SM PDF Chapter11

8/6/2019 SM PDF Chapter11

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297

Gravity, Planetary Orbits,and the Hydrogen Atom

CHAPTER OUTLINE

11.1 Newtons Law of UniversalGravitation Revisited

11.2 Structural Models11.3 Keplers Laws11.4 Energy Considerations in

Planetary and SatelliteMotion

11.5 Atomic Spectra and theBohr Theory of Hydrogen

11.6 Context ConnectionChanging from a Circularto an Elliptical Orbit

ANSWERS TO QUESTIONS

Q11.1 The Earth creates the same gravitational fieldgfor all objects nearthe Earths surface. The larger mass needs a larger force to give it just

the same acceleration.

Q11.2 To a good first approximation, your bathroom scale reading is unaffected because you, the Earth,and the scale are all in free fall in the Suns gravitational field, in orbit around the Sun. To a precisesecond approximation, you weigh slightly less at noon and at midnight than you do at sunrise orsunset. The Suns gravitational field is a little weaker at the center of the Earth than at the surfacesubsolar point, and a little weaker still on the far side of the planet. When the Sun is high in yoursky, its gravity pulls up on you a little more strongly than on the Earth as a whole. At midnight theSun pulls down on you a little less strongly than it does on the Earth below you. So you can haveanother doughnut with lunch, and your bedsprings will still last a little longer.

Q11.3 Because both the Earth and Moon are moving in orbit about the Sun. As described byF magravitational centripetal= , the gravitational force of the Sun merely keeps the Moon (and Earth) in a

nearly circular orbit of radius 150 million kilometers. Because of its velocity, the Moon is kept in itsorbit about the Earth by the gravitational force of the Earth. There is no imbalance of these forces, atnew moon or full moon.

Q11.4 Air resistance causes a decrease in the energy of the satellite-Earth system. This reduces the diameterof the orbit, bringing the satellite closer to the surface of the Earth. A satellite in a smaller orbit,however, must travel faster. Thus, the effect of air resistance is to speed up the satellite!

Q11.5 The escape speed from the Earth is 11.2 km/s and that from the Moon is 2.3 km/s, smaller by a factor

of 5. The energy requiredand fuelwould be proportional to v2 , or 25 times more fuel is requiredto leave the Earth versus leaving the Moon.

Q11.6 In a circular orbit each increment of displacement is perpendicular to the force applied. The dotproduct of force and displacement is zero. The work done by the gravitational force on a planet in anelliptical orbit speeds up the planet at closest approach, but negative work is done by gravity andthe planet slows as it sweeps out to its farthest distance from the Sun. Therefore, net work in onecomplete orbit is zero.


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298 Gravity, Planetary Orbits, and the Hydrogen Atom

Q11.7 For a satellite in orbit, one focus of an elliptical orbit, or the center of a circular orbit, must be locatedat the center of the Earth. If the satellite is over the northern hemisphere for half of its orbit, it mustbe over the southern hemisphere for the other half. We could share with Easter Island a satellite thatwould look straight down on Arizona each morning and vertically down on Easter Island eachevening.

Q11.8 The gravitational force of the Earth on an extra particle at its center must be zero, not infinite as oneinterpretation of Equation 11.1 would suggest. All the bits of matter that make up the Earth will pullin different outward directions on the extra particle.

Q11.9 Speed is maximum at closest approach. Speed is minimum at farthest distance.

Q11.10 The gravitational force is conservative. An encounter with a stationary mass cannot permanentlyspeed up a spacecraft. Jupiter is moving. A spacecraft flying across its orbit just behind the planetwill gain kinetic energy as the planets gravity does net positive work on it.

Q11.11 Cavendish determined G. Then from gGM

R=

2, one may determine the mass of the Earth.

Q11.12 The spacecraft did not have enough fuel to stop dead in its high-speed course for the Moon.

Q11.13 If an electron moved like a hockey puck, it could have any arbitrary frequency of revolution aroundan atomic nucleus. If it behaved like a charge in a radio antenna, it would radiate light withfrequency equal to its own frequency of oscillation. Thus, the electron in hydrogen atoms wouldemit a continuous spectrum, electromagnetic waves of all frequencies smeared together.

Q11.14 (a) Yesprovided that the energy of the photon is precisely enough to put the electron into oneof the allowed energy states. Strangelymore precisely non-classicallyenough, if theenergy of the photon is not sufficient to put the electron into a particular excited energylevel, the photon will not interact with the atom at all!

(b) Yesa photon of any energy greater than 13.6 eV will ionize the atom. Any extra energywill go into kinetic energy of the newly liberated electron.

Q11.15 An atomic electron does not possess enough kinetic energy to escape from its electrical attraction tothe nucleus. Positive ionization energy must be injected to pull the electron out to a very largeseparation from the nucleus, a condition for which we define the energy of the atom to be zero. Theatom is a bound system. All this is summarized by saying that the total energy of an atom isnegative.

Q11.16 From Equations 11.19, 11.20 and 11.21, we have = = + = +Ek e

r

k e

r

k e

rK Ue e e e

2 2 2

2 2. Then K E=

and U Ee = 2 .


3/34

Chapter 11 299

SOLUTIONS TO PROBLEMS

Section 11.1 Newtons Law of Universal Gravitation Revisited

P11.1 F m gGm m

r

= =11 2

2

gGm

r= =

=

22

11 4 3

27

6 67 10 4 00 10 10

1002 67 10

. ..

N m kg kg

mm s

2 2

2e je ja f

P11.2 (a) At the midpoint between the two objects, the forces exerted by the 200-kg and 500-kg objectsare oppositely directed,

and from FGm m

rg =

1 22

we have FG

=

=

50 0 500 200

0 200

2 50 102

5.

.

.kg kg kg

m

Nb gb g

a f

toward the 500-kg object.

(b) At a point between the two objects at a distance d from the 500-kg objects, the net force onthe 50.0-kg object will be zero when

G

d

G

d

50 0 200

0 400

50 0 5002 2

.

.

.kg kg

m

kg kg b gb ga f

b gb g

=

or d = 0 245. m

P11.3 FGMm

r= =

=

211

3

2 2106 67 10

1 50 15 0 10

4 50 10

7 41 10.. .

.

.N m kg kg kg

m

N2 2e jb ge j

e j

P11.4 The force exerted on the 4.00-kg mass by the 2.00-kg mass isdirected upward and given by

rF j j

j

244 2

242

112

11

6 67 104 00 2 00

3 00

5 93 10

= =

=

Gm m

r$ .

. .

.

$

. $

N m kg kg kg

m

N

2 2e jb gb g

a f

The force exerted on the 4.00-kg mass by the 6.00-kg mass isdirected to the left

rF i i

i

644 6

642

112

11

6 67 104 00 6 00

4 00

10 0 10

= =

=

Gm m

r$ .

. .

.

$

. $

e j e j b gb ga fN m kg

kg kg

m

N

2 2

r

r

FIG. P11.4

Therefore, the resultant force on the 4.00-kg mass isr r rF F F i j4 24 64

1110 0 5 93 10= + = + . $ . $e j N .


4/34


P11.5 (a) The Sun-Earth distance is 1 496 1011. m and the Earth-Moon distance is 3 84 108. m , so thedistance from the Sun to the Moon during a solar eclipse is

1 496 10 3 84 10 1 492 1011 8 11. . . = m m m

The mass of the Sun, Earth, and Moon are MS = 1 99 1030. kg

ME = 5 98 1024. kg

and MM = 7 36 1022. kg

We have FGm m

rSM = =

=

1 22

11 30 22

11 220

6 67 10 1 99 10 7 36 10

1 492 104 39 10

. . .

