gravity, planetary orbits, and the hydrogen...

297

Gravity, Planetary Orbits,and the Hydrogen Atom

CHAPTER OUTLINE 11.1 Newton’s Law of Universal

Gravitation Revisited 11.2 Structural Models 11.3 Kepler’s Laws 11.4 Energy Considerations in

Planetary and Satellite Motion

11.5 Atomic Spectra and the Bohr Theory of Hydrogen

11.6 Context Connection Changing from a Circular to an Elliptical Orbit

ANSWERS TO QUESTIONS Q11.1 The Earth creates the same gravitational field g for all objects near

the Earth’s 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 Sun’s gravitational field, in orbit around the Sun. To a precise second approximation, you weigh slightly less at noon and at midnight than you do at sunrise or sunset. The Sun’s gravitational field is a little weaker at the center of the Earth than at the surface subsolar point, and a little weaker still on the far side of the planet. When the Sun is high in your sky, its gravity pulls up on you a little more strongly than on the Earth as a whole. At midnight the Sun pulls down on you a little less strongly than it does on the Earth below you. So you can have another 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 by

F 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 its orbit about the Earth by the gravitational force of the Earth. There is no imbalance of these forces, at new moon or full moon.

Q11.4 Air resistance causes a decrease in the energy of the satellite-Earth system. This reduces the diameter

of 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 required—and fuel—would be proportional to v2 , or 25 times more fuel is required to leave the Earth versus leaving the Moon.

Q11.6 In a circular orbit each increment of displacement is perpendicular to the force applied. The dot

product of force and displacement is zero. The work done by the gravitational force on a planet in an elliptical orbit speeds up the planet at closest approach, but negative work is done by gravity and the planet slows as it sweeps out to its farthest distance from the Sun. Therefore, net work in one complete orbit is zero.

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 located at the center of the Earth. If the satellite is over the northern hemisphere for half of its orbit, it must be over the southern hemisphere for the other half. We could share with Easter Island a satellite that would look straight down on Arizona each morning and vertically down on Easter Island each evening.

Q11.8 The gravitational force of the Earth on an extra particle at its center must be zero, not infinite as one

interpretation of Equation 11.1 would suggest. All the bits of matter that make up the Earth will pull in 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 permanently

speed up a spacecraft. Jupiter is moving. A spacecraft flying across its orbit just behind the planet will gain kinetic energy as the planet’s gravity does net positive work on it.

Q11.11 Cavendish determined G. Then from gGMR

= 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 around

an atomic nucleus. If it behaved like a charge in a radio antenna, it would radiate light with frequency equal to its own frequency of oscillation. Thus, the electron in hydrogen atoms would emit a continuous spectrum, electromagnetic waves of all frequencies smeared together.

Q11.14 (a) Yes—provided that the energy of the photon is precisely enough to put the electron into one

of the allowed energy states. Strangely—more precisely non-classically—enough, if the energy of the photon is not sufficient to put the electron into a particular excited energy level, the photon will not interact with the atom at all!

(b) Yes—a photon of any energy greater than 13.6 eV will ionize the atom. Any “extra” energy

will 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 to

the nucleus. Positive ionization energy must be injected to pull the electron out to a very large separation from the nucleus, a condition for which we define the energy of the atom to be zero. The atom is a bound system. All this is summarized by saying that the total energy of an atom is negative.

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

rk e

rk e

rK Ue e e

e

2 2 2

2 2. Then K E=

and U Ee = −2 .

Chapter 11 299

SOLUTIONS TO PROBLEMS Section 11.1 Newton’s Law of Universal Gravitation Revisited

P11.1 F m gGm m

r= =1

1 22

gGm

r= =

× ⋅ × ×= ×

−−2

2

11 4 3

27

6 67 10 4 00 10 10

1002 67 10

. ..

N m kg kg

m m s

2 22e je j

a f

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

are oppositely directed,

and from FGm m

rg = 1 22

we have FG

∑ =−

= × −50 0 500 200

0 2002 50 102

5.

..

kg kg kg

m N

b gb ga 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 on

the 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 kgb gb ga f

b gb g−

=

or d = 0 245. m

P11.3 FGMm

r= = × ⋅

×

×= ×−

−

−

−2

113

2 2106 67 10

1 50 15 0 10

4 50 107 41 10.

. .

.. N m kg

kg kg

m N2 2e j

b ge je j

P11.4 The force exerted on the 4.00-kg mass by the 2.00-kg mass is

directed 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 j b gb ga f

The force exerted on the 4.00-kg mass by the 6.00-kg mass is

directed 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 f N m kg kg kg

m

N

2 2

r

r

FIG. P11.4

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

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


P11.5 (a) The Sun-Earth distance is 1 496 1011. × m and the Earth-Moon distance is 3 84 108. × m , so the distance 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 je j

N

(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 je 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 je j

Note that the force exerted by the Sun on the Moon is much stronger than the force of the Earth on the Moon. In a sense, the Moon orbits the Sun more than it orbits the Earth. The Moon’s path is everywhere concave toward the Sun. Only by subtracting out the solar orbital motion of the Earth-Moon system do we see the Moon orbiting the center of mass of this system.

