Ebook Physics for scientists and engineers Part 2

(BQ) Part 2 book Physics for scientists and engineers has contents: Electric charges and electric field; gauss’s law; capacitance, dielectrics, electric energy storage; electric currents and resistance; magnetism; electromagnetic induction and faraday’s law;...and other contents.

CHAPTER 21: Electric Charges and Electric Field Responses to Questions Rub a glass rod with silk and use it to charge an electroscope The electroscope will end up with a net positive charge Bring the pocket comb close to the electroscope If the electroscope leaves move farther apart, then the charge on the comb is positive, the same as the charge on the electroscope If the leaves move together, then the charge on the comb is negative, opposite the charge on the electroscope The shirt or blouse becomes charged as a result of being tossed about in the dryer and rubbing against the dryer sides and other clothes When you put on the charged object (shirt), it causes charge separation within the molecules of your skin (see Figure 21-9), which results in attraction between the shirt and your skin Fog or rain droplets tend to form around ions because water is a polar molecule, with a positive region and a negative region The charge centers on the water molecule will be attracted to the ions (positive to negative) See also Figure 21-9 in the text The negatively charged electrons in the paper are attracted to the positively charged rod and move towards it within their molecules The attraction occurs because the negative charges in the paper are closer to the positive rod than are the positive charges in the paper, and therefore the attraction between the unlike charges is greater than the repulsion between the like charges - + +++++++ - + - + - + A plastic ruler that has been rubbed with a cloth is charged When brought near small pieces of paper, it will cause separation of charge in the bits of paper, which will cause the paper to be attracted to the ruler On a humid day, polar water molecules will be attracted to the ruler and to the separated charge on the bits of paper, neutralizing the charges and thus eliminating the attraction The net charge on a conductor is the difference between the total positive charge and the total negative charge in the conductor The “free charges” in a conductor are the electrons that can move about freely within the material because they are only loosely bound to their atoms The “free electrons” are also referred to as “conduction electrons.” A conductor may have a zero net charge but still have substantial free charges Most of the electrons are strongly bound to nuclei in the metal ions Only a few electrons per atom (usually one or two) are free to move about throughout the metal These are called the “conduction electrons.” The rest are bound more tightly to the nucleus and are not free to move Furthermore, in the cases shown in Figures 21-7 and 21-8, not all of the conduction electrons will move In Figure 21-7, electrons will move until the attractive force on the remaining conduction electrons due to the incoming charged rod is balanced by the repulsive force from electrons that have already gathered at the left end of the neutral rod In Figure 21-8, conduction electrons will be repelled by the incoming rod and will leave the stationary rod through the ground connection until the repulsive force on the remaining conduction electrons due to the incoming charged rod is balanced by the attractive force from the net positive charge on the stationary rod © 2009 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 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual The electroscope leaves are connected together at the top The horizontal component of this tension force balances the electric force of repulsion (Note: The vertical component of the tension force balances the weight of the leaves.) Coulomb’s law and Newton’s law are very similar in form The electrostatic force can be either attractive or repulsive; the gravitational force can only be attractive The electrostatic force constant is also much larger than the gravitational force constant Both the electric charge and the gravitational mass are properties of the material Charge can be positive or negative, but the gravitational mass only has one form 10 The gravitational force between everyday objects on the surface of the Earth is extremely small (Recall the value of G: 6.67 x 10-11 Nm2/kg2.) Consider two objects sitting on the floor near each other They are attracted to each other, but the force of static fiction for each is much greater than the gravitational force each experiences from the other Even in an absolutely frictionless environment, the acceleration resulting from the gravitational force would be so small that it would not be noticeable in a short time frame We are aware of the gravitational force between objects if at least one of them is very massive, as in the case of the Earth and satellites or the Earth and you The electric force between two objects is typically zero or close to zero because ordinary objects are typically neutral or close to neutral We are aware of electric forces between objects when the objects are charged An example is the electrostatic force (static cling) between pieces of clothing when you pull the clothes out of the dryer 11 Yes, the electric force is a conservative force Energy is conserved when a particle moves under the influence of the electric force, and the work done by the electric force in moving an object between two points in space is independent of the path taken 12 Coulomb observed experimentally that the force between two charged objects is directly proportional to the charge on each one For example, if the charge on either object is tripled, then the force is tripled This is not in agreement with a force that is proportional to the sum of the charges instead of to the product of the charges