16.X.1

E = mc2 11 0.9882 (9 10

31kg)

(3 108 m

s

)2 5.2 1013 J

Er

= mc2 8.1 1014 JK = E E

r 4.4 1013 J

6.X.2

K 12m |~v|2

12

(1.4 105 kg)(

9 105 m3600 s2

) 4.4 109 J

6.X.3

|~v|

2Km

21.1 105 J1.5 103 kg

12 m/s

6.X.4

mc2 = K

m Kc2 1.1 10

5J(

3 108 ms)2

1.2 1012 kg

6.X.5

~r = ~rf~r

i 18.2,8,6 m

W = ~F ~rW 0.03,0.04,0.09 N 18.2,8,6 mW 1.41 J

26.X.6

W = ~F ~rW

0,

(9.8

Nkg

)(2 kg), 0

N 25,30, 0 m

W 585 J

6.X.7

(a) K decreases and W is negative

(b) K increases and W is positive

(c) K decreases and W is negative

(d) K increases and W is positive

6.X.8

W = ~F ~rW 300, 0, 0 N 2, 0, 0 mW 600 J

6.X.9 A gravitational force acts on the jar.

W = ~F ~r 0,4.9, 0 N 0,1, 0 m 4.9 N

6.X.10

(a) No work is done because Earths displacement and Suns gravitational force on Earth are perpendicular.

(b) |~p| =~p

f ~p

i

= |2P | = 2P6.X.11 Doubling the final speed requires quadrupling the work done. Work, in this case, is proportional to speed squared.Doubling the speed therefore quadruples the work.

6.X.12

3E 40.8mc2K 39.8mc2

Here, K is large compared to mc2, in fact its about 40 times larger than mc2. At this speed, most of the energy is kineticenergy. About 40/41 of the total energy is kinetic energy. The remaining 1/41 of the total energy is rest energy.

At low speeds, K is small compared to mc2.

6.X.13

0.8 MeV939.6 MeV

8.51 104

6.X.14

W1 30, 0, 0 N 2.3, 0, 0 m 69 J

W2 15, 0, 0 N 8, 0, 0 m 120 J

Wnet

= W1

+W2 189 J

6.X.15 Since the force and displacement are in the same direction, the dot product reduces to a simple scalar product.

W (

5 1019 N)

(0.5 m)

2.5 1019 J

6.X.16

Eparticle

+ U = 0

50 J + U = 0 U = 50 J

6.X.17 There are six interaction energy pairs: U12, U

13, U

14, U

23, U

24, and U

34. Note that U

12and U

21are the same since

they describe the same pairwise interaction. Dont count such terms twice!

6.X.18

4Ug GMm

R

(

6.7 1011 Nm/kg2) (6 1024 kg)(1 kg)

(6.4 106 m) 6.3 107 J

6.X.19

(1rf

1ri

)=

11 107 m +

11 108 m

= 9 108 m1

Yes, it is negative.

6.X.20

(1rf

1ri

)=

11 108 m +

11 107 m

= 9 108 m1

6.X.21

By choosing your system appropriately, W = 0 and E = 0, so Ei

= Ef.

6.X.22

r is the distance from the spacecraft to the center of the asteroid.

6.X.23

K decreases because U increases. You can see that the vertical separation between E and U decreases in going from r1to

r2.

6.X.24

A is bound because E < 0. C is unbound because E > 0. B is trapped because there are turning points where E = U andK = 0.

6.X.25

C is circular motion since r=constant. B is an elliptical orbit. A is linear motion where an object is lauched vertically froma star, rises until K = 0, and falls back into the star along a straight line.

56.X.26

In the problem in Chapter 3, the initial speed of the spacecraft is the speed necessary to just barely get it to the Moon andhave v = 0 when it gets there. The escape speed is the speed necessary for the object to go to r = and have zero speedwhen it gets there. It takes slightly more initial speed to reach r = than to reach the Moon.

6.X.27

If an object falls from rest from a very large distance from Earth (r =), then its speed when striking Earth is the escapespeed, 1.12 104 m/s.

6.X.28

You calculate the same final speed because

Kf

+ Uf

= Ki

+ Ui

Kf

= Ki

+ Ui U

f

= Ki

+ U= K

i+mgy

The final kinetic energy depends on y = yf y

i. Thus, initial height and final height are not important; it is the change

in height y that determines the final speed.

6.X.29

vi

= 100 mph = 44.7 m/s. Define the system to be the ball and Earth. At the initial state, the ball leaves the hand withspeed v

iand height y

i= 0. At the final state, it is at its peak and v

f= 0. There is no work done on the ball.

7

0Ui

+Ki

= Uf

+0

Kf

+>0

W

12mv2

i= mgyf

yf

=v2i

2g= 102 m

6.X.30

Define the system to be the ball and Earth. Its initial height is yiand its initial speed is vi =

v2xi

+ v2yi. Its height at the

peak is yfand speed at the peak is

vxf

. Since Fnet,x

= 0, vxis constant is v

xf= v

xi. There is no work done on the ball.

6Ui

+Ki

= Uf

+Kf

+>0

W

Uf

= Ui

+KiK

f

mgyf

= mgyi

+12mv2

i 1

2mv2

f

mgyf

= mgyi

+12m(v2

i v2

f)

mgyf

= mgyi

+12m(v2xi

+ v2yiv2xf

)

mgyf

= mgyi

+12mv2

yi

yf

= yi

+v2yi

2g

6.X.31

Convert the kinetic energy of a proton to J. Ki,proton

= 1.6 1014 J. Define the system to be the two protons, so Ki

=2(1.6 1014 J) = 1.6 1014 J. Assume that they are initially very far apart, so U

i 0. At the final state, their speeds are

zero at the turning point, which is the point of closest approach. The work done on the system is zero and their rest energiesare constant. Thus, the total initial kinetic energy is completely converted to the final electric potential energy.

Ki

= Uf

Ki

=1

4pi0

q1q2

r

q1

= q2

= 1.6 1019 C, thus

r = 7.2 1015 m

Note that the radius of a hydrogen atom is about 1 1010 m, so at these energies, the protons distance of closest approachis within the atom, and close to the diameter of the nucleus.

6.X.32

The rest energy of O2 is about

Erest = 16(940 MeV)= 1.5 1010 eV

The binding energy of O2 is 5 eV. Thus, the ratio is

5 eV1.5 1010 eV = 3.3 10

10

71 mol of O2 molecules has a mass of about 16 g. The best analytical balances measure to about 1 106 g. This gives aratio of 110

6 g16 g = 6 108. Thus, even the most precise analytical balances can not measure the mass difference between

1 mol of O2 molecules and 2 mol of atomic oxygen (individual oxygen atoms).

