chapter 44 pdf solutions to physics engineering

8/13/2019 Chapter 44 PDF Solutions to Physics Engineering

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CHAPTER 44: Astrophysics and Cosmology

Responses to Questions

1. The Milky Way appears murky or milky to the naked eye, and so before telesopes !ere used it

!as thou"ht to be loud#like. When $ie!ed !ith a telesope, muh of the murkiness is resol$edinto stars and star lusters, so !e no lon"er onsider the Milky Way to be milky.

2. %f a star "enerates more ener"y in its interior than it radiates a!ay, its temperature !ill inrease.

&onse'uently, there !ill be "reater out!ard pressure opposin" the "ra$itational fore direted

in!ard. To re"ain e'uilibrium, the star !ill e(pand. %f a star "enerates less ener"y than it radiatesa!ay, then its temperature !ill derease. There !ill be a smaller out!ard pressure opposin" the

"ra$itational fore direted in!ard, and, in order to re"ain e'uilibrium, the star !ill ontrat.

). *ed "iants are e(tremely lar"e stars !ith relati$ely ool surfae temperatures, resultin" in theirreddish olors. These stars are $ery luminous beause they are so lar"e. When the +un beomes a

red "iant, for instane, its radius !ill be on the order of the distane from the arth to the +un. - red

"iant has run out of hydro"en in its inner ore and is fusin" hydro"en to helium in a shellsurroundin" the ore. *ed "iants ha$e left their main se'uene positions on the /* dia"ram and

mo$ed up 0more luminous and to the ri"ht 0ooler.

4. - star mo$in" alon" arro! 1 !ould inrease in luminosity

!hile maintainin" the same surfae temperature. %t !ould

therefore also ha$e to inrease in si3e, sine eah s'uare

meter of its surfae !ould ha$e the same olor and therefore

same ener"y output as before. - star mo$in" alon" arro! 2

!ould inrease in luminosity and derease in temperature. %t

!ould also inrease in si3e, sine it !ould need to produe a

"reater luminosity e$en thou"h eah unit area of its surfae

!ould no! be produin" less ener"y. - star mo$in" alon"arro! ) maintains the same luminosity !hile inreasin" its

surfae temperature. %t !ill beome smaller, sine a unit area

of this star !ill inrease its ener"y and therefore a smaller

o$erall area !ill be needed to maintain the same luminosity.

- star mo$in" alon" arro! 4 dereases in both surfae temperature and luminosity. inally, a star

mo$in" alon" arro! 5 !ill derease in luminosity !hile maintainin" the same surfae temperature

and dereasin" in si3e. ote that these arro!s do not neessarily represent natural paths for stars

on the /* dia"ram.

5. The /* dia"ram is a plot of luminosity $ersus surfae temperature of a star and therefore does not

diretly tell us anythin" about the ore of a star. o!e$er, !hen onsidered in on6untion !ith

theories of stellar e$olution, the /* dia"ram does relate to the interior of a star. or instane, allmain se'uene stars are fusin" hydro"en to helium in their ores, so the loation of a partiular star

on the main se'uene does "i$e us that information.

7. The fate of a star depends on the mass of the star remainin" after the red "iant phase. %f the mass is

less than about 1.4 solar masses, the star !ill beome a !hite d!arf. %f the mass is "reater than thislimit, than the e(lusion priniple applied to eletrons is not enou"h to hold the star up a"ainst its

o!n "ra$ity and it ontinues to ontrat, e$entually beomin" a neutron star or, if its mass at this

sta"e is more than t!o or three solar masses, a blak hole.

2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 4thEdition Instructor Solutions Manual

9. :es. otter stars are found on the main se'uene abo$e and to the left of ooler stars. %f /*

dia"rams of lusters of stars are ompared, it is found that older lusters are missin" the upper left

portions of their main se'uenes. -ll the stars in a "i$en luster are formed at about the same time,

and the absene of the hotter main se'uene stars in a luster indiates that they ha$e shorter li$es

and ha$e already used up their ore hydro"en and beome red "iants. %n fat, the turn#off point, orpoint at !hih the upper end of the main se'uene stops, an be used to determine the a"es of

lusters.

;. The baseline used in measurin" paralla(es from the arth is the distane from the arth to the +un.

0+ee i"ure 44#11. %f you !ere measurin" paralla(es from the Moon instead, you !ould need to

make a sli"ht orretion based on the position of the Moon !ith respet to the arth/+un line at the

time of the measurement. %f you !ere measurin" paralla(es from Mars, you !ould need to use the

distane from Mars to the +un as the baseline. %n addition, you !ould need to !ait half a Martian

year bet!een measurements instead of half an arth year.

8. Wath the star o$er a period of se$eral days and determine its period throu"h obser$ation. eterminin" the natureof a "eodesi, for instane by obser$in" the motion of a body or li"ht near a lar"e mass, !ill help

determine the nature of the ur$ature of spae/time there.