Also, a charged object is not attracted to or repelled from a neutral object, which would be the case if the numerator in Coulomb’s law were proportional to the sum of the charges 13 When a charged ruler attracts small pieces of paper, the charge on the ruler causes a separation of charge in the paper For example, if the ruler is negatively charged, it will force the electrons in the paper to the edge of the paper farthest from the ruler, leaving the near edge positively charged If the paper touches the ruler, electrons will be transferred from the ruler to the paper, neutralizing the positive charge This action leaves the paper with a net negative charge, which will cause it to be repelled by the negatively charged ruler 14 The test charges used to measure electric fields are small in order to minimize their contribution to the field Large test charges would substantially change the field being investigated 15 When determining an electric field, it is best, but not required, to use a positive test charge A negative test charge would be fine for determining the magnitude of the field But the direction of the electrostatic force on a negative test charge will be opposite to the direction of the electric field The electrostatic force on a positive test charge will be in the same direction as the electric field In order to avoid confusion, it is better to use a positive test charge © 2009 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 Chapter 21 Electric Charges and Electric Field 16 See Figure 21-34b A diagram of the electric field lines around two negative charges would be just like this diagram except that the arrows on the field lines would point towards the charges instead of away from them The distance between the charges is l 17 The electric field will be strongest to the right of the positive charge (between the two charges) and weakest to the left of the positive charge To the right of the positive charge, the contributions to the field from the two charges point in the same direction, and therefore add To the left of the positive charge, the contributions to the field from the two charges point in opposite directions, and therefore subtract Note that this is confirmed by the density of field lines in Figure 21-34a 18 At point C, the positive test charge would experience zero net force At points A and B, the direction of the force on the positive test charge would be the same as the direction of the field This direction is indicated by the arrows on the field lines The strongest field is at point A, followed (in order of decreasing field strength) by B and then C 19 Electric field lines can never cross because they give the direction of the electrostatic force on a positive test charge If they were to cross, then the force on a test charge at a given location would be in more than one direction This is not possible 20 The field lines must be directed radially toward or away from the point charge (see rule 1) The spacing of the lines indicates the strength of the field (see rule 2) Since the magnitude of the field due to the point charge depends only on the distance from the point charge, the lines must be distributed symmetrically 21 The two charges are located along a line as shown in the 2Q diagram Q (a) If the signs of the charges are opposite then the point on the line where E = will lie to the left of Q In that region ℓ the electric fields from the two charges will point in opposite directions, and the point will be closer to the smaller charge (b) If the two charges have the same sign, then the point on the line where E = will lie between the two charges, closer to the smaller charge In this region, the electric fields from the two charges will point in opposite directions 22 The electric field at point P would point in the negative x-direction The magnitude of the field would be the same as that calculated for a positive distribution of charge on the ring: E Qx 4 o  x  a 3/ 23 The velocity of the test charge will depend on its initial velocity The field line gives the direction of the change in velocity, not the direction of the velocity The acceleration of the test charge will be along the electric field line 24 The value measured will be slightly less than the electric field value at that point before the test charge was introduced The test charge will repel charges on the surface of the conductor and these charges will move along the surface to increase their distances from the test charge Since they will then be at greater distances from the point being tested, they will contribute a smaller amount to the field © 2009 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 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual 25 The motion of the electron in Example 21-16 is projectile motion In the case of the gravitational force, the acceleration of the projectile is in the same direction as the field and has a value of g; in the case of an electron in an electric field, the direction of the acceleration of the electron and the field direction are opposite, and the value of the acceleration varies 26 Initially, the dipole will spin clockwise It will “overshoot” the equilibrium position (parallel to the field lines), come momentarily to rest and then spin counterclockwise The dipole will continue to oscillate back and forth if no damping forces are present If there are damping forces, the amplitude will decrease with each oscillation until the dipole comes to rest aligned with the field 27 If an electric dipole is placed in a nonuniform electric field, the charges of the dipole will experience forces of different magnitudes whose directions also may not be exactly opposite The addition of these forces will leave a net force on the dipole Solutions to Problems Use Coulomb’s law to calculate the magnitude of the force 1.602 1019 C 26 1.602 1019 C Q1Q2 2 F  k  8.988 10 N  m C  2.7 103 N  12 r 1.5 10 m       Use Coulomb’s law to