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 rwright@andrews.edu

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


 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



 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.

 𝑭 𝑛𝑒𝑡

= 𝑚𝒂

 Net force is the vector sum of all the forces.





 Free-body diagram

 Draw only forces acting on the object

 Represent the forces are vector arrows


 Weight

 Measure of force of gravity

 𝐹 = 𝑚𝑎

 𝑤 = 𝑚𝑔

 Unit: N

 Depends on local gravity

 Mass

 Not a force

 Measure of inertia or amount of matter

 Unit: kg

 Constant


 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/s 2 forward. What are W and F?

 W = F = 1160 N

R

W

F

T


 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.


 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?

 a d

= 1.07 m/s 2

 a b

= -0.15 m/s 2

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/s 2

 4a) 56.0 kg

 6) 4.12×10 5 N

 8) 253 m/s 2

 9) 0.130 m/s 2 , 0 m/s 2

 13) 1.5×10 3 N, 150 kg

 15) 2.64×10 7 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

F

N



 Weight pushes down

 So the table pushes up

 Called Normal force

 Newton’s 3 rd Law

 Normal force doesn’t always = weight

 Draw a freebody diagram to find equation



 A box is sitting on a ramp angled at 20°. If the box weighs 50 N, what is the normal force on the box?

F

N

20°

 47 N

20° w



 Tension

 Pulling force from rope, chain, etc.

 Everywhere the rope connects to something, there is an identical tension



 Problems-Solving Strategy

1.

Identify the principles involved and draw a picture

2.

List your knows and Draw a free-body diagram

3.

Apply 𝐹 𝑛𝑒𝑡

= 𝑚𝑎

4.

Check your answer for reasonableness



 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/s 2 , what is the apparent weight (normal force) of the bananas?

 F

N

= 15.1 N


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.

T

1

116

° x

T

2

 104 N 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

 4CQ23, 25-27

 Answers

 18) 779 N

 19) 7.84×10 -4 N, 1.89×10 -3 N, 2.41 time tension in vert strand

 20) 588 N, 678 N

 23) 6.20 m/s 2

 24) 1.03×10 3 kg

 25) 3.43×10 3 N

 26) 3.14×10 3 N

 27a) 4.41×10 5 N

 28) 9.89×10 4 kg, 1.70×10 4 N

 34) 1.42×10 3 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



 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.



 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.



 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



 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/s 2 @ 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/s 2 , 4.31×10 3 N

 52) 10 -13 , 10 -11

 53) 10 -38 , 10 -25 , 10 -36 , gravity has small nuclear influence

 54) 10 2

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



𝑓

𝑆

≤ 𝜇

𝑆

𝐹

𝑁

 More pushing down (F

N

), more friction

 𝜇

𝑆 is coefficient of static friction (0.01 to 1.5)

5.1 Friction

 Kinetic Friction

 Once motion happens

 𝑓 𝑘

= 𝜇 𝑘

𝐹

𝑁

 𝑓 𝑘 is usually less than 𝑓 𝑠

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 f k

W

F

N

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

𝑁

1.59 𝑚/𝑠 2 downhill 𝑓

15° 𝑤

Day 17 Homework

 Don’t let these problems cause friction between us

 5P1-5b, 8-11, 17-19

 Read 5.2

 5CQ6-8

 Answers

 1) 5.00 N

 2) 1.00×10 3 N, 30.0 N

 3) 10 N, 97.0 N

 4) 588 N, 1.96 m/s 2

 5a) 4.90 m/s 2 , 5b) will not slip

 8) work

 9) work

 10) 1.83 m/s 2

 11) 0.737 m/s 2 , 5.71°

 17) 272 N, 512 N, 0.268

 18) 51.0 N, 0.720 m/s 2

 19) 46.5 N, 0.655 m/s 2

5.2 Drag Forces

 Drag

 Resistive force from moving through a fluid

 Size depends on area, speed, and properties of the fluid

 For large objects

 𝐹

𝐷

=

1

2

𝐶𝜌𝐴𝑣

2

 𝐶 = drag coefficient

 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 𝑚 2 , the density of air is

1.21 𝑘𝑔/𝑚 3 , 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

 𝑚𝑔 =

1

2

𝐶𝜌𝐴𝑣 2

 𝑣 =

2𝑚𝑔 𝜌𝐶𝐴

5.2 Drag Forces

 Find the terminal velocity of a falling mouse in air ( 𝐴 = 0.004 𝑚 2 , 𝑚 = 0.02 𝑘𝑔, 𝐶 = 0.5

) and a falling human falling flat ( 𝐴 = 0.7 𝑚 2 , 𝑚 = 85 𝑘𝑔, 𝐶 = 1.0

). The density of air is 1.21 𝑘𝑔/𝑚 3 .