P11.6 Let θ represent the angle each cable makes with the vertical, L the

cable 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 of the balls. We have

Fy∑ = 0 : T mgcosθ − = 0

Fx∑ = 0 : TGmm

rsinθ − =2 0

r

r

r

FIG. P11.6

Then tanθ = Gmmr mg2

x

L x

Gm

g d x2 2 22−=

−a f x d xGm

gL x− = −2 2 2 2a f .

The factor Gm

gis 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 8452

11

m N m kg kg

m s m

2 2

2a f e jb g

e j=

× ⋅−.

. x = × −3 06 10 8. m.

continued on next page

Chapter 11 301

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

Possibility two: If d 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 kg m m

2 2

a f e jb gb g a 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 the nucleus of atom.

P11.7 gGMR

G

RG R

R

= = =2

43

2

3

43

ρπ ρ

πe j

If gg

G RG R

M

E

M M

E E= =1

64

33

4π ρ

π ρ

then ρρ

M

E

M

E

E

M

gg

RR

=FHGIKJFHGIKJ =FHGIKJ =1

64

23

a f .

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

rGmM

rE

E

M

M2 2=

so r rMM

rr

M EM

EE

E= = ××

=7 36 105 98 10 9 01

22

24.. .

r r r

r

r

E M EE

E

+ = × = +

= × = ×

3 84 109 01

3 84 103 46 10

8

88

..

..

m

m1.11

m

(b) At this distance the acceleration due to the Earth’s gravity is

g

GMr

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 je 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

2 N m kg

kg m m s

=

= =× ⋅ ×

×=

−6 67 10 6 68 10

242 100 076 1

11 19

3 2

. ..

e je j



(b) We ignore the difference (of about 4%) in g between the lip and the base of the cliff. For the vertical motion of the athlete we have

y y v a t

t

t

f i yi y= + +

− = + + −

=FHG

IKJ =

12

5 000 0 012

0 076 1

2 5 000

0 076363

2

2

2 1 2

m m s

m s

1 m s

2.

.

e jb g

(c) x x v t a tf i xi x= + + = + + = ×12

0 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 athlete’s trajectory. (d) v vxf xi= = 8 50. m s

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

Thus rv i jf = − = +8 50 27 6 8 5 27 62 2. $ . $ . .e j m s m s at tan

..

−1 27 68 5

below the x axis.

rv f = °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 m N

2 2e j e je je j

(b) ∆FGMmr

GMmr

= −front2

back2

∆ ∆g

Fm

GM r r

r r= =

−back2

front2

front2

back2

e j

FIG. P11.10

∆

∆

g

g

=× × × − ×L

NMOQP

× ×

= ×

−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 je j e j

m m

m m

N kg

Chapter 11 303

P11.11 g gMG

r a1 2 2 2= =+

g gy y1 2= − g g gy y y= +1 2

g g gx x1 2 2= = cosθ cosθ =+

r

a r2 2 1 2e j

rg i= −2 2g x

$e j

or rg =

+

22 2 3 2

MGr

r ae j toward the center of mass

r

r

FIG. P11.11

P11.12 (a) We require that GM m

rmv

rE

2

2

= , but gM GR

E

E

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

RE4 2

2

= or

vgRE= =

×= ×

2

9 80 6 37 10

25 59 10

63

. ..

m s m m s

2 e j.

(b) Tr

v= =

×

×=

2 2 2 6 37 10239

6π πa fe j. m

5.59 10 m s min3

(c) FGM m

R

mgE

E

= = = =2 4

300 9 807352b g

b ge j kg m s

4 N

2.

Section 11.2 Structural Models No problems in this section

Section 11.3 Kepler’s Laws

P11.13 (a) For the geosynchronous satellite, Σ =F mar r becomes GmM

rmv

rE

2

2

= and in turn

GMr

rT

E = FHGIKJ

2 2π or r

GM TE32

24=

π. Thus, the radius of the satellite orbit is

r =× ⋅ ×L

NMM

O

QPP = ×

−6 67 10 5 98 10 86 400

44 23 10

11 24 2

2

1 3

7. .

.

/ N m kg kg s

m2 2e je jb g

π.



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

d = × + × = ×6 37 10 4 23 10 4 27 106 2 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

tdc

= =×

×=2 2 4 27 10

3 00 100 285

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

vrr

p

a

a

p= =

+ ×+ ×

= =2 289 6 37 10

6 37 108 659

1 273

3 km km

459 km km km

6 829 km.

.. .

We do not need to know the period.

P11.15 Ta

GM2

2 34= π (Kepler’s third law with m M<< )

Ma

GT= =

×

× ⋅ ×= ×

−

4 4 4 22 10

6 67 10 1 77 86 4001 90 10

2 3

2

2 8 3

11 227π π .

. ..

m

N m kg s kg

2 2

e je jb g

(Approximately 316 Earth masses) P11.16 By Kepler’s Third Law, T ka2 3= (a = semi-major axis)

For any object orbiting the Sun, with T in years and a in A.U., k = 1 00. . Therefore, for Comet Halley

75 6 1 000 570

22

3

. ..a f a f=

+FHG

IKJ

y

The farthest distance the comet gets from the Sun is

y = − =2 75 6 0 570 35 22 3. . .a f A.U. (out around the orbit of Pluto)

FIG. P11.16

P11.17 Applying Newton’s 2nd Law, F ma∑ = yields F mag c= for each star:

GMM

r

Mvr2 2

2

a f = or Mv rG

= 4 2

.