calculate the magnitude of the force Q1Q2 r2 1.602  10 C C   4.0  10 m 19   8.988  10 N  m 2 15 2  14 N The charge on the plastic comb is negative, so the comb has gained electrons  3.0  10 C  1.6021e 10  6 m m   Use Coulomb’s law to calculate the magnitude of the force 25  106 C 2.5  103 C Q1Q2 2 F  k  8.988  10 Nm C  7200 N r  0.28m2 F k     Use the charge per electron to find the number of electrons  electron   2.37  1014 electrons 38.0  106 C   19  1.602  10 C       9.109  1031 kg  19  C  1e    4.9  1016  4.9  1014% 0.035kg Since the magnitude of the force is inversely proportional to the square of the separation distance, F  , if the distance is multiplied by a factor of 1/8, the force will be multiplied by a factor of 64 r   F  64F0  64 3.2 102 N  2.0 N © 2009 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 Chapter 21 Electric Charges and Electric Field Since the magnitude of the force is inversely proportional to the square of the separation distance, F  , if the force is tripled, the distance has been reduced by a factor of r r 8.45 cm r   4.88 cm 3 Use the charge per electron and the mass per electron  electron   2.871  1014  2.9  1014 electrons 46  106 C   19  1.602  10 C     2.871  10 e   9.1091e10 14 kg  16   2.6  10 kg    31  To find the number of electrons, convert the mass to moles, the moles to atoms, and then multiply by the number of electrons in an atom to find the total electrons Then convert to charge  1mole Al   6.022  1023 atoms   79 electrons   1.602  1019 C  15kg Au  15kg Au      1molecule    mole electron  0.197 kg       5.8  108 C The net charge of the bar is 0C , since there are equal numbers of protons and electrons 10 Take the ratio of the electric force divided by the gravitational force QQ 2 19 k 12 8.988  10 N  m C 1.602  10 C FE kQ Q r     2.3 1039 11 31 27 2 mm FG Gm1m2 6.67 10 N  m kg 9.1110 kg 1.67 10 kg G 12 r The electric force is about 2.3 1039 times stronger than the gravitational force for the given scenario        11 (a) Let one of the charges be q , and then the other charge is QT  q The force between the charges is FE  k q  QT  q  d FE dq  2k r2  k   qQT  q2 To find the maximum and minimum force, set the r r2 first derivative equal to Use the second derivative test as well k dFE k   QT  2q    q  12 QT FE   qQT  q  ; r dq r 2   q  12 QT gives  FE  max So q1  q2  12 QT gives the maximum force (b) If one of the charges has all of the charge, and the other has no charge, then the force between them will be 0, which is the minimum possible force So q1  0, q2  QT gives the minimum force © 2009 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 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual 12 Let the right be the positive direction on the line of charges Use the fact that like charges repel and unlike charges attract to determine the direction of the forces In the following expressions, k  8.988  109 N  m2 C2   75C 48C ˆ  75C85C ˆ F75  k ik i  147.2 N ˆi  150 N ˆi  0.35m2  0.70 m2   75C 48C ˆ  48C85C ˆ F48  k ik i  563.5 N ˆi  560 N ˆi  0.35m2  0.35m2  85C 75C ˆ 85C 48C ˆ F85  k ik i  416.3 N ˆi  420 N ˆi 2 0.70 m 0.35m     13 The forces on each charge lie along a line connecting the charges Let the variable d represent the length of a side of the triangle Since the triangle is equilateral, each angle is 60o First calculate the magnitude of each individual force F12  k Q1Q2 d2   8.988  10 N  m C 2   7.0  10 C8.0  10 C 6  F13 1.20 m   F23 Q1Q3 d2   8.988  109 N  m2 C2  Q2 d Q3 d  F21  0.3495 N F13  k Q1 d 6  F12  F32  F31  7.0  10 C 6.0  10 C 6 6 1.20 m2  0.2622 N F23  k Q2Q3 d2   8.988 10 N  m C 2  8.0 10 C 6.0 10 C  0.2996 N  F 6 6 1.20 m 2 32 Now calculate the net force on each charge and the direction of that net force, using components F1 x  F12 x  F13 x    0.3495 N  cos 60o   0.2622 N  cos 60o  4.365  102 N F1 y  F12 y  F13 y    0.3495 N  sin 60o   0.2622 N  sin 60o  5.297  101 N F1  F12x  F12y  0.53N 1  tan 1 F1 y  tan 1 F1x 5.297  101 N  265 4.365  102 N F2 x  F21x  F23 x   0.3495 N  cos 60o   0.2996 N   1.249  101 N F2 y  F21 y  F23 y   0.3495 N  sin 60o   3.027  101 N F2  F22x  F22y  0.33 N 2  tan 1 F2 y F2 x  tan 1 3.027  101 N 1.249  101 N  112 F3 x  F31 x  F32 x    0.2622 N  cos 60o   0.2996 N   1.685  101 N F3 y  F31 y  F32 y   0.2622 N  sin 60o   2.271  101 N F3  F32x  F32y  0.26 N 3  tan 1 F3 y F3 x  tan 1 2.271  101 N 1.685  101 N  53 © 2009 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 Chapter 21 Electric Charges and Electric Field 14 (a) If the force is repulsive, both charges must be positive since the total charge is positive Call the total charge Q kQ  Q  Q  kQ Q Fd Q1  Q2  Q F  12   Q12  QQ1  0 d d k Q  Q2  Q1  Fd k Q  Q2       90.0  106 C    Fd k  90.0  10 C  6 2 4   8.988  109 N  m2 C2   12.0N 1.16 m 2  60.1  106 C , 29.9  10 6 C (b) If the force is attractive, then the charges are of opposite sign The value used for F must then be negative Other than that, the solution method is the same as for part (a) kQ  Q  Q  kQ Q Fd  Q12  QQ1  0 Q1  Q2  Q F  12  d d k Q  Q2  Q1  Fd k    Q  Q2    k   12  90.0  106 C  Fd 90.0  106 C  4  12.0N 1.16 m   8.988  10   N  m C2    106.8  106 C ,  16.8  106 C 15 Determine the force on the upper right charge, and then use the symmetry of the configuration to determine the force on the other three charges The force at the upper right corner of the square is the vector sum of the forces due to the other three charges Let the variable d represent the 0.100 m length of a side of the square, and let the variable Q represent the 4.15 mC charge at each corner F41  k F42  k F43  k Q2 d2 Q2 2d Q2  F41x  k Q2 d2  F42 x  k , F41 y  Q2 2d  F43 x  , F43 y  k Q1 2Q 4d , F42 y  k Q4  F41 d Q2 cos45o  k  F42  F43 Q3 2Q 4d Q2 d2 d2 Add the x and y components together to find the total force, noting that F4 x  F4 y F4 x  F41x  F42 x  F43 x  k F4  F42x  F42y  k Q2  Q2 d2 1  d  k 2Q 4d 0  k 2 Q2    k Q2  2     F4 y d2   1  2  d  2 © 2009 