6.X.33

The binding energy of a deuteron is 2.2 MeV. The total energy of a deuteron is

E = mproton

c2 +mneutron

c2 + Uatom

The electrons rest energy is negligible. Also, the potential energy U of the proton and the electron is small compared to therest energy. Thus,

Erest

mproton

c2 +mneutron

c2

= 938 MeV + 940 MeV= 1878 MeV

The ratio of binding energy to rest energy is

2.2 MeV1878 MeV

= 0.1%

6.X.34

Fy

= Uy

= y

(GMmy

)Fy

=GMmy2

Fx

= Ux

= x

(GMmy

)= 0

86.X.35

[p2

m

]=

[kg m/s]2

[kg]

= [kgm2

s2]

= [N m]= [J]

[mc2] = [kgm2

s2]

= [N m]= [J]

6.X.36|~v| = 0.99c and m = 9.11 1031 kg

E = mc2

=1

1 0.992 (9.11 1031 kg)(3 108 m/s)2

= 7.08(8.20 1014 J)= 5.80 1013 J= 3.63 MeV

Erest

= mc2

= 8.20 1014 J= 0.512 MeV

K = E Erest

= 3.12 MeV. Thus, KErest 6.

6.X.37|~v| = 100 mph = 44.7 m/s. K 12m |~v|2 so

K

Erest

=12m |~v|2mc2

=12 |~v|2c2

= 1.11 1014

9This is very small! A baseballs rest energy is much greater than its kinetic energy at typical speeds for a baseball.

6.X.38

For a baseball, |~v| 90 mph 40 m/s and m 0.15 kg, so

K =12m |~v|2

= 120 J

For a basketball, |~v| 10 m/s and m 0.6 kg, so K = 30 J

6.X.39

Because the energy of the electron is so large compared to its rest energy, the electrons speed will be very close to thespeed of light. To achieve at least 7 significant figures in your answer, you need to use melectron = 9.10938215 1031 kg,1.672621638 1027 kg and c = 2.99792458 108 m/s.

Erest,proton

= mc2

= (1.672621638 1027 kg)(2.99792458 108m/s)2= 1.50327736 1010 J

Eelectron

= 1.50327736 1010 J and m = 9.10938215 1031 kg

Eelectron

= mc2

=1.50327736 1010 J

(9.10938215 1031 kg)(2.99792458 108 m/s)2= 1836.1527

=1

1 |~v|2c2

Solve for |~v|. Because of the very high value of , the speed of the electron is very close to c. In fact, the precision of yourcalculator may be so small, that it calculates |~v| to be exactly equal to c, which is not correct. Its possible to use a computerprogram such as Excel which has greater precision than your calculator. The result is |~v| /c = 0.99999985, though the 5 isnot a significant figure.

Note that most of the electrons energy is kinetic energy.

K = ( 1)mc2= (1836.1527 1)(9.10938215 1031 kg)(2.99792458 108 m/s)2= 1.502459 1010 J

6.X.40

10

(a)

Erest

= mc2

= (0.145 kg)(3 108 m/s)2= 1.305 1016 J

(b) yes, because |~v|

11

(a)

K =12m |~v|2

= 1470 J

(b) It is the same. K is a scalar and thus does not depend on the direction of ~v

6.X.43

K =12m |~v|2

=12

(0.144 kg)(222 + 232 + 112 m2/s2)

= 81.6 J

6.X.44

mHe

=(

4 gmol

)(1 kg

1000 g

)(1 mol

6.02 1023 atoms)

= 6.64 1027 kgK =

12m |~v|2

|~v| =

2Km

= 658 m/s

6.X.45

K = KfK

i= 0 because

~vf

= 19 m/s and ~vi

= 19 m/s.6.X.46

(a)

~p = ~Fnet

t= (0.3, 0, 0 N)(1.5 s)= 0.45, 0, 0 kg m/s

12

(b)

~vi

= 0.9, 0, 0 m/s

~v =pm

=0.45, 0, 0 kg m/s

0.8 kg= 0.5625, 0, 0 m/s

~vf

= ~vi

+ ~v= 0.3375, 0, 0 m/s

K = KfK

i

=12mv2

f 1

2mv2

i

= 0.081 J

6.X.47

W = ~F ~r=

~F |~r| cos = (40 N)(3 m)(30)= 104 J

6.X.48

W = ~F ~r=

~F |~r| cos = (30 N)(50 m)(cos 35

)

= 1230 J

6.X.49

(a) Zero. The displacement through which the force acts is zero.

13

(b) Zero. The force vector is perpendicular to the displacement d~r, and cos 90 = 0.

(c) Wby spring

= 12ks2i 12ks2f which is 0 for one cycle since si = sf . For one-half cycle, Wby spring = 0 because, if the spring

is stretched a distance sithen one half-cycle later it will be compressed the same distance, s

f= s

iand s2

f= s2

i.

6.X.50

(a)

W = ~F ~r= F

xx+ F

yy + F

zz

= (22 N)(25 (17)) m= 176 J

(b) K is negative, so K decreased.

(c)

W = (22 N)(17 (25)) m= 176 J

(d) K is positive, so K increased.

6.X.51

(a)

W = ~F ~r= F

yy

= mg(yf y

i)

= (0.7 kg)(

9.8Nkg

)(0 4.5 m)

= 30.9 J

(b) K is positive, so K increased.

(c) W = 30.9 J since ~F is opposite ~r.(d) K is negative, so K decreased.

6.X.52

Erest

= mc2

= 4.59 1010 J

14

E = mc2 where = 1q1 |~v|2

c2

= 5.61.

E = (5.61)Erest

= 2.58 109 J

K = E Erest

= 2.12 109 J

If W = 4.7 109 J, then E = W = 4.7 109 J.

Ef

= Ei

+ E

= 2.58 109 J + 4.7 109 J= 7.28 109 J

Erest

stays the same, 4.59 1010 J.

K = E Erest

= 7.28 109 J 4.59 1010 J= 6.82 109 J

Since Erest

stays the same, you can also calculate K using Kf

= Ki

+W = 6.82 109 J

6.X.53

False, W is not a vector

True

False, W is not the magnitude of a vector.

False, K 12m |~v|2 for |~v|

15

W = ~F ~r= (< 250, 440,220 > N) (< 6,4,5 > m)= 840 J

W = E

= K + *0E

rest

Ki

= KfW

=12mv2

fW

= 3745 J 840 J= 2905 J

Ki

=12mv2

i

vi

= 6.96 m/s

6.P.55

W = ~F ~r= 250, 490,160 N 3,9,5 m= 2860 J

Kf

= Ki

+W

=12mv2

i+W

= 8640 J + (2860 J)= 5780 J

Ki

=12mv2

f

vf

= 9.81 m/s

16

6.X.56You do the same work on each block. According to the Energy Principle, each block will have the same K. Since theyboth start from rest, then K

fis the same for each block.

Kf

=p2f

2m therefore the larger mass block will have a larger final momentum.

If you pull the blocks for the same amount of time, then according to the Momentum Principle, ~p = ~Fnet

t will be the

same for the two blocks. Since they both start from rest, ~pfwill be the same. Since ~v

f=

~pfm , then

~pf

is greater for thesmaller block and the smaller block will have a greater final kinetic energy.