11. %f the redshift of spetral lines of "ala(ies !ere diso$ered to be due to somethin" other than

e(pansion of the uni$erse, then the ?i" ?an" theory and the idea that the uni$erse is e(pandin"

!ould be alled into 'uestion. o!e$er, the e$idene of the osmi bak"round miro!a$e

radiation !ould onflit !ith this $ie!, unless it too !as determined to result from some ause other

than e(pansion.

12. o. %n an e(pandin" uni$erse, all "ala(ies are mo$in" a!ay from all other "ala(ies on a lar"e sale.

0@n a small sale, nei"hborin" "ala(ies may be "ra$itationally bound to eah other. Therefore, the

$ie! from any "ala(y !ould be the same. @ur obser$ations do not indiate that !e are at the enter.

0+ee i"ure 44#2).

1). They !ould appear to be reedin". %n an e(pandin" uni$erse, the distanes bet!een "ala(ies are

inreasin", and so the $ie! from any "ala(y is that all other "ala(ies are mo$in" a!ay.

14. -n e(plosion on arth !ould be affeted by the arthAs "ra$ity and air resistane. ah piee ofdebris !ould at like a pro6etile, !ith its indi$idual initial $eloity. More distant partiles !ould not

spread at a hi"her speed. This orresponds some!hat to a losed uni$erse, in !hih the "ala(iese$entually stop and then all ome bak to"ether a"ain. %n the ase of the e(plosion on arth, most of

the partiles !ould e$entually stop. Most !ould land on the "round. +ome mi"ht esape into spae.

The partiles !ould not all reassemble, as in the bi" runh.

15. ?lak holes ha$e tremendous "ra$ity, so !e an detet them by the "ra$itational defletion of other

ob6ets in their $iinity. -lso, matter aeleratin" to!ard a blak hole "i$es off (#rays, !hih an be

deteted. %n addition, "ra$itational lensin", the bendin" of li"ht omin" from stars and "ala(ies

loated behind the blak hole, an indiate that the blak hole is present.


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&hapter 44 Astrophysics and Cosmology

17. R !"M#c$, soM Rc$#!".

MB 05.28 ( 1=#110).== ( 1=;CDE207.79 ( 1=#11F B ).59 ( 1=17k"

19. ?oth the formation of the arth and the time durin" !hih people ha$e li$ed on arth are on the far

ri"ht ed"e of i"ure 44#)=, in the era of dark ener"y.

1;. The 2.9 G osmi miro!a$e bak"round radiation is the remnant radiation of the ?i" ?an". -s the

uni$erse e(panded, the !a$elen"ths of the ?i" ?an" radiation len"thened and beame redshifted.

The 2.9 G blakbody ur$e peaks at a !a$elen"th of about 9.)5 m, in the miro!a$e re"ion. The

temperature of this radiation is lo! beause the ener"y spread out o$er an inreasin"ly lar"e $olume

as the uni$erse e(panded.

18. The early uni$erse !as too hot for atoms to e(ist. The a$era"e kineti ener"ies of partiles !ere hi"h

and fre'uent ollisions pre$ented eletrons from remainin" !ith nulei.

2=. 0a Type %a superno$ae ha$e a ran"e of luminosities that an be e(trated from their obser$able

harateristis and an be deri$ed from the rate at !hih they bri"hten and fade a!ay.

0% The distane to a superno$a an be determined by omparin" the relati$e intensity to theluminosity.

21. The initial ?i" ?an" !as not perfetly symmetri. >e$iations in the symmetry enabled the

de$elopment of "ala(ies and other strutures.

22. %f the a$era"e mass density of the uni$erse is abo$e the ritial density, then the uni$erse !ill

e$entually stop its e(pansion and ontrat, ollapsin" on itself and endin" finally in a bi" runh.

This senario orresponds to a losed uni$erse, or one !ith positi$e ur$ature.

2). %f there !ere 9 protons for e$ery neutron, and it takes t!o protons and t!o neutrons to reate a

sin"le helium nuleus, then for e$ery helium nuleus there !ould be 12 hydro"en nulei. +ine the

mass of helium is four times the mass of hydro"en, the ratio of the total mass of hydro"en to the

total mass of helium should be 12H4, or )H1.

24. 0a Ira$ity bet!een "ala(ies should be pullin" the "ala(ies bak to"ether, slo!in" the e(pansion ofthe uni$erse.

0% -stronomers ould measure the redshift of li"ht from distant superno$ae and dedue the

reession $eloities of the "ala(ies in !hih they lie. ?y obtainin" data from a lar"e number of

superno$ae, they ould establish a history of the reessional $eloity of the uni$erse, and

perhaps tell !hether the e(pansion of the uni$erse is slo!in" do!n.