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 Physics for Scientists & Engineers with Modern Physics, 4th Edition   8.988  10 N  m C   tan 1 2   4.15  10 C 3  0.100 m  2 Instructor Solutions Manual     2.96  107 N   2  F4 y  45o above the x-direction F4 x For each charge, the net force will be the magnitude determined above, and will lie along the line from the center of the square out towards the charge 16 Determine the force on the upper right charge, and then use the symmetry of the configuration to determine the force on the other charges The force at the upper right corner of the square is the vector sum of the forces due to the other three charges Let the variable d represent the 0.100 m length of a side of the square, and let the variable Q represent the 4.15 mC charge at each corner Q2 Q2 F41  k  F41x  k , F41 y  d d F42  k F43  k Q2  F42 x  k 2d Q2 Q2 2d 2Q cos45  k o  F43 x  , F43 y  k 4d , F42 y  k  F41 Q1 Q4  F43 d Q2 2Q2  F42 Q3 4d Q2 d2 d2 Add the x and y components together to find the total force, noting that F4 x  F4 y Q2 F4 x  F41x  F42 x  F43 x  k F4  F42x  F42y  k d2 k Q2 d  0.64645 2Q 4d 2 k 0  k Q2 d2   8.988  10 N  m   tan 1 F4 y 2 Q2 d2   d2  1    0.64645k  F4 y  0.9142   4.15  10 C  0.9142  1.42  10 N C  3 Q2   0.100 m   225o from the x-direction, or exactly towards the center of the square F4 x For each charge, there are two forces that point towards the adjacent corners, and one force that points away from the center of the square Thus for each charge, the net force will be the magnitude of 1.42  107 N and will lie along the line from the charge inwards towards the center of the square 17 The spheres can be treated as point charges since they are spherical, and so Coulomb’s law may be used to relate the amount of charge to the force of attraction Each sphere will have a magnitude Q of charge, since that amount was removed from one sphere and added to the other, being initially uncharged F k Q1Q2 r2 k Q2 r2  Qr F k   0.12 m  1.7 102 N 8.988 109 N  m2 C2  electron  12   1.0 10 electrons 19  1.602 10 C   1.650 107 C  © 2009 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 Chapter 21 Electric Charges and Electric Field 18 The negative charges will repel each other, and so the third charge Q Q0 4Q0 must put an opposite force on each of the original charges Consideration of the various possible configurations leads to the x l–x conclusion that the third charge must be positive and must be between the other two charges See the diagram for the definition of variables l For each negative charge, equate the magnitudes of the two forces on the charge Also note that  x  l left: k k Q0Q k Q0Q x x 2 Q0Q k x2 k k 4Q02 l2 4Q0Q  l  x 4Q02 l 4Q0Q right: k  l  x 2 k 4Q02  l2  x  13 l  Q  4Q0 x2 l  Q0 Thus the charge should be of magnitude  3  94 Q0 Q0 , and a distance l from  Q0 towards  4Q0 19 (a) The charge will experience a force that is always pointing q Q Q towards the origin In the diagram, there is a greater force of dx dx Qq Qq to the left, and a lesser force of to 2 4  d  x  4  d  x  the right So the net force is towards the origin The same would be true if the mass were to the left of the origin Calculate the net force Qq Qq Qq  d  x    d  x   Fnet    2 2   4  d  x  4  d  x  4  d  x   d  x   4Qqd 4  d  x   d  x  2 x Qqd   d  x   d  x  2 x We assume that x  d Qqd Qq Fnet  x x 2  0d   d  x   d  x  This has the form of a simple harmonic oscillator, where the “spring constant” is kelastic  Qq  0d The spring constant can be used to find the period See Eq 14-7b T  2 m kelastic m Qq  2  0d  2 m 0d Qq (b) Sodium has an atomic mass of 23 T  2 m 0d Qq  2  29 1.66  1027 kg   8.85  1012 C2 1.60  10 19 C  N m2   10 10 m   1012 ps    0.24 ps  0.2 ps  1s   2.4  1013 s  © 2009 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 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual 20 If all of the angles to the vertical (in both cases) are assumed to   be small, then the spheres only have horizontal displacement, FT1 FT2 1 2 and so the electric force of repulsion is always horizontal Likewise, the small angle condition leads to tan   sin      for all small angles See the free-body diagram for each sphere,  FE1 m g F E2 m2g showing the three forces of gravity, tension, and the electrostatic force Take to the right to be the positive horizontal direction, and up to be the positive vertical direction Since the spheres are in equilibrium, the net force in each direction is zero (a)  F1x  FT1 sin 1  FE1   FE1  FT1 sin 1 F 1y  FT1 cos 1  m1 g  FT1  m1 g cos 1  FE1  m1 g cos 1 sin 1  m1 g tan 1  m1 g1 A completely parallel analysis would give FE2  m2 g Since the electric forces are a Newton’s third law pair, they can be set equal to each other in magnitude FE1  FE2  m1 g1  m2 g  1   m2 m1  (b) The same analysis can be done for this case FE1  FE2  m1 g1  m2 g  1   m1 m1  (c) The horizontal distance from one sphere to the other is s by the small angle approximation See the diagram Use the relationship derived above that FE  mg to solve for the distance Case 1: d  l 1  2   2l1  1  m1 g1  FE1  Case 2: kQ  2Q  d2 d  l 1  2   m1 g1  FE1  kQ  2Q  d2 d 2l 1/  4lkQ   mg  d   2l  mg  d l1  1   mg 2d 3l l 1 2 l l sin  l sin  2d 3l 1/  3lkQ    mg   d  21 Use Eq 21–3 to calculate the force Take east to be the positive x direction   F   E  F  qE  1.602  1019 C 1920 N C ˆi  3.08  1016 N ˆi  3.08  1016 N west q    22 Use Eq 21–3 to calculate the electric field Take north to be the positive y direction   F 2.18  1014 N ˆj E   1.36  105 N C ˆj  1.36  105 N C south 19 q 1.602  10 C 23 Use Eq 21–4a to calculate the electric field due to a point charge Q 33.0  106 C E  k  8.988  109 N  m2 C2  1.10  107 N C up r  0.164 m Note that the electric field points away from the positive charge   © 2009 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 10 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual Use the angle to calculate the distance in parsecs, and then convert to light years  