..

e je je j

e jN

(b) FEM =

=

6 67 10 5 98 10 7 36 10

3 84 101 99 10

11 24 22

8 220

. . .

..

N m kg N

2 2e je je j

e j

(c) FSE =

=

6 67 10 1 99 10 5 98 10

1 496 103 55 10

11 30 24

11 222

. . .

..

N m kg N

2 2e je je j

e j

Note that the force exerted by the Sun on the Moon is much stronger than the force of theEarth on the Moon. In a sense, the Moon orbits the Sun more than it orbits the Earth. TheMoons path is everywhere concave toward the Sun. Only by subtracting out the solarorbital motion of the Earth-Moon system do we see the Moon orbiting the center of mass ofthis system.

P11.6 Let represent the angle each cable makes with the vertical, L thecable length, x the distance each ball scrunches in, and d = 1 m the

original distance between them. Then r d x= 2 is the separation ofthe balls. We have

Fy = 0 : T mgcos = 0

Fx = 0 : TGmm

rsin =

20

r

r

r

FIG. P11.6

Then tan =Gmm

r mg2

x

L x

Gm

g d x2 22

2=

a f x d x

Gm

gL x = 2

2 2 2a f .

The factorGm

g

is numerically small. There are two possibilities: either x is small or else d x 2 is

small.

Possibility one: We can ignore x in comparison to d and L, obtaining

x 16 67 10 100

9 845

211

mN m kg kg

m sm

2 2

2a f e j

b g

e j=

.

. x = 3 06 10 8. m.

continued on next page


5/34

Chapter 11 301

The separation distance is r = = 1 2 3 06 10 1 000 61 38m m m nm. . .e j .

Possibility two: Ifd x 2 is small, x 0 5. m and the equation becomes

0 56 67 10 100

9 845 0 52

112 2

..

..m

N m kg kg

N kgm m

2 2

a f e jb g

b ga f a fr =

r = 2 74 10 4. m .

For this answer to apply, the spheres would have to be compressed to a density like that of thenucleus of atom.

P11.7 gGM

R

G

RG R

R

= = =2

43

2

3

4

3

e j

Ifg

g

G R

G RM

E

M M

E E

= =1

6

4

33

4

then

M

E

M

E

E

M

g

g

R

R=

FHG

IKJFHG

IKJ

=FHG

IKJ =

1

64

2

3a f .

P11.8 (a) At the zero-total field point,GmM

r

GmM

rE

E

M

M2 2

=

so r rM

Mr

rM E

M

EE

E= =

=

7 36 10

5 98 10 9 01

22

24

.

. .

r r rr

r

E M EE

E

+ = = +

=

=

3 84 109 01

3 84 103 46 10

8

88

..

..

m

m

1.11m

(b) At this distance the acceleration due to the Earths gravity is

gGM

r

g

EE

E

E

= =

=

2

11 24

8 2

3

6 67 10 5 98 10

3 46 10

3 34 10

. .

.

.

N m kg kg

m

m s directed toward the Earth

2 2

2

e je j

e j

*P11.9 (a) For the gravitational force on an object in the neighborhood of Miranda we have

m gGm m

r

gGm

r

objobj Miranda

Miranda2

Miranda

Miranda2

2

2

2N m kg

kg mm s

=

= =

=

6 67 10 6 68 10

242 100 076 1

11 19

3 2

. ..

e j

e j



6/34


(b) We ignore the difference (of about 4%) ingbetween the lip and the base of the cliff. For thevertical motion of the athlete we have

y y v a t

t

t

f i yi y= + +

= + +

=

FHG

IKJ

=

1

2

5 000 0 0

1

2 0 0761

2 5 000

0 076363

2

2

2 1 2m m s

m s

1 ms

2

.

.

e jb g

(c) x x v t a t f i xi x= + + = + + = 1

20 8 5 363 0 3 08 102 3. .m s s mb ga f

We ignore the curvature of the surface (of about 0.7) over the athletes trajectory.

(d) v vxf xi= = 8 50. m s

v v a t yf yi y= + = = 0 0 076 1 363 27 6. .m s s m s2e ja f

Thusrv i jf = = +8 50 27 6 8 5 27 6

2 2. $ . $ . .e j m s m s at tan.

.1 27 6

8 5below the x axis.

rvf = 28 9. m s at 72.9 below the horizontal

P11.10 (a) FGMm

r= =

+

=

2

11 30 3

4 217

6 67 10 100 1 99 10 10

1 00 10 50 01 31 10

. .

. ..

N m kg kg kg

m mN

2 2e j e je j

e j

(b) FGMm

r

GMm

r

=

front2

back2

gF

m

GM r r

r r= =

back2

front2

front2

back2

e j

FIG. P11.10

g

g

=

LNM

OQP

=

6 67 10 100 1 99 10 1 01 10 1 00 10

1 00 10 1 01 10

2 62 10

11 30 4 2 4 2

4 2 4 2

12

. . . .

. .

.

e j e j e j e j

e j e j

m m

m m

N kg


7/34

Chapter 11 303

P11.11 g gMG

r a1 2 2 2

= =+

g gy y1 2= g g g y y y= +1 2

g g g x x1 2 2= = cos cos =

+

r

a r2 2 1 2

e j

rg i= 2 2g x

$e j

orrg =

+

2

2 2 3 2

MGr

r ae jtoward the center of mass

r

r

FIG. P11.11

P11.12 (a) We require thatGM m

r

mv

rE

2

2

= , but gM G

RE

E

=2

. In this case r RE= 2 therefore,g v

RE4 2

2

= or

vgRE

= =

=

2

9 80 6 37 10

2

5 59 10

6

3. .

.m s m

m s

2 e j.

(b) Tr

v= =

=

2 2 2 6 37 10239

6 a fe j. m

5.59 10 m smin

3

(c) FGM m

R

mgE

E

= = = =

2 4

300 9 80735

2b gb ge jkg m s

4N

2.

Section 11.2 Structural Models

No problems in this section

Section 11.3 Keplers Laws

P11.13 (a) For the geosynchronous satellite, =F mar r becomesGmM

r

mv

rE

2

2

= and in turn

GM

r

r

TE

=FHG

IKJ

22

or r

GM TE32

24=

. Thus, the radius of the satellite orbit is

r = L

NMM

OQPP =

6 67 10 5 98 10 86 400

44 23 10

11 24 2

2

1 3

7. .

.

/

N m kg kg sm

2 2e je jb g

.



8/34


(b) The satellite is so far out that its distance from the north pole,

d = + = 6 37 10 4 23 10 4 27 1062 7 2 7. . .m m me j e j is nearly the same as its orbital radius.

The travel time for the radio signal is

t

d

c= =

=2 2 4 27 10

3 00 10 0 285

7

8

.

. .

m

m s s

e j.

P11.14 By conservation of angular momentum for the satellite,

r v r vp p a a= v

v

r

r

p

a

a

p

= =+

+ = =

2 289 6 37 10

6 37 10

8 6591 27

3

3

km km

459 km km

km

6 829 km

.

.. .

We do not need to know the period.

