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Physics 231 Lecture 22
•
•
Main points of today’s lecture:
Newton’s law of universal
gravitation:
GMm
F=
•
•
Kepler’s laws and the relation
between the orbital period and
orbital radius.
 4π 2 
2
r 3
T = 
 GM 
Gravitation potential energy
PE grav = −
•
r2
GM1M 2
r12
Tensile stress and strain
ΔF
ΔL
=Y
A
L0
•
Bulk stress and strain:
ΔF
ΔV
= ΔP = B
A
V
A uniform10 kg beam is support at one end by a hinge and at the other by
a cable. The cable from the wall is attached to the beam at an angle of 30°.
What is the tension in the cable?
a) 9.8 N
b) 19.6 N
c) 29.4 N
d) 39.2 N
d) 49 N
e) 58.8 N
f) 68.6 N
e) 98 N

Fhinge
T 
30o L T

Wbar
1. Draw the forces.
2. Choose the axis for calculating the
torques so as to avoid calculating torques
you don’ t want to know.
Don’t want to know Fhinge, so pput axis at hinge.
g
3. Set the net torque about this axis to zero and solve for the tension.
b g L / 2
 τ i = 0 = TL sin ( 30°) − m bar
i
TL sin ( 30°) = m bar g L / 2
T sin ( 30°) = m bar g / 2
 1
T   = m bar g / 2
T = m bar g
T = 98N
 2


To get Fhinge , use  Fi = 0.
i
Newton’s law of universal gravitation
•
All objects (even light photons) feel a gravitational force attracting
them to other objects. This force is proportional to the two masses and
inversely proportional to the square of the distance between them.
m1m2
F12 = G 2
r12
G=6
6.673
673 x 10-11 N m²
m /kg
/kg²
Example
•
A spaceship
hi iis on a journey
j
to
t the
th moon. The
Th masses off the
th earth
th andd
moon are, respectively, 5.98x1024 kg and 7.36x1022 kg. The distance
between the centers of the earth and the moon is 3.85x106m. At what
point as measured from the center of the earth,
point,
earth does the gravitational
force exerted on the craft by the earth balance the gravitational force
exerted by the moon? This point lies on a line between the centers of
GM moon mship
the earth and the moon.
GM earth m ship
Fmoon =
2
=
F
=
earth
rmoon _ ship
Moon
r2
earth _ ship
GM moon m ship

rmoon _ ship
2
moon _ ship
r
M moon
2
moon _ ship
r
rearth _ ship

rearth _ ship
rmoon _ ship
=
=
=
GM earth m ship
2
rearth
_ ship
M earth
2
earth _ ship
r
M earth
M moon

2
rearth
_ ship
2
rmoon
_ ship
5.98x1024 kg
=
= 9.01
22
7.36x10 kg
rearth _ ship + rmoon _ ship = 9.01rmoon _ ship + rmoon _ ship = 10.01r
Earth
rmoon _ ship = 3.85x105 m
M earth
=
M moon
 rearth _ ship
= 3.85x106 m
= 3.47x106 m
moon _ ship
Example
•
Mp
George weighs
G
i h 100 N on the
th Earth.
E th What
Wh t would
ld his
hi weight
i ht be
b on the
th
surface of another planet that has a planetary mass, which is 3 times
the mass of Earth and mass and a planetary radius, which is 2 times the
radius of Earth?
GM Earth m George
WEarth =
– a) 75 N
R 2Earth
– b) 100 N
G ( 3M earth ) m George
GM planet m George
=
Wplanet =
– c) 125 N
R 2planet
( 2R earth )2
– d) 150 N
3GM earth m George
3 GM earth m George
=
=
2
2
4R earth
4
R earth
3ME
Rp
2Re
WE
100 N
Wp
?
Wplanet =
Hint: This is a ratio problem.
3
3
Wearth = 100N = 75N
4
4
Conceptual question
•
An astronaut
A
t
t is
i on a spacecraft
ft floating
fl ti ““weightlessly”
i htl
l ” in
i orbit.
bit
While her feet are attached securely to the spacecrafte, she
shakes a large iron anvil rapidly back and forth. She reports
back to Earth that
– a) the shaking costs her no effort because the anvil has no
inertial mass in space.
– b) th
the shaking
h ki costs
t h
her some effort
ff t b
butt considerably
id bl lless
than on Earth.
– c) although weightless, the inertial mass of the anvil is the
same as on Earth.
E th
3
Kepler’s laws
•
Johannes Kepler proposed three laws of planetary
motion:
1. All planets moved in elliptical orbits with the
Sun at one of the focal points.
2 Planetary
2.
l
orbits
bi sweep out equall areas in
i equall
times. (This is a consequence of angular
momentum conservation.)
area =
1
1 Δθ
1
Δt ≈ r 2ωΔt
r ⋅ rΔθ = r 2
2
2 Δt
2
2a
Δθ
r
 r 2ω = const .
3 Th
3.
The square off the
th orbital
bit l period
i d off any planet
l t is
i proportional
ti l to
t the
th cube
b
of the average distance from the planet to the Sun. For a circular orbit:
GM star m
/ planet
m
/ planet v 2
GM star v 2
2
Fgrav =
=
m
a
=

