ = 8.6 ly φ ′′ 0.38′′ ⎝ 1pc ⎠ Convert the angle to seconds of arc, reciprocate to find the distance in parsecs, and then convert to light years o ⎛ 3600′′ ⎞ φ = 1.9 × 10−4 ⎜ o ⎟ = 0.684′′ ⎝ ⎠ ( d ( pc ) = ) ⎛ 3.26 ly ⎞ 1 = = 1.46 pc ⎜ ⎟ = 4.8 ly φ ′′ 0.684 ⎝ 1pc ⎠ The reciprocal of the distance in parsecs is the angle in seconds of arc 1 (a) φ ′′ = = = 0.02778′′ ≈ 2.8 × 10−2 ′′ d ( pc ) 36 pc ( ) ⎛ 1o o o ⎞ = ( 7.717 × 10−6 ) ≈ ( 7.7 × 10−6 ) ⎟ ⎝ 3600′′ ⎠ (b) 0.02778′′ ⎜ Convert the light years to parsecs, and then reciprocate the parsecs to find the parallax angle in seconds of arc ⎛ 1pc ⎞ 55 ly ⎜ φ= = 16.87 pc ≈ 17 pc = 0.059′′ ⎟ 16.87 pc ⎝ 3.26 ly ⎠ The parallax angle is smaller for the further star Since tan φ = d , as the distance D to the star D increases, the tangent decreases, so the angle decreases And since for small angles, tan φ ≈ φ , we © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 398 www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli have that φ ≈ d D Thus if the distance D is doubled, the angle φ will be smaller by a factor of Find the distance in light years That value is also the time for light to reach us ⎛ 3.26 ly ⎞ 35 pc ⎜ ⎟ = 114 ly ≈ 110 ly → It takes light 110 years to reach us ⎝ 1pc ⎠ (a) The apparent brightness is the solar constant, 1.3 × 103 W m (b) Use Eq 33-1 to find the absolute luminosity L l= → L = 4π d 2l = 4π 1.496 × 1011 m 2 4π d ( ) (1.3 × 10 ) W m = 3.7 × 1026 W The apparent brightness of an object is inversely proportional to the observer’s distance from the L object, given by l = To find the relative brightness at one location as compared to another, 4π d take a ratio of the apparent brightness at each location L lJupiter lEarth = 4π d Jupiter L 4π d Earth = d Earth d Jupiter ⎛d = ⎜ Earth ⎜d ⎝ Jupiter ⎞ ⎛ ⎞2 −2 ⎟⎟ = ⎜ ⎟ = 3.7 × 10 ⎠ ⎝ 5.2 ⎠ 10 The angular width is the inverse tangent of the diameter of our Galaxy divided by the distance to the nearest galaxy According to Figure 33-2, our Galaxy is about 100,000 ly in diameter Galaxy diameter 1.0 × 105 ly φ = tan −1 = tan −1 = 4.2 × 10−2 rad ≈ 2.4o Distance to nearest galaxy 2.4 × 10 ly φMoon = tan −1 Moon diameter Distance to Moon = tan −1 3.48 × 106 m 3.84 × 10 m ( = 9.1 × 10−3 rad ≈ 0.52o ) The galaxy width is about 4.5 times the moon width 11 The density is the mass divided by the volume M M 1.99 × 1030 kg = = = 1.4 × 10−4 kg m ρ= 3 11 V R π π 1.5 × 10 m 3 ( ) 12 The angular width is the inverse tangent of the diameter of the Moon divided by the distance to the Sun o −1 Moon diameter −1 3.48 × 10 m = tan = 2.33 × 10−5 rad ≈ 1.33 × 10−3 ≈ 4.79′′ φ = tan 11 Distance to Sun 1.496 × 10 m ( ) 13 The density is the mass divided by the volume M M 1.99 × 1030 kg = Sun3 = = 1.83 × 109 kg