Tài liệu Physics exercises_solution: Chapter 12 pptx

E 2 2 2 12.1: The ratio will be the product of the ratio of the mass of the sun to the mass of the earth and the square of the ratio of the earth-moon radius to the sun-moon radius... 1

Note: to obtain the numerical results given in this chapter, the following numerical values

of certain physical quantities have been used;

kg

1097.5and

sm80.9 ,kgmN10673

E 2

2 2

12.1: The ratio will be the product of the ratio of the mass of the sun to the mass of the

earth and the square of the ratio of the earth-moon radius to the sun-moon radius Using the earth-sun radius as an average for the sun-moon radius, the ratio of the forces is

.18.2kg105.97

kg101.99m

101.50

m1084.3

24

30 2

kg)2150)(

kg1097.5()kgmN10673.6

2 6 5

24 2

2 11 2

kg2150(

)N1067.1(

Gm mg

r m Gm

yielding the same result

12.3:

)(

))(

(

12 2

12

2 1 2

nm nm

12.4: The separation of the centers of the spheres is 2R, so the magnitude of the

gravitational attraction is GM2 (2R)2 GM2 4R2

12.5: a) Denoting the earth-sun separation as R and the distance from the earth as x,

the distance for which the forces balance is obtained from

,)

E 2

S

x

m GM x

R

m GM



which is solved for

m

1059.21

8

E S

R x

b) The ship could not be at equilibrium for long, in that the point where the forces balance is moving in a circle, and to move in that circle requires some force The spaceship could continue toward the sun with a good navigator on board

12.6: a) Taking force components to be positive to the right, use of Eq (12.1) twice

2 2 11

1032.2

,m0.600

kg0.10m

0.400

kg00.5kg

100.0kgm10

673.6

g

F

with the minus sign indicating a net force to the left

b) No, the force found in part (a) is the net force due to the other two spheres.

12.7:         2.4 10

m1078.3

kg1035.7kg70kgm.10673

2 8

22 2

,23

000,

12.9: Denote the earth-sun separation asr and the earth-moon separation as 1 r 2

a)   6.30 10 ,

)(

20 2

2

E 2 2 1

r

m Gm

toward the sun b)The moon distance is sufficiently small compared to the

earth-sun distance (r2 << r2) that the vector from the earth to the moon can be taken to be perpendicular to the vector from the sun to the moon The components of the

gravitational force are then

,1099.1,

1034

2 2

E M 20

2 1

S

r

m Gm r

m Gm

and so the force has magnitude 4.771020 and is directed 24.6 from the direction toward the sun

r

20 2

2

E 2 2 1

m Gm

toward the sun

12.10:

square theofcenter the

towardN,

102.8

m)10.0(

kg)800()kgNm1067.6(

m)(0.10

45coskg)800)(

kgNm1067.6(2

45cos2

F45cos2

3

2

2 2

2 11 2

2 2

2 11

2 AD

D A 2

AB

B A

D B

m Gm

F F

left the toN,106.1

N10043.1

N10668.1

m40.0m;

10

0

kg500

3 3

1

4 2

23

3 2 3

3 2

12

2 1 1

23 12

3 2

F

r

m m

G

F

r

m m

G

F

r r

m m

4)2()2( 2 2 1 2 d22 1

m m Gm d

m Gm d

m

12.13: For convenience of calculation, recognize that the mass of the small sphere will

cancel The acceleration is then

,sm101.20.10

0.6m)10(10.0

kg)260.0(

2 2

2 6

22 2

2 11

12.15: To decrease the acceleration due to gravity by one-tenth, the distance from the

earth must be increased by a factor of 10,and so the distance above the surface of the earth is

)905)(

sm80.9(

949

1)815(

1m

g

2 2

2 E

2

v

E E

v 2 E E

2 E

2

v

E E E

v 2

v

v v

R m

m R Gm

R R

R m

m G R Gm

where the subscripts v refer to the quantities pertinent to Venus b) (8.87m s2)(5.00kg)N

8)sm80.9

r

r m

E ( 8 ) 1700)

1 (

r

r m

m

ρ

  1656kg m ,)

ρ

12.19: 2E

r

mm G

F 

r600103mRE so F 610N

At the surface of the earth, w  mg735N

The gravity force is not zero in orbit The satellite and the astronaut have the same acceleration so the astronaut’s apparent weight is zero