Solutions to Problems

1. &on$ert the an"le to seonds of ar, reiproate to find the distane in parses, and then on$ert to

li"ht years.

( )

( )

o4

o

)7==2.8 1= 1.=44

1

1 1 ).27 lyp =.85;p ).1ly

1.=44 1pd

= =

= = = =

2.


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( )1 1 ).27 ly

p ).9=4p ).9=4 p 12 ly=.29 1p

d

= = = =

). &on$ert the li"ht years to parses, and then take the reiproal of the number of parses to find the

paralla( an"le in seonds of ar.

( )1p 1

75ly 18.84 p 2= p 2 si". fi". =.=5=).27ly 18.84 p

= = =

4. The reiproal of the distane in parses is the an"le in seonds of ar.

0a( )

1 1=.=19;7 =.=1;

p 57 pd = = =

0% ( ) ( )o

o o7 71

=.=19;7 4.871 1= 5.= 1=)7==

=

5. The paralla( an"le is smaller for the further star. +ine tan ,d = as the distaneto the starinreases, the tan"ent dereases, so the an"le dereases. -nd sine for small an"les, tan , !eha$e that .d Thus if the distaneis doubled, the an"le !ill be smaller by a fator of 2 .

7. ind the distane in li"ht years. That $alue is also the time for li"ht to reah us.

).27ly;5 p 299 ly 2;= ly %t takes li"ht 2;= years to reah us.

1p=

9. The apparent bri"htness of an ob6et is in$ersely proportional to the s'uare of the obser$erAs distane

from the ob6et, "i$en by '. 44#1. To find the relati$e bri"htness at one loation as ompared to

another, take a ratio of the apparent bri"htness at eah loation.

222 2

Jupiter Jupiter arth arth

2

arth Jupiter Jupiter 2

arth

4 1=.=)9

5.2

4

'

% d d d

'% d d

d

= = = = =

;. 0a The apparent bri"htness is the solar onstant,) 2

1.) 1= W m .

0%


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5

1 1 o

7

7

1 1 ) o

Moon ;

Iala(y diameter 1.= 1= lytan tan =.=42 rad 2.4

>istane to nearest "ala(y 2.4 1= ly

Moon diameter ).4; 1= mtan tan 8.1 1= rad =.52

>istane to Moon ).;4 1= m

= = =

= = =

The Iala(y !idth is about 4.5 times the Moon !idth.

11. The )#$alue is the mass ener"y of the reatants minus the mass ener"y of the produts. The masses

are found in -ppendi( .

( )[ ] ( )

[ ] ( )

4 4 ;

2 2 4

2 2 2 2

e ?e

4 ; 12

2 4 7

2 2 2 2 2

?e e &

e e ?e

2 2 4.==27=)u ;.==5)=5 u 8)1.5 Me$ =.=82 MeK

e ?e &

4.==27=)u ;.==5)=5u 12.====== 8)1.5Me$

9.)77 MeK

) m c m c c c

) m c m c m c c c

+

= = =

+

= + = +

=

12. The an"ular !idth is the in$erse tan"ent of the diameter of the Moon di$ided by the distane to the

+un.

( )7

o1 1 5 )

11

Moon diameter ).4; 1= mtan tan 2.)) 1= rad 1.)) 1= 4.98

>istane to +un 1.487 1= m

= = =

1). The density is the mass di$ided by the $olume.

( )

)=

8 )+un

))4 74arth) )

1.88 1= k"1.;) 1= k" m

7.); 1= m

MM

( R

= = = =

+ine the $olumes are the same, the ratio of the densities is the same as the ratio of the masses.)=

5

24

arth arth

1.88 1= k").)) 1= times lar"er

5.8; 1= k"

M

M

= = =

14. The density of the neutron star is its mass di$ided by its $olume.


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!here is a onstant, andAis the radiatin" area. ThePin the +tefan/?olt3mann e'uation is the

same as the luminosity'in this hapter. The luminosity'is related to the apparent bri"htness %by

'. 44#1. %t is "i$en that 1 2 =.=81% % = , 1 2d d= , L1 49=nm, = and L2 92=nm. =

2 L1 1 2

L1 1 L2 2 1 2 2 2

1 L2 1 2

=.=81 =.=814 4

* ' '

* * % %* d d

= = = =

( ) 2 42 4 4 22 22 2 2 2 2 2 22 4 2 4 4 2

1 1 1 1 1 1 1 1 1

2 2 2

1 2 2

2 1 1

=.=81 4=.=81 =.=81 =.=811 =.=81

4

49=nm=.=81 =.=81 =.=81 =.12;5

92=nm

r *d ' P A * * r

d ' P A* r * * r

r *

r *

= = = = = =

= = = =

The ratio of the diameters is the same as the ratio of radii, so1

2

=.1) .