m ρ= V π REarth π 6.38 × 10 m 3 ( ) Since the volumes are the same, the ratio of the densities is the same as the ratio of the masses © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 399 www.elsolucionario.net Chapter 33 Astrophysics and Cosmology ρ ρ Earth = M M Earth 1.99 × 1030 kg = 5.98 × 10 kg 24 = 3.33 × 105 times larger 14 The density of the neutron star is its mass divided by its volume Use the proton to calculate the density of nuclear matter ρ neutron = star ρ neutron star ρ white = M V = M Sun = πR Earth ( 1.5 1.99 × 1030 kg 5.354 × 1017 kg m 1.83 × 109 kg m3 π (11000 m ) ) = 5.354 ×10 17 ρ neutron star = 2.9 × 10 ρ nuclear dwarf = kg m ≈ 5.4 × 1017 kg m 5.354 × 1017 kg m matter 1.673 × 10−27 kg π (1 × 10−15 m ) = 1.3 15 From Equation 30-2, the Q-value is the mass energy of the reactants minus the mass energy of the products He + 42 He → 48 Be ( ) Q = mBe c − 2mHe c = [ ( 4.002603 u ) − 8.005305 u ] c 931.5 Mev c = −0.092 MeV He + 48 Be → 12 C ( Q = mBe c − 2mHe c = [ 4.002603 u + 8.005305 u − 12.000000] c 931.5 Mev c ) = 7.366 MeV 16 Wien’s law says that the λPT = α , where α is a constant, and so λP1T1 = λP2T2 The StefanBoltzmann equation says that the power output of a star is given by P = β AT , where β is a constant, and A is the radiating area The P in the Stefan-Boltzmann equation is the same as the luminosity L in this chapter It is given that l1 = l2 and r1 = r2 We assign λP1 = 800 nm and λP2 = 400 nm λP1T1 = λP2T2 → L1 l1 = l2 → 4π d12 T2 T1 = = λP1 =2 λP2 L2 4π d 22 β A2T24 4π r22T24 T24 ⎛ T2 ⎞ → = = = = = =⎜ ⎟ d12 L1 P1 β AT 4π r12T14 T14 ⎝ T1 ⎠ 1 d 22 L2 P2 → ⎛T ⎞ = ⎜ ⎟ = ( 2) = d1 ⎝ T1 ⎠ d2 The star with the peak at 400 nm is times further away than the star with the peak at 800 nm 17 Wien’s law says that the λPT = α , where α is a constant, and so λP1T1 = λP2T2 or The StefanBoltzmann equation says that the power output of a star is given by P = β AT , where β is a constant, and A is the radiating area The P in the Stefan-Boltzmann equation is the same as the luminosity L in this chapter It is given that l1 l2 = 0.091 , d1 = d , λP1 = 500 nm and λP2 = 700 nm © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 400 www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli T2 λP1T1 = λP2T2 → 1= d 22 = d12 0.091L2 L1 T1 = = λP1 = λP2 0.091P2 P1 = 0.091A2T24 = AT 1 L1 l1 = 0.091l2 → 4π d ( 0.091) 4π r22T24 4π r12T14 = 0.091 = 0.091 L2 → 4π d 22 T24 r22 T14 r12 → ⎛T ⎞ ⎛5⎞ = 0.091 ⎜ ⎟ = 0.091 ⎜ ⎟ = 0.15 r2 ⎝7⎠ ⎝ T1 ⎠ r1 The ratio of the diameters is the same as the ratio of radii, so 18 The Schwarzschild radius is given by R = 2GM Sun (a) RSun = c (b) REarth = = 2GM Earth c R= c = D2 ( ( )( ) ( 6.67 × 10 N im kg )( 5.98 × 10 = ( 3.00 ×10 m s ) 3.00 × 108 m s −11 2 ( ( 3.00 ×10 24 ) = 2950 m = kg 2.95 km ) = 8.86 × 10 −3 m ≈ 8.9 mm 2GM c2 × 1041 kg )( m s) 6.67 × 10−11 N i m kg = 0.15 2GM c2 6.67 × 10 −11 N im kg 1.99 × 1030 kg 19 The Schwarzschild radius is given by R = 2GM D1 )= × 1014 m 20 (a) For the vertices of the triangle we choose the North pole and two points on a latitude line on nearly opposite sides of the Earth, as shown on the diagram Let the angle at the North pole be 179° 180° 179o 90° 90o 90°o 90 (b) If the triangle is drawn on a small enough portion of the sphere that the portion is flat, then the sum of the angles will be 180° 21 The limiting value for the angles in a triangle on a sphere is 540o Imagine drawing an equilateral triangle near the north pole, enclosing the north pole If that triangle were small, the surface would be approximately flat, and the each angle in the triangle would be 60o Then imagine “stretching” each side of that triangle down towards the equator, while keeping sure that the north pole stayed inside the triangle The angle at each vertex of the triangle would expand, with a limiting value of 180o The three 180o angles in the triangle would add up to 540o 22 Use Eq 33-6, Hubble’s law v = Hd → d = v H = ( 0.010 ) ( 3.00 × 108 m s ) 2.2 × 10 m s Mly = 140 Mly = 1.4 ì 108 ly â 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 401 www.elsolucionario.net Chapter 33 Astrophysics and Cosmology 23 Use Eq 33-6, Hubble’s law v 3500 km s v = Hd → d = = = 160 Mly = 1.6 × 108 ly H 22 km s Mly 24 Use Eq 33-6, Hubble’s law ⎛ ⎞ ⎟ = 0.88 c ⎝ 3.00 × 10 m s ⎠ v = Hd = ( 22000 m s Mly )(12000 Mly ) = 2.64 × 108 m s ⎜ c 25 We find the velocity from Hubble’s law (Eq 33-6), and the observed wavelength from the Doppler shift, Eq 33-3 v Hd ( 22000 m s Mly )(1.0 Mly ) (a) = = = 7.33 × 10−5 c c 3.00 × 108 m s λ = λ0 (b) v c = Hd c λ = λ0 (c) v c = Hd c λ = λ0 1+ v c 1− v c = ( 22000 m s + 7.33 × 10−5 ( Mly ) 1.0 × 102 Mly 3.00 × 10 m s 1− v c = ( 656 nm ) ( 22000 m s 1+ v c 1− v c ) = 7.33 × 10 + 7.33 × 10−3 ( 3.00 × 10 m s + 0.733 − 0.733 −3 = 660.83 nm ≈ 661nm − 7.33 × 10−3 Mly ) 1.0 × 10 Mly = ( 656 nm ) = 656.05 nm ≈ 656 nm − 7.33 × 10−5 1+ v c = = ( 656 nm ) ) = 0.733 = 1671.3 nm ≈ 1670 nm 26 Use Eqs 33-4 and 33-5a to solve for the speed of the galaxy ∆λ = 610 nm − 434 nm = 176 nm 2 ⎛ ∆λ ⎞ ⎛ 176 nm ⎞ + 1⎟ − ⎜ λ + 1⎟ − ⎜ ∆λ 1+ v c v ⎝ v ⎝ 434 nm ⎠ 0.9755 ⎠ = −1 → = = = = 0.327c → z= 2 λ0 1− v c c ⎛ ∆λ c ⎛ 176 nm 2.9755 ⎞ ⎞ ⎜ λ + 1⎟ + ⎜ 434 nm + 1⎟ + ⎝ ⎠ ⎝ ⎠ v ≈ 0.33c Use Hubble’s law, Eq 33-6, to solve for the distance v v 0.3278 v = Hd → d = = c = = 4.47 × 103 Mly ≈ 4.5 × 109 ly H 22000 m s Mly ( ) H c 3.00 ì 108 m s â 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 402 www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli 27 Use Eq 33-35a to solve for the speed of the galaxy z= 1+ v c 1− v c −1 → v c ( z + 1) − 1.62 − = = 0.44 ( z + 1) + 1.62 + = → v = 0.44c 28 Use Eq 33-35a to solve for the redshift parameter z= 1+ v c 1− v c 29 Eq 33-3 states λ = λ0 λ = λ0 + 0.50 −1 = − 0.50 1+ v c 1− v c − = 0.73 1/ 1+ v c ⎛ v⎞ ⎛ v⎞ = λ0 ⎜ + ⎟ ⎜ − ⎟ 1− v c ⎝ c⎠ ⎝ c⎠ ⎛ ⎛ λ ≈ λ0 ⎜ + ⎜ 12 ⎝ ⎝ −1/ v ⎞⎛ v⎞ v⎞ ⎛ ⎛ ≈ λ0 ⎜ + 12 ⎟ ⎜ − ( − 12 ) ⎟ = λ0 ⎜ + 12 ⎟ c ⎠⎝ c⎠ c⎠ ⎝ ⎝ v ⎞⎞ v v ⎛ v⎞ → λ − λ0 = ∆λ = λ0 → ⎟ ⎟ = λ0 ⎜ + ⎟ = λ0 + λ0 c ⎠⎠ c c ⎝ c⎠ ∆λ λ0 = v c 30 Wien’s law is given as Eq 27-2 in chapter 27 2.90 × 10−3 mi K 2.90 × 10−3 mi K = = 1.1× 10−3 m λPT = 2.90 × 10−3 miK → λP = T 2.7 K 31 We use the proton as typical nuclear matter kg 1nucleon 10−26 × = nucleons m3 m 1.67 × 10−27 kg 32 If the scale of the universe is inversely proportional to the temperature, then the scale times the temperature should be constant If we call the current scale “1”, and knowing the current temperature to be about K, then the product of scale and temperature should be about Use Figure 33-25 to estimate the temperature at various times For purposes of illustration, we assume the universe has a current size of about 1010 ly (a) At t = 106 yr , the temperature is about 1000 K Thus the scale is found as follows 3 = = × 10−3 ≈ × 107 ly ( Scale )( Temperature ) = → Scale = Temperature 1000 (b) At t = s , the temperature is about 1010 K 3 Scale = = 10 = × 10−10 ≈ 3ly Temperature 10 (c) At t = 10−6 s , the temperature is about 1013 K 3 Scale = = 13 = × 10−13 ≈ × 10−3 ly ≈ × 1013 m Temperature 10 (d) At t = 10−35 s , the temperature is about 1027 K 3 Scale = = 27 = × 10−27 ≈ × 10−17 ly ≈ 0.3 m Temperature 10 © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 403 www.elsolucionario.net Chapter 33 Astrophysics and Cosmology 33 The temperature corresponding to the average kinetic energy needed to produce a particle of mass m mc is given by 32 kT = mc → T = 23 k 2 −19 mc 2 500 × 10 ev c c 1.60 × 10 J ev (a) T = =3 = 3.86 × 1012 K −23 k 1.38 × 10 J K ( ) ( ) From Figure 33-25, this corresponds to a time of ∼ 10−5 s (b) T = mc k = ( 9500 ×10 ) ( ev c c 1.60 × 10−19 J ev 1.38 × 10 −23 J K ) = 7.34 × 10 13 K From Figure 33-25, this corresponds to a time of ∼ 10−7 s (c) T = mc k = (100 × 10 ) ( ev c c 1.60 × 10−19 J ev 1.38 × 10 −23 J K ) = 7.73 × 10 11 K From Figure 33-25, this corresponds to a time of ∼ 10−4 s 34 A: temperature increases, luminosity stays the same, size decreases B: temperature stays the same, luminosity decreases, size decreases C: temperature decreases, luminosity increases, size increases 35 The apparent luminosity is given by Eq 33-1 Use that relationship to derive an expression for the absolute luminosity, and equate