02-Forces and Uniform Circular Motion
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Transcript 02-Forces and Uniform Circular Motion
Forces and Uniform
Circular Motion
Physics
Unit 2
This Slideshow was developed to accompany the textbook
OpenStax Physics
Available for free at https://openstaxcollege.org/textbooks/college-physics
By OpenStax College and Rice University
2013 edition
Some examples and diagrams are taken from the textbook.
Slides created by
Richard Wright, Andrews Academy
[email protected]
4.1-4.4 Newton’s Laws of Motion
Kinematics
How things move
Dynamics
Why things move
Force
A push or a pull
Is a vector
Unit: Newton (N)
Measured by a spring scale
4.1-4.4 Newton’s Laws of Motion
A body at rest remains at rest, or, if in motion, remains in motion at a constant
velocity unless acted on by a net external force.
Inertia
Property of objects to remain in constant motion or rest.
Mass is a measure of inertia
Watch Eureka! 01
Watch Eureka! 02
4.1-4.4 Newton’s Laws of Motion
Acceleration of a system is directly proportional to and in the same direction as
the net force acting on the system, and inversely proportional to its mass.
𝑭𝑛𝑒𝑡 = 𝑚𝒂
Net force is the vector sum of all the forces.
Watch Eureka! 03
Watch Eureka! 04
Watch Eureka! 05
4.1-4.4 Newton’s Laws of Motion
Free-body diagram
Draw only forces acting
on the object
Represent the forces are
vector arrows
4.1-4.4 Newton’s Laws of Motion
Weight
Mass
Measure of force of gravity
Not a force
𝐹 = 𝑚𝑎
𝑤 = 𝑚𝑔
Measure of inertia or
amount of matter
Unit: N
Unit: kg
Depends on local gravity
Constant
4.1-4.4 Newton’s Laws of Motion
A 1000 kg sailboat’s engine pushes forward with 5000 N thrust.
The wind blows the boat at an angle of 30° starboard with a force
of W. Another boat is pulling it at 30° port with force of F. There is
a resistance force of 2000 N. There is an acceleration of 5.0 m/s2
forward. What are W and F?
F
R
W = F = 1160 N
T
W
4.1-4.4 Newton’s Laws of Motion
Whenever one body exerts a force on a second body, the first body
experiences a force that is equal in magnitude and opposite in
direction to the force that it exerts.
Every force has an equal and opposite reaction force.
You push down on your chair, so the chair pushed back up on you.
4.1-4.4 Newton’s Laws of Motion
A 70-kg diver jumps off a boat on a smooth lake. He applies 75 N
horizontally against the 500-kg boat. What are the accelerations
of the diver and the boat?
ad = 1.07 m/s2
ab = -0.15 m/s2
Day 14 Homework
Force yourself to do these problems
4P1-4a, 6, 8-9, 13, 15-16
Read 4.5-4.6
4CQ21-22
Answers
1) 265 N
2) 9.26 s
3) 13.3 m/s2
4a) 56.0 kg
6) 4.12×105 N
8) 253 m/s2
9) 0.130 m/s2, 0 m/s2
13) 1.5×103 N, 150 kg
15) 2.64×107 N
16) 692 N, 932 N
4.5 Normal, Tension, and Other Forces
4.6 Problem-Solving Strategies
When two objects touch there is often a force
Normal Force
Perpendicular component of the contact force between two
objects
FN
4.5 Normal, Tension, and Other Forces
4.6 Problem-Solving Strategies
Weight pushes down
So the table pushes up
Called Normal force
Newton’s 3rd Law
Normal force doesn’t always = weight
Draw a freebody diagram to find
equation
4.5 Normal, Tension, and Other Forces
4.6 Problem-Solving Strategies
A box is sitting on a ramp
angled at 20°. If the box
weighs 50 N, what is the
normal force on the box?
47 N
FN
20°
20°
w
4.5 Normal, Tension, and Other Forces
4.6 Problem-Solving Strategies
Tension
Pulling force from rope, chain, etc.