=

17. WienAs la! 0'. )9#1 says that the * = , !here is a onstant, and so 1 1 2 2.* * = The

+tefan/?olt3mann e'uation 0'. 18#19 says that the po!er output of a star is "i$en by

4

P A*= ,!here is a onstant, andAis the radiatin" area. ThePin the +tefan/?olt3mann e'uation is the

same as the luminosity'in this hapter. The luminosity'is related to the apparent bri"htness %by

'. 44#1. %t is "i$en that 1 2 ,% %= 1 2 ,r r= L1 95=nm, = and L2 45=nm. =

2 1

1 1 2 2

1 2

42 4 2 4 4

1 2 2 2 2 2 2 2 2 2 2

1 2 2 2 2 4 2 4 4

1 2 1 1 1 1 1 1 1 1 1

2 2 2

2 2 1

1 1 2

4

4 4 4

95=2.;

45=

** *

*

' ' d ' P A * r * * *% %

d d d ' P A* r * * *

d *

d *

= =

= = = = = = = =

= = = =

The star !ith the peak at 45= nm is 2.; times further a!ay than the star !ith the peak at 95= nm.

19. The +h!ar3shild radius is2

2.

"M

c

( ) ( )

( )

11 2 2 24

),arth

,arth 22 ;

2 7.79 1= m k" 5.8; 1= k"2;.;7 1= m ;.8 mm

).== 1= m s

"MR

c

= = =

g

1;. The +h!ar3shild radius is "i$en by2

2.

"MR

c= -n appro(imate mass for our Iala(y is

alulated in (ample 44#1.

( ) ( )

( )

11 2 2 41

14

22 ;

2 7.79 1= m k" 2 1= k"2) 1= m

).== 1= m s

"MR

c

= = =

g

18. The limitin" $alue for the an"les in a trian"le on a sphere is o54= . %ma"ine dra!in" an e'uilateral

trian"le near the north pole, enlosin" the north pole. %f that trian"le !ere small, the surfae !ould


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be appro(imately flat, and eah an"le in the trian"le !ould be o7= . Then ima"ine strethin" eah

side of that trian"le do!n to!ards the e'uator, !hile keepin" sure that the north pole stayed inside

the trian"le. The an"le at eah $erte( of the trian"le !ould e(pand, !ith a limitin" $alue of o1;= .

The three o1;= an"les in the trian"le !ould sum too

54= .

2=. To 6ust esape from an ob6et, the kineti ener"y of the body at the surfae of the body must be e'ualto the ma"nitude of the "ra$itational potential ener"y at the surfae. oppler

shift, '. 44#).

0a( ) ( ) 4

;

22=== m s Mly 9.= Mly5.1)) 1=

).== 1= m s

+ -d

c c

= = =

( )4

= 4

1 1 5.1)) 1=757 nm 757.)4 nm 757 nm

1 1 5.1)) 1=

+ c

+ c

+ + = = =

0%( ) ( ) )

;

22=== m s Mly 9= Mly5.1)) 1=

).== 1= m s

+ -d

c c

= = =


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( ))

= )

1 1 5.1)) 1=757 nm 758.);nm 758 nm

1 1 5.1)) 1=

+ c

+ c

+ + = = =

27.


9/18


rest rest

;

rest

).== 1= m s =.1=m75Mly

22,=== m s Mly 21m

++ c + -d

c

+ cd

- -

= = =

= = = =

)1. WienAs la! is "i$en in '. )9#1.) )

) )

2.8= 1= m G 2.8= 1= m G 2.8= 1= m G 1.1 1= m

2.9 G*

*

= = = =

g gg

)2. We use WienAs la!, '. )9#1. rom i"ure 44#)=, the temperature is about 1=1= G.) )

) 1)

1=

2.8= 1= m G 2.8= 1= m G 2.8= 1= m G ) 1= m

1= G*

*

= = = =

g gg f

rom i"ure )1#12, that !a$elen"th is in the "amma ray re"ion of the M spetrum.

)). We use the proton as typial nulear matter.

27 )

) 29

k" 1nuleon1= 7 nuleons m

m 1.79 1= k"

=

)4. %f the uni$erseAs sale is in$ersely proportional to the temperature, the sale times the temperature

should be onstant. %f !e all the urrent sale 1, and kno!in" the urrent temperature to be about

) G, then the produt of sale and temperature should be about ).


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2

2

mc/* mc *

/= = .

0a( ) ( )2 2 1)2 12

2)

5== MeK 1.7= 1= J MeK7 1= G

1.); 1= J G

c cmc*

/

= = =

rom i"ure 44#)=, this orresponds to a time of 51= s: .

0%( ) ( )2 2 1)2

14

2)

85== MeK 1.7= 1= J MeK1 1= G

1.); 1= J G

c cmc*

/

= = =

rom i"ure 44#)=, this orresponds to a time of 91= s

: .