that for two stars L l= → L = 4π d 2l 4π d 2 Ldistant = LSun → 4π d distant ldistant = 4π dSun lSun → star star d distant = dSun star lSun ldistant star ( = 1.5 × 1011 m ) ⎛ 1ly ⎞ ⎟ = ly 15 −11 ⎜ 10 ⎝ 9.461 × 10 m ⎠ star 36 The angular momentum is the product of the rotational inertia and the angular velocity ( I ω )initial = ( I ω )final → ωfinal ⎛ MRinitial ⎞ ⎛I ⎞ ⎛ Rinitial ⎞ ⎛ × 108 m ⎞ = ωinitial ⎜ initial ⎟ = ωinitial ⎜ 52 = ω = 1rev month )⎜ ⎟ ⎟ ( initial ⎜ ⎟ ⎝ × 10 m ⎠ ⎝ I final ⎠ ⎝ Rfinal ⎠ ⎝ MRfinal ⎠ = 4.9 × 109 rev month = 4.9 × 109 37 The rotational kinetic energy is given by rev month × 1month 30 d × 1d 24 h × 1h 3600 s = 1900 rev s I ω The final angular velocity, from problem 36, is about 4.9 × 109 rev month KE final KE initial = 2 I finalωfinal I initialωinitial = 5 2 ωfinal MRfinal 2 ωinitial MRinitial ⎛ R ω ⎞ = ⎜ final final ⎟ ⎝ Rinitialωinitial ⎠ © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 404 www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli ⎛ (1 × 104 m )( 4.9 × 109 rev month ) ⎞ =⎜ ⎟ = 4.9 × 10 ⎜ ⎟ × 10 m 1rev month ) ⎠ ( )( ⎝ 38 The power output is the energy loss divided by the elapsed time 2 ∆KE 12 I ω ( fraction lost ) 12 52 MR ω ( fraction lost ) = = P= ∆t ∆t ∆t = ( )( 30 (1.5 ) 1.99 × 10 kg 1.0 × 10 m ) ( 2π rad s ) (1×10 ) = 2 (1d )( 24 h d )( 3600 s h ) 39 Use Newton’s law of universal gravitation F =G m1m2 r −9 ( = 6.67 × 10 −11 N i m kg ) ( ×10 41 kg ) 2.7 × 1025 W ⎡ ⎛ 9.46 × 10 m ⎞ ⎤ ⎢( × 10 ly ) ⎜ ⎟⎥ 1ly ⎝ ⎠⎦ ⎣ 15 = 1.68 × 1028 N ≈ × 1028 N 40 (a) Assume that the nucleons make up only 2% of the critical mass density nucleon mass density = 0.02 10−26 kg m3 ( nucleon number density = ) ( 0.02 10−26 kg m 1.67 × 10 −27 ) kg nucleon = 0.12 nucleon m neutrino number density = 109 ( nucleon number density ) = 1.2 × 108 neutrino m3 ( 0.98 10 −26 kg m ) kg = 8.17 × 10−35 × 9.315 × 108 eV c −27 = 46 eV c 1.2 × 10 neutrino m neutrino 1.66 × 10 kg (b) Assume that the nucleons make up only 5% of the critical mass density nucleon mass density = 0.05 10−26 kg m ( nucleon number density = ) ( 0.05 10−26 kg m3 1.67 × 10 −27 ) kg nucleon = 0.30 nucleon m3 neutrino number density = 109 ( nucleon number density ) = 3.0 × 108 neutrino m3 ( 0.95 10−26 kg m ) 3.0 × 108 neutrino m3 = 3.17 × 10 −35 kg neutrino × 9.315 × 108 eV c 1.66 × 10−27 kg = 18 eV c 41 The temperature of each star can be found from Wien’s law λPT = 2.90 × 10−3 miK → T600 = 2.90 × 10 −3 mi K = 4830 K T400 = 2.90 × 10 −3 mi K 600 × 10−9 m 400 × 10−9 m The luminosity of each star can be found from the H-R diagram L600 ∼ 1026 W L400 ∼ 1027 W = 7250 K © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 405 www.elsolucionario.net Chapter 33 Astrophysics and Cosmology The Stefan-Boltzmann