Everywhere the rope connects to something, there is an
identical tension
4.5 Normal, Tension, and Other Forces
4.6 Problem-Solving Strategies
Problems-Solving Strategy
1. Identify the principles involved and draw a picture
2. List your knows and Draw a free-body diagram
3. Apply 𝐹𝑛𝑒𝑡 = 𝑚𝑎
4. Check your answer for reasonableness
4.5 Normal, Tension, and Other Forces
4.6 Problem-Solving Strategies
A lady is weighing some bananas in a grocery store when the floor
collapses. If the bananas mass is 2 kg and the floor is accelerating
at -2.25 m/s2, what is the apparent weight (normal force) of the
bananas?
FN = 15.1 N
4.5 Normal, Tension, and Other Forces
4.6 Problem-Solving Strategies
y
A stoplight is suspended by two cables over a
street. Weight of the light is 110 N and the
cables make a 116° angle with each other.
Find the tension in each cable.
104 N
116°
T1
x T2
W
Day 15 Homework
The tension is mounting… I can’t wait to see
what’s next!
4P18-20, 23-27a, 28, 34-35
Read 4.7-4.8
24) 1.03×103 kg
26) 3.14×103 N
27a) 4.41×105 N
Answers
28) 9.89×104 kg, 1.70×104 N
18) 779 N
19)
N,
in vert strand
23) 6.20 m/s2
25) 3.43×103 N
4CQ23, 25-27
7.84×10-4
20) 588 N, 678 N
1.89×10-3
N, 2.41 time tension
34) 1.42×103 N, 539 N
35) 73 N
4.7 Further Applications of Newton’s Laws of Motion
4.8 The Four Basic Forces
More fun problems!
The helicopter in the drawing is
moving horizontally to the right at a
constant velocity. The weight of the
helicopter is 53,800 N. The lift force
L generated by the rotating blade
makes an angle of 21.0° with respect
to the vertical. What is the
magnitude of the lift force?
57600 N
4.7 Further Applications of Newton’s Laws of Motion
4.8 The Four Basic Forces
A 1380-kg car is moving due east with an initial speed of 27.0
m/s. After 8.00 s the car has slowed down to 17.0 m/s. Find the
magnitude and direction of the net force that produces the
deceleration.
4.7 Further Applications of Newton’s Laws of Motion
4.8 The Four Basic Forces
A mountain climber, in the process of crossing between two cliffs
by a rope, pauses to rest. She weighs 535 N. Find the tensions in
the rope to the left and to the right of the mountain climber.
4.7 Further Applications of Newton’s Laws of Motion
4.8 The Four Basic Forces
Four Basic Forces
All forces are made up of only 4 forces
Gravitational - gravity
Electromagnetic – static electricity, magnetism
Weak Nuclear - radioactivity
Strong Nuclear – keeps nucleus of atoms together
4.7 Further Applications of Newton’s Laws of Motion
4.8 The Four Basic Forces
All occur because particles with that force property play catch with a different particle
Electromagnetic uses photons
Scientists are trying to combine all forces together in Grand Unified Theory
Have combined electric, magnetic, weak nuclear
Gravity is the weakest
We feel it because the electromagnetic cancels out over large areas
Nuclear forces are strong but only over short distance
Day 16 Homework
Electromagnetic forces are responsible for
doing homework.
4P40-44, 46, 52-54
Read 5.1
5CQ1, 3
Answers
40) 10.23 m/s2 @ 85.33° above horizontal
41) 376 N
42) 736 N, 194 N
43) -68.5 N
44) 7.43 m/s, 2.97 m
46) 4.20 m/s, 29.4 m/s2, 4.31×103 N
52) 10-13, 10-11
53) 10-38, 10-25, 10-36, gravity has small
nuclear influence
54) 102
5.1 Friction
Normal force – perpendicular to surface
Friction force – parallel to surface, and opposes motion
Comes from rough surface
Not well understood
5.1 Friction
Static Friction
Keeps things from moving.