3.26 ly  1 d  pc     3.704 pc  3.704 pc    12 ly   0.27  1pc  Convert the light years to parsecs, and then take the reciprocal of the number of parsecs to find the parallax angle in seconds of arc  1pc  65ly    19.94 pc  20 pc  sig fig   0.050  19.94 pc  3.26 ly  The reciprocal of the distance in parsecs is the angle in seconds of arc 1   0.01786  0.018 (a)    d  pc  56 pc o o  1o    4.961  106    5.0  106    3600  (b) 0.01786  The parallax angle is smaller for the further star Since tan   d D , as the distance D to the star increases, the tangent decreases, so the angle decreases And since for small angles, tan    , we have that   d D Thus if the distance D is doubled, the angle  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  85 pc    277 ly  280 ly  It takes light 280 years to reach us  1pc  The apparent brightness of an object is inversely proportional to the square of the observer’s distance from the object, given by Eq 44-1 To find the relative brightness at one location as compared to another, take a ratio of the apparent brightness at each location L bJupiter bEarth 2  d Earth    4 d Jupiter d Earth        0.037 L d Jupiter  d Jupiter   5.2  4 d Earth (a) The apparent brightness is the solar constant, 1.3  103 W m (b) Use Eq 44-1 to find the intrinsic luminosity L b  L  4 d 2b  4 1.496  1011 m 2 4 d   1.3  10  W m  3.7  1026 W The density is the mass divided by the volume M M 1.99  1030 kg     103 kg m  3 10 V r    10 m 3   © 2009 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 634 Chapter 44 Astrophysics and Cosmology 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 44-2, our Galaxy is about 100,000 ly in diameter Galaxy diameter 1.0  105 ly  tan 1  0.042 rad  2.4 o   tan 1 Distance to nearest galaxy 2.4  106ly Moon  tan 1 Moon diameter Distance to Moon  tan 1 3.48  106 m 3.84  108 m  9.1  103 rad  0.52 o The Galaxy width is about 4.5 times the Moon width 11 The Q-value is the mass energy of the reactants minus the mass energy of the products The masses are found in Appendix F He  42 He  48 Be   Q  2mHe c  mBe c    4.002603 u   8.005305 u  c 931.5 Mev c  0.092 MeV He  48 Be  12 C  Q  mBe c  mHe c  mC c   4.002603 u  8.005305 u  12.000000  c 931.5 Mev c   7.366 MeV 12 The angular width is the inverse tangent of the diameter of the Moon divided by the distance to the Sun o Moon diameter 3.48  106 m   tan 1  tan 1  2.33  105 rad  1.33  103  4.79 11 Distance to Sun 1.496  10 m   13 The density is the mass divided by the volume M 1.99  1030 kg M   Sun3   1.83  109 kg m 3  REarth  6.38  10 m V 3   Since the volumes are the same, the ratio of the densities is the same as the ratio of the masses  1.99  1030 kg M    3.33  105 times larger  Earth M Earth 5.98  1024 kg 14 The density of the neutron star is its mass divided by its volume Use the proton to calculate the density of nuclear matter The radius of the proton is taken from Eq 41-1 30 M 1.5 1.99  10 kg   5.354  1017 kg m3  5.4  1017 kg m  neutron  3 V star  11  10 m   neutron star  white   5.354  1017 kg m3 1.83  109 kg m    neutron  2.9  10 star  nuclear dwarf matter  5.354  1017 kg m 1.673  1027 kg  1.2  1015 m   2.3 © 2009 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 635 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual 15 Wien’s law (Eq 37-1) says that the PT   , where  is a constant, and so P1T1  P2T2 The Stefan–Boltzmann equation (Eq 19-17) says that the power output of a star is given by P   AT , where  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 The luminosity L is related to the apparent brightness b by Eq 44-1 It is given that b1 b2  0.091 , d1  d , P1  470 nm, and P2  720 nm T2 P1T1  P2T2  1 d 22 d12  0.091L2 L1 T1  P1 L1 L ; b1  0.091 b2   0.091 2 4 d1 4 d P2  0.091P2 P1  0.091 A2T24 4 r12T14 A1T14 0.091 4 r22T24    0.091  T24 r22 T14 r12  T     470 nm   0.091    0.091  P2   0.091    0.1285 r2  720 nm   T1   P1  r1 The ratio of the diameters is the same as the ratio of radii, so D1 D2  0.13 16 Wien’s law (Eq 37-1) says that the PT   , where  is a constant, and so P1T1  P2T2 The Stefan–Boltzmann equation (Eq 19-17) says that the power output of a star is given by P   AT , where  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 The luminosity L is related to the apparent brightness b by Eq 44-1 It is given that b1  b2 , r1  r2 , P1  750 nm, and P2  450 nm T2 P1T1  P2T2  T1 L1 b1  b2  4 d12   P1 P2 L2 4 d 22  A2T24 4 r22T24 T24  T2          d12 L1 P1  AT 4 r12T14 T14  T1  1 d 22 L2 P2   T      750       P1      2.8 d1  T1   P2   450  d2 The star with the peak at 450 nm is 2.8 times further away than the star with the peak at 750 nm 17 The Schwarzschild radius is REarth  2GM Earth c  2GM c2 6.67  1011 N m kg   3.00 10 18 The Schwarzschild radius is given by R   5.98 10 m s 2GM c2 24 kg   8.86 10 3 m  8.9 mm An approximate mass for our Galaxy is calculated in Example 44-1 11 2 41 2GM 6.67  10 N  m kg  10 kg    1014 m R c2 3.00  10 m s      © 2009 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 636 Chapter 44 Astrophysics and Cosmology 19 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 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 sum to 540o 20 To just escape from an object, the kinetic energy of the body at the surface of the body must be equal to the magnitude of the gravitational potential energy at the surface Use Eq 8-19 vesc  2GM RSchwarzchild 2GM  2GM c  c 21 We find the time for the light to cross the elevator, and then find how far the elevator moves during that time due to its acceleration 9.80 m s  2.4 m  g  x  x ; y  g  t   t    1.3  10 16 m 2 2c c 3.00  10 m s     Note that this is smaller than the size of a proton 22 Use Eq 44-4, Hubble’s law v 1850 km s v  Hd  d    84 Mly  8.4  107 ly H 22 km s Mly 23 Use Eq 44-4, Hubble’s law v  Hd  d  v H   0.015   3.00  108 m s  2.2  10 m s Mly  204.5 Mly  2.0  10 Mly  2.0  108 ly 24 (a) Use Eq 44-6 to solve for the speed of the galaxy z obs  rest rest   455 nm  434 nm  v  