P11.15 Ta

GM2

2 34=

(Keplers third law with m M


9/34

Chapter 11 305

P11.18 F ma = :Gm M

r

m v

r

planet star planet

2

2

=

GM

rv r

GM r r r

r

r

x x y y

y xx

yy

star

star

yr yr

= =

= = =

=

FHG

IKJ

=F

HGIKJ

=

2 2 2

3 3 3 2 3 2

3 2

3 290 0

5 003

468

5 00

.

. .

So planet has turned through 1.30 revolutionsY .

FIG. P11.18

P11.19 The speed of a planet in a circular orbit is given by

F ma = :GM m

r

mv

rsun2

2

= vGM

r=

sun .

For Mercury the speed is vM =

=

6 67 10 1 99 10

5 79 104 79 10

11 30

10

4. .

..

e je je j

m

sm s

2

2

and for Pluto, vP =

=

6 67 10 1 99 10

5 91 104 74 10

11 30

12

3. .

..

e je je j

m

sm s

2

2.

With greater speed, Mercury will eventually move farther from the Sun than Pluto. With originaldistances rP and rM perpendicular to their lines of motion, they will be equally far from the Sun

after time t where

r v t r v t

r r v v t

t

P P M M

P M M P

2 2 2 2 2 2

2 2 2 2 2

12 2 10 2

4 2 3 2

25

98

5 91 10 5 79 10

4 79 10 4 74 10

3 49 10

2 27 101 24 10 3 93

+ = +

=

=

=

= =

e j

e j e j

e j e j

. .

. .

.

.. . .

m m

m s m s

m

m ss yr

2

2 2


10/34


P11.20 For the Earth, F ma = :GM m

r

mv

r

m

r

r

Ts

2

2 22= =

FHG

IKJ

.

Then GM T rs2 2 34= .

Also the angular momentum L mvr mr

T

r= =2

is a constant for the Earth.

We eliminate rLT

m=

2between the equations:

GM TLT

ms

2 23 2

42

=FHG

IKJ GM T

L

ms

1 2 23 2

42

=FHG

IKJ .

Now the rate of change is described by

GM TdT

dtG

dM

dtTs

s1

21 01 2 1 2

FHG

IKJ+

FHG

IKJ=

dT

dt

dM

dt

T

M

T

ts

s

= FHG

IKJ

2

T t dMdt

TM

T

s

s

FHG IKJ= FHG IKJ

FHG IKJ

=

2 5 000 3 16 10 3 64 10 2 1

1 82 10

7

9

2

yr s1 yr

kg s yr1.991 10 kg

s

30. .

.

e j

Section 11.4 Energy Considerations in Planetary and Satellite Motion

P11.21 (a)

= =

= M

rS

E43

2

30

6 39

3 1 99 10

4 6 37 101 84 10

.

..

kg

mkg m3

e j

e j

(b) gGM

rS

E

= =

=

2

11 30

6 26

6 67 10 1 99 10

6 37 103 27 10

. .

..

N m kg kg

mm s

2 2

2e je j

e j

(c) UGM m

rg

S

E

= =

=

6 67 10 1 99 10 1 00

6 37 102 08 10

11 30

613

. . .

..

N m kg kg kg

mJ

2 2e je jb g

P11.22 W UGm m

r= =

FHG

IKJ

1 2 0

W=

+

=

6 67 10 7 36 10 1 00 10

1 74 10 2 82 10

11 22 3

69

. . .

. .

N m kg kg kg

m J

2 2e je je j


11/34

Chapter 11 307

P11.23 For her jump on earth,1

22mv mgyi f= (1)

v gyi f= = =2 2 9 8 0 5 3 13. . .m s m m s2e ja f

We assume that she has the same takeoff speed on the asteroid. Here

1

20 02mv

GM m

RiA

A

= + (2)

The equality of densities between planet and asteroid,

= =M

R

M

R

E

E

A

A4

3

2 4

3

2implies M

R

RMA

A

EE=

FHG

IKJ

3

(3)

Note also at Earths surface gGM

R

E

E

=2

(4)

Combining the equations (2), (1), (3) and (4) by substitution gives

1

22v

GM

Ri

A

A

= 1

22

3

gyG

R

R

RMf

A

A

EEd i =

FHG

IKJ

GM

Ry

GM R

RE

Ef

E A

E2

2

3= R y R A f E

2 0 5= = . m 6.37 10 m6a fe j

RA = 1 78 103. m

P11.24 (a) The work must provide the increase in gravitational energy

W U U U

GM M

r

GM M

r

GM M

R y

GM M

R

GM MR R y

W

g gf gi

E p

f

E p

i

E p

E

E p

E

E pE E

= =

= +

= +

+

= +

FHG

IKJ

= F

HG

I

KJ

F

HG I

KJ

=

1 1

6 67 105 98 10 100

1

6 37 10

1

7 37 10

850

1124

6 6

..

. .

N m

kgkg kg

m m

MJ

2

2 e jb g



12/34


(b) In a circular orbit, gravity supplies the centripetal force:

GM M

R y

M v

R y

E p

E

p

E+

=+b g b g2

2

Then, 12

12

2M v GM MR y

pE p

E

=+b g

So, additional work= kinetic energy required

=

=

1

2

6 67 10 5 98 10 100

7 37 10

2 71 10

11 24

6

9

. .

.

.

N m kg kg

kg m

J

2

2

e je jb ge je j

W

P11.251

2

1 1 1

2

2 2mv GM m

r r

mvi Ef i

f+

F

HG

I

KJ=

1

2

01 1

2

2 2v GM

R

vi EE

f+ F

HG

I

KJ=

or v vGM

Rf

E

E

212 2

=

and v vGM

Rf

E

E

= FHG

IKJ1

2

1 22

vf = LNM

OQP = 2 00 10 1 25 10 1 66 10

4 2 81 2

4. . .e j m s

P11.26 (a) vM G

REsolar escape

Sun

Sun

km s= =

242 1.

(b) Let r R xE S= represent variable distance from the Sun, with x in astronomical units.

vM G

R x xE S= =

2 42 1Sun .

Ifv =125 000 km

3 600 s, then x = = 1 47 2 20 1011. .A.U. m

(at or beyond the orbit of Mars, 125 000 km/h is sufficient for escape).


13/34

Chapter 11 309

P11.27 To obtain the orbital velocity, we use FmMG

R

mv

R = =2

2

or vMG

R=

We can obtain the escape velocity from 12 mv mMGResc2 =

or vMG

Rvesc = =

22

*P11.28 (a) Energy conservation of the object-Earth system from release to radius r:

K U K U

GM m

R hmv

GM m

r

v GM r R h

dr

dt

gh

gr

E

E

E

EE

+ = +

+

=

= +

FHG

IKJ

FHG

IKJ =

e j e jaltitude radius

01

2

2

1 1

2

1 2

(b) dtdr

v

dr

vi

f

i

f

f

i

z z z= = . The time of fall is

t GMr R h

dr

tr

dr

EER

R h

E

E

= +

FHG

IKJ

FHG

IKJ

=

FHG

IKJ

LNM

OQP

+

z

z

21 1

2 6 67 10 5 98 101 1

6 87 10

1 2

11 246

1 2

6 37 106

. ..

.m

m

6.87 10 m6

We can enter this expression directly into a mathematical calculation program.

Alternatively, to save typing we can change variables to ur

=106

. Then

tu

duu

du=

FHG

IKJ =

FHG

IKJ

z z7 977 10 110

1

6 87 1010 3 541 10

10

10

1 1

6 8714 1 2

6 6

1 26

6 37

6 878

6

6 1 2

1 2

6 37

6 87

..