Cancels out applied force
until the applied force gets
too big.
Depends on force pushing
down and roughness of
surface
5.1 Friction
Static Friction
Depends on force pushing down and roughness of surface
𝑓𝑆 ≤ 𝜇𝑆 𝐹𝑁
More pushing down (FN), more friction
𝜇𝑆 is coefficient of static friction (0.01 to 1.5)
5.1 Friction
Kinetic Friction
Once motion happens
𝑓𝑘 = 𝜇𝑘 𝐹𝑁
𝑓𝑘 is usually less than 𝑓𝑠
5.1 Friction
A car skids to a stop after initially going 30.0
m/s. k = 0.800. How far does the car go
before stopping?
57.3 m
W
fk
FN
5.1 Friction
A 65-kg skier is coasting downhill on a 15° slope. Assuming the
coefficient of friction is that of waxed wood on snow, what is the
skier’s acceleration?
𝐹
1.59 𝑚/𝑠 2 downhill
𝑁
𝑓
15°
𝑤
Day 17 Homework
Don’t let these problems cause friction between
us
5P1-5b, 8-11, 17-19
Read 5.2
8) work
10) 1.83 m/s2
11) 0.737 m/s2, 5.71°
Answers
17) 272 N, 512 N, 0.268
1) 5.00 N
2)
5a) 4.90 m/s2, 5b) will not slip
9) work
5CQ6-8
1.00×103
4) 588 N, 1.96 m/s2
N, 30.0 N
3) 10 N, 97.0 N
18) 51.0 N, 0.720 m/s2
19) 46.5 N, 0.655 m/s2
5.2 Drag Forces
Drag
For large objects
Resistive force from moving
through a fluid
𝐹𝐷 = 2 𝐶𝜌𝐴𝑣 2
Size depends on area, speed, and
properties of the fluid
𝐶 = drag coefficient
1
𝜌 = density of fluid
𝐴 = area of object
𝑣 = speed of object relative to the
fluid
5.2 Drag Forces
The drag coefficient of a car is measured in a wind tunnel. If the
wind is blowing at 25 𝑚/𝑠, the area is 24 𝑚2 , the density of air is
1.21 𝑘𝑔/𝑚3 , and the drag force is measured at 3630 𝑁. What is
the drag coefficient of the car?
5.2 Drag Forces
Terminal Velocity
Falling objects will
accelerate until the
downward force of gravity
= drag force
1
2
𝑚𝑔 = 𝐶𝜌𝐴𝑣 2
𝑣 =
2𝑚𝑔
𝜌𝐶𝐴
5.2 Drag Forces
Find the terminal velocity of a falling mouse in air (𝐴 = 0.004 𝑚2 , 𝑚 = 0.02 𝑘𝑔, 𝐶 = 0.5) and
a falling human falling flat (𝐴 = 0.7 𝑚2 , 𝑚 = 85 𝑘𝑔, 𝐶 = 1.0). The density of air is 1.21 𝑘𝑔/𝑚3 .
Mouse: 12.7 m/s
Human: 44.4 m/s
5.2 Drag Forces
Terminal Velocity of very small objects
(like pollen)
Stoke’s Law
𝐹𝑆 = 6𝜋𝑟𝜂𝑣
𝑟 = radius of object
𝜂 = viscosity of fluid (kg/m·s) Table 12.1
𝑣 = velocity of object
What is terminal velocity for a Ragweed
pollen grain? 𝑑 = 17 𝜇𝑚, density of pollen
is 1320 𝑘𝑔/𝑚3 , and viscosity of air is 1.81
× 10−5 𝑘𝑔/𝑚 ⋅ 𝑠.