v  c   0.04839c  0.048 c 434 nm c   (b) Use Hubble’s law, Eq 44-4, to solve for the distance v 0.04839 3.00  10 m s   660 Mly  6.6  108 ly v  Hd  d  H 22000 m s Mly   25 We find the velocity from Hubble’s law, Eq 44-4, and the observed wavelength from the Doppler shift, Eq 44-3 v Hd  22000 m s Mly  7.0 Mly     5.133  104 (a) c c 3.00  10 m s   0 1 v c 1 v c   656 nm   5.133  104  5.133  104  656.34 nm  656 nm © 2009 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 637 Physics for Scientists & Engineers with Modern Physics, 4th Edition (b) v c  Hd c   0   22000 m s Mly  70 Mly  3.00  10 m s 1 v c 1 v c  5.133  103  5.133  103   656 nm  Instructor Solutions Manual  5.133  103  659.38 nm  659 nm 26 Use Eqs 44-3 and 44-4 to solve for the distance to the galaxy 2 obs  rest 1 v c  vc obs  rest 1 v c obs  rest     d H H  v 2 c obs  rest obs  rest      3.00  10 m s   423.4 nm    393.4 nm     2.2  10 m s Mly   423.4 nm    393.4 nm   2 2  1.0  103 Mly  1.0  109 ly 27 Use Eqs 44-3 and 44-5a to solve for the speed of the galaxy z obs  rest obs 1 v c  1  1  rest rest 1 v c  z  1  1.0602     0.05820 c  z  1  1.0602  v  v  0.058 c The approximation of Eq 44-6 gives v  zc  0.060 c 28 Use Eqs 44-3 and 44-5a to solve for the redshift parameter z obs  rest obs 1 v c  0.075  1  1    0.078 1 v c  0.075 rest rest Or, we use the approximation given in Eq 44-6 v z   0.075 c 29 Eq 44-3 states   rest   rest 1 v c 1 v c 1 v c 1/  v  rest    1 v c  c 1  v     c 1/    rest   v   v   v   rest    12    rest     rest  rest c  c   c  v  v  v         rest    c  c c     rest    rest v c   rest  v c © 2009 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 638 Chapter 44 Astrophysics and Cosmology 30 For small relative wavelength shifts, we may use Eq 44-6 to find the speed We use Eq 44-4 to find the distance v   c rest d  vc  rest ; v  Hd  v c   3.00  108 m s  0.10 cm       65 Mly H H rest  22, 000 m s Mly   21cm  31 Wien’s law is given in Eq 37-1 PT  2.90  10 mK  P  3 2.90  103 mK T  2.90  103 m K 2.7 K  1.1  103 m 32 We use Wien’s law, Eq 37-1 From Figure 44-30, the temperature is about 1010 K 2.90  103 mK 2.90  103 m K PT  2.90  103 mK  P     1013 m  10 10 K T From Figure 31-12, that wavelength is in the gamma ray region of the EM spectrum 33 We use the proton as typical nuclear matter  26 kg   1nucleon   nucleons m3  10   27 m   1.67  10 kg   34 If the universe’s scale is inversely proportional to the temperature, 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 44-30 to estimate the temperature at various times For purposes of illustration, we assume the universe has a current size of about 1010 ly There will be some variation in the answer due to reading the figure (a) At t  106 yr , the temperature is about 1000 K Thus the scale is found as follows  Scale  Temperature      Scale  Temperature  1000   103   Size   10 3 1010 ly   107 ly (b) At t  s , the temperature is about 1010 K 3 Scale   10   1010 Temperature 10     Size   1010 1010 ly  3ly (c) At t  106 s , the temperature is about 1013 K 3 Scale   13  1013  Temperature 10    Size   1013 1010 ly   103 ly   1013 m 35 (d) At t  10 s , the temperature is about 1027 K 3 Scale   27   1027 Temperature 10    Size   10 27 1010 ly   10 17 ly  0.3 m © 2009 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 639 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual 35 We approximate the temperature–energy relationship by kT  E  mc as suggested on page 1217 mc kT  mc  T  (a) T  mc k k  500 MeV c  c 1.60  10  2 1.38  10 23 13 J MeV J K    10 12 K From Figure 44-30, this corresponds to a time of  105 s (b) T  mc k  9500 MeV c  c 1.60  10  1.38  10 23 13 J MeV J K    10 14 K From Figure 44-30, this corresponds to a time of  107 s (c) T  mc k 100 MeV c  c 1.60  10  2 1.38  10 23 13 J MeV J K    10 12 K From Figure 44-30, this corresponds to a time of  104 s There will be some variation in the answers due to reading the figure 36 (a) According to the text, near Figure 44-33, the visible matter makes up about one-tenth of the total baryonic matter The average baryonic density is therefore 10 times the density of visible matter M  baryon  10  visible  10 visible3 R 10  10 11    14  10 ly  9.46  10 galaxies 1011 stars galaxy 2.0  1030 kg star 15  m ly    2.055 1026 kg m3  2.1  1026 kg m3 (b) Again, according to the text, dark matter is about times more plentiful than normal matter  dark   baryon   2.055  1026 kg m   8.2  1026 kg m3 37 (a) From page 1201, a white dwarf with a mass equal to that of the Sun has a radius about the size of the Earth’s radius, 6380 km From page 1202, a neutron star with a mass equal to 1.5 solar masses has a radius of about 20 km For the black hole, we use the Schwarzschild radius formula R 2GM c     m s 6.67  1011 N  m kg 1.99  1030 kg  3.00  10   8849m  8.85 km (b) The ratio is 6380 : 20 : 8.85  721 : 2.26 :  700 : : © 2009 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 640 Chapter 44 Astrophysics and Cosmology 38 The angular momentum is the product of the rotational inertia and the angular velocity  I  initial   I  final  final  MRinitial  I   Rinitial    108 m  1rev month   initial  initial   initial     initial        10 m   I final   Rfinal   MRfinal   7.66  109 rev month  7.66  109 rev month  1month 30 d  1d 24 h  1h 3600s  2953 rev s  3000 rev s 39 The rotational kinetic energy is given by I  The final angular velocity, from problem 43, is 7.66  109 rev month K final K initial  2 I finalfinal I initialinitial  2 MRfinal final 5 2 initial MRinitial  R     final final   Rinitialinitial     103 m  7.66  109 rev month       10  10 m 1rev month        40 The apparent luminosity is given by Eq 44-1 Use that relationship to derive an expression for the absolute luminosity, and equate that for two stars L  L  4 d 2b b 4 d 2 bdistant  4 d Sun bSun  Ldistant  LSun  4 d distant star star d distant  d Sun star lSun ldistant star   1.5  1011 m   1ly   5ly  15 11  10  9.461  10 m  star 41 A: The temperature