Tài liệu Physics exercises_solution: Chapter 12 pptx

Note: to obtain the numerical results given in this chapter, the following numerical values
of certain physical quantities have been used;
kg.1097.5andsm80.9 ,kgmN10673.6
24
E
22211


mgG
Use of other tabulated values for these quantities may result in an answer that differs in
the third significant figure.
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
.18.2
kg105.97
kg101.99
m101.50
m1084.3
24
30
2
11
8






















12.2: Use of Eq. (12.1) gives
N.1067.1
m)106.38m10(7.8
kg)2150)(kg1097.5(
)kgmN10673.6(
4
265
24
2211
2
21
g






r
mm
GF
The ratio of this force to the satellite’s weight at the surface of the earth is
%.7979.0
)sm80.9)(kg2150(
)N1067.1(
2
4


(This numerical result requires keeping one extra significant figure in the intermediate
calculation.) The ratio, which is independent of the satellite mass, can be obtained
directly as
,
2
E
2
E
2
E







r

R
gr
Gm
mg
rmGm
yielding the same result.
12.3:
.
)(
))((
12
2
12
21
2
12
21
F
r
mm
G
nr
nmnm
G

12.4: The separation of the centers of the spheres is 2R, so the magnitude of the
gravitational attraction is
.4)2(
2222
RGMRGM 

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
,
)(
2
E
2
S
x
mGM
xR
mGM


which is solved for
m.1059.2
1
8
E
S



M
M
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
gives
 
 
 
 
 
 










11
22
2211
1032.2
,
m0.600
kg0.10
m0.400
kg00.5
kg100.0kgm10673.6
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:
 
 
 
 
.104.2
m1078.3
kg1035.7kg70
kgm.10673.6
3
28
22
2211





12.8:
 
 
4
2
1003.6
500,23
000,333



12.9: Denote the earth-sun separation as
1
r
and the earth-moon separation as
2
r
.
a)
 
,1030.6
)(
20
2
2
E
2
21
S
M









r
m

rr
m
Gm
toward the sun. b)The earth-moon distance is sufficiently small compared to the earth-
sun distance (r
2
<< r
2
) 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
,1099.1,1034.4
20
2
2
EM
20
2
1
SM

r
mGm
r
mGm
and so the force has magnitude

20
1077.4
and is directed

6.24
from the direction
toward the sun.
c)
 
 
,1037.2
r
20
2
2
E
2
21
S
M









r
m
r
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
45cos
2
F45cos2
3
2
22211
2
22211
2
AD
DA
2
AB
BA
DBonA













r
mGm
r
mGm
FF

12.11:


left the toN,106.1
N10043.1
N10668.1
m40.0m;10.0
kg500
3
31
4
2
23
32
3
3
2

12
21
1
2312
321








FFF
r
mm
GF
r
mm
GF
rr
mmm


12.12: The direction of the force will be toward the larger mass, and the magnitude
will be
.
)(4
)2()2(
2

12
2
1
2
2
d
mmGm
d
mGm
d
mGm


12.13: For convenience of calculation, recognize that the mass of the small sphere will
cancel. The acceleration is then
,sm101.2
0.10
0.6
m)10(10.0
kg)260.0(2
29
22




G
directed down.
12.14: Equation (12.4) gives
  

 
.sm757.0
m1015.1
kg105.1kgmN10763.6
2
2
6
222211





g
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
 
m.1038.1R110
7
E

12.16: a) Using
,sm80.9g
2
E

Eq

 
4.12
gives
,sm87.8
)905)(.sm80.9(
949.
1
)815(.
1
mg
2
2
2
E
2
v
E
E
v
2
E
E
2
E
2
v
E
E
E
v

2
v
v
v




















































g
R
R
m
m

R
Gm
RR
R
m
m
G
R
Gm
where the subscripts v refer to the quantities pertinent to Venus. b)
kg)00.5()sm87.8(
2
N.3.44
12.17: a) See Exercise 12.16;
 
.sm369.0
1700
8
)sm80.9(
2
2
2
Titania











g
b)
3
T
3
E
E
T
E
T
.
r
r
m
m
ρ
ρ

, or rearranging and solving for density,

3
E
3
E
E
)81(
1700)1(

E
..
r
r
m
m
T
E
ρρ
 
,mkg1656)mkg5500(
3
1700
512
3

or about
E.0.39

12.18:
kg1044.2
21
2

G
gR
M
and
 
.mKg101.30

33
34
3

Rπ
M
ρ
12.19:
2
E
r
mm
GF


E
3
m10600 Rr 
so
N610F
At the surface of the earth,
N.735g  mw
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