..

.

.

.

.

e je j

A mathematics program returns the value 9.596 for this integral, giving for the time of fall

t = = =3 541 10 10 9 596 339 8 3408 9. . . s .


14/34


P11.29v

R h

GM

R h

i

E

E

E

2

2+

=

+b g

K mvGM m

R hi i

E

E

= =+

FHG

IKJ

=

+

L

N

MM

O

Q

PP =

1

2

1

2

1

2

6 67 10 5 98 10 500

6 37 10 0 500 101 45 102

11 24

6 6

10. .

. ..

N m kg kg kg

m mJ

2 2e je jb g

e j e j

The change in gravitational potential energy of the satellite-Earth system is

UGM m

R

GM m

RGM m

R RE

i

E

fE

i f

= =

FHG

IKJ

= =

1 1

6 67 10 5 98 10 500 1 14 10 2 27 1011 24 8 1 9. . . .N m kg kg kg m J2 2e je jb ge j

Also, K mvf f= = = 1

2

1

2500 2 00 10 1 00 102 3

2 9kg m s Jb ge j. . .

The energy transformed due to friction is

E K K U i fint J J

= = + = 14 5 1 00 2 27 10 1 58 10

9 10

. . . .a f .

P11.30 The gravitational force supplies the needed centripetal acceleration.

Thus,GM m

R h

mv

R hE

E E+

=+b g b g22

or vGM

R hE

E

2=

+

(a) Tr

v

R hEGM

R hE

E

= =+

+

2 2 b g

b g

TR h

GM

E

E

=+

2

3

b g

(b) v GMR h

E

E

=+

(c) Minimum energy input is E K U K U f gf i gimin = + e j e j.

It is simplest to

launch the satellite from a location on the equator, and launch it toward the east.

This choice has the object starting with energy K mvi i=1

22

with vR R

iE E

= =2

1 00

2

86 400

. day s

and UGM m

R

giE

E

= .

Thus, E mGM

R h

GM m

R hm

R GM m

RE

E

E

E

E E

Emin =

+

FHG

IKJ

+

L

NMM

O

QPP

+1

2

1

2

4

86 400

2 2

2

sb g

or E GM mR h

R R h

R mE

E

E E

Emin =

+

+

LNMM

OQPP

2

2

2

86 400

2 2

2b g b g

s


15/34

Chapter 11 311

*P11.31 (a) Energy conservation for the object-Earth system from firing to apex:

K U K U

mvGmM

R

GmM

R h

gi

gf

iE

E

E

E

+ = +

= +

e j e j1

202

where1

2mv

GmM

RE

Eesc2

= . Then

1

2

1

2

1

2

1

2

2

2

2

2

2

2

2

v v vR

R h

v vv R

R h

v v

R h

v R

h v Rv v R v R v R v Rv v

hR v

v v

iE

E

iE

E

i

E

E

E

iE E E i E

i

E i

i

= +

=+

=

+

=

=

+

=

esc2

esc2

esc2 esc

2

esc2

esc2

esc

2

esc2 esc

2

esc

2

esc2

esc2

(b) h =

= 6 37 10

11 2 8 761 00 10

6 2

2 27.

. ..

m 8.76m

a fa f a f

(c) The fall of the meteorite is the time-reversal of the upward flight of the projectile, so it isdescribed by the same energy equation

v v RR h

v hR h

v

iE

E E

i

2 2 2 3 2 7

6 7

8

4

1 11 2 10 2 51 106 37 10 2 51 10

1 00 10

1 00 10

= +

FHG IKJ=

+FHG IKJ

=

+ FHG IKJ

=

=

esc esc

2 2

m s mm m

m s

m s

. .. .

.

.

e j

(d) With v vi


16/34


P11.33 (a) The major axis of the orbit is 2 50 5a = . AU so a = 25 25. AU

Further, in Figure 11.5, a c+ = 50 AU so c = 2475. AU

Then ec

a= = =

2475

25 250 980

.

..

(b) In T K as2 3= for objects in solar orbit, the Earth gives us

1 12 3

yr AUb g a f= Ks Ks =1

1

2

3

yr

AU

b ga f

Then T22

3

31

125 25=

yr

AUAU

b ga f

a f. T= 127 yr

(c) UGMm

r= =

=

6 67 10 1 991 10 1 2 10

50 1 496 102 13 10

11 30 10

11

17. . .

..

N m kg kg kg

mJ

2 2e je je je j

Section 11.5 Atomic Spectra and the Bohr Theory of Hydrogen

P11.34 (a) The energy of the photon is found as E E En n

i fi f

= =

13 606 13 606

2 2

. .eV eVa f

En nf i

=

F

HG

I

KJ13 606

1 12 2

. eV

Thus, for n = 3 to n = 2 transition E = F

HG I

KJ=13 606 1

9

1 89. .eV1

4

eV

(b) Ehc

=

and =

=

6 626 10

1 89656

34

19

.

.

J s 2.998 10 m s

eV 1.602 10 J eVnm

8e je j

(c) fc

=

f =

=

3 10

6 56 104 57 10

8

714m s

mHz

..


17/34

Chapter 11 313

P11.35 (a) Lyman series1

112

= FHG

IKJ

Rni

ni = 2 3 4, , , K

1 1

94 96 101 097 10 1

19

72

=

= FHG

IKJ.

.e jni

ni = 5

(b) Paschen series:1 1

3

12 2

= FHG

IKJ

Rni

ni = 4 5 6, , , K

The shortest wavelength for this series corresponds to ni = for ionization

11 097 10

1

9

172

= FHG

IKJ

.ni

For ni = , this gives = 820 nm

This is larger than 94.96 nm, so this wave length

cannot be associated with the Paschen series .

Balmer series:1 1

2

12 2

= FHG

IKJ

Rni

ni = 3 4 5, , , K

11 097 10

1

4

172

= FHG

IKJ

.ni

with ni = for ionization, min = 365 nm

Once again the shorter given wavelength cannot be associated with the Balmer series .

P11.36 (a) vk e

m re

e1

2

1

=

where r a12

0111 0 005 29 5 29 10= = = a f . .nm m

v1

9 19 2

31 11

68 99 10 1 60 10

9 11 10 5 29 102 19 10=

=

. .

. ..

N m C C

kg mm s

2 2e je je je j

(b) K m ve1 12 31 6 2 181

2

1

29 11 10 2 19 10 2 18 10 13 6= = = = . . . .kg m s J eVe je j

(c) Uk e

r

e1

2

1

9 19 2

1118

8 99 10 1 60 10

5 29 10

4 35 10 27 2= =

= =

. .

.

. .N m C C

m

J eV

2 2e je j


18/34


P11.37 En ni f

=

F

HG

I

KJ13 6

1 12 2

. eVa f

Where for E > 0 we have absorption and for E < 0 we have emission.

(i) for ni

= 2 and nf

= 5 , E = 2 86. eV (absorption)

(ii) for ni = 5 and nf = 3 , E = 0 967. eV (emission)

(iii) for ni = 7 and nf = 4 , E = 0 572. eV (emission)

(iv) for ni = 4 and nf = 7 , E = 0 572. eV (absorption)

(a) Ehc

=

so the shortest wavelength is emitted in transition ii .

(b) The atom gains most energy in transition i .

(c) The atom loses energy in transitions ii and iii .

P11.38 We use En

n =13 6

2

. eV.