12.20: Get g on the neutron star

2

ns ns

2

ns ns

R

GM g

R

GmM mg

R

mGM mg

30 2

2 11

kg)1099.1)(

kgNm10(6.67s

m8.9

E 2

2

2 E 2 1

Fr

R m gm m

12.22: a) From Example 12.4 the mass of the lander is 4000 kg Assuming Phobos to

be spherical, its mass in terms of its density ρandradiusRis (4 3)ρR3, and so the gravitational force is

N.27)m1012)(

mkg2000)(

kg4000)(

34()

kg4000)(

R G

b) The force calculated in part (a) is much less than the force exerted by Mars in Example 12.4

12.23: 2GM R  2(6.67310 11 Nm2 kg2 (3.61012kg) (700m)

0.83m s

One could certainly walk that fast

12.24: a) F Gm E m r2andU GmEm r, so the altitude above the surface of the earth is 9.36 105m

 R

F

U

b) Either of Eq (12.1) or Eq (12.9) can be used with the

result of part (a ) to find m, or noting that U2 G2M E2m2 r2, mU2 FGME

12.26: a) The kinetic energy is ,or (629kg)(3.33 103m s)2,

2 1 2

or KE   9

m1087.2

)kg629)(

kg1097.5)(

kgmN10673.6(GM

9

24 2

2 11

05.1

sm6200

kg1097.5kgmN10673.622

4

3

24 2

2 11 3

Gm

b) 2 3.71m s2

T πv

12.28: Substitution into Eq (12.14) gives T 6.96103s, or 116 minutes

12.29: Using Eq (12.12),

m1080.7m1038.6

kg1097.5kgmN10673

5 6

24 2

2 11

12.30: Applying Newton’s second law to the Earth

 

kg1001.2

)]

(3.365[(

)kgNm1067.6(

m)1050.1(44

2and

;

30

2 10 64 8 2

2 11

3 11 2

2 E

3 2 2 2

Earth 2

2 2

s

E s E

d

GT

r G

r m

T

r v G

rv m

r

v m r

m Gm

ma F

12.31: F mac for thebaseball

The net force is the gravity force exerted on the baseball by Deimos, so

sm7.4m)100.6(kg)100.2(kgmN1067.6

2 2

12.32: Apply Newton’s second law to Vulcan

days9.47s400,86

ds1014.4

kg)1099.1)(

kgNm1067.6(

m)1079.5(44

22:

6

30 2

2 11

3 10 3

2 2 s

3 2

2 s

2 v

2 vs

T

r r

Gm

T

r v

r

v m r

m Gm ma F

8

))11.0m)(10((1.50kg)

1099.185.0)(

kgmN10673.6(

4

11 30

2 2 11

v

b) 2r v1.25106s(about two weeks)

12.34: From either Eq (12.14) or Eq (12.19),

kg

1098.1

d))s1064.8d)(

7.224()kgmN10673.6(

m)1008.1(44

30

2 4 2

2 11

3 11 2

2

3 2 S

12.36: a)  1 2 7.071010m.

F

m Gm r

b) From Eq (12.19), using the result of part (a),

days

121s1005.1kg)1090.1)(

kgmN10673.6(

m)1007.7(

30 2

2 11

2 10

kgmN10673.6

The speed doesn’t have this value, so the orbit is not circular b) The escape speed for any object at this radius is 2(56km s)79km s, so the spacecraft must be in a closed elliptical orbit

12.38: a) Divide the rod into differential masses dm at position l, measured from the right end of the rod Then , dm = dl ( M L), and

x l

dl L

GmM x

l

dm Gm

GmM x

l

dl L

GmM

.1ln

For x >> L, the natural logarithm is ~ L x , and U Gm M x. b) The x-component

of the gravitational force on the sphere is

,)(

))(1(

)(

2

2

Lx x

GmM x

L

x L L

GmM δx

with the minus sign indicating an attractive force As x >> L, the denominator in the

above expression approaches x , and 2 F x Gm M x2, as expected The derivative may also be taken by expressing

x L x x

L

ln)ln(

12.39: a) Refer to the derivation of Eq (12.26) and Fig (12.22) In this case, the red

ring in Fig (12.22) has mass M and the common distance s is x2 a2.Then,

.2

2 a x GMm

U   b) When x >> a, the term in the square root approaches x2and U GMm x, as expected

GMmx





with the minus sign indicating an attractive force d) when x >> a, the term inside the

parentheses in the above expression approaches x and 2 F x GMmx (x2) 2

,2

0



x This makes sense because the mass at the center is a constant distance a from the

mass in the ring The result of part (c) indicates that F x 0when x0 At the center of the ring, all mass elements that comprise the ring attract the particle toward the respective parts of the ring, and the net force is zero

12.40: At the equator, the gravitational field and the radial acceleration are parallel,

and taking the magnitude of the weight as given in Eq (12.30) gives

.rad

0 ma mg

0

rad

w a

g

a w ma

m

12.42: a)   .