Day 18 Homework
Even though homework can be a
drag, it’s good for you
5P20-23, 25-28
Read 5.3
5CQ9, 11, 14-16
21) Assume A=0.14 m2 49.2 s
22) 25.1 m/s, 9.9 m/s
23) 44.8 N, 91.5 N; 357 N, 729 N
25) 313 m/s, 9.8 m/s
26) work
Answers
20) 115 m/s, 414 km/h
27) 𝜂 for water 1.005 × 10−3 ,
𝑚⋅𝑠
2.38×10-6 m/s
28) 0.76 kg/m·s
𝑘𝑔
5.3 Elasticity: Stress and Strain
Forces that deform
Change shape
For small deformations
Object returns to original shape
Deformation proportional to force
Hooke’s Law
𝐹 = 𝑘Δ𝐿
5.3 Elasticity: Stress and Strain
Hooke’s Law
The amount of stretching (𝑘) depends on several things
Tension and Compression
𝑘 depends
applied force
cross-sectional area
property called elastic modulus or Young’s modulus
5.3 Elasticity: Stress and Strain
For Tension and Compression Hooke’s Law becomes
𝐹
Δ𝐿 =
𝐿
𝑌𝐴 0
Where
Δ𝐿 is change in length
𝐹 is applied force
𝑌 is Young’s modulus (see Table 5.3)
𝐴 is cross-sectional area
𝐿0 is the original length
5.3 Elasticity: Stress and Strain
Rearranging produces
𝐹
Δ𝐿
=𝑌
𝐴
𝐿0
𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑌 × 𝑠𝑡𝑟𝑎𝑖𝑛
5.3 Elasticity: Stress and Strain
What is the Young’s Modulus for nylon rope? It’s radius is 0.005 m
and a 1 meter length stretches 0.0016 m with 50 kg is hung on it’s
end.
5.3 Elasticity: Stress and Strain
Sideways Stress: Shear Modulus
Force perpendicular to length of material
𝐹
Δ𝑥 =
𝐿0
𝑆𝐴
𝑆 is shear modulus
5.3 Elasticity: Stress and Strain
Changes in Volume: Bulk Modulus
Deforming material in all directions
Gases very easy
Solids and liquids more difficult
𝐹
Δ𝑉 =
𝑉
𝐵𝐴 0
𝐵 is bulk modulus
Depends on molecular arrangement within
material and overcome electromagnetic forces
5.3 Elasticity: Stress and Strain
If it didn’t break, how much would the end of a 2 m long wooden
pole, 3 cm in diameter, bend if a 80 kg person hung from the end
of it?
Day 19 Homework
Let these problems stress and strain your mind.
33) 9 𝑐𝑚
5P30-39, 41-42
34) 0.57 𝑚𝑚
Read 6.1-6.2
35) 8.59 𝑚𝑚
6CQ1-2
36) 706 𝑁
37) 1.49 × 10−7 𝑚
Answers
38) 3.34 × 10−6 𝑚
30) 4.5 × 10−5 𝑚
39) 3.99 × 10−7 𝑚, 9.67 × 10−8 𝑚
31) 1 𝑚𝑚
41) 4 × 102 𝑁/𝑐𝑚2
32) 0.190 𝑚𝑚
42) 2.0 × 104 𝑁/𝑐𝑚2
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
Newton’s Laws of motion primarily
relate to straight-line motion.
Uniform Circular Motion
Motion in circle with constant
speed
Rotation Angle (Δ𝜃)
Angle through which an object
rotates
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
Arc Length is the distance around part of
circle
Δ𝒔
Δ𝜃 =
𝒓
Angle Units:
Revolutions: 1 circle = 1 rev
Degrees: 1 circle = 360°
Radians: 1 circle = 2𝜋
Arc Length formula must use radians and
angle unit
2𝜋 = 360° = 1 𝑟𝑒𝑣
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
Convert 60° to radians
Angular Velocity
How fast an object rotates
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
Angular Velocity (𝜔)
𝜔=
Δ𝜃
Δ𝑡
𝑣=
Δ𝑠
Δ𝑡
Unit: rad/s
CCW +, CW –
Δ𝑠
→ Δs = rΔθ
𝑟
𝑟Δ𝜃
𝑣=
= 𝑟𝜔
Δ𝑡
Δ𝜃 =
A CD rotates 320 times in 2.4 s. What is its
angular velocity in rad/s? What is the linear
velocity of a point 5 cm from the center?