We can write r in terms of the period, T, by considering the time and

distance of one complete cycle. The distance traveled in one orbit is the circumference of the stars’ common orbit, so 2πr vT= . Therefore

Mv rG

vG

vT= = FHGIKJ

4 42

2 2

π

FIG. P11.17

so, Mv TG

= =×

× ⋅= × =

−

2 2 220 10 14 4 86 400

6 67 101 26 10 63 3

3 3 3

1132

π π

m s d s d

N m kg kg solar masses

2 2

e j a fb ge j

.

.. .

Chapter 11 305

P11.18 F ma∑ = : Gm M

r

m v

rplanet star planet

2

2

=

GMr

v r

GM r r r

rr

x x y y

y xx

yy

star

star

yr yr

= =

= = =

=FHGIKJ = °F

HGIKJ = °

2 2 2

3 3 3 2 3 2

3 2

3 290 05 00

3468

5 00

ω

ω ω ω

ω ω ω .. .

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

rmv

rsun2

2

= vGM

r= sun .

For Mercury the speed is vM =× ×

×= ×

−6 67 10 1 99 10

5 79 104 79 10

11 30

104

. .

..

e je je j

m

s m s

2

2

and for Pluto, vP =× ×

×= ×

−6 67 10 1 99 10

5 91 104 74 10

11 30

123

. .

..

e je je j

m

s m s

2

2.

With greater speed, Mercury will eventually move farther from the Sun than Pluto. With original distances 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 49 102 27 10

1 24 10 3 93

+ = +

− = −

=× − ×

× − ×= ×

×= × =

e je j e je j e j

. .

. .

..

. . . m m

m s m s

m m s

s yr2

2 2


P11.20 For the Earth, F ma∑ = : GM m

rmv

rmr

rT

s2

2 22= = FHGIKJ

π.

Then GM T rs2 2 34= π .

Also the angular momentum L mvr mr

Tr= = 2π

is a constant for the Earth.

We eliminate rLT

m=

2π between the equations:

GM TLT

ms2 2

3 2

42

= FHGIKJπ

π GM T

Lms

1 2 23 2

42

= FHGIKJπ

π.

Now the rate of change is described by

GM TdTdt

GdM

dtTs

s12

1 01 2 1 2−FHG

IKJ +FHG

IKJ =

dTdt

dMdt

TM

Tt

s

s= −

FHGIKJ ≈2

∆∆

∆ ∆

∆

T tdM

dtT

M

T

s

s≈ −

FHGIKJ = − ×F

HGIKJ − ×

×

FHG

IKJ

= × −

2 5 0003 16 10

3 64 10 21

1 82 10

79

2

yr s

1 yr kg s

yr1.991 10 kg

s

30.

.

.

e j

Section 11.4 Energy Considerations in Planetary and Satellite Motion

P11.21 (a) ρπ π

= =×

×= ×

MrS

E43

2

30

6 39

3 1 99 10

4 6 37 101 84 10

.

..

kg

m kg m3e 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

m m s

2 22e je j

e j

(c) UGM m

rgS

E= − = −

× ⋅ ×

×= − ×

−6 67 10 1 99 10 1 00

6 37 102 08 10

11 30

613

. . .

..

N m kg kg kg

m J

2 2e je jb g

P11.22 W UGm m

r= − = −

−−F

HGIKJ∆ 1 2 0

W =+ × ⋅ × ×

×= ×

−6 67 10 7 36 10 1 00 10

1 74 102 82 10

11 22 3

69

. . .

..

N m kg kg kg

m J

2 2e je je j

Chapter 11 307

P11.23 For her jump on earth, 12

2mv 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

12

0 02mvGM m

RiA

A− = + (2)

The equality of densities between planet and asteroid,

ρπ π

= =M

R

M

RE

E

A

A43

2 43

2 implies M

RR

MAA

EE=

FHGIKJ

3

(3)

Note also at Earth’s surface gGMR

E

E= 2 (4)

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

12

2vGM

RiA

A=

12

23

gyG

RRR

MfA

A

EEd i = F

HGIKJ

GMR

yGM R

RE

Ef

E A

E2

2

3= R y RA f E2 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

= × ⋅FHG

IKJ ×

×−

×FHG

IKJ

=

−

∆

1 1

6 67 105 98 10 100

16 37 10

17 37 10

850

1124

6 6.

.. .

N mkg

kg kg m m

MJ

2

2 e jb g



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

GM M

R y

M v

R yE p

E

p

E+=

+b g b g2

2

Then, 12

12

2M vGM M

R ypE p

E=

+b g

So, additional work = kinetic energy required

=

× ⋅ ×

×

= ×

−12

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

P11.25 12

1 1 12

2 2mv GM mr r

mvi Ef i

f+ −FHG

IKJ =

12

01 1

22 2v GM

Rvi E

Ef+ −

FHG

IKJ =

or v vGMRf

E

E

212 2

= −

and v vGMRf

E

E= −FHG

IKJ1

21 2

2

v f = × − ×LNM

OQP = ×2 00 10 1 25 10 1 66 104 2 8

1 24. . .e j m s

P11.26 (a) vM GRE

solar escapeSun

Sun km s= =

⋅

242 1.