0c( ) ( )2 2 1)2

12

2)

1== MeK 1.7= 1= J MeK1 1= G

1.); 1= J G

c cmc*

/

= = =

rom i"ure 44#)=, this orresponds to a time of 41= s

: .

There !ill be some $ariation in the ans!ers due to readin" the fi"ure.

)7. 0a -ordin" to the te(t, near i"ure 44#)), the $isible matter makes up about one#tenth of the

total baryoni matter. The a$era"e baryoni density is therefore 1= times the density of $isible

matter.

( ) ( ) ( )

( ) ( )

$isible

baryon $isible )4

)

11 11 )=

)8 154

)

27 )27 )

1= 1=

1= "ala(ies 1= stars "ala(y 2.= 1= k" star 1=

14 1= ly 8.47 1= m ly

2.1 1= k" m2.=55 1= k" m

M

R

= =

=

=

0% -"ain, aordin" to the te(t, dark matter is about 4 times more plentiful than normal matter.

( )27 ) 27 )dark baryon4 4 2.=55 1= k" m ;.2 1= k" m = =

)9. 0a rom pa"e 12=1, a !hite d!arf !ith a mass e'ual to that of the +un has a radius about the si3e

of the arthAs radius, 7);=km . rom pa"e 12=2, a neutron star !ith a mass e'ual to 1.5 solar

masses has a radius of about 2=km . or the blak hole, !e use the +h!ar3shild radius

formula.

( ) ( )

( )

11 2 2 )=

22 ;

2 7.79 1= m k" ) 1.88 1= k"2;;48m ;.;5 km

).== 1= m s

"MR

c

= = =

g

0% The ratio is 7);= H 2= H ;.;5 921H 2.27 H1 9== H 2 H1 .=

);. The an"ular momentum is the produt of the rotational inertia and the an"ular $eloity.


12;


11/18


( ) ( )

( )

initial final

2 22 ;2initial initial initial5

final initial initial initial2 )2final final final5

8 8

9 1= m1re$ month

; 1= m

re$ 1month

9.77 1= re$ month 9.77 1= month )

I I

I MR R

I MR R

=

= = = =

= =

1d 1h

285)=d 24 h )7==s

)===re$ s

re$ s =

)8. The rotational kineti ener"y is "i$en by21

2I . The final an"ular $eloity, from problem 4), is

89.77 1= re$ month .

( ) ( )

( )( )

22 2 21 2

final final final final final final final2 5

2 2 21 2initial initial initial initial initial initial initial2 5

2) 8

8

;

; 1= m 9.77 1= re$ month

; 1=9 1= m 1re$ month

0 I MR R

0 I MR R

= = =

= =

4=. The apparent luminosity is "i$en by '. 44#1.


12/18


( )

( )

( )

( )

27 )

27 )

)

29

8 ; )

27 )

; )

nuleon mass density =.=2 1= k" m

=.=2 1= k" mnuleon number density =.12 nuleon m

1.79 1= k" nuleon

neutrino number density 1= nuleon number density 1.2 1= neutrino m

=.8; 1= k" m

1.2 1= neutrino m

=

= =

= =

; 2

)5 2

29

k" 8.)15 1= eK;.19 1= 47 eK

neutrino 1.77 1= k"

cc

= =

0% -ssume that the nuleons make up only 5N of the ritial mass density.

( )

( )

( )

( )

27 )

27 )

)

29

8 ; )

27 )

; )

nuleon mass density =.=5 1= k" m

=.=5 1= k" mnuleon number density =.)= nuleon m

1.79 1= k" nuleon

neutrino number density 1= nuleon number density ).= 1= neutrino m

=.85 1= k" m

).= 1= neutrino m

=

= =

= =

; 2

)5 229

k" 8.)15 1= eK).19 1= 1;eK

neutrino 1.77 1= k"

cc

= =

45. The temperature of eah star an be found from WienAs la!.)

) )

77= 4;=8 8

2.8= 1= m G

2.8= 1= m G 2.8= 1= m G 4)8=G 7=4=G

77= 1= m 4;= 1= m

*

* *

=

= = = =

g

g g

The luminosity of eah star an be found from the /* dia"ram.25 27

77= 4;=) 1= W ) 1= W' '

The +tefan/?olt3mann e'uation says that the po!er output of a star is "i$en by4

P A*= , !here

is a onstant, andAis the radiatin" area. ThePin the +tefan/?olt3mann e'uation is the same asthe luminosity'"i$en in '. 44#1. orm the ratio of the t!o luminosities.

( )

( )

24 2 4 2 27

4;= 4;= 4;= 4;= 4;= 4;= 4;= 77=

24 2 4 2 25

77= 77= 77= 77= 77= 77= 77= 4;=

4)8=G4 ) 1= W 1.79

4 ) 1= W 7=4=G

' A * r * r ' *

' A * r * r ' *

= = = = =

The diameters are in the same ratio as the radii.

