Download s 37, 44, 53, 55, 56, 61, 69, 74, 75, 82, 92

Ch 11 HW Day 4, p 362 – 369 #’s 37, 44, 53, 55, 56, 61, 69, 74, 75, 82, 92
37 •
Picture the Problem Newton’s 2nd law of motion relates the
masses and accelerations of these objects to their common
accelerating force.
(a) Apply Newton’s 2nd
law to the standard object:
F  m1a1
Apply Newton’s 2nd law
to the object of
unknown mass:
F  m2 a2
Eliminate F between
these two equations
and solve for m2:
m2 
Substitute numerical
values and evaluate m2:
2.6587 m/s 2
1 kg   2.27 kg
m2 
1.1705 m/s 2
(b)
a1
m1
a2
It is the inertial mass of m2 .
44 ••
Picture the Problem Let the zero of gravitational potential
energy be at infinity, m represent the mass of the object, and h
the maximum height reached by the object. We’ll use
conservation of energy to relate the initial potential and kinetic
energies of the object-earth system to the final potential
energy.
Use conservation of energy
to relate the initial
potential energy of the
system to its energy as the
object is about to strike the
earth:
K f  Ki  U f  U i  0
or, because Kf = 0, (move all else to the
other side),
K RE   U RE   U RE  h  0
(1)
where h is the initial height above the
earth’s surface.
Express the potential
energy of the object-earth
system when the object is
at a distance r from the
surface of the earth:
Substitute in equation (1) to
obtain:
U r   
Solve for h:
Little m cancels out of all 3 terms. Move
the first two terms to the other side of
the eq. Get a common denominator of
2RE. Then, you’ll have a single fraction
on both the left and right. This way, you
can cross-multiply, and then subtract RE
from both sides to isolate h. Don’t forget
that ALL units need to be SI.
1
2
GM E m
r
GM E m GM E m
mv 

0
RE
RE  h
h
2
2GM E RE
 RE
2GM E  RE v 2
Or, you can use their derivation below.
h
RE
2 gRE
1
v2
Substitute numerical values and
h
evaluate h:
6.37 106 m
2 9.81m/s 2 6.37 106 m


4 10 m
3
2
 1
 935 km
53 ••
Picture the Problem We can use its definition to express the
period of the spacecraft’s motion and apply Newton’s 2nd law
to the spacecraft to determine its orbital velocity. We can then
use this orbital velocity to calculate the kinetic energy of the
spacecraft. We can relate the spacecraft’s angular momentum
to its kinetic energy and moment of inertia.
(a) Express the period of
the spacecraft’s orbit about
the earth:
Use Newton’s 2nd law to
relate the gravitational
force acting on the
spacecraft to its orbital
speed:
Solve for v to obtain:
T
2 R 2 3RE  6 RE


v
v
v
where v is the orbital speed of the
spacecraft.
NOTE: 3RE because you have to
use 1 RE + a height of 2 RE for a
total of 3 RE.
F = ma, where a = v2/r, so,
Fradial 
v
GM E m
3RE 2
v2
m
3 RE
gRE
3
NOTE: normally, v =  gRE , but
since this orbits at a height of 3RE
(total), after you divide both sides
by 3 and take the square root, you
end up with the above expression.
Substitute for v in our
expression for T to obtain:
RE
T  6 3
g
Substitute numerical values
and evaluate T:
6.37  106 m
T  6 3π
9.81 m/s 2
1h
 7.31 h
3600 s
NOTE: I would have left it as 2.6 x
104 s, since that’s SI.
 2.631  10 4 s 
(b) Using its definition,
express the spacecraft’s
kinetic energy:
K  12 mv 2  12 m 13 gRE 
100 kg 9.81m/s 2 6.37 106 m
Substitute numerical values
and evaluate K:
K
(c) Express the kinetic
energy of the spacecraft in
terms of its angular
momentum:
L2
K
2I
Solve for L:
L  2IK
1
6
 1.04 GJ
Express the moment of
inertia of the spacecraft
with respect to an axis
through the center of the
earth:
I  m3RE 
Substitute and solve for L:
L  18mRE2 K  3RE 2mK
2
 9mRE2
Substitute numerical values and evaluate L:




L  3 6.37 106 m 2100 kg  1.04 109 J  8.72 1012 J  s
55 ••
Picture the Problem We can express the energy difference
between these two orbits in terms of the total energy of a
satellite at each elevation. The application of Newton’s 2nd law
to the force acting on a satellite will allow us to express the
total energy of each satellite as function of its mass, the radius
of the earth, and its orbital radius.
Express the energy difference:
E  Egeo  E1000
(1)
Express the total energy of
an orbiting satellite:
E tot  K  U
 12 mv 2 
GM E m
R
where R is the orbital radius.
(2)
Apply Newton’s 2nd law to
a satellite to relate the
gravitational force to the
orbital speed:
Fradial
GM E m
v2

m
2
R
R
Which is N’s Law of Gravity, a
centripetal force, and N2, F = ma,
where a = v2/R. Also, since
GME = gRE2, and little m cancels
out, this can be simplified to:
gRE2 v 2