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

vM GR x xE S

= =⋅

2 42 1Sun .

If v =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).

Chapter 11 309

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

Rmv

R∑ = =2

2

or vMG

R=

We can obtain the escape velocity from 12

mvmMG

Resc2 =

or vMGR

vesc = =22

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

K U K U

GM mR h

mvGM m

r

v GMr R h

drdt

g h g r

E

E

E

EE

+ = +

−+

= −

= −+

FHG

IKJ

FHG

IKJ = −

e j e jaltitude radius

012

21 1

2

1 2

(b) dtdrv

drvi

f

i

f

f

iz 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 977 101

101

6 87 1010 3 541 10

10

10

1 16 87

14 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 .


P11.29 v

R hGM

R hi

E

E

E

2

2+=

+b g

K mvGM mR hi i

E

E= =

+FHG

IKJ =

× ⋅ ×

× + ×

L

NMM

O

QPP = ×

−12

12

12

6 67 10 5 98 10 500

6 37 10 0 500 101 45 102

11 24

6 610

. .

. ..

N m kg kg kg

m m J

2 2e je jb ge j e j

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

∆U

GM mR

GM mR

GM mR R

E

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= = × = ×12

12

500 2 00 10 1 00 102 3 2 9 kg m s Jb ge j. . .

The energy transformed due to friction is

∆ ∆E K K Ui fint J J= − − = − + × = ×14 5 1 00 2 27 10 1 58 109 10. . . .a f .

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

Thus, GM m

R h

mvR h

E

E E+=

+b g b g2

2

or vGMR h

E

E

2 =+

(a) Tr

vR hE

GMR h

E

E

= =+

+

2 2π πb gb g

TR hGME

E=

+2

3

πb g

(b) vGMR h

E

E=

+

(c) Minimum energy input is ∆E K U K Uf 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= 12

2

with vR R

iE E= =

21 00

286 400

π π. day s

and UGM m

RgiE

E= − .

Thus, ∆E mGMR h

GM mR h

mR GM m

RE

E

E

E

E E

Emin =

+FHG

IKJ − +

−LNMM

OQPP +

12

12

4

86 400

2 2

2π

sb g

or ∆E GM mR h

R R hR m

EE

E E

Emin =

++

LNMM

OQPP

−2

22

86 400

2 2

2b g b gπ

s

Chapter 11 311

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

K U K U

mvGmM

RGmMR h

g i g f

iE

E

E

E

+ = +

− = −+

e j e j12

02

where 12

mvGmM

RE

Eesc2 = . Then

12

12

12

1

2

2

2

2

2

2

2

2

v v vR

R h

v vv RR h

v vR hv R

hv R

v vR

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

esc2

esc2

esc2

esc2

esc2

esc2

(b) h =×

−= ×

6 37 10

11 2 8 761 00 10

6 2

2 27.

. ..

m 8.76 m

a fa f a f

(c) The fall of the meteorite is the time-reversal of the upward flight of the projectile, so it is

described by the same energy equation

v vR

R hv

hR h

v

iE

E E

i

2 2 2 3 2 7

6 7

8

4

1 11 2 102 51 10

6 37 10 2 51 10

1 00 10

1 00 10

= −+

FHG

IKJ = +FHG

IKJ = × ×

× + ×FHG

IKJ

= ×

= ×

esc esc

2 2

m s m

m m

m s

m s

..

. .

.

.

e j

(d) With v vi << esc , hR vv

R v RGM

E i E i E

E≈ =

2

2

2

2esc

. But gGM

RE

E

= 2 , so hv

gi=2

2, in agreement with

0 2 02 2= + − −v g hi b ga f . P11.32 For a satellite in an orbit of radius r around the Earth, the total energy of the satellite-Earth system is

EGM

rE= −

2. Thus, in changing from a circular orbit of radius r RE= 2 to one of radius r RE= 3 , the

required work is

W EGM m

rGM m

rGM m

R RGM m

RE

f

E

iE

E E

E

E= = − + = −

LNM

OQP

=∆2 2

14

16 12

.


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 = 24 75. AU

Then eca

= = =24 7525 25

0 980..

.

(b) In T K as

2 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 T 22

331

125 25=

yr

AU AU

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

1117

. . .

..

N m kg kg kg

m J

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 ni f

i f= − = − −

−13 606 13 6062 2

. . eV eVa f

En nf i

= −FHG

IKJ

13 6061 1

2 2. eV

Thus, for n = 3 to n = 2 transition E = −FHGIKJ =13 606

19

1 89. . eV14

eV

(b) Ehc=λ

and λ =× ⋅ ×

×=

−

−

6 626 10

1 89656

34

19

.

.

J s 2.998 10 m s

eV 1.602 10 J eV nm

8e je j

(c) fc=λ

f =×

×= ×−

3 106 56 10

4 57 108

714 m s

m Hz

..

Chapter 11 313

P11.35 (a) Lyman series 1

112λ

= −FHGIKJR

ni

ni = 2 3 4, , , K

1 1

94 96 101 097 10 1

19

72λ

=×

= × −FHGIKJ−.