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
Uniform Circular Motion
Speed is constant
Velocity is not constant
Velocity is always changing
This acceleration is “centripetal” acceleration
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
Object moves in circular
path
At time t0 it is at point O
with a velocity tangent to
the circle
At time t, it is at point P
with a velocity tangent to
the circle
The radius has moved
through angle
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
Draw the two velocity vectors so
that they have the same tails.
The vector connecting the heads
is v
Draw the triangle made by the
change in position and you get
the triangle in (b)
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
Since the triangles have the same
angle are isosceles, they are
similar.
Δ𝑣 𝑣Δ𝑡
=
𝑣
𝑟
Δ𝑣 𝑣 2
=
Δ𝑡
𝑟
𝑣2
𝑎𝐶 =
= 𝑟𝜔2
𝑟
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
At any given moment
v is pointing tangent to the circle
ac is pointing towards the center of the circle
If the object suddenly broke from circular motion would travel in line tangent
to circle
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
Two identical cars are going around two corners at 30 m/s. Each
car can handle up to 1 g. The radius of the first curve is 50m and
the radius of the second is 100 m. Do either of the cars make the
curve? (hint find the ac)
50 m
100 m
Day 20 Homework
Rotating too fast can make you sick, but these
problems won’t.
6P2-6, 10-11, 13, 16, 18-19
Read 6.3, 6.4
6CQ4-6, 9, 11, 14,
5) 117 rad/s
6) 39.0 m/s
10) 12.9 rev/min
11) 2.5 m/s2
13) 126 rad/s, 145 m/s, 1.82 × 104 m/s 2 , 1.85
× 103 g
Answers
16) 31.4 rad/s, 118 m/s2, 384 m/s2, comments
2) 0.1 rps, 0.63 rad/s
18) 0.524 km/s, 29.7 km/s
3) 5 × 107 rotations
19) 0.313 rad/s
4) 86400 s, 7.3 × 10−5 rad/s, 470 m/s
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
Newton’s 2nd Law
Whenever there is acceleration there is a force to cause it
F = ma
Fc = mac
𝑚𝑣 2
𝐹𝐶 =
= 𝑚𝑟𝜔2
𝑟
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
Centripetal Force is not a new, separate force created by nature!
Some other force creates centripetal force
Swinging something from a string tension
Satellite in orbit gravity
Car going around curve friction
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
A 1.25-kg toy airplane is attached to a string and swung in a circle
with radius = 0.50 m. What was the centripetal force for a speed
of 20 m/s? What provides the Fc?
Fc = 1000 N
Tension in the string
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
What affects Fc more: a change in mass, a change in radius, or a
change in speed?
A change in speed since it is squared and the others aren’t.
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
When a car travels around an unbanked curve, static friction
provides the centripetal force.
By banking a curve, this reliance on friction can be eliminated for
a given speed.
Derivation of Banked Curves
A car travels around a friction free banked curve
Normal Force is perpendicular to road
x component (towards center of circle) gives centripetal force
𝑚𝑣 2
𝐹𝑁 sin 𝜃 =
𝑟
y component (up) cancels the weight of the car
𝐹𝑁 cos 𝜃 = 𝑚𝑔
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
Divide the x by the y
𝑚𝑣 2
𝐹𝑁 sin 𝜃 =
𝑟
𝐹𝑁 cos 𝜃 = 𝑚𝑔
Gives
𝑣2
tan 𝜃 =
𝑟𝑔
Notice mass is not involved
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
In the Daytona International Speedway, the corner is banked at 31 and
r = 316 m. What is the speed that this corner was designed for?
v = 43 m/s = 96 mph
Cars go 195 mph around the curve. How?