4;=

77=

1.79 1.9d

d=

The luminosities are fairly sub6eti$e, sine they are read from the /* dia"ram. >ifferent ans!ers

may arise from different readin"s of the /* dia"ram.

47. 0a The number of parses is the reiproal of the an"ular resolution in seonds of ar.

( ) ( ) ( )o

o771 1 1

1== parse =.=1 ) 1=7= 7=

2.9; 1=

= =

=

0% We use the *aylei"h riterion, '. )5#1=, !hih relates the an"ular resolution to the diameter of


1)=


13/18


the optial element. We hoose a !a$elen"th of 55= nm, in the middle of the $isible ran"e.

( )

( ) ( )

8

7

1.22 55= 1= m1.22 1.22 B B1).;)m 14 m

2.9; 1= rad 1;=

= =

The lar"est optial telesopes urrently in use are about 1= m in diameter.

49. We appro(imate the temperature/kineti ener"y relationship by /* 0= as "i$en on pa"e 1219.

( ) ( )12 1817

2)

1.87 1= eK 1.7= 1= J eK 2 1= G

1.); 1= J G

0/* 0 *

/

= = = =

rom i"ure 44#)=, this is in the hadron era .

4;. We assume that "ra$ity auses a entripetal fore on the "as. +ol$e for the speed of the rotatin" "as,

and use '. 44#7.

( )( ) ( )

( )

2"as blak hole "as "as

"ra$ity entripetal 2

8 )=blakhole 11 2 2 5

"as 15

5

) )

;

2 1= 1.88 1= k"7.79 1= m k" 7.42 1= m s

8.47 1= m7;ly

1ly

7.42 1= m s2.14 1= 2 1=

).== 1= m s

m mm +

1 1 "r r

m+ "

r

+.

c

= =

= = =

= =

g

48. 0a To find the ener"y released in the reation, !e alulate the )#$alue for this reation. rom '.

42#2a, the )#$alue is the mass ener"y of the reatants minus the mass ener"y of the produts.

The masses are found in -ppendi( .

( )[ ] ( )2 2 2 2

& M"2 2 12.====== u 2).8;5=42 u 8)1.5 Me$ 1).8)MeK) m c m c c c= = =0% The total kineti ener"y should be e'ual to the eletrial potential ener"y of the t!o nulei !hen

they are 6ust touhin". The distane bet!een the t!o nulei !ill be t!ie the nulear radius,

from '. 41#1. ah nuleus !ill ha$e half the total kineti ener"y.

( )( ) ( )( )

( ) ( ) ( )

( )( )

21D ) 1D )15 15 nuleus

=

2

nuleus1 1

2 2

=

22 18

8 2 21

2 1D ) 1)15

11.2 1= m 1.2 1= m 12

4 2

1

4 2

7 1.7= 1= & 1MeK;.8;; 1= n & 4.911MeK

1.7= 1= J2 1.2 1= m 12

4.9 MeK

2r A 3

r

20 3

r

= = =

= =

=

=

g

0c We appro(imate the temperature/kineti ener"y relationship by /* 0= as "i$en on pa"e1219.

( )( )1) 1=2)

4.911MeK 1.7= 1= J MeK 5.5 1= G

1.); 1= J G

0/* 0 *

/

= = = =

5=. 0a ind the )#$alue for this reation. rom '. 42#2a, the )#$alue is the mass ener"y of the


1)1


14/18


reatants minus the mass ener"y of the produts.

( )[ ] ( )

17 17 2; 4

; ; 14 2

2 2 2 2 2

@ +i e

@ @ +i e

2 2 15.884815 u 29.897829 u 4.==27=) 8)1.5 Me$

8.584 MeK

) m c m c m c c c

+ +

= =

=

0% The total kineti ener"y should be e'ual to the eletrial potential ener"y of the t!o nulei !henthey are 6ust touhin". The distane bet!een the t!o nulei !ill be t!ie the nulear radius,from '. 41#1. ah nuleus !ill ha$e half the total kineti ener"y.

( )( ) ( )( )

( ) ( ) ( )

( )( )

21D) 1D)15 15 nuleus

=

2

nuleus1 1

nuleus 2 2

=

22 18

8 2 21

2 1D) 1815

11.2 1= m 1.2 1= m 17

4 2

1

4 2

; 1.7= 1= & 1eK ;.8;; 1= n & 9.7=8 MeK

1.7= 1= J2 1.2 1= m 17

9.7MeK

2r A 3

r

20 3

r

= = =

= =

= =

g

0c We appro(imate the temperature/kineti ener"y relationship by /* 0= as "i$en on pa"e1219.

( )18

7

1=

2)

1.7= 1= J9.7=8 1= eK

1eK ;.; 1= G

1.); 1= J G

0/* 0 *

/

= = = =

51. We treat the ener"y of the photon as a rest mass, and so2

photon photonO .Om E c= To 6ust esape

from a spherial massMof radiusR, the ener"y of the photon must be e'ual to the ma"nitude of the

"ra$itational potential ener"y at the surfae.