.e jni

ni = 5

(b) Paschen series: 1 1

31

2 2λ= −FHG

IKJR

ni

ni = 4 5 6, , , K

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

1

1 097 1019

172λ

= × −FHG

IKJ.

ni

For ni = ∞ , this gives λ = 820 nm

This is larger than 94.96 nm, so this wave length cannot be associated with the Paschen series .

Balmer series: 1 1

21

2 2λ= −FHG

IKJR

ni

ni = 3 4 5, , , K

1

1 097 1014

172λ

= × −FHG

IKJ.

ni

with ni = ∞ for ionization, λ min = 365 nm

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

P11.36 (a) vk em r

e

e1

2

1=

where r a12

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

v1

9 19 2

31 116

8 99 10 1 60 10

9 11 10 5 29 102 19 10=

× ⋅ ×

× ×= ×

−

− −

. .

. ..

N m C C

kg m m s

2 2e je je je j

(b) K m ve1 12 31 6 2 181

212

9 11 10 2 19 10 2 18 10 13 6= = × × = × =− −. . . . kg m s J eVe je j

(c) Uk ere

1

2

1

9 19 2

1118

8 99 10 1 60 10

5 29 104 35 10 27 2= − = −

× ⋅ ×

×= − × = −

−

−−

. .

.. .

N m C C

m J eV

2 2e je j


P11.37 ∆En ni f

= −FHG

IKJ

13 61 12 2. eVa f

Where for ∆E > 0 we have absorption and for ∆E < 0 we have emission. (i) for ni = 2 and n f = 5 , ∆E = 2 86. eV (absorption)

(ii) for ni = 5 and n f = 3 , ∆E = −0 967. eV (emission)

(iii) for ni = 7 and n f = 4 , ∆E = −0 572. eV (emission)

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

(a) Ehc=λ

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 Enn = −13 6

2. eV

.

To ionize the atom when the electron is in the n th level,

it is necessary to add an amount of energy given by E Enn= − = 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 61 51

..

eV9

eV .

P11.39 (a) r2

2 20 052 9 2 0 212= =. . nm nmb ga f

(b) m vm k e

ree e

2

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

mee

e2 2

2 22 25 2

31191

2 2

9 95 10

2 9 11 105 43 10 3 40= = =

× ⋅

×= × =

−

−−b g e j

e j.

.. .

kg m s

kg J eV


Chapter 11 315

(e) Uk ere

2

2

2

9 19 2

918

8 99 10 1 60 10

0 212 101 09 10 6 80= − = −

× ⋅ ×

×= − × = −

−

−−

. .

.. .

N m C C

m J eV

2 2e je j

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

P11.40 Starting with 12 2

22

m vk e

ree=

we have vk em r

e

e

22

=

and using rn

m k ene e

=2 2

2h

gives vk e

m n m k en

e

e e e

22

2 2 2=

he j

or vk enne=

2

h.

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

so 12

6 626 10 3 00 10

1 216 101 63 102

34 8

718mv

hc= =× ⋅ ×

×= ×

−

−−

λ

. .

..

J s m s

m J

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

611e je je j

e 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

EGMm

af = − = −× ⋅ ×

×= − ×

−

2

6 67 10 5 98 10 10

2 69 101 48 10

11 24 4

711

. .

..

N m kg kg kg

m J

2 2e 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 .


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

0 0

12

2

2+ = −

=

mvGM m

r

vGM

r

r

r

Mars

Mars

After the engine burn, for a circular orbit we have

Σ = =

=

F maGM m

rmv

r

vGM

r

: Mars

Mars

20

2

0

The percentage reduction from the original speed is

v v

vv v

vr

r

−=

−= − × =0 0 0

0

22

2 12

100% 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 Earth’s orbit, and x the distance from

Earth to the spacecraft.

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

r xs

s

E

=−b g2

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

xEE= 2

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

Thus, F FGM m

r x

GM mx

mvr x

mr x

r xTS E

S

E

E

E E

E− =−

− =−

=−

−LNMM

OQPPb g b g b g

b g2 2

2 22π

which reduces to GM

r x

GMx

r x

TS

E

E E

−− =

−

b gb g

2 2

2

2

4π. (1)

Cleared of fractions, this equation would contain powers of x 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 1024. kg , rE = ×1 496 1011. m, and T = = ×1 000 3 156 107. . yr s .


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 hand side, 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 ACE satellites are at one. Another is beyond the far side of the Sun. Another is beyond the night side of the Earth. Two more are on the Earth’s 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 due

to the gravitational force of the Moon is given by

a GM

d= Moon

2

At the point A nearest the Moon, a GM

d rM

+ =−a f2

At the point B farthest from the Moon, a GM

d rM

− =+a f2

FIG. P11.45

∆a a a GMd r dM= − =

−−

LNMM

OQPP+

1 12 2a f

For d r>> , ∆aGM r

dM= = × −2

1 11 1036. m s2

Across the planet, ∆ ∆gg

ag

= =×

= ×−

−2 2 22 109 80

2 26 106

7..

. m s

m s

2

2


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

− = − + +

−FHGIKJ = = −LNM

OQP

FHG

IKJ

GmmR

Gmmr

mv mv

Gmr R

v v Gmr R

212

12

12

1 12

1

2 2

21 2

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

mv m Gmr R

Gmr R

= −LNMOQP

FHG

IKJ = −FHG

IKJ

LNM

OQP

2 21 2

31 2

12

1 12

1.

