Download 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
 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
 Answers
 33) 5.979 × 1024 kg
 41) 1.66 × 10−10 m/s2, 2.17 × 105
m/s
 37) 3.42 × 10−5 m/s2, 3.34 × 10−5
m/s2
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