To ionize the atom when the electron is in the nth level,

it is necessary to add an amount of energy given by E En

n= =13 6

2

. eV.

(a) Thus, in the ground state where n = 1, we have E = 13 6. eV .

(b) In the n = 3 level, E = =13 6

1 51.

.eV

9eV .

P11.39 (a) r22 20 052 9 2 0 212= =. .nm nmb ga f

(b) m vm k e

re

e e2

2

2

31 9 19 2

9

9 11 10 8 99 10 1 60 10

0 212 10= =

. . .

.

kg N m C C

m

2 2e je je j

m ve 2259 95 10= . kg m s

(c) L m v re2 2 225 9 349 95 10 0 212 10 2 11 10= = = . . .kg m s m kg m s2e je j

(d) K m vm v

me

e

e2 2

2 22 25

2

31

191

2 2

9 95 10

2 9 11 105 43 10 3 40= = =

= =

b g e je j

.

.. .

kg m s

kgJ eV



19/34

Chapter 11 315

(e) Uk e

re

2

2

2

9 19 2

918

8 99 10 1 60 10

0 212 101 09 10 6 80= =

= =

. .

.. .

N m C C

mJ eV

2 2e je j

(f) E K U2 2 2 3 40 6 80 3 40= + = = . . .eV eV eV

P11.40 Starting with1

2 22

2

m vk e

re

e=

we have vk e

m re

e

22

=

and using rn

m k en

e e

=

2 2

2

h

gives vk e

m n m k en

e

e e e

22

2 2 2=

he j

or vk e

nn

e=

2

h.

P11.41 Each atom gives up its kinetic energy in emitting a photon,

so1

2

6 626 10 3 00 10

1 216 101 63 102

34 8

7

18mvhc

= =

=

. .

..

J s m s

mJ

e je je j

v = 4 42 104. m s .

Section 11.6 Context ConnectionChanging from a Circular to an Elliptical Orbit

P11.42 The original orbit radius is r a= = + 6 37 10 500 106 3. m m=6.87 10 m6 . The original energy is

EGMm

ai = =

=

2

6 67 10 5 98 10 102 90

11 24 4. ..

N m kg kg kg

2 6.87 10 m10 J

2 2

6

11e je je je j

.

We assume that the perigee distance in the new orbit is 6 87 106. m. Then the major axis is

2 6 87 10 2 00 10 2 69 106 7 7a = + = . . .m m m and the final energy is

E GMma

f = =

=

26 67 10 5 98 10 10

2 69 101 48 10

11 24 4

711. .

..N m kg kg kg

mJ

2 2

e je je j .

The energy input required from the engine is E Ef i = = 1 48 10 2 90 10 1 42 1011 11 11. . .J J Je j .


20/34


P11.43 (a) Energy of the spacecraft-Mars system is conserved as the spacecraft moves between a verydistant point and the point of closest approach:

0 01

2

2

2+ =

=

mvGM m

r

vGM

r

r

r

Mars

Mars

After the engine burn, for a circular orbit we have

= =

=

F maGM m

r

mv

r

vGM

r

: Mars

Mars

20

2

0

The percentage reduction from the original speed is

v v

v

v v

vr

r

=

=

=

0 0 0

0

2

2

2 1

2100% 29 3%.

(b) The answer to part (a) applies with no changes , as the solution to part (a) shows.

Additional Problems

P11.44 Let m represent the mass of the spacecraft, rE the radius of the Earths orbit, and x the distance from

Earth to the spacecraft.

The Sun exerts on the spacecraft a radial inward force of FGM m

r x

ss

E

=

b g2

while the Earth exerts on it a radial outward force of FGM m

xE

E=

2

The net force on the spacecraft must produce the correct centripetal acceleration for it to have anorbital period of 1.000 year.

Thus, F FGM m

r x

GM m

x

mv

r x

m

r x

r x

TS E

S

E

E

E E

E =

=

=

LNMM

OQPPb g b g b g

b g2 2

22

2

which reduces toGM

r x

GM

x

r x

T

S

E

E E

=

b g

b g2 2

2

2

4. (1)

Cleared of fractions, this equation would contain powers ofx ranging from the fifth to the zeroth.

We do not solve it algebraically. We may test the assertion that x is between 1 47 109. m and

1 48 109. m by substituting both of these as trial solutions, along with the following data:

MS = 1 991 1030. kg , ME = 5 983 10

24. kg , rE = 1 496 1011. m, and T= = 1 000 3 156 107. .yr s .



21/34

Chapter 11 317

With x = 1 47 109. m substituted into equation (1), we obtain

6 052 10 1 85 10 5 871 103 3 3. . . m s m s m s2 2 2

or 5 868 10 5 871 103 3. . m s m s2 2

With x = 1 48 109

. m substituted into the same equation, the result is

6 053 10 1 82 10 5 870 8 103 3 3. . . m s m s m s2 2 2

or 5 870 9 10 5 870 8 103 3. . m s m s2 2 .

Since the first trial solution makes the left-hand side of equation (1) slightly less than the right handside, and the second trial solution does the opposite, the true solution is determined as between the

trial values. To three-digit precision, it is 1 48 109. m.As an equation of fifth degree, equation (1) has five roots. The Sun-Earth system has five

Lagrange points, all revolving around the Sun synchronously with the Earth. The SOHO and ACEsatellites are at one. Another is beyond the far side of the Sun. Another is beyond the night side of theEarth. Two more are on the Earths orbit, ahead of the planet and behind it by 60. Plans are under way

to gain perspective on the Sun by placing a spacecraft at one of these two co-orbital Lagrange points.The Greek and Trojan asteroids are at the co-orbital Lagrange points of the Jupiter-Sun system.

P11.45 The acceleration of an object at the center of the Earth dueto the gravitational force of the Moon is given by

a GM

d=

Moon2

At the point A nearest the Moon, a GM

d r

M+

=

a f2

At the point B farthest from the Moon, a GM

d r

M =

+af

2

FIG. P11.45

a a a GMd r d

M= =

L

NMM

O

QPP+

1 12 2a f

For d r>> , aGM r

dM

= = 2 1 11 10

36. m s2

Across the planet, g

g

a

g= =

=

2 2 22 10

9 802 26 10

67.

..

m s

m s

2

2


22/34


*P11.46 Energy conservation for the two-sphere system from release to contact:

= + +

FHG

IKJ= =

L

NM

O

QP

F

H

GI

K

J

Gmm

R

Gmm

rmv mv

Gmr R

v v Gmr R

2

1

2

1

2

1

2

1 1

2

1

2 2

21 2

(a) The injected impulse is the final momentum of each sphere,

mv m Gmr R

Gmr R

= LNM

OQP

FHG

IKJ =

FHG

IKJ

LNM

OQP

2 21 2

31 2

1

2

1 1

2

1.

(b) If they now collide elastically each sphere reverses its velocity to receive impulse

mv mv mv Gmr R

= = FHG

IKJ

LNM

OQP

a f 2 2 12

131 2

P11.47 (a) The free-fall acceleration produced by the Earth is g GMr

GM rE E= = 2 2 (directed downward)

Its rate of change isdg

drGM r GM rE E= =

2 23 3a f .

The minus sign indicates thatgdecreases with increasing height.

At the Earths surface,dg

dr

GM

RE

E

= 2

3.

(b) For small differences,

g

r

g

h

GM

RE

E= =

23 Thus, g GM hR

E

E= 2 3

(c) g =

=

2 6 67 10 5 98 10 6 00

6 37 101 85 10

11 2 24

6 35

. . .