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

mv mv mv Gmr R

− − = = −FHGIKJ

LNM

OQPa f 2 2

12

131 2

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

rGM rE

E= = −2

2 (directed downward)

Its rate of change is dgdr

GM r GM rE E= − = −− −2 23 3a f .

The minus sign indicates that g decreases with increasing height.

At the Earth’s surface, dgdr

GMR

E

E

= −2

3 .

(b) For small differences,

∆∆

∆g

r

g

hGMR

E

E

= =2

3 Thus, ∆gGM 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

m m s

22e 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 − Gmdmr

to the system energy add up to −GmM

rring . When the object is at A,

this is

− × ⋅ ×

× + ×= − ×

−6 67 10 1 000

1 10 2 107 04 10

11

8 2 8 2

4..

N m kg 2.36 10 kg

kg m m J

2 20

2 e j e j.


Chapter 11 319

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

−× ⋅ ×

×= − ×

−6 67 101 10

1 57 1011

85.

. N m 1 000 kg 2.36 10 kg

kg m J

2 20

2 .

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

K U K U

v

v

g A g B

B

B

+ = +

− × = − ×

= × ×FHG

IKJ =

e j e j0 7 04 10

12

1 000 1 57 10

2 8 70 1013 2

4 2 5

4 1 2

. .

..

J kg J

J1 000 kg

m s

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

mv

mg hIo

2

2= h = 70 000 m g

GMrIo = =2 1 79. m s2

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

A more precise answer is given by

12

2

1 2mv

GMmr

GMmr

− = −

12

6 67 10 8 90 101

1 82 101

1 89 102 11 22

6 6v = × ××

−×

FHG

IKJ

−. .. .

e je j v = 492 m s

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

3 98 103 m2

mπ

.

From the drop: ∆y gt g= = =12

12

29 2 1 402 2. . s ma f

so, gMGr

= = × =−2 1 40

29 23 28 102

32

.

..

m

s m s2a f

a f ∴ = ×M 7 79 1014. kg


P11.51 For both circular orbits,

F ma∑ = : GM m

rmv

rE

2

2

=

vGM

rE=

r r

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 53

. .

..

N m kg kg

m m m 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

63

. .

..

N m kg kg

m m s

2 2e je je j

.

The energy of the satellite-Earth system is

K U mvGM m

rm

GMr

GMr

GM mrg

E E E E+ = − = − = −12

12 2

2

(c) Originally Ei = −× ⋅ ×

×= − ×

−6 67 10 5 98 10 100

2 6 57 103 04 10

11 24

69

. .

..

N m kg kg kg

m J

2 2e je jb ge j

.

(d) Finally E f = −× ⋅ ×

×= − ×

−6 67 10 5 98 10 100

2 6 47 103 08 10

11 24

69

. .

..

N m kg kg kg

m J

2 2e je jb ge j

.

(e) Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy is

so 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 very

small 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 pp

a=FHGIKJ = × F

HGIKJ = ×3 027 10

1 4711 521

2 93 104 4...

. m s m se j


Chapter 11 321

(b) K mvp p= = × × = ×12

12

5 98 10 3 027 10 2 74 102 24 4 2 33. . .e je j J

UGmM

rpp

= − = −× × ×

×= − ×

−6 673 10 5 98 10 1 99 10

1 471 105 40 10

11 24 30

1133

. . .

..

e je je j J

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

Compare to find that K Up p+ = − ×2 66 1033. J and K Ua a+ = − ×2 65 1033. J . They agree.

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

conserved we have,

012

121 1

22 2

2 1 2= + −m v m vGm 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 21 2

2=+b g and v m

Gd 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

v141 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 3112

2 67 10= = ×. J

*P11.54 (a) Let R represent the radius of the asteroid. Then its volume is 43

3π R and its mass is ρ π43

3R .

For your orbital motion, F ma∑ = , Gm m

Rm v

R1 22

22

= , G R

RvR

ρ π43

3

2

2

=

Rv

G=FHGIKJ =

× ⋅

FHGG

IKJJ = ×−

34

3 8 5

6 67 10 1 100 41 53 10

2 1 2 2

11

1 2

4

ρ π π.

..

m s kg m

N m kg m

2 3

2

b gb g

(b) ρ π π43

1 10043

1 53 10 1 66 103 4 3 16R = × = × kg m m kg3e j e j. .

(c) vR

T=

2π T

Rv

= =×

= × =2 2 1 53 10

8 51 13 10 3 15

44π π .

.. .

m

m s s h

e j



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

L L

m vR I

m vR m RT

m vm R

T

Tm R

m v

i f∑ ∑=

= −

= −

=

= =× ×

= × =

0

025

2

45

45

4 1 66 10 1 53 10

5 90 8 58 37 10 26 5

2

2 12

21

1

2

16 417

ωπ

π

π π

asteroid

asteroid

asteroid

kg m

kg m s s billion years

. .

.. .

e je jb gb g

This problem is realistic. Many asteroids, such as Ida and Eros, are roughly 30 km in

diameter. 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 a visitor the same experience of easy orbital motion. So would many Kuiper-belt objects.