6.P.57(a), (b) as follows:

WJack

400, 0, 200 N 2, 0,1 m 1000 J

WJill

150, 0, 300 N 2, 0,1 m 0 J

(c) Jill does no work, so her force must be perpendicular to the boats displacement.(d)

Kf K

i+W~v

f

~vi

2 + 2Wm

(1.3 m/s)2 +2(1000 J)

3000 kg

1.0 m/s

6.P.58

K = ~F ~r = ~F |~r|~F = K|~r|

K =(f

i

)mc2

K 1

1 0.93c2c2

11 0.99c2

c2

(9 1031 kg)(3 108 ms

)2

K 3.5 1013 J~F 3.5 1013 J3 m

1 1013 N

6.P.59

17

(a)

W = ~F ~r 0,mg, 0 0,y, 0 (0.12 kg)(9.8 N

kg)(0.07 m) 0.082 J

(b)

K = Wspring

+WEarth

Wspring

= K WEarth

12m~v

f

2 12m~v

i

2 + 0.082 J 0.288 J

6.P.60 Steps 1, 2, 5, and 6 must be included, but not necessarily in that order.

W = ~F ~r

1.6 1013 , 0, 0

N 2, 0, 0 m

3.2 1013 JEf

= Ei

+W

11 0.91c2

c2

(9 1031 kg)(3 108 ms

)2 + 3.2 1013 J

5.15 1013 J

Solve for speed as a function of energy (final energy, that is).

E = mc2

=E

mc2=

11 |~v|

2

c2

|~v|c

=

1 mc

2

E

|~v|c

1 (9 10

31 kg)(3 108 ms )25.15 1013 J

|~v|c 0.99

Significant figures are important in this problem.

6.P.61 Significant figures are important in this problem.

18

(a)

W = ~F ~r (2 1012 N)(3.2 103 m) 6.4 109 J

Ef

=7

0Ei

+W 6.4 109 J|~v|c

=

1 mc

2

E

1 (9 1031 kg)(3 108 ms )26.4 109 J

0.99999999992

(b) Since the electron experiences a constant force, its acceleration is constant, and therefore its average speed is thearithmetic mean of its initial and final speeds, or 0.49999999996c. So the electron takes about 3.210

3m

0.49999999996c 0.00002 sto go the distance.

6.X.62

si

= 0 and sf

= 0.1 m. The average force exerted between si

= 0 and sf

= 0.1 is

~Favg = 12(ksf + ksi )=

12

(20

Nm

)(0.1 m)

= 1.0 N

W =~Favg |~r|

= (1.0 N)(0.1 m)W = 0.1 J

6.X.63

Assume that ~F is parallel ~r.

(a)

W = ~F ~r= F

xx

= (130 N)(6 m)= 780 J

(b)

W = Fxx

= (40 N)(5 m)= 200 J

19

(c)

Wtotal

= 780 J +200 J= 580 J

6.X.64

W = W1

+W2

= ~F1~r

1+ ~F

2~r

2

= 90, 150, 195 N 4, 6,4 m + 90,90,585 N 4, 6,4 m= ((90 N)(4 m) + (150 N)(6 m) + (195 N)(4 m)) + ((90 N)(4 m) + (90 N)(6 m) + (585 N)(4 m))= 480 J + 2160 J= 2640 J

W = E= E

f E

i

Ef

= Ei

+WK

f= K

i+W

=12mv2

i+W

= 4.67 104 J + 2.64 103 J= 4.93 104 J

Kf

=12mv2

f

vf

= 29.8 m/s

6.P.65

W1 250, 400,170 N 6, 7,4 m 4980 J

W2 140, 250, 150 N 4,7, 5 m 440 J

W = W1

+W2 4540 J

Kf

= Ki

+W~vf

= ~vi

2 + 2Wm

(3.5 m/s)2 +2(4540 J)

100 kg

10.2 m/s

20

6.P.66

Wme

400, 310,250 N 3.6, 4.2,1.2 m 162 JW = K

Wme

+Wfriend

= KW

friend= K

fK

iW

me

12

(700 kg)(4.01 m/s)2 12

(700 kg)(4 m/s)2 162 J 134 J

6.P.67

(a) W1 (130 N)(7 m) 910 J

(b) W2 (40 N)(7 m) 200 J

(c) W = W1

+W2 710 J

6.P.68

Assume that the force by you on the crate is parallel to the crates displacement.

W = ~F ~r= ~F

1~r

1+ ~F

2~r

2+ ~F

3~r

3

= F1x

x1

+ F2x

x2

+ F3x

x3

= (34 N)(2 m) + (13 N)(6 m) + (40 N)(2 m)= 66 J

6.P.69

W = (180 N)(6 m) + (170 N)(4 m)= 400 J

W = KK

f= W +K

i

= W +12mv2

i

= 400 J + 613 J= 1013 J

21

Kf

=12mv2

f

vf

= 4.5 m/s

6.X.70

Ei

= Ef

Eelectron

+ Epositron

= 2Ephoton

2Eelectron

= 2Ephoton

Eelectron

= Ephoton

Ephoton

= Eelectron

= mc2

= 8 1014 J= 0.5 MeV

6.P.71

Ei

= Ef

(a)

Ei,nuc

= E + Ef,nuc= Erest, + Erest,f,nuc +K

K = Ei,nuc Erest, + Erest,f,nuc

= mic2 mc2 +mf c2

= 3.145285 108 J 5.968326 1010 J 3.085536 108 J= 6.58 1013 J

(b)

1 eV = 1.6 1019 JK = 4.11 106 eV

= 4.11 MeV

6.P.72

22

(a)

Ei,rest

= mc2

= (3.894028 1025 kg)(2.99792 108 m/s)2= 3.499767 108 J

(b)

Erest,alpha + Erest, new nucleus = mc2 +m

nucleusc2

= (6.640678 1027 kg)(2.99792 108 m/s)2 + (3.827555 1025 kg)(2.99792 108 m/s)2= 5.968326 1010 J + 3.440024 108 J= 3.499708 108 J

(c) The rest energy decreased.

(d)

K = Erest,i E

rest,f

= 5.95 1012 J= 3.72 MeV

6.X.73

Define the system to be the ball and Earth.

Ei

= Ef

Ui

+Ki

= Uf

+Kf

:0mgy

i+

12mv2

i= mgv

f+

12*0mvf

vi

=

2gyf

= 19.8 m/s

6.X.74

Define the system to be the object and the Earth. Very far away, Uf

= 0 and Kf

= 0.

23

W = E= E

f E

i

= 0(GmM

ri

)

=GmM

ri

=

(6.7 1011 Nm2

kg2

)(1 kg)(6 1024 kg)

6.4 106 m= 6.28 107 J

6.X.75

Note that the gravitational force is in the negative radial direction (toward the center of Earth). This is important for gettingyour signs correct.