Friction provides the rest of the centripetal force
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
Inertial reference frame – nonaccelerating
Non-inertial reference frames produce seemingly magic forces
Stomp on the brakes, everything flies forward
Car was reference frame, it was accelerating
Fictitious force pushed stuff forward
Really just Newton’s first law
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
Remember the good old days when cars were big, the seats
were vinyl bench seats, and there were no seat belts? Well
when a guy would take a girl out on a date and he wanted to get
cozy, he would put his arm on the back of the seat then make a
right hand turn. The car and the guy would turn since the tires
and steering wheel provided the centripetal force. The friction
between the seat and the girl was not enough, so the girl would
continue in a straight path while the car turned underneath her.
She would end up in the guy’s arms.
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
According to Newton’s first law, objects will travel in straight line
If the reference frame is rotating, it will appear to move in an arc
Show Coriolis video
If rotation is counterclockwise, the path will bend to the right
If rotation is cw, the path will bend left
6.3 Centripetal Force
6.4 Fictitious Forces: The Coriolis Effect
The earth rotates
When viewed from north pole, it rotates counterclockwise
Storms, ocean current, etc in northern hemisphere turn to
right
Day 21 Homework
There is a real force to make you do these
problems.
6P23-27, 29, 30a
Read 6.5
6CQ19-21
Answers
23) 483 N, 17.4 N, 2.24 times, 0.0807 times
24) 3.9 × 103 N
25) 4.14°
26) 18.9 m/s
27) 24.6 m, 36.6 m/s2, 3.73 g
29) 2.56 rad/s, 5.71°
30a) 16.2 m/s
6.5 Newton’s Universal Law of Gravitation
Why does g = 9.80 m/s2?
It’s related to Newton’s Law of Universal Gravitation
Watch Eureka 6
6.5 Newton’s Universal Law of Gravitation
Every particle in the universe exerts a force on every other particle
𝑚𝑀
𝐹𝑔 = 𝐺 2
𝑟
G = 6.673 x 10-11 N m2/kg2
m and M are the masses of the particles
r = distance between the particles
6.5 Newton’s Universal Law of Gravitation
For bodies
Using calculus – apply universal gravitation for bodies
Estimate (quite precisely)
Assume bodies are particles based at their center of mass
For spheres assume they are particles located at the center
6.5 Newton’s Universal Law of Gravitation
What is the gravitational attraction between a 75-kg boy (165 lbs) and
the 50-kg girl (110 lbs) seated 1 m away in the next desk?
Fg = 2.5 x 10-7 N
= 2.6 x 10-8 lbs of force
6.5 Newton’s Universal Law of Gravitation
Weight is Gravitational Force the earth exerts on an object
Unit: Newton (N)
Remember!!!
Weight is a Force
Watch Eureka 7
6.5 Newton’s Universal Law of Gravitation
Weight
𝑚𝑀
𝑊=𝐺 2
𝑟
𝑊 = 𝑚𝑔
𝑀
𝑔=𝐺 2
𝑟
r is usually RE
So g = 9.80 m/s2
6.5 Newton’s Universal Law of Gravitation
The gravitational pull from the moon and sun causes tides
Water is pulled in the direction of the moon and sun
Gravitational pull from satellites causes the main body to move slightly
Moon causes earth to move
Planets cause sun/star to move
6.5 Newton’s Universal Law of Gravitation
Astronauts in the space shuttles and
international space station seem to float
They appear weightless
They are really falling
Acceleration is about g towards earth
Watch Mir Collision Video (8 min)
(MIR_Space_Station_collision.mp4)
… they were finally able to close and repressurize
the hatch. Several months later a new team of
cosmonauts returned and found the hatch
impossible to permanently repair. Instead they
attached a set of clamps to secure it in place.
It is this set of clamps that Linenger and Tsibliyev
are staring at uneasily seven years later. To his relief,
the commander opens the hatch Without incident
and crawls outside onto an adjoining ladder just
after nine o’clock. Linenger begins to follow. Outside
the Sun is rising. The Russians have planned the EVA
at a sunrise so as to get the longest period of light.
But because of that, Linenger’s first view of space is
straight into the blazing Sun. “The first view I got
was just blinding rays coming at me,” Linenger told
his postflight debriefing session. “Even with my gold
visor down, it was just blinding. [I] was basically
unable to see for the first three or four minutes
going out the hatch.”