increases, the luminosity stays the same, and the size decreases B: The temperature stays the same, the luminosity decreases, and the size decreases C: The temperature decreases, the luminosity increases, and the size increases 42 The power output is the energy loss divided by the elapsed time 2 K K initial  fraction lost  21 I   fraction lost  12 25 MR   fraction lost  P    t t t t    30 1.5 1.99  10 kg 8.0  10 m 43 Use Newton’s law of universal gravitation F G r2 2 9 1d  24 h d  3600 s h  m1m2   2 rad s  1  10   1.74610   6.67  1011 N m kg  10 kg     10 ly  9.46  10 41 25 W  1.7  1025 W 15  m ly   1.68  1028 N   1028 N © 2009 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 641 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual 44 (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   0.12 nucleon m kg nucleon 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  46 eV c neutrino 1.66  1027 kg 1.2  108 neutrino m3 (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   0.30 nucleon m3 kg nucleon neutrino number density  10  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 45 The temperature of each star can be found from Wien’s law PT  2.90  103 m K  T660  2.90  103 m K  4390 K T480  2.90  103 m K 660  109 m 480  109 m The luminosity of each star can be found from the H–R diagram L660   1025 W L480   1026 W  6040 K The Stefan–Boltzmann equation says that the power output of a star is given by P   AT , where  is a constant, and A is the radiating area The P in the Stefan–Boltzmann equation is the same as the luminosity L given in Eq 44-1 Form the ratio of the two luminosities L480 L660  4 T480  A480T480 4 r480  4 4 r660 T660  A660T660  r480 r660  L480 T660 L660 T480   1026 W  4390 K   1025 W  6040 K   1.67 The diameters are in the same ratio as the radii d 480  1.67  1.7 d 660 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 46 (a) The number of parsecs is the reciprocal of the angular resolution in seconds of arc o  1     2.78  106    106 o     0.01   100 parsec       60   60      © 2009 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 642 Chapter 44 Astrophysics and Cosmology (b) We use the Rayleigh criterion, Eq 35-10, which relates the angular resolution to the diameter of the optical element We choose a wavelength of 550 nm, in the middle of the visible range 1.22 550  109 m 1.22 1.22  D = =13.83m  14 m  D   2.78  106    rad 180      The largest optical telescopes currently in use are about 10 m in diameter 47 We approximate the temperature–kinetic energy relationship by kT  K as given on page 1217 kT  K  T  K 1.96  10 12   eV 1.60  1019 J eV 1.38  10 k From Figure 44-30, this is in the hadron era 23 J K   1016 K 48 We assume that gravity causes a centripetal force on the gas Solve for the speed of the rotating gas, and use Eq 44-6 mgas mblack mgas vgas hole Fgravity  Fcentripetal  G   r2 r mblack hole vgas  G z v c  r   6.67  10 6.42  105 m s 11 N  m kg    10 1.99  10 30 kg   6.42  10  9.46  10 m   68ly    1ly   15 m s  2.14  103   103 3.00  10 m s 49 (a) To find the energy released in the reaction, we calculate the Q-value for this reaction From Eq 42-2a, the Q-value is the mass energy of the reactants minus the mass energy of the products The masses are found in Appendix F   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 41-1 Each nucleus will have half the total kinetic energy qnucleus 1/ 1/ r  1.2  10 15 m  A   1.2  1015 m 12  U 4 r   K  12 U   2  qnucleus 4 2r 8.988  10   2 1.60  1019 C   1MeV   1.2  1015 m 12 1/  1.60  1013J   4.711MeV    Nn C  4.7 MeV (c) We approximate the temperature–kinetic energy relationship by kT  K as given on page 1217 13 K  4.711MeV  1.60  10 J MeV kT  K  T    5.5  1010 K 23 1.38  10 J K k   © 2009 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 643 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual 50 (a) Find the Q-value for this reaction From Eq 42-2a, the Q-value is the mass energy of the reactants minus the mass energy of the products 16 28 O  168 O  14 Si  24 He  Q  2mO c  mSi c  mHe c   15.994915 u   27.976927 u  4.002603 c 931.5 Mev c   9.594 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 41-1 Each nucleus will have half the total kinetic energy qnucleus 1/3 1/3 15 15 r  1.2  10 m  A   1.2  10 m 16  U 4 2r   K nucleus  12 U   2  qnucleus 4 2r 8.988 10 Nn  1.60 10 C  C  1.2  10 m  16  8 2 19 1/3 15  1eV 1.60  1019 J  7.609 MeV  7.6 MeV (c) We approximate the temperature–kinetic energy relationship by kT  K as given on page 1217  1.60  1019 J  7.609  106 eV   1eV K    8.8  1010 K kT  K  T   23 k 1.38  10 J K   51 We treat the energy of the photon as a “rest mass,” and so mphoton "  " Ephoton c To just escape from a spherical mass M of radius R, the energy of the photon must be equal to the magnitude of the gravitational potential energy at the surface GMmphoton GMmphoton GM Ephoton c GM  R   Ephoton  R Ephoton Ephoton c   52 We use the Sun’s mass and given density to calculate the size of the Sun M M    V  rSun 1/  3M  rSun     4  rSun d Earth-Sun   1.99  1030 kg    26   4 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 d galaxy  382 ly 100,000 ly   103 © 2009 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 644 Chapter 44 Astrophysics and Cosmology 53 We approximate the temperature–kinetic energy relationship by kT  K as given on page 1217 kT  K  T  K k 14  10  12  eV 1.60  1019 J eV 1.38  10 23 J K   1.6  10 17 K From Figure 44-30, this might correspond to a time around 1015 s Note that this is just a very rough estimate due to the qualitative nature of Figure 44-30 54 (a) We consider the photon as entering from the left, grazing the Sun, and moving off in a new direction The deflection is assumed to be very small In particular, we consider a small part of the motion in which the photon moves a horizontal distance dx  cdt while located at (x,y) relative to the center of the Sun Note that y  R and r  x  y If the photon has energy E, it will