=
=
ω
planet c
3
r2
r
r2
2r
GM star  2π 
2
2π

 3
4
π

=
2
ωT = 2π  ω =
 
3
r
 T = 
r
T
T 
GM star 
G
2 

2  4π
a 3

=
T
4. For an elliptical orbit:
 GM star 
4


Conceptual question
•
Two satellites
T
t llit A and
d B off the
th same mass are going
i around
dE
Earth
th
in concentric circular orbits.The distance of satellite B from
Earth’s center is twice that of satellite A.What is the ratio of the
centripetal force acting on B to that acting on A?
– a) 1/8
GM E ms
GM E ms
F
=
F
=
c,A
c.B
– b) 1/4
rA 2
rB 2
– c) 1/2
– d) 1 2
GM E ms
GM E ms
1
=
Fc.B
=
– e)) 1
cB
= F
2
2
RB
2RA
Fc,B/Fc,A
?
( 2rA )
Fc,B
Fc,A
=
4A
4r
4
c,A
A
1
4
Hint: This is a ratio problem.
5
Example
•
One satellite
O
t llit is
i in
i an orbit
bit about
b t Jupiter
J it off radius
di r1 and
d a period
i d off 100
days. If a second satellite is placed in an orbit with 4 times the radius
(i.e. r2=4r1), what is the period for the orbit of the second satellite?
– a)) 200 d
2

 3
4
π
2
T1 = 
r1
– b) 400 d

 GM Jupiter 
– c) 800 d
– d) 1600 d
2
2




4π
3
4π
3
=
4r
T22 = 
r
 GM
 ( 1)
 2
GM


Jupiter 
Jupiter 
r2
4r1
T1
100 d
T2
?
 4π 2  3
= 64 
r1 = 64T 2

1
 GM Jupiter
J it 
d = 800 days
d
Hint:
Hi
t this
thi can also
l
 T22 = 64T12  T2 = 8T1 = 8 ⋅100 days
be solved as a ratio problem
6
Conceptual quiz
•
Suppose E
S
Earth
th h
had
d no atmosphere
t
h
and
dab
ballll were fi
fired
d ffrom
the top of Mt. Everest in a direction tangent to the ground. If the
initial speed were high enough to cause the ball to travel in a
circular trajectory around Earth
Earth, the ball’s
ball s acceleration would
– a) be much less than g (because the ball doesn’t fall to the
ground).
Note: someone in a space
– b) b
be approximately
i t l g.
ship in the same orbit would
– c) depend on the ball’s speed.
feel “weightless” and many
would call this a “zero g”
environment.
“weightlessness”
weightlessness is a
sensation that occurs when
one feels no effects of gravity.
It happens
h
when
h g=00 or when
h
someone is falling freely.
7
Example
•
The Earth
Th
E th orbits
bit the
th sun iin an circular
i l orbit
bit off radius
di rE=1.5x10
1 5 1011 m.
What is the mass of the sun?
2
2



 3
4
π
4
π
2
3
2
T =
r  M sun T = 
r


 GM sun 
 G 
 M sun
 4π 2  r 3
=
 G  T 2
(
)
3
1.50x 10 m
4 ⋅ ( 3.14)
30
2.0x10
kg
=
=
2
-11
11
6.673 x 10 N m² /kg²   365d   24h   3600s 
1y  y   d   d 


2
11
8
Example
•
A uniform steel beam of length 5.00 m has a weight of 4.5x103 N. One end of the beam is
bolted to a vertical wall. The beam is held in a horizontal pposition by
y a cable attached
between the other end of the beam and a point on the wall. The cable makes an angle of 25o
above the horizontal. A load whose weight is 12.0x103 N is hung from the beam at a point
that is 3.5 m from the wall. Find (a) the magnitude of the tension in the supporting cable and
(b) the magnitude of the force exerted on the end of the beam by the bolt that attaches to the
wall.
Choose pivot to be at hinge:
 τ i = 0 = T(5m) sin ( 25o ) − Wload ( 3.5m ) − Wbeam ( 2.5m)
i
(3.5m) + W ⋅ ( 2.5m)
T=W ⋅
( )
load

R
25o

Wbeam
load

Wload

T
beam
( )
(5m) sin 25o
(5m) sin 25o
 T = 2.54x10 4 N

Hinge forces:  Fi = 0; x-comp: R x -Tcos 25o = 0
( )
i
R x = Tcos 25o = 2.29x104 N
( )
y comp: R y − Wload − Wbeam + T sin(25o ) = 0
y-comp:
R y = Wload + Wbeam − T sin(25o ) = 12.0x103 N + 4.5x103 N − 10.7x103 N
R y = 5.8x103 N  R = R 2x + R 2y = 23.6x103 N
5.8x103 N
0
tan(θ R ) =
.25
14
above horizontal
=

θ
=
R
4
2.3x10 N