The situation only gets worse once his eyes
clear. Exiting the airlock, Linenger climbs out
onto a horizontal ladder that stretches out
along the side of the module into the darkness.
Glancing about, trying in vain to get his
bearings, he is suddenly hit by an
overwhelming sense that he is falling, as if
from a cliff. Clamping his tethers onto the
handrail, he fights back a wave of panic and
tightens his grip on the ladder. But he still
can’t shake the feeling that he is plummeting
through space at eighteen thousand miles an
hour. His mind races.
You’re okay. You’re okay. You’re not going to
fall. The bottom is way far away.
And now a second, even more intense
feeling washes over him: He’s not just
plunging off a cliff. The entire cliff is crumbling
away. “It wasn’t just me falling, but everything
was falling, which gave [me] even a more
unsettling feeling,” Linenger told his
debriefers. “So, it was like you had to
overcome forty years or whatever of life
experiences that [you] don’t let go when
everything falls. It was a very strong, almost
overwhelming sensation that you just had to
control. And I was able to control it, and I was
glad I was able to control it. But I could see
where it could have put me over the edge.”
The disorientation is paralyzing. There is no
up, no down, no side. There is only threedimensional space. It is an entirely different
sensation from spacewalking on the shuttle,
where the astronauts are surrounded on three
sides by a cargo bay. And it feels nothing—
nothing—like the Star City pool. Linenger is an
ant on the side of a falling apple, hurtling
through space at eighteen thousand miles an
hour, acutely aware what will happen if his
Russian-made tethers break. As he clings to
the thin railing, he tries not to think about the
handrail on Kvant that came apart during a
cosmonaut’s spacewalk in the early days of
Mir. Loose bolts, the Russians said.
Loose bolts.
Day 22 Homework
Just like gravity, these problems are
attractive.
6P33-37, 41
Read 6.6
6CQ23
Answers
33) 5.979 × 1024 kg
34) 3.33 × 10−5 m/s2, 5.93 × 10−3
m/s2, 178
35) 1.62 m/s2, 3.75 m/s2
36) 274 m/s2, 28.0 times
37) 3.42 × 10−5 m/s2, 3.34 × 10−5
m/s2
41) 1.66 × 10−10 m/s2, 2.17 × 105
m/s
6.6 Satellites and Kepler’s Laws
After studying motion of planets, Kepler came up with his laws of
planetary motion
Newton then proved them all using his Universal Law of Gravitation
Assumptions:
A small mass, m, orbits much larger mass, M, so we can use M as an
approximate inertia reference frame
The system is isolated
6.6 Satellites and Kepler’s Laws
1. The orbit of each planet
about the Sun is an ellipse
with the sun at one focus.
6.6 Satellites and Kepler’s Laws
2. Each planet moves so that an
imaginary line drawn from
the sun to the planet sweeps
out equal areas in equal
times.
6.6 Satellites and Kepler’s Laws
3. The ratio of the squares of the
periods of any two planets
about the sun is equal to the
ratio of the cubes of their
average distances from the sun.
2
𝑇1
𝑇22
=
3
𝑟1
𝑟23
These laws work for all satellites
For circular orbits
𝑇2 =
4𝜋2 3
𝑟
𝐺𝑀
𝑇2
𝑟3
=
4𝜋2
𝐺𝑀
Table 6.2 gives data about the
planets and moons
6.6 Satellites and Kepler’s Laws
Use the data of Mars in Table 6.2 to find the mass of sun.
Mars, 𝑟 = 2.279 × 108 km, 𝑇 = 1.881 y
Day 23 Homework
Draw an ellipse around your
answers to these problems
44) 1.98 × 1030 kg
6P43-47, 48a-b
46) 316
Answers
47) 3 × 108 y, 2 × 1013 solar
masses
43) 4.23 × 104 km
48) 7401 m/s, 1.05 × 104 m/s
45) 1.89 × 1027 kg