( )2photonphoton photonphoton 2

photon photon

"M E c"Mm "Mm "M

E RR E E c

= = = =

52. We use the +unAs mass and "i$en density to alulate the si3e of the +un.

( )( )

)4+un)

1D )1D ) )=

1;

+un 1527 )

1;

9 )+un +un

11

arth#+un "ala(y

) 1.88 1= k") 1ly).72 1= m );2ly 4==ly

4 8.47 1= m4 1= k" m

).72 1= m );2ly2 1= 4 1=

1.5= 1= m 1==,===ly

M M

( r

Mr

r r

d d

= =

= = = =

= =

5). We appro(imate the temperature/kineti ener"y relationship by /* 0= as "i$en on pa"e 1219.


1)2


15/18


( ) ( )12 1819

2)

14 1= eK 1.7= 1= J eK 1.7 1= G

1.); 1= J G

0/* 0 *

/

= = = =

rom i"ure 44#)=, this mi"ht orrespond to a time around 151= s . ote that this is 6ust a $ery

rou"h estimate due to the 'ualitati$e nature of i"ure 44#)=.

54. 0a We onsider the photon as

enterin" from the left, "ra3in" the

+un, and mo$in" off in a ne!

diretion. The defletion is

assumed to be $ery small. %n

partiular, !e onsider a small part

of the motion in !hih the photon

mo$es a hori3ontal distaned cdt = !hile loated at 0,yrelati$e to the enter of the +un.

ote that y R and 2 2 2 .r y= +

%f the photon has ener"yE, it !ill

ha$e a mass of2

,m E c= and amomentum of ma"nitude

.p E c mc= = To find the han"eof momentum in they#diretion, !e

use the impulse produed by they#omponent of the "ra$itational fore.

( )2 2 )D2

2 2os

y y

"Mm d "Mm d "MmR ddp 1 dt

r c r c

R

r c R= = = =

+

To find the total han"e in they#momentum, !e inte"rate o$er all0the entire path of the

photon. We use an inte"ral from -ppendi( ?#4.

( ) ( ))D2 1D2

2 2 2 2 2 2

2 2y

"MmR d, "MmR , "Mm "Mpp

c c cR c R, R R , R

= = = = + +

The total ma"nitude of defletion is the han"e in momentum di$ided by the ori"inal

momentum.

2

2

2

2y

"Mpp "Mc R

c Rp p

= = =

0% We use data for the +un.

( )( ) ( )

2

11 )=

2

22 ; ;

7

m2 7.79 1= 1.88 1= k"

k"2

).== 1= m s 7.87 1= m

1;= )7==4.2); 1= rad =.;9

rad 1

"M

c R

= =

= =

g


1))

,

y R=

Ep mc

c= =

2

"Mm1

r=

Rr


16/18


55. ?eause Kenus has a more ne"ati$e apparent ma"nitude, Kenus is bri"hter. We !rite the

lo"arithmi relationship as follo!s, lettin" mrepresent the ma"nitude and %the bri"htness.

( ) ( )

( ) ( )

( )Kenus +irius2 1

2 1 2 1 2 1

2 1

2 1

4.4 1.4

Kenus2 2.5 2.52 1 2 1

1 +irius

lo" lo" lo" lo"

52.5

lo" lo" =.=1

lo" 1= 1= 1= 17m mm m

/

m / % m

m

m

m / % % / % %

m/

% %

%%m / % %

% %

+

= = =

+= = =

= = = = =

57. %f there arenuleons, !e assume that there are appro(imately1

2neutrons and

1

2protons.

Thus, for the star to be neutral, there !ould also be 12eletrons.

0a rom '. 4=#12 and 4=#1), !e find that if all eletron le$els are filled up to the ermi ener"y

,E the a$era"e eletron ener"y is

)5

.E

( ) ( )e2D) 2D )2 2

e) )1e 5 2 5

e e

) ) 1 )

; 5 2 ; 2

Eh h

E

m ( m (

= = =

0% The ermi ener"y for nuleons !ould be a similar e(pression, but the mass !ould be the mass

of a nuleon instead of the mass of the eletron. uleons are about 2=== times hea$ier than

eletrons, so the ermi ener"y for the nuleons !ould be on the order of 1D1=== the ermi

ener"y for the eletrons. We !ill i"nore that small orretion.