For r < R,

Fr

= dUdr

= grR

U = gr

Rdr

=g

R

r2

2+ C

where C is an integration constant. At r = R,

U(r = R) = gR

2+ C

=GM

R2R

2+ C

=12GM

R+ C

For r > R, U(r > R) = GMr . At r = R, U(r = R) = GMR . The expression for U(r < R) must match the expression forU(r > R) at r = R. Thus,

12GM

R+ C = GM

R

C = 32GM

R

= 32gR

24

And therefore, U(r R, the potential energy is proportional to 1/r.

-1e+08

-9e+07

-8e+07

-7e+07

-6e+07

-5e+07

-4e+07

-3e+07

0 2e+06 4e+06 6e+06 8e+06 1e+07 1.2e+07

U (J

)

r (m)

UEarth vs. distance from center of Earth

Figure 1: U(r) for Earth.

6.X.76

In nuclear energy production, there is a change in rest energy due to splitting or joining of nuclei. In the case of fusion, forexample, the change in rest energy results in gamma rays (i.e. high energy photons). In a chemical reaction used for chemicalenergy production, the number of nuclei (of various types) stays the same. This is called stoichiometry in chemistry. Nonuclei split or join, but rather molecules split and atoms can join together to make new molecules. There is a change in thebinding energy (electric potential energy) which can result in an increase of thermal energy (in the case of an exothermicreaction).

Nuclear energy production and chemical energy production are alike in the sense that something splits or joins to produceenergy. The difference is that, in the case of nuclear energy, nuclei split or join, and in the case of chemical energy, moleculessplit or are formed (i.e. bonds are broken or formed between atoms).

6.X.77

(a) Momentum increases as the comet gets closer to Sun; therefore,~p

E

> ~pD

> ~pC

> ~pB

> ~pA

25

(b) K increases as |~p| increases; therefore, KE> K

D> K

C> K

B> K

A.

(c) false

false

false

true

true

(d) false

true

false

true

true

(e) U increases as r increases; thus, UA> U

B> U

C> U

D> U

E.

6.X.78

If E = EA, then

false

true

true

false

false

true

If E = EB, then

true

true

false

false

6.X.79

Fx

= Ux

= x

(bx2

)=2bx3

26

Fy

= Uy

= y

(bx2

)= 0

6.X.80

vescape

=

2GMR

= 2320 m/s

The escape speed from Earth is 1.1 104 m/s, which is 4.8 times larger than the Moon. Thus, a thruster on the Moon mustdo approximately 1/(4.8)2 = 1/23 times less work than a thruster on Earth to cause an equal mass object to escape.

6.P.81

(a)

Ui

+Ki

=7

0Uf

+Kf

Ki

= Kf U

i

12mv2

i=

12mv2

f GMm

ri

vi

=

v2f

+2GM

ri

=

10002 +

2(6.7 1011 Nm2kg2

)(0.6 1024 kg)3.4 106 m

vi

= 4960 m/s

(b) Ufand K

f= 0, so

Ui

+Ki

= 0K

i= U

i

12mv2

i=

(GMm

ri

)

vi

=

2GM

ri

= 4860 m/s

27

6.P.82

Ui

+Ki

=70

Uf

+Kf

(a)

Ki

= Kf U

i

12mv2

i=

12mv2

f GMm

ri

vi

=

v2f

+2GM

ri

=

20002 +

2(6.7 1011 Nm2kg2

)(0.6 1024 kg)3.4 106 m

vi

= 5260 m/s

(b) Ufand K

f= 0, so

Ui

+Ki

= 0K

i= U

i

12mv2

i=

(GMm

ri

)

vi

=

2GM

ri

= 4860 m/s

6.P.83

Use escape speed to get M for the asteroid.

vesc

=

2GMR

M =12

v2RG

= 7.46 1015 kg

If vi

= 20 m/s, vfis found from E

i= E

f.

28

Ui

+Ki

=7

0Uf

+Kf

GMmri

+12mv2

i=

12mv2

f

vf

=

v2i 2GM

Rvf

= 17.3 m/s

6.P.84

vesc

=

2GMR

v2esc

=2GMR

Use the Energy Principle.

Ei

= Ef

Ui

+Ki

=7

0Uf

+Kf

GMmri

+12mv2

i=

12mv2

f

v2f

= v2i 2GM

R

v2f

= v2i v2

esc

vf

=

v2i v2

esc

=

(35 m/s)2 (24 m/s)2vf

= 25.5 m/s

6.P.85

The Earth-Moon distance is 3.84 105 km = 3.84 108 m. The Energy Principle for the Earth-Moon-spacecraft systemgives

Ei

= Ef

UE,i

+ UM,i

+Ki

= UE,f

+ UM,f

+Kf

Kf

= UE,i U

E,f+ U

M,i U

M,f+K

i

12

v2f

= GME

(1

rE,i

1rE,f

)+GM

M

(1

rM,i

1rM,f

)+

12

v2i

29

rE,i

= RE

= 6.4 106 mrE,f

= dRM

= 3.823 108 mrM,i

= dRE

= 3.776 108 mrM,f

= RM

= 1.75 106 m

Substitute and solve for vf

12

v2f

= 6.176 107 Jkg

+ 2.80 106 Jkg

+ 8.45 107 Jkg

12

v2f

= 2.55 107 Jkg

vf

= 7.15 103 m/s

The sample program below simulates the motion of the spacecraft launched along a straight line between Earth and Moon.The speed of impact at Moon as determined by the simulation is 7.12 103 m/s which is similar to the speed calculatedfrom The Energy Principle of 7.15 103 m/s.

1 from __future__ import d i v i s i o n2 from v i s u a l import 3

4 RE = 6.4 e6 #rad ius o f Earth5 RM = 1.75 e6 #rad ius o f Moon6

7 s p a c e c r a f t = sphere ( pos=(RE, 0 , 0) , c o l o r=co l o r . cyan , rad iu s =0.25RE)8 Earth = sphere ( c o l o r=co l o r . blue , r ad iu s=RE)9 Moon = sphere ( pos=(3.84 e8 , 0 , 0 ) , c o l o r=co l o r . white , r ad iu s =0.5RE)10

11 m=1.5 e4 #mass o f s p a c e c r a f t12 ME = 6e24 #mass o f Earth13 MM = 7.35 e22 #mass o f Moon14 G = 6.67 e1115

16 v=vecto r ( 1 . 3 e4 , 0 , 0 ) #i n i t i a l v e l o c i t y o f s p a c e c r a f t17 p=mv #i n i t i a l momentum of s p a c e c r a f t18

19 t=020 #dt =0.013600 #time s t ep21 dt=1022

23 rmag=mag( spa c e c r a f t . pos ) ; #di s t ance o f s p a c e c r a f t from Earth24 rrelmoonmag=mag( spa c e c r a f t . posMoon . pos ) #di s t ance o f s p a c e c r a f t from Moon25

