12-22
Chapter 12
EXECUTE:
(a)
2
E
g
2
E
(10.0 kg)(9.80 m/s )
98.0 N
mm
FG
m
g
R
==
=
=
.
(b)
The mass of the inner core is
33
4
3
63
2
2
44
inner
inner
2
1
(
)
(1.3 10 kg/m )
(1.2 10 m)
9.4 10 kg
mR
R
ρπ
π
=−
=
×
×
=
×
.
The
mass of the outer core is
43
6
3
6
3
2
4
4
outer
3
(1.1 10 kg/m )
([3.6 10 m]
[1.2 10 m] )
2.1 10 kg
m
=×
×
−×
=
×
.
Only the inner
and outer cores contribute to the force.
22
24
11
2
2
g
62
(9.4 10 kg
2.1 10 kg)(10.0 kg)
(6.67 10
N m /kg )
110 N
(3.6 10 m)
F
−
×+
×
⋅
=
×
.
(c)
Only the inner core contributes to the force and
22
11
2
2
g
(9.4 10 kg)(10.0 kg)
(6.67 10
N m /kg )
44 N
(1.2 10 m)
F
−
×
⋅
=
×
.
(d)
At
0
r
=
,
g
0
F
=
.
EVALUATE:
In this model the earth is spherically symmetric but not uniform, so the result of Example 12.10
doesn’t apply.
In particular, the force at the surface of the outer core is greater than the force at the surface of the
earth.
12.71.
IDENTIFY:
Eq.(12.12) relates orbital period and orbital radius for a circular orbit.
SET UP:
The mass of the sun is
30
1.99 10 kg
M
.
EXECUTE:
(a)
The period of the asteroid is
32
11
2
. Inserting 3 10 m
a
T
GM
for
a
gives
11
2.84 y and 5 10 m
×
gives a period of 6.11 y.
(b)
If the period is
11
5.93 y, then
4.90 10 m.
a
(c)
This happens because
0.4
2 5,
=
another ratio of integers. So once every 5 orbits of the asteroid and 2 orbits of
Jupiter, the asteroid is at its perijove distance. Solving when
11
4.74 ,
4.22 10 m.
Ty
a
×
EVALUATE:
The orbit radius for Jupiter is
11
7.78 10 m
×
and for Mars it is
11
2.21 10 m
×
.
The asteroid belt lies
between Mars and Jupiter.
The mass of Jupiter is about 3000 times that of Mars, so the effect of Jupiter on the
asteroids is much larger.
12.72.
IDENTIFY:
Apply the work-energy relation in the form
WE
= Δ
, where
EKU
=
+
.
The speed
v
is related to the
orbit radius by Eq.(12.10).
SET UP:
24
E
5.97 10 kg
m
EXECUTE:
(a)
In moving to a lower orbit by whatever means, gravity does positive work, and so the speed does increase.
(b)
()
1/2
E
vG
m r
−
=
, so
3/2
E
E
3
22
rr
G
m
m
r
r
−
−Δ
Δ
⎛⎞
⎛
⎞
Δ=
−
=
⋅
⎜⎟
⎜
⎟
⎝⎠
⎝
⎠
Note that a positive
r
Δ
is given as a
decrease in radius. Similarly, the kinetic energy is
( ) ( )
2
E
1/2
1/2
/
K
mv
Gm m r
, and so
( )
2
E
1/2
/
K
Gm m r
r
Δ
and
( )
2
E
/
UG
m
m
r
r
−
Δ
.
( )
2
E
/2
WUK
G
m
m
=Δ +Δ =−
Δ
(c)
3
E
/7
.
7
2
1
0
m
/
s
,
m
r
×
\
3
E
/
28.9 m/s,
vrG
m
r
Δ=Δ
=
10
E
8.95 10 J
EG
m
m
r
×
(from Eq.(12.15)),
( )
28
E
6.70 10 J,
KG
m
m
×
9
21
.
3
4
1
0
J
UK
−Δ=
− ×
, and
8
6.70 10 J.
WK
=−Δ =−
×
(d)
As the term &burns up± suggests, the energy is converted to heat or is dissipated in the collisions of the debris
with the ground.
EVALUATE:
When
r
decreases,
K
increases and
U
decreases (becomes more negative).
12.73.
IDENTIFY:
Use Eq.(12.2) to calculate
g
.
F
Apply Newton²s 2nd law to circular motion of each star to find the
orbital speed and period. Apply the conservation of energy expression, Eq.(7.13), to calculate the energy input
(work) required to separate the two stars to infinity.
(a) SET UP:
The cm is midway between the two stars since they have equal masses. Let
R
be the orbit radius for
each star, as sketched in Figure 12.73.