..

N m kg kg m

mm s

2

2e je ja f

e j

*P11.48 (a) Each bit of mass dm in the ring is at the same distance from the object at A. The separate

contributions Gmdm

rto the system energy add up to

GmM

r

ring. When the object is at A,

this is

+

=

6 67 10 1 000

1 10 2 10

7 04 1011

8 2 8 2

4. .N m kg 2.36 10 kg

kg m m

J2 20

2 e j e j.



23/34

Chapter 11 319

(b) When the object is at the center of the ring, the potential energy is

=

6 67 10

1 101 57 10

11

85. .

N m 1 000 kg 2.36 10 kg

kg mJ

2 20

2.

(c) Total energy of the object-ring system is conserved:

K U K U

v

v

gA

gB

B

B

+ = +

=

= F

HGIKJ

=

e j e j

0 7 04 101

21 000 1 57 10

2 8 70 1013 2

4 2 5

4 1 2

. .

..

J kg J

J

1 000 kgm s

P11.49 To approximate the height of the sulfur, set

mvmg hIo

2

2

= h = 70 000 m gGM

rIo = =2

1 79. m s2

v g hIo= 2 v = 2 1 79 70 000 500.a fb g b gm s over 1 000 mi h

A more precise answer is given by

1

22

1 2

mvGMm

r

GMm

r =

1

26 67 10 8 90 10

1

1 82 10

1

1 89 10

2 11 226 6

v =

FHG

IKJ

. .. .

e je j v = 492 m s

P11.50 From the walk, 2 25 000r = m. Thus, the radius of the planet is r = = 25 000

3 98 103m

2m

.

From the drop: y gt g = = =1

2

1

229 2 1 402

2. .s ma f

so, gMG

r= = =

2 1 40

29 23 28 10

23

2

.

..

m

sm s2

a fa f

= M 7 79 1014. kg


24/34


P11.51 For both circular orbits,

F ma = :GM m

r

mv

rE

2

2

=

vGM

r

E=

rr

r

FIG. P11.51

(a) The original speed is vi =

+

=

6 67 10 5 98 10

6 37 10 2 107 79 10

11 24

6 5

3. .

..

N m kg kg

m mm s

2 2e je je j

.

(b) The final speed is vi =

=

6 67 10 5 98 10

6 47 107 85 10

11 24

6

3. .

..

N m kg kg

mm s

2 2e je je j

.

The energy of the satellite-Earth system is

K U mvGM m

rm

GM

r

GM

r

GM m

rg

E E E E+ = = =

1

2

1

2 22

(c) Originally Ei =

=

6 67 10 5 98 10 100

2 6 57 103 04 10

11 24

6

9. .

..

N m kg kg kg

mJ

2 2e je jb ge j

.

(d) Finally Ef =

=

6 67 10 5 98 10 100

2 6 47 103 08 10

11 24

6

9. .

..

N m kg kg kg

mJ

2 2e je jb g

e j.

(e) Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy isso large that the total energy decreases by

E Ei f = = 3 04 10 3 08 10 4 69 109 9 7. . .J J Je j .

(f) The only forces on the object are the backward force of air resistance R, comparatively verysmall in magnitude, and the force of gravity. Because the spiral path of the satellite is not

perpendicular to the gravitational force, one component of the gravitational force pulls

forward on the satellite to do positive work and make its speed increase.

P11.52 (a) The net torque exerted on the Earth is zero. Therefore, the angular momentum of the Earth

is conserved;

mr v mr va a p p= and v vr

ra p

p

a

=FHG

IKJ

= FHG

IKJ= 3 027 10

1 471

1 5212 93 104 4.

.

..m s m se j



25/34

Chapter 11 321

(b) K mvp p= = = 1

2

1

25 98 10 3 027 10 2 74 102 24 4

2 33. . .e je j J

UGmM

rp p= =

=

6 673 10 5 98 10 1 99 10

1 471 105 40 10

11 24 30

1133

. . .

..

e je je jJ

(c) Using the same form as in part (b), Ka = 2 57 1033. J and Ua = 5 22 10

33. J .

Compare to find that K Up p+ = 2 66 1033. J and K Ua a+ = 2 65 10

33. J . They agree.

P11.53 (a) At infinite separation U= 0 and at rest K= 0 . Since energy of the two-planet system isconserved we have,

01

2

1

21 1

22 2

2 1 2= + m v m v

Gm m

d(1)

The initial momentum of the system is zero and momentum is conserved.

Therefore, 0 1 1 2 2= m v m v (2)

Combine equations (1) and (2): v mG

d m m1 2

1 2

2=

+b gand v m

G

d m m2 1

1 2

2=

+b g

Relative velocity v v vG m m

dr= =

+

1 21 22b g b g

(b) Substitute given numerical values into the equation found for v1 and v2 in part (a) to find

v1

41 03 10= . m s and v2

32 58 10= . m s

Therefore, K m v1 1 12 321

21 07 10= = . J and K m v2 2 2

2 311

22 67 10= = . J

*P11.54 (a) Let R represent the radius of the asteroid. Then its volume is4

33R and its mass is

4

33R .

For your orbital motion, F ma = ,Gm m

R

m v

R1 22

22

= ,G R

R

v

R

4

3

3

2

2

=

Rv

G=

FHG

IKJ

=

F

HGG

I

KJJ =

3

4

3 8 5

6 67 10 1 100 41 53 10

2 1 2 2

11

1 2

4

.

..

m s kg m

N m kg m

2 3

2

b gb g

(b) 4

31100

4

31 53 10 1 66 103 4

3 16R = = kg m m kg 3e j e j. .

(c) vR

T=

2 T

R

v= =

= =2 2 1 53 10

8 51 13 10 3 15

4

4 .

.. .

m

m ss h

e j



26/34


(d) Angular momentum is conserved for the asteroid-you system:

L L

m vR I

m vR m R

T

m vm R

T

Tm R

m v

i f =

=

=

=

= =

= =

0

02

5

2

4

5

4

5

4 1 66 10 1 53 10

5 90 8 58 37 10 26 5

2

2 12

21

1

2

16 4

17

asteroid

asteroid

asteroid

kg m

kg m ss billion years

. .

.. .

e je jb gb g

This problem is realistic. Many asteroids, such as Ida and Eros, are roughly 30 km indiameter. They are typically irregular in shape and not spherical. Satellites such as Phobos(of Mars), Adrastea (of Jupiter), Calypso (of Saturn), and Ophelia (of Uranus) would allow avisitor the same experience of easy orbital motion. So would many Kuiper-belt objects.

P11.55 (a) T rv

= =

= =

2 2 30 000 9 46 10

2 50 107 10 2 10

15

515 8 .

.

m

m ss yre j

(b) Ma

GT= =

=

4 4 30 000 9 46 10

6 67 10 7 13 102 66 10

2 3

2

2 15 3

11 15 241

.

. ..

m

N m kg skg

2 2

e j

e je j

M = 1 34 10 1011 11. ~solar masses solar masses

The number of stars is on the order of 1011 .

P11.56 (a) From the data about perigee, the energy of the satellite-Earth system is

E mvGM m

rp

E

p

= =

1

2

1

21 60 8 23 10

6 67 10 5 98 10 1 60

7 02 10

2 3 211 24

6. .

. . .

.a fe j

e je ja f

or E = 3 67 107. J

(b) L mvr mv rp p= = =

=

sin sin . . . .