have a “mass” of m  E c , and a p E  mc c x  r R yR  F GMm r2 momentum of magnitude p  E c  mc To find the change of momentum in the y-direction, we use the impulse produced by the y-component of the gravitational force GMm dx GMm R dx GMmR dx dp y  Fy dt   cos    3/2 r c r r c c x2  R2   To find the total change in the y-momentum, we integrate over all x (the entire path of the photon) We use an integral from Appendix B-4   p y    GMmR dx x c R  3/  GMmR c x R x R   1/ 2GMm cR  2GMp c2 R  The total magnitude of deflection is the change in momentum divided by the original momentum 2GMp p y 2GM    c R  c R p p (b) We use data for the Sun  Nm   6.67  10 11 1.99  1030 kg  kg  2GM     c R 3.00  108 m s 6.96  108 m       180   3600     0.87   rad   1   4.238  10 6 rad  © 2009 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 645 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual 55 Because Venus has a more negative apparent magnitude, Venus is brighter We write the logarithmic relationship as follows, letting m represent the magnitude and b the brightness m  k log b ; m2  m1  k  log b2  log b1   k log  b2 b1   k m2  m1 log  b2 b1   5  2.5 log  0.01 b   10 b1 m2  m1  k log  b2 b1  m2  m1 k b  Venus  10 bSirius 56 If there are N nucleons, we assume that there are approximately mVenus  mSirius 2.5  10 N neutrons and 4.4 1.4 2.5  16 N protons Thus, for the star to be neutral, there would also be N electrons (a) From Eq 40-12 and 40-13, we find that if all electron levels are filled up to the Fermi energy EF , the average electron energy is 53 EF Ee  N e  EF    N  5 h2  Ne  8me   V  2/3   h2  N    N     8me   2V  2/3 (b) The Fermi energy for nucleons would be a similar expression, but the mass would be the mass of a nucleon instead of the mass of the electron Nucleons are about 2000 times heavier than electrons, so the Fermi energy for the nucleons would be on the order of 1/1000 the Fermi energy for the electrons We will ignore that small correction To calculate the potential energy of the star, think about the mass in terms of shells Consider the inner portion of the star with radius r < R and mass m, surrounded by a shell of thickness dr and mass dM See the diagram From Gauss’s law applied to gravity, the gravitational effects of the inner portion of the star on the shell are the same as if all of its mass were at the geometric center Likewise, the spherically-symmetric outer portion of the star has no gravitational effect on the shell Thus the gravitational energy of the inner portion–shel) combination is given by a form of Eq 8-17, dU  G is given by   dr R r mdM The density of the star r M We use that density to calculate the masses, and then integrate over the  R3 full radius of the star to find the total gravitational energy of the star M r3 m   43  r  r M   R3  R3      dM   4 r dr  dU  G mdM r  M  G  R3 M  4 r dr   3Mr R r 3Mr R3 R3 r dr  dr 3GM R r dr © 2009 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 646 Chapter 44 Astrophysics and Cosmology R  3GM U    R 3GM 3GM R   GM R6 R 0 R6 5 R  (c) The total energy is the sum of the two terms calculated above The mass of the star is primarily due to the nucleons, and so M  Nmnucleon  Etotal  Ee  U   3h  r  dr   3  5 M  80me  mnucleon 92/33h M 5/3 Let a   h 3 N     8me   2V  5/3 nucleon 2/3  N   M     2mnucleon  R  and b  320 me m the equilibrium radius 4/3 r dr   GM R 2/3  GM  R a GM , so Etotal    R  b R 92/33h M 5/3 5/3 R2 320 4/3 me mnucleon  GM R We set dEtotal dR  to find 92/33h M 5/3 5/3 320 4/3 me mnucleon 92/3 h  5/3 dR R3 R b 32 4/3GM 1/3 me mnucleon GM We evaluate the equilibrium radius using the Sun’s mass 92/3 h Req  5/3 32 4/3GM 1/3 me mnucleon dEtotal 2a   b   Req  2a   92/3 6.63  1034 J s   Nm   32 4/3  6.67  1011 kg 30   2.0  10 kg   1/3   9.11 10 31  kg 1.67  1027 kg  5/3  7.178  106 m  7.2  103 km 57 There are N neutrons The mass of the star is due only to neutrons, and so M  Nmn From Eqs 40-12 and 40-13, we find that if all energy levels are filled up to the Fermi energy EF , the average energy is 53 EF We follow the same procedure as in Problem 56 The expression for the gravitational energy does not change En  N n  EF    N  Etotal  En  U  Let a  18  2/3 160 18  8/3 n m h2  N    8mn   V  2/3 h M 5/3 160 4/3 mn8/3 R h M 5/3 4/3 and b   2/3 M   mn  3h   40mn 3 M       R mn  2/3  18  2/3 h M 5/3 160 4/3 mn8/3 R GM R GM , so Etotal  a R  b R We set dEtotal dR  to find the equilibrium radius © 2009 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 647 Physics for Scientists & Engineers with Modern Physics, 4th Edition 18  2/3 Instructor Solutions Manual h M 5/3 160 4/3 mn8/3 18  h  3 dR R R b 16 4/3GM 1/3 mn8/3 GM We evaluate the equilibrium radius for a mass of 1.5 solar masses dEtotal Req   2a  18  2/3 b   Req  2a 2/3  h2 16 4/3GM 1/3 mn8/3  182/3 6.63  1034 J s   Nm   16 4/3  6.67  1011 kg  30  1.5  2.0  10 kg    1/3 1.67 10 27 kg  8/3  1.086  104 m  11km 58 We must find a combination of c, G, and  that has the dimensions of time The dimensions of c  L3  L  ML2  are   , the dimensions of G are  , and the dimensions of  are   2 T   MT   T      3   2  L   L   ML  t P  c G   T        L  M    T         T   MT   T    3  2  ;     ;           5  ;   1  3     5  1  3    tP  c 5/ 1/ 1/ G   G c5  ;  ;    52  6.67  10 11  21  6.63  10  3.00 10 m s  N m kg 34 J s   5.38  10 44 s © 2009 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 648 ...  k F43  k Q2 d2 Q2 2d Q2  F41x  k Q2 d2  F 42 x  k , F41 y  Q2 2d  F43 x  , F43 y  k Q1 2Q 4d , F 42 y  k Q4  F41 d Q2 cos45o  k  F 42  F43 Q3 2Q 4d Q2 d2 d2 Add the x and y components...  sin 60o   3. 027  101 N F2  F22x  F22y  0.33 N 2  tan 1 F2 y F2 x  tan 1 3. 027  101 N 1 .24 9  101 N  1 12 F3 x  F31 x  F 32 x    0 .26 22 N  cos 60o   0 .29 96 N   1.685... k , F41 y  d d F 42  k F43  k Q2  F 42 x  k 2d Q2 Q2 2d 2Q cos45  k o  F43 x  , F43 y  k 4d , F 42 y  k  F41 Q1 Q4  F43 d Q2 2Q2  F 42 Q3 4d Q2 d2 d2 Add the x and y components together