P11.55 (a) Tr

v= =

× ×

×= × = ×2 2 30 000 9 46 10

2 50 107 10 2 10

15

515 8π π .

.

m

m s s yr

e 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 241π π .

. ..

m

N m kg s kg

2 2

e je 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

rpE

p= − = × −

× ×

×

−12

12

1 60 8 23 106 67 10 5 98 10 1 60

7 02 102 3 2

11 24

6. .. . .

.a fe j e je ja f

or E = − ×3 67 107. J

(b) L mvr mv rp p= = ° = × ×

= × ⋅

sin sin . . . .

.

θ 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 Earth

are conserved,

at apogee we must have 12

2mvGMm

rEa

a− =

and mv r La a sin .90 0° = .


Chapter 11 323

Thus, 12

1 606 67 10 5 98 10 1 60

3 67 10211 24

7.. . .

.a f e je ja fv

raa

−× ×

= − ×−

J

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

Solving simultaneously, 12

1 606 67 10 5 98 10 1 60 1 60

9 24 103 67 102

11 24

107.

. . . .

..a f e je ja fa f

vv

aa

−× ×

×= − ×

−

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

mvaa

= =× ⋅

×= ×

9 24 10

1 60 5 58 101 04 10

10

37.

. ..

kg m s

kg m s m

2

b ge j.

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

a = × + × = ×12

7 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 6 3

11 24

π π .

. .

m

N m kg kg2 2

e je je j

T = =8 060 134 s 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 as it moves between the distant point and the point where it grazes the Earth, moving perpendicular to the radius:

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

r r r rr v r v× = ×

m R v mR v

v vE i E f

f i

3

3

b g ==

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

K U K Ug i g f+ = +e j e j :

12

012

2 2mv mvGM m

Ri fE

E+ = −

12

12

92 2v vGM

Ri iE

E= −e j

GM

RvE

Ei= 4 2 : v

GMRi

E

E=

4


P11.58 From Kepler’s third law, minimum period means minimum orbit size. The “treetop satellite” in Figure P11.27 has minimum period. The radius of the satellite’s circular orbit is essentially equal to the radius R of the planet.

F ma∑ = : GMm

Rmv

RmR

RT2

2 22= = FHGIKJ

π

G V

R R

RT

G RR

T

ρπ

ρ π π

=

FHGIKJ =

2 2 2

2

32 3

2

4

43

4

e j

The radius divides out: T G2 3ρ π= TG

= 3πρ

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

then

0 2 1=−+

Mr mrM mb g

and Mr mr2 1=

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

mrMGm

d1 12

2ω = and MrMGm

d2 22

2ω =

r r

FIG. P11.59

Adding these two equations gives r rM m G

d1 22

2+ =+b g a f

ω

with ω ω ω1 2= = . Now using d r r= +1 2

and T = 2πω

we find Td

G M m2

2 34=+

πa f .

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

m gGm m

r2 21 22= . The equal magnitude force exerted on the Earth by m2 produces negligible

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

gGmr2

12

11 24

7 2

6 67 10 5 98 10

1 20 102 77= =

× ⋅ ×

×=

−. .

..

N m kg kg

m m s

2 22e 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


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

m m s

2 22e je j

e j

The distance between the masses closes with relative acceleration of

g g grel2 2 2 m 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

radius r2

around 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 to k er

mvr

e2

2

2

2= .

We can eliminate vnmr

= h to find

k er

mnm r

e2 2 2

2 22= h

.

So the separation distances are rn

mk ea n n

e

= = = × −22 1 06 10

2 2

2 02 10 2h

. me j .

The orbital radii are r

a n2 0

2= , the same as for the electron in hydrogen.

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

re= + = + −1

212

2 22

.

Since mvk e

re2

2

2= , E

k er

k er

k er

k ea n n

e 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 from

the 500 kg mass

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

P11.6 see the solution, either 1 000 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


P11.14 1.27 P11.16 35.2 AU P11.18 planet Y has turned through

1.30 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) 23

πR hGME

E

+b g; (b)

GMR h

E

E +;

(c) GM mR h

R R hR m

EE

E E

E++

LNMM

OQPP

−2

22

86 400

2 2

2b g b gπ

s,

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

P11.32 GM m

RE

E12

P11.34 (a) 1.89 eV; (b) 656 nm; (c) 4 57 1014. × Hz P11.36 (a) 2 19 106. × 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 P11.44 see the solution P11.46 (a) see the solution;

(b) 212

131 2

Gmr R

−FHGIKJ

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 1033. J , U = − ×5 22 1033. J ,

K Ua a+ = − ×2 65 1033. 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) 5 580 m s , 1 04 107. × m ;

(d) 8 69 106. × m ; (e) 134 min P11.58 see the solution 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 from

now. Your spacecraft can be here then by leaving the Earth’s orbit and slowing down relative to the Sun, to fall into an elliptical orbit around the Sun. This point in the Earth’s orbit is the aphelion of the spacecraft’s new orbit. For its period to be 1/2 year, we want

TGM

a a

a

a

S

22

3 19 3

3

12

7 2

1932

10

42 97 10

3 156 10

2 97 108 38 10

9 43 10

=FHGIKJ = ×

=×

×= ×

= ×

−

−

π.