To alulate the potential ener"y of the star, think about the

mass in terms of shells. &onsider the inner portion of the

star !ith radius rPRand mass m, surrounded by a shell of

thikness drand mass dM. +ee the dia"ram. rom IaussAs

la! applied to "ra$ity, the "ra$itational effets of the inner

portion of the star on the shell are the same as if all of its

mass !ere at the "eometri enter. Like!ise, thespherially#symmetri outer portion of the star has no

"ra$itational effet on the shell. Thus the "ra$itational

ener"y of the inner portion/shel ombination is "i$en by a

form of '. ;#19, .d3 "mdM

r= The density of the star

is "i$en by )4)

.M

R

= We use that density to alulate the masses, and then inte"rate o$er the

full radius of the star to find the total "ra$itational ener"y of the star.

( ) ( )

( ) ( )

)

) )4 4

) )) )4

)

2

2 2

) )4

)

) 2

2) )

4

7

)4 4

)

)

M rm r r M

R R

M MrdM r dr r dr dr

R R

r MrM dr

mdM "M R Rd3 " " r dr r r R

= = =

= = =

= = =


1)4

r

Rdr


17/18


( )2 2 2 5 2

4 4

7 7 7

= =

) ) ) )

5 5

R R"M "M "M R "M

3 r dr r dr R R R R

= = = =

0c The total ener"y is the sum of the t!o terms alulated abo$e. The mass of the star is primarily

due to the nuleons, and so nuleon.M m=2D )2 2

total e

e

2D )2 2 2D) 2 5D) 2

) 4D) 5D) 24

e nuleon nuleon e nuleon)

) 1 ) )5 2 ; 2 5

) ) ) 8 ) )

;= 2 5 )2= 5

h "M E E 3 m ( R

h M M "M h M "M

m m m R R m m R R

= + =

= =

Let

2D) 2 5D)

4D) 5D)

e nuleon

8 )

)2=

h Ma

m m= and 2

),

5% "M= so total 2 .

a %E

R R= We set total =dE dR= to find

the e'uilibrium radius.2D) 2 5D)

4D) 5D) 2D) 2

total e nuleon

e') 2 4D) 1D) 5D)2

e nuleon

8 )2

)2=2 2 8

= ) )2

5

h M

dE m ma % a h

RdR R R % "M m m"M

= + = = = =

We e$aluate the e'uilibrium radius usin" the +unAs mass.

( )

( ) ( ) ( )

2D) 2

e' 4D) 1D) 5D)

e nuleon

22D) )4

21D) 5D)

4D) 11 )= )1 29

2

7 )

8

)2

8 7.7) 1= J s

m)2 7.79 1= 2.= 1= k" 8.11 1= k" 1.79 1= k"

k"

9.19; 1= m 9.2 1= km

hR

"M m m

=

=

=

g

g

59. There areneutrons. The mass of the star is due only to neutrons, and so n .M m= rom 's.

4=#12 and 4=#1), !e find that if all ener"y le$els are filled up to the ermi ener"y ,E the a$era"e

ener"y is )5

.E We follo! the same proedure as in roblem 57. The e(pression for the

"ra$itational ener"y does not han"e.

( ) ( ) ( )

2D ) 2D )2D ) 2 5D)2 2

) )

n n 5 5 ) 4D) ;D) 24

n n n n n)

) 1;) ) )

; 4= 17=

h Mh M h M E E

m ( m m R m m R = = = =

( ) 2total n

2D ) 2 5D)

4D) ;D) 2n

)

5

) 1;

17=

"ME E 3

R

h M

m R= + =

Let( )

2D) 2 5D)

4D) ;D)

n

) 1;

17=a

h M

m= and 2

),

5% "M= so

total 2.

a %E

R R= We set total =dE dR= to find the

e'uilibrium radius.


1)5


18/18


( )

( )

2D ) 2 5D)

2D ) 24D) ;D)

total n

e') 2 4D) 1D) ;D)2

n

) 1;2

1;17=2 2=

) 17

5

h M

hdE ma % aR

dR R R % "M m"M

= + = = = =

We e$aluate the e'uilibrium radius for a mass of 1.5 solar masses.

( )

( )

( ) ( )

2D ) 2

e' 4D) 1D) ;D)

n

22D) )4

21D) ;D)

4D) 11 )= 29

2

4

1;

17

1; 7.7) 1= J s

m17 7.79 1= 1.5 2.= 1= k" 1.79 1= k"

k"

1.=;7 1= m 11km

hR

"M m

=

=

=

g

g

5;. We must find a ombination of c, ", and h that has the dimensions of time. The dimensions of c

are ,'*

the dimensions of "are

)

2,'

M*

and the dimensions of h are2

.M'*

[ ] [ ] [ ] [ ]

( )

) 2) 2 2

2

51 1

2 2 2

11 2 2

5D2 1D 2 1D2

5

) 2 = = 2 1 5 = 1 )

5 1 )

17.79 1= m k" 7.7

2

P

P

' ' M't c " * ' M *

* M* *

"t c "

c

+ +

= = =

+ + = = = + = =

= = = =

= = =

h

gh

h

( )

( )

)4

44

5;

) 1= J s

5.); 1= s

).== 1= m s

=

g


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