Your weight would be

2
ns
nsns
R
mGM
mgw


24
302211
2
m)10(
kg)1099.1)(kgNm10(6.67
sm8.9
N675












N101.9
13

12.21: From eq. (12.1),
; (12.4), Eq.fromand,G
2
E
2
1
2
RGmgmmFr
E

combining
and solving for
E
R
,
kg.1098.5
24
2
2
E21

E

Fr
Rmgm
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
,)34( isradiusand
3
ρRRρ 
and so the
gravitational force is
N.27)m1012)(mkg2000)(kg4000)(34(
)kg4000)(34(
33
2
3



G
R
RG
b) The force calculated in part (a) is much less than the force exerted by Mars in Example
12.4.
12.23:
)m700()kg106.3()kgmN10673.6(22
122211



RGM

s.m83.0
One could certainly walk that fast.
12.24: a)
so ,and
E
2
rmGmUrmGmF
E

the altitude above the surface of the
earth is
m.1036.9
5
E
 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
,
22
2
22
rmMGU
E

E
2

FGMUm 
kg.1055.2
3

12.25: The escape speed, from the results of Example 12.5, is
.2 RGM
a)
.sm1002.5)m1040.3()kg1042.6()kgmN10673.6(2
36232211


b)
s.m1006.6)m1091.6()kg1090.1()kgmN10673.6(2
47272211


c) Both the kinetic energy and the gravitational potential energy are proportional to
the mass.
12.26: a) The kinetic energy is
,)sm1033.3)(kg629(or ,
23
2
1
2
2
1
 KmvK
J.103.49or
9
KE

b)
,
m1087.2
)kg629)(kg1097.5)(kgmN10673.6(GM
9
242211




r
m
U
or
J.1073.8
7
U
12.27: a) Eliminating the orbit radius r between Equations (12.12) and (12.14) gives
  
 
min.1751005.1
sm6200
kg1097.5kgmN10673.622
4
3
242211
3





s
v
Gm
T
E

b)
.sm71.3
2
2

T
πv
12.28: Substitution into Eq. (12.14) gives
s,1096.6
3
T
or 116 minutes.
12.29: Using Eq. (12.12),
  
 
s.m1046.7
m1080.7m1038.6
kg1097.5kgmN10673.6
3
56
242211






v
12.30: Applying Newton’s second law to the Earth

 
kg1001.2
)]()3.365[()kgNm1067.6(
m)1050.1(4
4
2
and
;
30
2
1064.8
2211
3112
2
E
32
2
2
Earth
2
2
2
4











d
s
T
r
s
s
E
sE
d
GT
r
G
r
m
T
r
v
G
rv
m
r

v
m
r
mGm
maF
E






12.31:
baseball.for the
c
maF 
The net force is the gravity force exerted on the baseball by Deimos, so
sm7.4m)100.6(kg)100.2()kgmN1067.6(
3152211
2
2



DD
DD
D
RGmv
R
v

m
R
mm
G
A world-class sprinter runs 100 m in 10 s so have
sm4.7s;m10  vv
for a thrown
baseball is very achieveable.
12.32: Apply Newton’s second law to Vulcan.
 
days9.47
s400,86
d1
s1014.4
kg)1099.1)(kgNm1067.6(
m)1079.5(4
4
2
2
:
6
302211
310
3
2
2
s
32
2
s

2
v
2
vs



























Gm
r
T
T
r
r
Gm
T
r
v
r
v
m
r
mGm
maF


12.33: a)

s.m1027.8
))11.0m)(10((1.50kg)1099.185.0)(kgmN10673.6(
4
11302211




rGmv

b)
weeks).(about twos1025.12
6
vr

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
242211
3112
2
32
S






GT
r
m

12.35: a) The result follows directly from Fig. 12.18. b)
m)1092.5)(248.01(
12

m,1045.4

12
 y.248c) m.104.55m)1050.4)(010.01(
1212
 T
12.36: a)
m.1007.7
10
21

F
mGm
r
b) From Eq. (12.19), using the result of part (a),
days.121s1005.1
kg)1090.1)(kgmN10673.6(
m)1007.7(2
7
302211
2310