26 t r a i l=curve ( c o l o r=spa c e c r a f t . c o l o r )27

28 while rmag>0.99999999RE and rrelmoonmag>RM: #stop i f rmag < RE or rrelmoonmag < RM29 r a t e (100)30 #ca l c u l a t e Fgrav on s pa c e c r a f t by Earth31 r=spa c e c r a f t . pos32 rmag=mag( r )

30

33 run i t=r /rmag34 FgravE=GmME/rmag2 run i t35

36 #ca l c u l a t e Fgrav on s pa c e c r a f t by Moon37 rrelmoon=spa c e c r a f t . pos Moon . pos38 rrelmoonmag=mag( rrelmoon )39 rre lmoonunit=rrelmoon/rrelmoonmag40 FgravM=GmMM/rrelmoonmag 2 r re lmoonunit41

42 #ca l c u l a t e net f o r c e43 Fnet=FgravE + FgravM44

45 #update momentum and po s i t i o n46 p = p + Fnetdt47 v = p/m48 s p a c e c r a f t . pos = spa c e c r a f t . pos + vdt49

50 t r a i l . append ( pos=spa c e c r a f t . pos )51

52 t = t + dt53

54 print mag(v )

6.P.86

The system is Mars and the spacecraft.

Ei

= Ef

Ui

+Ki

= Uf

+Kf

Kf

= Ui U

f+K

i

= GMm(

1ri

1rf

)+K

i

12mv2

f= GMm

(1ri

1rf

)+

12mv2

i

12

v2f

= GM(

1ri

1rf

)+

12

v2i

= GM(

17 106

14 106

)+

12

v2i

12

v2f

= 4.59 106 Jkg

+ 4.5 106 Jkg

vf

= 4.26 103 m/s

6.P.87

31

vcircular/orbit

=

GM

R= 7925 m/s 7900 m/s

Define the system to be the satellite and Earth.

W = E= E

f E

i

= Uf

+Kf U

iK

i

=>

0GMm

rf

+12>

0mv2

f GMm

ri

12mv2

i

=GMm

R 1

2mv2

i

=GMm

R 1

2mGM

R

=12GMm

R

= 6.3 109 J

6.P.88

Ei

= Ef

Ui

+Ki

= Uf

+Kf

Kf

= Ui U

f+K

i

12mv2

f=GMm

ri

GMmrf

+12mv2

i

12

v2f

= GMSun

(1ri

1rf

)+

12

v2i

= 3.33 109 Jkg

+ 3.34 109 Jkg

vf

= 4420 m/s

6.P.89

W =~F ~r cos = (1.6 1013 N)(2 m) = 3.2 1013 J

32

W = E= E

f E

i

Ef

= W + Ei

fmc2 = W +

imc2

= 3.2 1013 J + 11 (0.95)2 (9.11 10

31 kg)(3 108 m/s)2

= 3.2 1013 J + 2.63 1013 J= 5.83 1012 J

fmc2 = 5.83 1013 J

f

=5.83 1013 J

(9.11 1031 kg)(3 108 m/s)2f

= 7.11

f

=1

1 v2c2

Solve for v which gives vf

= 0.99c.

6.P.90

(a)

Ui

+Ki

= Uf

+Kf

Ki

=7

0Uf U

i+K

f

12mv2

i= GMm

R+

12mv2

f

12

v2i

=GM

R+

12

v2f

= 134Jkg

+ 4.5Jkg

vi

= 16.6 m/s

6.P.91

ri and v

i 0. r

f,Jrelative to Jupiter is Jupiters radius. r

f,Srelative to the Sun is the orbital radius of Jupiter plus

Jupiters radius (which is negligibly small in comparison to the distance from the Sun.) rf,S 8 1011 m. Define the system

to be the rock, Sun, and Jupiter.

33

Ei

= Ef

0

Ui,S,rock

+0

Ui,J,rock

+0

Ki

= Uf,S,rock

+ Uf,J,rock

+Kf

0 =GM

Sm

rf,s

+GM

Jm

rf,J

+12mv2

f

12

v2f

=GM

S

rf,S

+GM

J

RJ

= 1.68 108 Jkg

+ 9.57 108 Jkg

vf

= 4.7 104 m/s

6.X.92

The system is the ball and Earth.

Ui

+Ki

= Uf

+Kf

mgyi

+12>

0mv2

i= mg

0yf

+12mv2

f

mgyi =12mv2

f

vf

=

2gyi

= 6.3 m/s

6.X.93

The system is the bear and Earth.

(a)

Ui

+0

Ki

= Uf

+Kf

Kf

= Ui U

f

= mg(yi y

f)

= (0.5 kg)(

9.8Nkg

)(2 m)

= 9.8 J

Kf

=12mv2

f

vf

= 6.3 m/s

34

(b) The flower pot has twice the mass of the teddy bear, so it will have twice the kinetic energy when hitting the ground,K

f= 19.6 J. Since mass cancels out, its speed when hitting the ground will be the same as the teddy bear, 6.3 m/s.

6.X.94

(a)

~Fnet

=~pt

mg =m(v

fy v

iy)

t

mg =mv

iy

tviy

= gt

=(

9.8Nkg

)(1.55 s)

= 15 m/s

(b) The system is the ball and the Earth

7

0Ui

+Ki

= Uf

+0

Kf

12mv2

i= mgyf

yf

=12

v2i

g= 11.8 m

6.P.95

W =

r=Rr=0

mg

Rr dr

=mgR

2

To move the object from r = R to r = , W = Ugrav

= Uf U

iwhere U

f= 0. W = U

i= GMmR =

GMRR2 m = mgR. It

takes twice as much energy to move the object from R to as it takes to move the object from 0 to R.

6.X.96

For two protons,

35

U =(

9 109 Nm2

c2

)q1q2

r

Where q1

= q2

= 1.6 1019 C

U = 1.44 1011 J

If the two protons move closer together, U increases.

For an electron and proton, q1

= 1.6 1019 C and q2

= 1.6 1019 C, so U = 1.44 1011 J

As the electron and proton come closer together, U decreases because it becomes more negative.

false

false

true

6.X.97

Graph (a). Two electrons are an unbound system, so E > 0. Also, U > 0 and decreases to U = 0 at r =.

Graph (b). An electron and proton that are moving when very far apart are an unbound system, so E > 0. For oppositelycharged particles, U < 0. Also, K increases as r 0.

6.P.98

A sketch of the protons is shown in Figure 2. It is useful for calculating the distances between each pair of protons.

r1,4

=d2 + d2 =

2d

36

d

d d

d

1

2

3

4

Figure 2: Four protons at the corners of a square of length d.