2

2

2 S 2 2

2 S

R mc r

c R r

2 6

2 2

sm1000.3m1000.142

24 2

2 11

2 8 3

12.43: a) From Eq (12.12),

.102.1kg103.4

kgmN10673.6

sm10200lym10461.9ly5.7

S 7 37

2 2 11

2 3 15

2

M G

Rv M

c

R v c

GM R

which does fit

12.44: Using the mass of the sun for M in Eq (12.32) gives

3.00 10 m s 2.95km.

kg1099.1kgmN10673.62

2 8

30 2

2 11

sun sun

M c

Gm R

Using 3.0 km instead of 2.95 km is accurate to 1.7%

12.45:   

3 10 m s 6.38 10 m 1.4 10 .

kg1097.5kgmN1067.6

6 2

8

24 2

2 11 E

12.46: a) From symmetry, the net gravitational force will be in the direction

45from thex -axis (bisecting the x and y axes), with magnitude

)kg0.1(2))m50.0(2(

)kg0.2()kg0150.0)(

kgmN10673

)kg0.1(2m)(0.502

)kg0.2(2

Canceling the factor of m and solving for v, and using the numerical values gives

s

m10

3.02  5





12.47: The geometry of the 3-4-5 triangle is available to simplify some of the algebra,

The components of the gravitational force are

5

3)

m000.5(

)kg0.80)(

kg500.0)(

kgmN10673.6(

2

2 2

)kg0.80()m000.4(

)kg0.60()kg500.0)(

kgmN10673.6

x

F

2.10510 10 N,

so the magnitude is 2.2010 10 N and the direction of the net gravitational force is

163counterclockwisefrom thex-axis b)A at x0,y1.39m

12.48: a) The direction from the origin to the point midway between the two large

masses is arctan (0.2000 100mm)26.6, which isnot theangle(14.6) found in the example

b) The common lever arm is 0.100 m, and the force on the upper mass is at an angle of

isnet torqueThe

2 2 11

)m200.0(

)m100.0(m)

2(0.200

45m)sin (0.100kg)

kg)(0.500)(0.0100

kgmN10

12.49: a) The simplest way to approach this problem is to find the force between the

spacecraft and the center of mass of the earth-moon system, which is 4.67106 mfrom the center of the earth

The distance from the spacecraft to the center of mass of the earth-moon system is 3.82108m Using the Law of Gravitation, the force on the spacecraft is 3.4 N, an angle of 0.61from the earth-spacecraft line This equilateral triangle arrangement

of the earth, moon and spacecraft is a solution of the Lagrange Circular Restricted Three-Body Problem The spacecraft is at one of the earth-moon system Lagrange points The Trojan asteriods are found at the corresponding Jovian Lagrange points

b)Theworkis 6 673 10 N m / kg )( 53.84 97 1010 mkg 7 35 10 kg)(1250 kg),or

8

22 24

2 2 11

12.50: Denote the 25-kg sphere by a subscript 1 and the 100-kg sphere by a

subscript 2 a) Linear momentum is conserved because we are ignoring all other forces, that is, the net external force on the system is zero Hence, m1v1 m2v2.

This relationship is useful in solving part (b) of this problem b)From the energy theorem,

2 2

2 1 1 i

f 2 1

2

111

v m m m r

r m

i f 2 1

2 2 2

Gm v

with a similar expression for v2 Substitution of numerical values gives

s

m1008.4s,m10

a

a x

x    When the spheres finally make contact, their centers will be a distance of 2Rapart,or x1x2 2R40m,

m

402

,m10614.5

2

d)s400,86d)(

(27.3kg)1097.5)(

kgmN10673.6(

3 25

2 24

2 2 11

kg) 0 20 )(

kg m N 10 673 6 (

2 2 2



center of the sphere

12.53: a) From Eq (12.14),

,m10

10(5.97)kgmN10673.6(2

3 22

2 24

2 2 11 2

h Note that the period to use for the earth’s rotation

is the siderial day, not the solar day (see Section 12.7) b) For these observers, the satellite is below the horizon

12.54: Equation 12.14 in the text will give us the planet’s mass:

P

22

kgmN10673.6(

m)1080.4m1075.5(44

2.7311024 kg,or about half earth’s mass

Now we can find the astronaut’s weight on the surface (The landing on the north pole removes any need to account for centripetal acceleration):