.

90 0 1 60 8 23 10 7 02 10

9 24 10

3 6

10

kg m s m

kg m s2

b ge je j

(c) Since both the energy of the satellite-Earth system and the angular momentum of the Earthare conserved,

at apogee we must have1

22mv

GMm

rEa

a

=

and mv r La a sin .90 0 = .



27/34

Chapter 11 323

Thus,1

21 60

6 67 10 5 98 10 1 603 67 102

11 24

7.. . .

.a f e je ja f

vr

aa

=

J

and 1 60 9 24 1010. .kg kg m s2b gv ra a = .

Solving simultaneously,1

2 1 60

6 67 10 5 98 10 1 60 1 60

9 24 10 3 67 102

11 24

107

.

. . . .

. .a fe je ja fa f

v

v

a

a

=

which reduces to 0 800 11 046 3 672 3 10 02 7. .v va a + =

so va = 11 046 11 046 4 0 800 3 672 3 10

2 0 800

2 7b g a fe ja f

. .

..

This gives va = 8 230 m s or 5 580 m s . The smaller answer refers to the velocity at the

apogee while the larger refers to perigee.

Thus, rL

mva

a

= =

= 9 24 10

1 60 5 58 101 04 10

10

3

7.

. ..

kg m s

kg m sm

2

b ge j

.

(d) The major axis is 2a r rp a= + , so the semi-major axis is

a = + = 1

27 02 10 1 04 10 8 69 106 7 6. . .m m me j

(e) Ta

GME= =

4 4 8 69 10

6 67 10 5 98 10

2 3 2 63

11 24

.

. .

m

N m kg kg 2 2e j

e je j

T= =8 060 134s min

P11.57 Let m represent the mass of the meteoroid and vi its speed when far away.

No torque acts on the meteoroid, so its angular momentum is conserved asit moves between the distant point and the point where it grazes the Earth,moving perpendicular to the radius:

FIG. P11.57L Li f= : m mi i f f

r r r rr v r v =

m R v mR v

v v

E i E f

f i

3

3

b g ==

Now energy of the meteoroid-Earth system is also conserved:

K U K U gi

gf

+ = +e j e j :1

20

1

22 2mv mv

GM m

Ri f

E

E

+ =

1

2

1

292 2v v

GM

Ri i

E

E

= e j

GM

RvE

Ei= 42 : v

GM

Ri

E

E

=4


28/34


P11.58 From Keplers third law, minimum period means minimum orbit size. The treetop satellite inFigure P11.27 has minimum period. The radius of the satellites circular orbit is essentially equal tothe radius R of the planet.

F ma = :GMm

R

mv

R

m

R

R

T2

2 22= =

FHG

IKJ

G VR R

RT

G RR

T

=

FHG

IKJ=

2 2 2

2

32 3

2

4

4

3

4

e j

The radius divides out: T G2 3 = TG

=3

P11.59 If we choose the coordinate of the center of mass at the origin,then

0 2 1=

+

Mr mr

M mb g and Mr mr2 1=

(Note: this is equivalent to saying that the net torque must bezero and the two experience no angular acceleration.) For eachmass F ma= so

mrMGm

d1 1

22

= and MrMGm

d2 2

22

=

r r

FIG. P11.59

Adding these two equations gives r rM m G

d1 2

22

+ =+b g a f

with 1 2= = . Now usingd r r= +1 2

and T=2

we find Td

G M m2

2 34=

+

a f.

*P11.60 (a) The gravitational force exerted on m2 by the Earth (mass m1 ) accelerates m2 according to:

m gGm m

r2 2

1 22

= . The equal magnitude force exerted on the Earth by m2 produces negligible

acceleration of the Earth. The acceleration of relative approach is then

gGm

r2

12

11 24

7 2

6 67 10 5 98 10

1 20 102 77= =

=

. .

..

N m kg kg

mm s

2 2

2e je j

e j.

(b) Again, m2 accelerates toward the center of mass with g2 2 77= . m s2 . Now the Earth

accelerates toward m2 with an acceleration given as



29/34

Chapter 11 325

m gGm m

r

gGm

r

1 11 22

12

2

11 24

7 2

6 67 10 2 00 10

1 20 100 926

=

= =

=

. .

..

N m kg kg

mm s

2 2

2e je j

e j

The distance between the masses closes with relative acceleration of

g g g rel2 2 2m s m s m s= + = + =1 2 0 926 2 77 3 70. . . .

P11.61 Let r represent the distance between the electron and the positron. The two move in a circle of

radiusr

2around their center of mass with opposite velocities. The total angular momentum of the

electron-positron system is quantized according to

Lmvr mvr

nn = + =2 2

h

where n=

1 2 3, , , K .

For each particle, F ma = expands tok e

r

mv

re

2

2

2

2= .

We can eliminate vn

mr=

hto find

k e

r

mn

m re

2 2 2

2 2

2=

h.

So the separation distances are rn

mk ea n n

e

= = = 2 2 1 06 10

2 2

2 02 10 2h . me j .

The orbital radii arer

a n2

02

= , the same as for the electron in hydrogen.

The energy can be calculated from E K U mv mvk e

re

= + = + 1

2

1

22 2

2

.

Since mvk e

re2

2

2= , E

k e

r

k e

r

k e

r

k e

a n ne e e e

= = =

=

2 2 2 2

02 22 2 4

6 80. eV.

ANSWERS TO EVEN PROBLEMS

P11.2 (a) 2 50 10 5. N toward the 500-kg mass;(b) between the masses and 0.245 m fromthe 500 kg mass

P11.4 + 10 0 5 93 10 11. $ . $i je j N

P11.6 see the solution, either 1000 61 3. .m nm

or 2 74 10 4. m

P11.8 (a) 3 46 108. m;

(b) 3 34 10 3. m s2 toward the Earth

P11.10 (a) 1 31 1017. N toward the black hole;

(b) 2 62 1012. N kg

P11.12 (a) 5 59 103. m s ; (b) 239 min;

(c) 735 N downward


30/34


P11.14 1.27

P11.16 35.2 AU

P11.18 planet Yhas turned through1.30 revolutions

P11.20 18.2 ms

P11.22 2 82 109. J

P11.24 (a) 850 MJ ; (b) 2 71 109. J

P11.26 (a) 42.1 km/s; (b) 2 20 1011. m

P11.28 (a) see the solution; (b) 340 s

P11.30 (a) 2

3

R h

GM

E

E

+b g; (b)

GM

R h

E

E +

;

(c) GM mR h

R R h

R mE

E

E E

E+

+

LNMM

OQPP

2

2

2

86 400

2 2

2b g b g

s,

launch the satellite from a location on theequator, and launch it toward the east

P11.32GM m

RE

E12

P11.34 (a) 1.89 eV; (b) 656 nm; (c) 4 57 1014. Hz

P11.36 (a) 2 19 10

6

.

m s ; (b) 13.6 eV; (c) 27.2 eV

P11.38 (a) 13.6 eV; (b) 1.51 eV

P11.40 see the solution

P11.42 1 42 1011. J


P11.46 (a) see the solution;

(b) 21

2

131 2

Gmr R

FHG

IKJ

LNM

OQP

P11.48 (a) 7 04 104. J; (b) 1 57 105. J;(c) 13.2 m/s

P11.50 7 79 1014. kg

P11.52 (a) 2 93 104. m s; (b) K= 2 74 1033. J,

U= 5 40 1033. J ;