Ei

= Ef

Ui

+0

Ki

=7

0Uf

+Kf

Ui

= Kf

Kf

= U1,2

+ U1,3

+ U1,4

=1

4pi0

q2

d+

14pi

0

q2

d+

14pi

0

q22d

4Kproton

=1

4pi0

q2(

2d

+12d

)4(

12mv2

f

)=

14pi

0

q2

d

(4 +

22

)

v =

(1

4pi0

q2

md(1 +

18

)

) 12

6.P.99

(a) When very far apart and at rest, their potential energy is zero and their total kinetic energy is zero. Thus, their totalenergy is zero. Their potential energy at any distance r apart is

U =1

4pio

(+e)(e)r

= 14pi

o

(1.6 1019 C)2r

37

Since the total energy of the system is zero, then their total kinetic energy is K = U . As a result,

K = +1

4pio

(1.6 1019 C)2r

These quantities are shown on the graph in Figure 3.

-2.5e-13

-2e-13

-1.5e-13

-1e-13

-5e-14

0

5e-14

1e-13

1.5e-13

2e-13

2.5e-13

5e-15 1e-14 1.5e-14 2e-14 2.5e-14 3e-14 3.5e-14 4e-14 4.5e-14 5e-14

E (J

)

r (m)

E vs. r for a particle of charge +e and its antiparticle of charge -e

U (r)K (r)E (r)

Figure 3: Graphs of E, U , and K for a particle and antiparticle.

(b) U = 0 when very far apart.

38

Ei

= Ef

2Erest,M

= 2Erest,m

2Kf

2Mc2 = 2mc2 + 2Kf

Kf

= Mc2 mc2= (M m)c2

Kf

= (f)mc2

f

=K

f

mc2+ 1

=(M m)c2

mc2+ 1

=M mm

+ 1

=M

m 1 + 1

f

=M

m1

1 v2c2=

M

m

1 v2

c2=

m2

M2

v2

c2= 1 m

2

M2

v =

1 m

2

M2c

If m > 1 and v

fis close to c. The particles are definitely relativistic.

(c) mproton

= 1.67 1027 kg and mpion

= 2.5 1028 kg

v =

1(mprotonm

pion

)2c

=

1

(2.5 1028 kg1.67 1027 kg

)2c

v = 0.99c

(d)

Kf

= ( 1)mc2

=1

1 (0.99)2= 7.09

Kf

= 1.37 1010

39

Calculate the distance between the pions when their potential energy is 10% of the final kinetic energy.

U = 0.1Kf

= 1.37 1011 JU =

14pi

0

q1q2

r

r =(9 109 Nm2c2 )(1.6 1019 c

2)1.37 1011 J

= 1.7 1017 m

The radius of a proton is on the order of 1015 m. Thus, the pions are separated a distance of about 1100 of the radiusof a proton. They do not have to be very far apart for U to be negligible.

6.P.100

(a) The initial state is the original plutonium nucleus, which is (c) in the figure.

The final state is when the daughter nuclei are far apart and Uf 0. This is (b) in the figure.

Initially, the system only has rest energy. In the final state, it has rest energy and kinetic energy, since Uf 0.

The Energy Principle gives

Ef

= Ei

Kf

+ 2Erest,Ag

= EPu

Kf

= EPu 2E

rest,Ag

= mPuc2 2m

Agc2

= (242.007)(1.66054 1027 kg)(2.99792 108 m/s)2 2(120.894)(1.66054 1027 kg)(2.99792 108 m/s)2= 3.61175 108 J 3.60848 108 J

Kf

= 3.27 1011 J

If v

40

v is about 4% of the speed of light, so it is safe to assume that the nucleus is non-relativistic.

(b) 1 Pu-242 nucleus gives 3.27 1011 J of kinetic energy. 1 mol of Pu-242 nuclei is 6.02 1023 nuclei, which results in(6.02 1023)(3.27 1011 J) = 1.96 1013 J of kinetic energy, This is 10 million times more energy than the 106 Jderived from 1 mol of gasoline.

(c) In going from state (a) to state (b) in the figure, there is no change in rest energy.

From this initial state to the final state where the nuclei are far apart, the potential energy changes and kinetic energychanges.

Ei

= Ef

Ui

+0

Ki

=7

0Uf

+Kf

Ui

= Ki

14pi

0

q1q2

r= 3.27 1011 J

Where q1

= q2

= 47(1.60218 1019 C) = 7.53025 1018 C.Solving for r gives

r =(8.98755 109 Nm2C2 )(7.53025 1018 C)2

3.27 1011 J= 1.56 1014 m

The radius of 121 nucleons is about

R = (1.3 1015 m)(121) 13= 6.43 1015 m

A sketch is shown in Figure 4. Units are femtometers, where 1 fm = 1015 m. The diameter of a nucleus is 2(6.43 fm) =12.9 fm. This leaves a gap of 15.6 fm 12.9 fm = 2.7 fm. The gap is 2.712.9 = 21% of the center-to-center distance.

6.4 fm 6.4 fm

2.7 fm

15.6 fm

Figure 4: The Ag-121 nuclei immediately after fission.

41

6.P.101

(a) The initial state is shown in state (c), the last state in the figure.

The final state in this case is when the nuclei interact, which is state (b), the middle state shown in the figure.

The potential energy has increased and the kinetic energy has decreased in this process.

Ef

= Ei

Uf

+Kf

=7

0Ui

+Ki

Kf

= Ki U

f

= Kproton,i

+Kdeuterium,i

+1

4pi0

qproton

qdeuterium

rf

= 4.14 1013 J + 2.07 1013 J (8.98755 109 Nm2

C2)(1.60218 1019 C)2

2(0.9 1015 m)= 4.14 1013 J + 2.07 1013 J 1.28171 1013 J= 4.928 1013 J

Now this is the initial kinetic energy of the system and the final state is the 3He nucleus and the gamma ray, which isstate (c) (the top state in the figure).

(b) During this process, there is a change in the rest energy, a change in potential energy, and a change in kinetic energy.

Ef

= Ei

Erest,Helium

+Kf

= Ki

+ Erest,proton

+ Erest,deuterium

Kf

= Ki

+ Erest,proton

+ Erest,deuterium

Erest,Helium

= 4.928 1013 J +mproton

c2 +mdeuterium

c2 mHelium

c2

= 4.928 1013 J + 1.50331 1010 J + 3.00513 1010 J 4.50038 1010 J= 4.928 1013 J + 8.05904 1013 J

Kf

= 1.30 1012 J

(c)

K = KfK

i

= 1.30 1012 J 4.928 1013 J= 8.06 1013 J

(d) If 1 proton and 1 deuterium nucleus produces 8.06 1013 J of kinetic energy, then 1 mol of protons and 1 mol ofdeuterium nuclei will produce (6.02 1023)(8.06 1013 J) = 4.8 1011 J of energy. This is about half a million timesthe energy obtained by burning 1 mol of gasoline.

42

6.P.102

(a) The momentum of the Ra-220 nucleus is zero; therefore, the total momentum of the Ra-216 nucleus and alpha particlemust also be zero. Since they exert equal magnitude forces on each other in opposite directions, then they will havethe same magnitude momentum, but in opposite directions as shown in Figure 5. Thus,

~pf,Ra-216

= |~p|.

p p

Figure 5: The Ra-216 nucleus and alpha particle after decay.