.

..

.

s m

s

s m m

m

2 3

2 33

e j

e je j

FIG. QCC2.1


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 we will want some insulation, air conditioning and a highly reflective surface for the spacecraft. From its current speed in solar orbit,

v =×

×= ×

2 1 496 10

3 156 102 98 10

11

74

π .

..

m

s m s

e j,

the spacecraft needs to drop to a speed given by

E K U

GMma

mvGMm

r

v GMr a

v

g= +

− = −

= −FHGIKJ

= × ⋅ ××

−×

FHG

IKJ

= ×

−

212

21 1

2

2 6 67 10 1 991 101

1 496 101

1 89 10

1 92 10

2

2

11 3011 11

4

. .. .

.

N m kg kg m m

m 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 picnic

basket a very slow push (say at 0.5 m/s) tangent to the orbit. Both space shuttles and the basket are in free fall around the Earth. Since 0.5 m/s is negligible in comparison to orbital speed, the basket will maintain nearly the same orbit as the spacecrafts, to bump into yours or pass close to it. More precisely, 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 launching the basket with any convenient tangential speed toward you and with a small appropriately chosen radially outward speed component.

CONTEXT 2 CONCLUSION SOLUTIONS TO PROBLEMS CC2.1 The Hohmann transfer path is one half of an ellipse

with 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= π

gives for the

transfer time

12

12

4 1 29 10

6 67 10 1 991 10

1 26 10 146

2 11 3

11 30

7

T =×

× ⋅ ×

= × =

−

π .

. .

.

m

N m kg kg

s d

2 2

e je je j

FIG. CC2.1



During this time Venus will move around in its orbit by θ ω= = ° ××

FHG

IKJ = °t 360

1 26 101 94 10

2347

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 6 km m m m. . . . From this, as in Example 11.1, we can find the orbital speed of the space station, which is the initial

speed 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

m m s

2 2e je j.

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

rE= −

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:

TT

TT

b

s

b

s= →

FHGIKJ =2 00 4 00

2

. . .

According to Kepler’s third law, the ratio of the squares of the periods of the golf ball and the space

station should be the equal to the ratio of the cubes of the semimajor axes:

TT

aa

a a a rb

s

b

sb s s

FHGIKJ =FHGIKJ = → = = =

2 31 34 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 its

circular 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 energy must be equal to the change in kinetic energy of the golf ball:

∆ ∆E K

GM ma

GM mr

GM mr

GM mr

GM mr

GM mr

E

b

E E E

E E

= = −−

− −FHGIKJ = − +

= − +FHG

IKJ =

2 2 2 1 587 2

21

1 5871 0 185

.

..

a f

Setting this equal to ∆K mv mvf i= −12

12

2 2 , we find,


Chapter 11 329

12

12

0 185

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 vf 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 Earth’s distance is given by

Σ = =

= =× ⋅ ×

×

= ×

−

F maGM m

rmv

r

vGM

r

E E

E

:

. .

.

.

Sun

Sun2 2 N 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 mr

v GMr r r

v

v

gE

E E M

= + − = −

= −+

FHG

IKJ

= × ⋅ ××

−× + ×

FHGG

IKJJ

= ×

−

Sun Sun

Sun

2 2N m kg kgm m

m s

212

21 1

2 6 67 10 1 991 101

1 496 101

1 496 10 2 28 10

3 27 10

32

3

311 30

11 11 11

34

. .. . .

.

e je j e j

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

(b) The craft’s aphelion speed is vv rr

E

M4

34

43 27 10

2 15 10= =× ×

×= ×

..

m s 1.496 10 m

2.28 10 m m 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 28 102 41 10= =

× ⋅ ×

×= ×

−Sun

2 2 N 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 .



(c) Consider the Earth as a frame of reference, rather than the Sun. The speed of a launch point

on the equator is vRE

0

62

1

2 6 37 10

86 400463= =

×=

π π

d

m

s m s

.e j. Escape speed from the Earth’s

gravity is given by

12

2 2 6 67 10 5 98 10

6 37 101 12 10

12

1

11 24

64

mvGM m

R

vGMR

E

E

E

E

=

= =× ⋅ ×

×= ×

−. .

..

N m kg kg

m m s

2 2e je j

So the launch speed change should be v v1 0310 7 10− = ×. m s .

(d) In the reference frame of Mars, imagine a spacecraft at rest with respect to the planet and far

from it, then falling in to the surface and making a soft landing at the equator, headed east. The speed it would gain as it falls is given by

0 012

2 2 6 67 10 6 42 10

3 37 105 04 10

62

6

11 23

63

+ = −

= =× ⋅ ×

×= ×

−

mvGM m

R

vGMR

M

M

M

M

. .

..

N m kg kg

m m s

2 2e je j

It should land with speed v7

62 3 37 10

24 6239=

×=

π .

.

m

h 3 600 s 1 h m s

e jb g . So it must change speed by

v v6 734 80 10− = ×. m s . Comparison of our answers shows that the most fuel is used in

taking off from Earth.

ANSWERS TO EVEN CONTEXT 2 CONCLUSION PROBLEMS CC2.2 1 30 103. × m s

gravity, planetary orbits, and the hydrogen...

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