T
c) From Eq. (12.14) the radius is

32
)8(

four times that of the large planet’s orbit,
or
m.1083.2
11

12.37: a) For a circular orbit, Eq. (12.12) predicts a speed of
s.km56m)1043(kg)1099.1)(kgmN10673.6(
9302211


The speed doesn’t have this value, so the orbit is not circular. b) The escape speed for any
object at this radius is
skm79)skm56(2 
, 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 (
LM
), and
dU =
.
xl
dl
L
GmM
xl
dmGm





Integrating,
U =










L
O
x
L
L
GmM
xl
dl
L
GmM
.1ln
For x >> L, the natural logarithm is
 
xL~
, and U

.xMGm
b) The x-component
of the gravitational force on the sphere is
,
)())(1(
)(
2
2
Lxx
GmM
xL
xL
L
GmM
δx
δU
F
x





with the minus sign indicating an attractive force. As x >> L, the denominator in the
above expression approaches
2
x
, and
x
F

2
xMGm
, as expected. The derivative may
also be taken by expressing
xLx
x
L
ln)ln(1ln 







at the cost of a little more algebra.
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
.
22
ax 
Then,
.
22
axGMmU 
b) When x >> a, the term in the square root approaches
2
x
and
xGMmU 

, as expected.
c)
x
F
δx
δU

,2322
)( ax
GMmx


,
with the minus sign indicating an attractive force. d) when x >> a, the term inside the
parentheses in the above expression approaches
2
x
and
232
)(xGMmxF
x

,
2
xGMm
as expected. e) The result of part (a) indicates that
a
GMm
U



when
.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
0
x
F
when
.0x
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
.
rad0
mamgw 
The difference between the measured weight and the force of gravitational attraction is
the term
.
rad
ma
The mass
m
is found by solving the first relation for
.,
0 rad
ag
mm




Then,
 
.
1
rad0rad0
rad
rad




ag
w
ag
a
wma
Using either
2
0
sm80.9g
or calculating
0
g
from Eq. (12.4) gives
N.40.2
rad
ma

12.41: a)
 
 
N54or N,5.53kg5.00sm7.10
22
N
RmGm
to two figures.
b)
   
 
  
 
N.0.52
hs3600h16
m105.24
sm 7.10kg00.5
2
72
2
rad0













agm
12.42: a)
 
.
2
2
2
S
2
2
2
S
2
r
Rmc
r
cR
r
GMm

b)
 
   
 
N.350
m1000.32
m104.1sm1000.3kg00.5

2
6
2
2
8




c) Solving Eq. (12.32) for
,M
  
 
.kg1044.9
kgmN10673.62
sm1000.3m1000.14
2
24
2211
2
832
S






G
cR

M
12.43: a) From Eq. (12.12),
 
  
 
.102.1kg103.4
kgmN10673.6
sm10200lym10461.9ly5.7
S
737
2211
2
3152
M
G
Rv
M





b) It would seem not.
c)
m,1032.6
22
10
2
2
2

S

c
Rv
c
GM
R
which does fit.
12.44: Using the mass of the sun for
M
in Eq. (12.32) gives
  
 
km.95.2
sm1000.3
kg1099.1kgmN10673.62
2
8
302211
S





R
That is, Eq. (12.32) may be rewritten
.
M
km95.2

2
sunsun
2
sun
S


















mm
M
c
Gm
R
Using 3.0 km instead of 2.95 km is accurate to 1.7%.
12.45:

  
   
.104.1
m1038.6sm103
kg1097.5kgmN1067.62
9
6
2
8
242211
E
S






R
R
12.46: a) From symmetry, the net gravitational force will be in the direction
45
x thefrom
-axis (bisecting the x and y axes), with magnitude









45sin
)m50.0(
)kg0.1(
2
))m50.0(2(
)kg0.2(
)kg0150.0)(kgmN10673.6(
22
2211

N.1067.9
12

b) The initial displacement is so large that the initial potential may be taken to be zero.
From the work-energy theorem,
.
)m50.0(
)kg0.1(
2
m)(0.502
)kg0.2(
2
1
2








Gmmv
Canceling the factor of m and solving for v, and using the numerical values gives
s.m103.02
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
2211



y
F

N10406.6
11











5
4
)m000.5(
)kg0.80(
)m000.4(
)kg0.60(
)kg500.0)(kgmN10673.6(
22
2211
x
F

N,10105.2
10

so the magnitude is
N1020.2
10

and the direction of the net gravitational force is
163
m.1.39y0,at xA b) axis.- thefromckwisecounterclo  x