(b)

Ei

= Ef

Erest,Ra-220

= Erest,Ra-116

+ Erest, +KRa-216 +K

= Erest,Ra-116

+ Erest, +p2Ra216

2mRa-216

+p2

2m

Erest,Ra-220

= Erest,Ra-116

+ Erest, +p22

(1

mRa-216

+1m

)

Solve for p.

p22

(1

mRa-216

+1m

)= E

rest,Ra-220 E

rest,Ra-216 Erest,

= mRa-220

c2 mRa-216

c2 mc2= [(219.96274)(1.6603 1027 kg) (215.95308)(1.66 1027 kg) (4.00151)(1.6603 1027 kg)](2.99792 108 m/s)2= 3.28228 108 J 3.22245 108 J 5.97105 1010 J= 1.216 1012 J

43

p2 =2(1.216 1012 J)(

1mRa-216

1m)

p = 1.26 1019 kg m/sK =

p22m

= 1.194 1012 J

(c)

KRa-216

+K = Erest,Ra-220 Erest,Ra-216 E= 1.216 1012 J

KRa-216

= 1.216 1012 J 1.194 1012 J= 2.2 1014 J

Note that KRa-216

< K as expected. They have the same magnitude of momentum. Since K = p2

2m , the larger massnucleus has less kinetic energy.

(d)

K = 1.194 1012 J( 1)mc2 = 1.194 1012 J

= 1.002

Since 1, v

44

Note that mdeuteron

6= mproton

+mneutron

because of the small kinetic energy of the photon.

(b) For a photon, pc = E so pphoton

= 3.521013 J

2.99792108 m/s = 1.17 1021 kg m/s. Thus, pdeuteron = 1.17 1021 kg m/s dueto conservation of momentum.

Kdeuteron

=p2

2m

=(1.17 1021 kg m/s)2

2(3.34 1027 kg)= 2.0 1016 J

Note that Kdeuteron

45

(c)

R = (1.3 1015 m)(118) 13= 6.38 1015 m

The distance between the centers of the nuclei is 15.7 1015 m. Thus the gap between the surfaces is 15.72(6.38) =2.95 1015 m. A sketch is shown in Figure 6. Note that 1015 m = 1 fm.

15.7 fm

2.94 fm

6.4 fm 6.4 fm

Figure 6: Two palladium nuclei after fission.

6.P.105

(a) Consider a proton and deuteron initially very far apart. At the minimum energy needed to make contact, their speedwill be zero at contact. Apply the Energy Principle.

Ei

= Ef

7

0Ui

+Ki

= Uf

+0

Kf

Ki

= Uf

=1

4pi0

q1q2

rf

Substitute rf

= 2(1 1015 m) = 1 1015 m, q1

= q2

= 1.6 1019 C.

Ki

= 1.15 1013 J

46

(b) Upon production of the Helium-3 nucleus and gamma ray photon,

Ei

= Ef

Erest,proton

+ Erest,deuteron

+Ki

= Erest,He

+Kf

Kf

= Erest,proton

+ Erest,deuteron

+Ki E

rest,He

= mproton

c2 +mdeuteron

c2 +Kim

Hec2

= 1.50331 1010 J + 3.00513 1010 J + 1.15 1013 J 4.50038 1010 J= 8.06 1013 J + 1.15 1013 J= 9.21 1013 J

(c)

KfK

i= 9.21 1013 J 1.15 1013 J= 8.06 1013 J

This is the change in rest energy of the nuclei in this reaction.

(d) Graph 3. It correctly represents the nuclear interaction at short distances before fusion (the system is bound, thusU < 0), the coulomb interaction after fusion (repulsive force and unbound system, thus U>0), the coulomb repulsionat greater distances (U > 0, but decaying), and no interaction at great, great distances (U 0 as r ).

6.P.106

Let 1 be the original nucleus and 2 be the new nucleus after alpha decay. Use c = 2.992792 108 m/s.

Ei

= Ef

Erest,1

= Erest, + Erest,2 +KfK

f= E

rest,1 Erest, Erest,2

= m1c2 mc2 m2 c2

= 3.52065 108 J 5.96833 1010 J 3.46089 1008 J= 8.38 1013 J

6.P.107

(a) Define y = 0 to be at the lowest point of the pendulums motion ( = 0). Then the height of the pendulum isy = L L cos = L(1 cos ). Because it is near Earth, we can use

U = mgyU = mgL(1 cos )

47

0

2

4

6

8

10

12

14

16

18

20

-200 -150 -100 -50 0 50 100 150 200

U (J

)

e ()

U vs. e for a pendulum

Figure 7: U vs. for a pendulum.

(b) To sketch the graph, assume that m = 1 kg and L = 1 m. Then, U varies between the values of 0 and 2g. An examplegraph is shown in Figure 7.

(c) s = L, so = sL . Write U in terms of s.

U = mgL(1 cos sL

)

The tangential component of the net force on the mass is

Fnet,tan

= Us

= s

(mgL(1 cos sL

))

= mgL( sin sL

)(1L

)

= mg sin sL

= mg sin

The net force can also be found by sketching a free-body diagram for the pendulum and summing the forces to calculatethe net force on the pendulum. A free-body diagram is shown in Figure 8.

The gravitational force can be written in terms of its radial component and its tangential component. The tangentialcomponent of ~F

gravis

48

Frad

Fgrav , rad

Fgrav, tan

Fgrav

Figure 8: A free-body diagram for the pendulum.

Fgrav,tan

= Fgrav

sin = mg sin

The negative sign gives the direction, showing that the force is to the left and angent to the path for positive anglesand to the right and tangent to the path for negative angles.

(d) To barely make it to the top, vf

= 0 at the top. Apply the Energy Principle with the initial state at = 0 and thefinal state at = 180.

Ei

= Ef

Ki

= Uf

Ki

= mgL(1 cos 180)12mv2

i= 2mgL

vi

=

4gL

Use m = 1 kg and L = 1 m. The initial energy needed to barely make it to the top is Ei

= 2mgL = 2g, for m = 1 andL = 1. If E > 2g, then it easily goes all the way around. If E < 2g, then it never makes it to the top. An examplegraph is shown in Figure 9.

6.P.108

49

0

5

10

15

20

25

30

-200 -150 -100 -50 0 50 100 150 200

U (J

)

e ()

U vs. e for a pendulum

Uv > critical speedv = critical speedv < critical speed

Figure 9: Various energy levels for a pendulum.

E2particle

(pc)2 = (mc2)2 (m |~v| c)2

= 2(mc2)2 2m2 |~v|2 c2

=1(

1 |~v|2c2) (mc2)2 1(

1 |~v|2c2)m2 |~v|2 c2

=(mc2)2 m2 |~v|2 c2(

1 |~v|2c2)

=m2c4 m2 |~v|2 c2(

1 |~v|2c2)

=m2c4(1 m2|~v|2c2m2c4 )(

1 |~v|2c2)

=m2c4(1 |~v|2c2 )(

1 |~v|2c2)

= m2c4

= (mc2)2

6.P.109

A sample program is shown below.