(c) K=

2 57 10

33

. J , U=

5 22 10

33

. J ,K Ua a+ = 2 65 10

33. J , total energy is

constant

P11.54 (a) 1 53 104. m ; (b) 1 66 1016. kg ;

(c) 1 13 104. s ; (d) 8 37 1017. s

P11.56 (a) 3 67 107. J ; (b) 9 24 1010. kg m s2 ;

(c) 5580 m s , 1 04 107. m ;

(d) 8 69 106. m ; (e) 134 min


P11.60 (a) 2 77. m s2 ; (b) 3 70. m s2

CONTEXT 2 CONCLUSION ANSWERS TO QUESTIONS

QCC2.1 The hypothetical anti-Earth will be right here six months fromnow. Your spacecraft can be here then by leaving the Earthsorbit and slowing down relative to the Sun, to fall into anelliptical orbit around the Sun. This point in the Earths orbit isthe aphelion of the spacecrafts new orbit. For its period to be1/2 year, we want

TGM

a a

a

a

S

22

3 19 3

3

1

2

72

1932

10

42 97 10

3 156 10

2 97 108 38 10

9 43 10

=FHG

IKJ

=

=

=

=

.

.

..

.

s m

s

s mm

m

2 3

2 33

e j

e je j FIG. QCC2.1



31/34

Chapter 11 327

Then the major axis is 2 1 89 1011a = . m . The perihelion distance is

1 89 10 1 496 10 3 9 1011 11 10. . . = m m m , less than the distance from the Sun to Mercury, so wewill want some insulation, air conditioning and a highly reflective surface for the spacecraft. From itscurrent speed in solar orbit,

v =

=

2 1 496 10

3 156 102 98 10

11

74

.

..

m

sm se j ,

the spacecraft needs to drop to a speed given by

E K U

GMm

amv

GMm

r

v GMr a

v

g= +

=

= FHG

IKJ

=

F

HG I

KJ

=

2

1

2

21 1

2

2 6 67 10 1 991 101

1 496 10

1

1 89 101 92 10

2

2

11 3011 11

4

. .

. ..

N m kg kg

m mm s

2 2e je j

QCC2.2 If you can survive for an hour, the commander of the other shuttle can most simply give the picnicbasket a very slow push (say at 0.5 m/s) tangent to the orbit. Both space shuttles and the basket are infree fall around the Earth. Since 0.5 m/s is negligible in comparison to orbital speed, the basket willmaintain nearly the same orbit as the spacecrafts, to bump into yours or pass close to it. Moreprecisely, in this maneuver the speed of the basket would be not quite enough for the circular orbit.It would start at the apogee of a slightly eccentric eclipse and would pass below your spacecraft,perhaps just out of reach. Your benefactor could compensate by doing a calculation and launchingthe basket with any convenient tangential speed toward you and with a small appropriately chosenradially outward speed component.

CONTEXT 2 CONCLUSION SOLUTIONS TO PROBLEMS

CC2.1 The Hohmann transfer path is one half of an ellipsewith major axis:

r r aEarth Venus m m+ = + =1 496 10 1 08 10 211 11. .

Then a = 1 29 1011. m . Now Ta

GMS

22 34

=

gives for the

transfer time

1

2

1

2

4 1 29 10

6 67 10 1 991 10

1 26 10 146

2 11 3

11 30

7

T=

= =

.

. .

.

m

N m kg kg

s d

2 2

e je je j

FIG. CC2.1



32/34


During this time Venus will move around in its orbit by = =

FHG

IKJ

= t 3601 26 10

1 94 10234

7

7

.

.

s

s. The

spacecraft departure and rendezvous points are 180 apart, so the spacecraft should be sent off when

Venus is behind the Earth by 234 180 53 8 = . .

*CC2.2 The radius of the circular orbit of the space station is

r RE= + = + = 500 6 37 10 5 00 10 6 87 106 5 6km m m m. . . .

From this, as in Example 11.1, we can find the orbital speed of the space station, which is the initialspeed vi of the golf ball:

vGM

riE

= =

=

6 67 10 5 98 10

6 87 107 62 10

11 24

63

. .

..

N m kg kg

mm s

2 2e je j.

The initial energy of the golf ball-Earth system, using Equation 11.8, is EGM m

r

E=

2

where m is the

mass of the golf ball.After the golf ball is hit, it returns to precisely the same location after the space station has

made 2.00 orbits. Thus, the period of the golf ball is 2.00 times that of the space station:

T

T

T

Tb

s

b

s

= FHG

IKJ

=2 00 4 00

2

. . .

According to Keplers third law, the ratio of the squares of the periods of the golf ball and the spacestation should be the equal to the ratio of the cubes of the semimajor axes:

T

T

a

a a a a rb

s

b

sb s s

FHG

IKJ =

FHG

IKJ = = = =

2 31 3

4 00 4 00 1 587 1 587. . . .a f

where we have identified the semimajor axis of the orbit of the space station as the radius r of itscircular orbit. From the semimajor axis of the golf ball orbit, we can find the new energy of the golf

ball-Earth system, using Equation 11.9: EGM m

aE

b

= 2

. Just after the golf ball is hit, the golf ball-Earth

system has the same potential energy as it did before the ball was hit, so the difference in energymust be equal to the change in kinetic energy of the golf ball:

E KGM m

a

GM m

r

GM m

r

GM m

r

GM mr

GM mr

E

b

E E E

E E

= =

FHG

IKJ= +

= +FHG IKJ=

2 2 2 1 587 2

21

1 5871 0 185

.

..

a f

Setting this equal to K mv mvf i= 1

2

1

22 2 , we find,



33/34

Chapter 11 329

1

2

1

20 185

0 370 0 370 1 17

2 2

2 2 2

mv mvGM m

r

v vGM

rv v v

f iE

f iE

i i i

=

= + = + =

.

. . .

Thus, the speed of the golf ball relative to the space station is,

v v v v v v f i i i i= = = = = 1 17 0 17 0 17 7 62 10 1 30 103 3. . . . .m s m se j

CC2.3 (a) The speed of an object in a circular orbit at the Earths distance is given by

= =

= =

=

F maGM m

r

mv

r

vGM

r

E E

E

:

. .

.

.

Sun

Sun

2 2N m kg kg

m

m s

22

2

2

11 30

11

4

6 67 10 1 991 10

1 496 10

2 98 10

e je j

For the transfer orbit at perihelion, the speed is described by

E K UGM m

amv

GM m

r

v GMr r r

v

v

gE

E E M

= + =

= +

FHG

IKJ

=

+

F

HGG

I

KJJ

=

Sun Sun

Sun

2 2N m kg kg m m

m s

2

1

2

21 1

2 6 67 10 1 991 101

1 496 10

1

1 496 10 2 28 10

3 27 10

32

3

311 30

11 11 11

34

. .. . .

.

e je je j

The speed change required is v v3 232 95 10 = . m s .

(b) The crafts aphelion speed is vv r

rE

M4

3

4

43 27 10

2 15 10= =

=

..

m s 1.496 10 m

2.28 10 mm s

11

11

e je j

. The

speed for a circular orbit at the distance of Mars is

vGM

rM5

11 30

114

6 67 10 1 991 10

2 28 10

2 41 10= =

=

Sun

2 2N m kg kg

m

m s. .

.

.e je j

.

The change in speed required at rendezvous to fly along with Mars is

v v5 432 65 10 = . m s .



34/34

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