equation says that the power output of a star is given by P = β AT , where β is a constant, and A is the radiating area The P in the Stefan-Boltzmann equation is the same as the luminosity L Form the ratio of the two luminosities L400 L600 4 β A400T400 π r400 T400 = = 4 β A600T600 π r600T600 → r400 r600 = L400 T600 = L600 T400 1027 W ( 4830 K ) 1026 W ( 7250 K ) = 1.4 The diameters are in the same ratio as the radii d 400 = 1.4 d 600 The luminosities are fairly subjective, since they are read from the H-R diagram Different answers may arise from different readings of the H-R diagram 42 The number of parsecs is the reciprocal of the angular resolution in seconds of arc o o ⎛ 1′ ⎞ ⎛ ⎞ ′′ → φ = ( 0.01 ) ⎜ ≈ × 10−6 100 parsec = ⎟ ⎜ ⎟ φ ′′ ⎝ 60′′ ⎠ ⎝ 60′ ⎠ ( ) 43 From Eq 27-16, we see that the wavelengths from single-electron energy level transitions are inversely proportional to the square of the atomic number of the nucleus Thus the lines from singlyionized helium are usually one fourth the wavelength of the corresponding hydrogen lines Because of their red shift, the lines have four times their usual wavelength Use Eq 33-3 λ = λ0 1+ v c 1− v c → 4λ0 = λ0 1+ v c 1− v c → 16 = 1+ v c 1− v c v → c = 15 → v = 0.88 c 17 44 From section 33-7, we can approximate the temperature – kinetic energy relationship by kT = KE → T = KE (1.8 ×10 12 = )( eV 1.60 × 10 k 1.38 × 10 From Figure 23-25, this is in the hadron era −23 −19 J K J eV ) = 1.4 ×10 16 kT = KE K 45 We assume that gravity causes a centripetal force on the gas Solve for the speed of the rotating gas, and use Eq 33-5b to find the Doppler shift mgas mblack mgas vgas hole Fgravity = Fcentripetal → G = → r2 r mblack hole vgas = G ∆λ r v = ( 6.67 × 10−11 N im kg ) ( ×10 )(1.99 ×10 30 kg ) = 6.84 × 10 ⎛ 9.46 × 10 m ⎞ ( 60 ly ) ⎜ ⎟ 1ly ⎝ ⎠ 15 m s 6.84 × 105 m s ≈ × 10−3 c 3.00 × 108 m s This is the Doppler shift as compared to the light coming from the center of the galaxy z= λ0 ≈ = © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 406 www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli 46 (a) Use Eqs 33-4 and 33-5a to solve for the speed of the galaxy ∆λ = 650 nm − 434 nm = 216 nm z= ∆λ λ0 1+ v c = 1− v c −1 → 2 ⎛ ∆λ ⎞ ⎛ 216 nm ⎞ + 1⎟ − + 1⎟ − ⎜ ⎜ v ⎝ λ0 v ⎝ 434 nm ⎠ 1.243 ⎠ = = = = 0.383c → v ≈ 0.383 c 2 c ⎛ ∆λ c ⎛ 216 nm ⎞ 3.243 ⎞ + 1⎟ + ⎜ ⎜ 434 nm + 1⎟ + λ ⎝ ⎠ ⎝ ⎠ (b) Use Hubble’s law, Eq 33-6, to solve for the distance v v 0.383 v = Hd → d = = c = = 5.22 × 103 Mly ≈ 5.2 × 109 ly ( 22000 m s Mly ) H H c 3.00 × 108 m s 47 (a) Find the Q-value for this reaction From Equation 30-2, the Q-value is the mass energy of the reactants minus the mass energy of the products 12 C + 126 C → 24 Mg 12 ( ) Q = 2mC c − mMg c = [ (12.000000 u ) − 23.985042 