50

1 from __future__ import d i v i s i o n2 from v i s u a l import 3

4 RE = 6.4 e6 #rad ius o f Earth5 RM = 1.75 e6 #rad ius o f Moon6 h = 5e4 #i n i t i a l a l t i t u d e o f Ranger7

8 Earth = sphere ( pos =(0 ,0 ,0) , c o l o r=co l o r . blue , r ad iu s=5RE)9 Moon = sphere ( pos=(4e8 , 0 , 0 ) , c o l o r=co l o r . white , r ad iu s =0.5Earth . rad iu s )10 ranger = sphere ( pos=(RE+h , 0 , 0) , c o l o r=co l o r . cyan , rad iu s =0.25Earth . rad iu s )11

12 m=173 #mass o f ranger13 ME = 6e24 #mass o f Earth14 MM = 7e22 #mass o f Moon15 G = 6.67 e1116

17 v=vecto r ( 1 . 32 e4 , 0 , 0) #i n i t i a l v e l o c i t y o f ranger18 p=mv #i n i t i a l momentum of ranger19

20 t=021 dt=522

23 #i n i t i a l i z e work24 work = 025

26 rmag=mag( ranger . pos ) ; #di s t ance o f ranger from Earth27 rrelmoonmag=mag( ranger . posMoon . pos ) #di s t ance o f ranger from Moon28

29 t r a i l=curve ( c o l o r=ranger . c o l o r )30

31 while rmag>RE and rrelmoonmag>RM: #stop i f rmag < RE or rrelmoonmag < RM32 r a t e (1000)33

34 #ca l c u l a t e Fgrav on ranger by Earth35 r=ranger . pos36 rmag=mag( r )37 run i t=r /rmag38 FgravE=GmME/rmag2 run i t39

40 #ca l c u l a t e Fgrav on ranger by Moon41 rrelmoon=ranger . pos Moon . pos42 rrelmoonmag=mag( rrelmoon )43 rre lmoonunit=rrelmoon/rrelmoonmag44 FgravM=GmMM/rrelmoonmag 2 r re lmoonunit45

46 #ca l c u l a t e net f o r c e47 Fnet=FgravE + FgravM48

49 #update momentum and po s i t i o n50 p = p + Fnetdt51 v = p/m52 ranger . pos = ranger . pos + vdt53

54 #ca l c u l a t e work done by g r a v i t a t i o n a l f o r c e s

51

55 work = work + dot ( Fnet , vdt )56

57 t r a i l . append ( pos=ranger . pos )58

59 t = t + dt60

61 print " crashed at t=" , t62 print "= ( in hours ) " , t , ( t /3600)63 print "= ( in days ) " , t , ( t /3600/24)64 print " speed=" , mag(v )65 print "work=" , work66 print "change in K=" , 0 .5m(mag(v ) 2 ( 1 . 32 e4 ) 2)

6.P.110

A sample program is shown below. Note that the energy is calculated before the position and momentum update so thatthe speed in the kinetic energy calculation corresponds to the same clock reading when the distances (such as rmag) arecalculated.

1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from v i s u a l . graph import 4

5 RE = 6.4 e6 #rad ius o f Earth6 RM = 1.75 e6 #rad ius o f Moon7 h = 5e4 #i n i t i a l a l t i t u d e o f Ranger8

9 Earth = sphere ( pos =(0 ,0 ,0) , c o l o r=co l o r . blue , r ad iu s=5RE)10 Moon = sphere ( pos=(4e8 , 0 , 0 ) , c o l o r=co l o r . white , r ad iu s =0.5Earth . rad iu s )11 ranger = sphere ( pos=(RE+h , 0 , 0) , c o l o r=co l o r . cyan , rad iu s =0.25Earth . rad iu s )12

13 m=173 #mass o f ranger14 ME = 6e24 #mass o f Earth15 MM = 7e22 #mass o f Moon16 G = 6.67 e1117

18 v=vecto r ( 1 . 32 e4 , 0 , 0) #i n i t i a l v e l o c i t y o f ranger19 p=mv #i n i t i a l momentum of ranger20

21 t=022 dt=1023

24 rmag=mag( ranger . pos ) ; #di s t ance o f ranger from Earth25 rrelmoonmag=mag( ranger . posMoon . pos ) #di s t ance o f ranger from Moon26

27 UGraph=gd i sp l ay ( t i t l e="Evs . r " , x t i t l e= ' r (m) ' , y t i t l e= 'E (J ) ' , x=450 , y=0, width=400 ,he ight=300)

28 UPlot=gcurve ( c o l o r=co l o r . cyan )29 EPlot=gcurve ( c o l o r=co l o r . ye l low )30 KPlot=gcurve ( c o l o r=co l o r . white )31

32

33 t r a i l=curve ( c o l o r=ranger . c o l o r )34

52

35 while rmag>RE and rrelmoonmag>RM: #stop i f rmag < RE or rrelmoonmag < RM36 r a t e (1000)37

38 #ca l c u l a t e Fgrav on ranger by Earth39 r=ranger . pos40 rmag=mag( r )41 run i t=r /rmag42 FgravE=GmME/rmag2 run i t43

44 #ca l c u l a t e Fgrav on ranger by Moon45 rrelmoon=ranger . pos Moon . pos46 rrelmoonmag=mag( rrelmoon )47 rre lmoonunit=rrelmoon/rrelmoonmag48 FgravM=GmMM/rrelmoonmag 2 r re lmoonunit49

50 #ca l c u l a t e net f o r c e51 Fnet=FgravE + FgravM52

53 #ca l c u l a t e energy54 Umoon = GmMM/rrelmoonmag55 Uearth = GmME/rmag56 U = Umoon + Uearth57 K = 1/2mmag(v ) 258 E= K + U59

60 #update momentum and po s i t i o n61 p = p + Fnetdt62 v = p/m63 ranger . pos = ranger . pos + vdt64

65 t r a i l . append ( pos=ranger . pos )66

67 t = t + dt68

69 UPlot . p l o t ( pos=(rmag ,U) )70 EPlot . p l o t ( pos=(rmag ,E) )71 KPlot . p l o t ( pos=(rmag ,K) )

Viewing the graphs shows that the total energy indeed remains constant. If the time step is too large (dt > 100 s for example),the total energy will become inaccurate.

The escape speed from Earth is approximately 1.12 104 m/s. If you change the initial speed to be less than the escapespeed, you will notice that the total energy is negative and the spacecraft never makes it to Moon. If you launch it at theescape speed, you will notice that the total energy is zero, as expected. Since the escape speed would cause the spacecraft tojust barely reach r = , then it certainly has enough energy to get to Moon. Thus, even at the escape speed, with E = 0,it still reaches Moon as one would expect.