N677

m1080.4

kg6.85kg10731.2kgmN10673.6

2 6

24 2

2 11 2

p

a p

12.55: In terms of the density ,the ratio M Ris4 3R2, and so the escape speed is

v 8 3 6.67310 11Nm2 kg22500kg m3150103m2 177m s

12.56: a) Following the hint, use as the escape velocity v 2gh, where his the height one can jump from the surface of the earth Equating this to the expression for the escape speed found in Problem 12.55,

,4

3

or ,3

8

G

gh R

GR

π gh

.mkg1003.3

12.57: a) The satellite is revolving west to east, in the same direction the earth is

rotating If the angular speed of the satellite is ω s and the angular speed of the earth is ω E,the angular speed ωrel of the satellite relative to you is ωrel ωs ωE

1 E rel

mm G

2 2

E

ω

Gm r

r v r

Gm

This is the radius of the satellite’s orbit Its height habove the surface of the earth is hrR E 1.39107 m

b) Now the satellite is revolving opposite to the rotation of the earth

If west to east is positive, then  121 rev h

rel  

ω

 rev h 7.27 10-5rad s

241E

ω ω

ω s

m1059.3andm1022.4

12.58: (a) Get radius of X:41 2R 18,850km

kgNm1067.6(

)m1020.1)(

N915(

:

25 2

2 11

2 7 2

x x

x 2

mg R

GmM ma

F

Apply Newton’s second law to astronant on a scale at the equator of X

 

dayoneis which hr,

36.7s3600

hr1s1065.2

)m1020.1)(

kg2.96(4N850.0N

0.915

42

:

4 2

7 2

2

2 2

2 scale grav

2 scale grav

m F

F T

R v R

mv F

F ma F

r

T r r

v m r m

)kg1005.2)(

kgNm1067.6(

)m102m1020.1(44

3

25 2

2 11

3 6 7

2 x

3 2

1

R h R Gm

g Gmm

mgh

h R R

At this point, it is advantageous to use the algebraic expression for g as given in

Eq (12.4) instead of numerical values to obtain the fractional difference as

,)

(

1 RE h RE h RE so if the fractional difference is

m

104.6(0.01)

If the algebraic form for g in terms of the other parameters is not used, and the

numerical values from Appendix F are used along with g9.80m s2,

,10

12.60: (a) Get g on Mongo: It takes 4.00 s to reach the maximum height, where

v = 0 then v0gt012.0m sg(4.00s)

2sm00.3



g

Apply Newton’s second law to a falling object:

G gR M R

GmM mg

)sm00.3

2 2 11

2

210 m00 2 2 2

GmM ma

T

r v

3 24



3.0 10 m

2km000,

r

kg)1056.4)(

kgNm1067.6(

m)100.3(

4

25 2

2 11

3 7

210 m00 2

h1s1054

At the top of Mount Everest, a height of h8800mabove sea level, the gravity force on you is

2 E

2 E

E 2

E 2

)1

()

mm G

h R

mm G

,

21)

1

(

R

h F

F R

h R

28.02E 1

2

r

h F

F

F

%

12.62: a) The total gravitational potential energy in this model is

r R

m r

m Gm U

b) See Exercise 12.5 The point where the net gravitational field vanishes is

m

1046.31

8 E

R r

Using this value for r in the expression in part (a) and the work-energy theorem,

including the initial potential energy of Gm(mE RE mM (REMRE)) gives

theorem, the rocket impacts the moon with a speed of 2.9km s

12.63: One can solve this problem using energy conservation, units of J/kg for energy,

and basic concepts of orbits ,or 2 ,

E      where E,KandU are the

energies per unit mass, v is the circular orbital velocity of 1655 m/s at the lunicentric

distance of 1.79106 m The total energy at this distance is1.37106J Kg When the velocity of the spacecraft is reduced by 20 m/s, the total energy becomes

,m)

10(1.79

kg)1035.7(kg/mN10673.6(s)/m20s/m1655(2

1

6

22 2

2 11 2

m

10706.1

12.64: Combining Equations (12.13) and (3.28) and setting

 

so that 0in Eq 12.30 ,s

which is 84.5 min, or about an hour and a half

12.65: The change in gravitational potential energy is

E

E E

E

h R R

h m Gm R

m Gm h

R

m Gm U

E

E

2 f

GmM

E R

R   The work needed to put it in orbit is the difference between these: -

E

E

2 i

GmM

E E

b) The total energy of the satellite far away from the Earth is zero, so the additional work needed is 0  

E E E

E

2

GmM R

  c) The work needed to put the satellite into orbit was the same as the work needed

to put the satellite from orbit to the edge of the universe