u ] c 931.5 Mev c = 13.93 MeV (b) The total kinetic energy should be equal to the electrical potential energy of the two nuclei when they are just touching The distance between the two nuclei will be twice the nuclear radius, from Eq 30-1 Each nucleus will have half the total kinetic energy qnucleus 1/ 1/ −15 −15 PE = r = 1.2 × 10 m ( A ) = 1.2 × 10 m (12 ) 4πε 2r ( ) KE nucleus = 12 PE = = ( ) qnucleus 4πε 2r (8.988 × 10 Nin (1.60 ×10 C ) C ) (1.2 × 10 m ) (12 ) 2 ( 6) −19 −15 1/ × 1eV 1.60 × 10−19 J (c) We can approximate the temperature – kinetic energy relationship by × 10 ( 4.71×10 eV ) ⎛⎜ 1.601eV kT = KE → T = KE k = ⎝ 1.38 × 10−23 J K −19 = 4.71MeV kT = KE J⎞ ⎟ ⎠ = 3.6 × 1010 K 48 (a) Find the Q-value for this reaction From Equation 30-2, the Q-value is the mass energy of the reactants minus the mass energy of the products 16 28 C + 168 C → 14 Si + 24 He ( Q = 2mC c − mSi c − mHe c = [ (15.994915 u ) − 27.976927 u − 4.002603] c 931.5 Mev c = 9.954 MeV © 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 407 www.elsolucionario.net ) Chapter 33 Astrophysics and Cosmology (b) The total kinetic energy should be equal to the electrical potential energy of the two nuclei when they are just touching The distance between the two nuclei will be twice the nuclear radius, from Eq 30-1 Each nucleus will have half the total kinetic energy qnucleus 1/ 1/ r = 1.2 × 10−15 m ( A ) = 1.2 × 10−15 m (16 ) PE = 4πε 2r ( ) KE nucleus = 12 PE = = ( ) qnucleus 4πε 2r (8.988 × 10 (1.60 ×10 C ) C ) (1.2 × 10 m ) (16 ) Nin (8) 2 −19 1/ −15 × 1eV 1.60 × 10−19 J (c) We can approximate the temperature – kinetic energy relationship by kT = KE → T = KE k ( = = 7.61MeV kT = KE ⎛ 1.60 × 10−19 J ⎞ ) 1eV ⎟ ⎝ ⎠ = 5.9 × 1010 K −23 7.61 × 106 eV ⎜ 1.38 × 10 J K 49 We use the Sun’s mass and given density to calculate the size of the Sun M M ρ= = → π V r Sun 1/ rSun ⎛ 3M ⎞ =⎜ ⎟ ⎝ 4πρ ⎠ rSun d Earth-Sun = ⎡ (1.99 × 1030 kg ) ⎤ =⎢ ⎥ −26 ⎢⎣ 4π (10 kg m ) ⎥⎦ 3.62 × 1018 m 1.50 × 10 m 11 ≈ × 107 ; 1/ 1ly ⎛ ⎞ ⎟ = 382 ly ≈ 400 ly 15 9.46 × 10 m ⎝ ⎠ = 3.62 × 1018 m ⎜ rSun rgalaxy = 382 ly 50,000 ly ≈ ì 103 â 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 408 www.elsolucionario.net ... without permission in writing from the publisher www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli (a) 1.156 1.156 100 (b) 21.8 2.18 101 (c) 0.0068 2.7635 101 (d)... without permission in writing from the publisher www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli 15 (a) 93 million miles 93 10 miles 1610 m mile 11 (b) 1.5 10... without permission in writing from the publisher www.elsolucionario.net Physics: Principles with Applications, 6th Edition Giancoli 30 Estimate one side of a house to be about 40 feet long, and