2
R
R
Simplify and solve for v2:
Substitute in equation (2) to
obtain:
Substitute in equation (1)
and simplify to obtain:
2
gR
E
v2 
R
Etot
2
2
2
gR
gR
m
mgR
E
 12 m E  E  
R
R
2R
mgRE2 mgRE2
E  

2 Rgeo 2 R1000
mgRE2

2
 1
1 


R

 1000 Rgeo 
Substitute numerical values and evaluate E:

2
1
1
  11.1GJ
E  12 500 kg 9.81 N / kg  6.37 106 m 

6
7
7
.
37

10
m
4.22

10
m




56 ••
Picture the Problem. We can use Kepler’s 3rd law to relate the
periods of the moon and Earth, in their orbits about the earth and
the sun, to their mean distances from the objects about which they
are in orbit. We can solve these equations for the masses of the sun
and the earth and then divide one by the other to establish a value
for the ratio of the mass of the sun to the mass of the earth.
Using Kepler’s 3rd law,
relate the period of the
moon to its mean
distance from the
earth:
Using Kepler’s 3rd law,
relate the period of the
earth to its mean
distance from the sun:
Solve equation (1) for ME:
Solve equation (2) for Ms:
4 2 3
T 
rm
GM E
2
m
(1)
where rm is the distance between the
centers of the earth and the moon.
4 2 3
T 
rE
GM s
2
E
(2)
where rE is the distance between the
centers of the earth and the sun.
4 2 3
ME 
r
2 m
GTm
(3)
4 2 3
Ms 
r
2 E
GTE
(4)
Divide equation (4) by
equation (3) and
simplify to obtain:
M s  rE 
  
M E  rm 



2
Substitute numerical
values and evaluate
Ms/ME:
M s  1.5  1011 m 


M E  3.82  108 m 
3
3
 Tm

 TE
 3.38  105
 27.3 d 


365
.
24
d


2
Express the difference
between this value and
the measured value of
3.33105:
3.38  105  3.33  105
% diff 
3.33  105
 1.50 %
61 ••
Picture the Problem The
configuration of point masses is
shown to the right. The
gravitational field at any point
can be found by superimposing
the fields due to each of the point
masses.
(a) Express the gravitational
field at
x = 2 m as the sum of the fields
due to the point masses m1 and
m2:
  
g  g1  g2
(1)
Express g1 and g 2 :
N2 = N’s LoG, cxl m’s


Gm
Gm
g1   2 1 iˆ and g 2  2 2 iˆ
x2
x1
Substitute in equation (1) to
obtain:

Gm
Gm
g   2 1 iˆ  2 2 iˆ
x1
x2

Gm1 ˆ Gm2 ˆ
i
i
2
x12
2 x1 

G
m1  14 m2  iˆ
2
x1
Substitute numerical values
and evaluate g :

6.6726  10 11 N  m 2 /kg 2
g
2 m 2
 2 kg  1 4 kg  iˆ
4

(b) Express


g1 and g 2 :
Substitute in equation (1) to
obtain:
Substitute numerical values
and evaluate g :
 1.67 10
11

N/kg iˆ

Gm

Gm
g1   2 1 iˆ and g 2   2 2 iˆ
x1
x2

Gm
Gm
g   2 1 iˆ  2 2 iˆ
x1
x2

Gm1 ˆ Gm2 ˆ
i 2 i
2
x2
2 x2 

G 1
 m  m2  iˆ
2 4 1
x2

6.6726  10 11 N  m 2 /kg 2
g
6 m 2
  1 2 kg   4 kg  iˆ
4

(c) Express the condition that g = 0:
 8.34  10
12

N/kg iˆ
Gm1
Gm2

0
2
2
x
6  x 
or
2
4

0
2
2
x
6  x 
Express this quadratic
equation in standard form:
x 2  12 x  36  0 , where x is in
meters.
Solve the equation to obtain:
x  2.48 m and x  14.5 m
From the diagram it is clear
that the physically meaningful
root is the positive one at:
x  2.48 m
69 ••
Picture the Problem: The magnitude of the gravitational force is F = mg
where g inside a spherical shell is zero and outside is given
2
g

GM
r
. This is N2 set equal to N’s Law of Gravity (Eq 11by
29).
(a) At r = 3a, the masses of
both spheres contribute to
g:
F  mg  m

G M 1  M 2 
3a 2
GmM 1  M 2 
9a 2
(b) At r = 1.9a, g due to M2 =
0:
GM1
GmM1
F  mg  m

2
3.61a 2
1.9a 
(c) At r = 0.9a, g = 0:
F 0
74 •••
Picture the Problem The diagram
shows the portion of the solid
sphere in which the hollow sphere
is embedded. g1 is the field due to
the solid sphere of radius R and
density 0 and g 2 is the field due to
the sphere of radius ½R and
negative density 0 centered at ½R.
We can find the resultant field by
adding the x and y components of


g1 and g 2 .
Set N2 = N’s L.of G., cancel m; use givens and sub.
Use its definition to express g1 :

GM G
g1 
0V

r2
r2
40 rG

3
40 r 3G

3r 2
Find the x and y components of

g1 :
40Gx
x
g1x   g1 cos    g1    
3
r
and
40Gy
 y
g1 y   g1 sin    g1    
3
r
where the negative signs indicate that
the field points inward.
Use its definition to express g2 :

GM 2 G 0V2 40 r23G
g2  2 

r
r22
3r22

where
40 r2G
3
r2 
x  12 R 2  y 2
Express the x and y components
of g 2 :
 x  12 R  40 Gx  12 R 
 
 g 2 
r
3
2


g2x
g2 y
Add the x components to
obtain the x component of the
resultant field:
 y  40Gy
 g 2   
3
 r2 
g x  g1x  g 2 x
40Gx 40G x  12 R 


3
3
20GR

3
where the negative sign indicates that
the field points inward.
Add the y components to
obtain the y component of the
resultant field:
g y  g1 y  g 2 y
Express g in vector form and
evaluate g :

 20GR  ˆ
g  g x iˆ  G y ˆj   
i
3



40Gy 40Gy

0
3
3
and

20GR
g
3
75 •••
Picture the Problem The gravitational field will exert an inward radial
force on the objects in the tunnel. We can relate this force to the
angular velocity of the planet by using Newton’s 2nd law of motion.
Letting r be the distance from
the objects to the center of the
planet, use Newton’s 2nd law
to relate the gravitational
force acting on the objects to
their angular velocity:
Solve for  to obtain:
Use its definition to express g:
Fnet  Fg  mr 2
or
mg  mr

2
g
r
(1)
GM G 0V 40 r 3G
g 2 

2
r
r
3r 2
40 rG

3
Substitute in equation (1) and
simplify:

40 rG
3

r
40G
3
82 •
Picture the Problem Consider an object of mass m at the
surface of the earth. We can relate the weight of this object to
the gravitational field of the earth and to the mass of the earth.
This is N2 set equal to N’s Law of Gravity.
Using Newton’s 2nd law,
relate the weight of an
object at the surface of the
earth to the gravitational
force acting on it:
Solve for ME:
w  mg 
gRE2
ME 
G
GM E m
RE2
Substitute numerical
values and evaluate ME:
ME
2

9.81 N / kg 6.37 106 m 

6.6726 1011 N  m 2 /kg 2
 5.97 1024 kg
92 ••
Picture the Problem. Let r represent the separation of the particle from
the center of the earth and assume a uniform density for the earth. The
work required to lift the particle from the center of the earth to its
surface is the integral of the gravitational force function. This function
can be found from the law of gravity and by relating the mass of the
earth between the particle and the center of the earth to the earth’s
mass. We can use the work-kinetic energy theorem to find the speed with
which the particle, when released from the surface of the earth, will
strike the center of the earth. Finally, the energy required for the
particle to escape the earth from the center of the earth is the sum of the
energy required to get it to the surface of the earth and the kinetic
energy it must have to escape from the surface of the earth.
(a) Express the work
required to lift the
particle from the
center of the earth to
the earth’s surface:
Using the law of
gravity, express the
force acting on the
particle as a function of
its distance from the
center of the earth:
RE
W   Fdr
(1)
0
where F is the gravitational force acting on
the particle.
GmM
F
r2
(2)
where M is the mass of a sphere whose
radius is r.
Express the ratio of M to
ME:


 43  r 3
M
 4
M E  3  RE3



r3
M  ME 3
RE
Substitute for M in
equation (2) to obtain:
GmM E
mgRE2
mg
F
r

r

r
3
3
RE
RE
RE
Substitute for F in equation
(1) and evaluate the
integral:
mg E
gmRE
W
rdr 

RE 0
2
(b) Use the workkinetic energy theorem
to relate the kinetic
energy of the particle
as it reaches the center
of the earth to the work
done on it in moving it
to the surface of the
earth:
W  K  mv
Substitute for W and
solve for v:
v
(c) Express the total
energy required for the
particle to escape when
projected from the
center of the earth:
E esc  W  12 mve2
Substitute for W and
solve for vesc:
R
1
2
2
gRE
2
 12 mvesc
where ve is the escape speed from the
surface of the earth.
vesc  3gRE
Substitute numerical
values and evaluate
vesc:

vesc  39.81 N/kg  6.37 106 m
 13.7 km/s