 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

 𝐹

𝑆

= 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/𝑚 3 , 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

 Answers

 20) 115 m/s, 414 km/h

 21) Assume A=0.14 m 2 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

 27) 𝜂 for water 1.005 × 10 −3 𝑘𝑔

2.38×10 -6 m/s 𝑚⋅𝑠

,

 28) 0.76 kg/m·s

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


 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


 For Tension and Compression Hooke’s Law becomes

 Where

 Δ𝐿 is change in length

 𝐹 is applied force

 𝑌 is Young’s modulus (see Table 5.3)

 𝐴 is cross-sectional area

Δ𝐿 =

𝐹

𝑌𝐴

𝐿

0

 𝐿

0 is the original length


 Rearranging produces

𝐹

𝐴

= 𝑌

Δ𝐿

𝐿

0 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑌 × 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.


 Sideways Stress: Shear Modulus

 Force perpendicular to length of material

Δ𝑥 =

𝐹

𝑆𝐴

𝐿

0

 𝑆 is shear modulus


 Changes in Volume: Bulk Modulus

 Deforming material in all directions

 Gases very easy

 Solids and liquids more difficult

Δ𝑉 =

𝐹

𝐵𝐴

𝑉

0

 𝐵 is bulk modulus

 Depends on molecular arrangement within material and overcome electromagnetic forces


 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.

 5P30-39, 41-42

 Read 6.1-6.2

 6CQ1-2

 Answers

 30) 4.5 × 10 −5 𝑚

 31) 1 𝑚𝑚

 32) 0.190 𝑚𝑚

 33) 9 𝑐𝑚

 34) 0.57 𝑚𝑚

 35) 8.59 𝑚𝑚

 36) 706 𝑁

 37) 1.49 × 10

−7 𝑚

 38) 3.34 × 10

−6 𝑚

 39) 3.99 × 10

−7 𝑚 , 9.67 × 10

−8 𝑚

 41) 4 × 10

2

𝑁/𝑐𝑚

2

 42) 2.0 × 10

4

𝑁/𝑐𝑚

2

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



 Arc Length is the distance around part of circle

Δ𝜃 =

Δ𝒔 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



 Convert 60° to radians  Angular Velocity

 How fast an object rotates



 Angular Velocity ( 𝜔 ) 𝜔 =

Δ𝜃

Δ𝑡

 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?

 Unit: rad/s

 CCW +, CW –

Δ𝑠 𝑣 =

Δ𝑡

Δ𝑠

Δ𝜃 = 𝑣 =

→ Δs = rΔθ 𝑟 𝑟Δ𝜃

Δ𝑡

= 𝑟𝜔



 Uniform Circular Motion

 Speed is constant

 Velocity is not constant

 Velocity is always changing

 This acceleration is “centripetal” acceleration



 Object moves in circular path

 At time t the circle

0 it is at point O with a velocity tangent to

 At time t, it is at point P with a velocity tangent to the circle

 The radius has moved through angle 



 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)



 Since the triangles have the same angle are isosceles, they are similar.

𝑎

𝐶

Δ𝑣 𝑣Δ𝑡

= 𝑣 𝑟

Δ𝑣

=

Δ𝑡 𝑣 2

= 𝑟 𝑣 𝑟

2

= 𝑟𝜔 2



 At any given moment

 v is pointing tangent to the circle

 a c is pointing towards the center of the circle

 If the object suddenly broke from circular motion would travel in line tangent to circle



 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 a c

)

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,

 Answers

 2) 0.1 rps, 0.63 rad/s

 3) 5 × 10 7 rotations

 4) 86400 s, 7.3 × 10

−5 rad/s, 470 m/s

 5) 117 rad/s

 6) 39.0 m/s

 10) 12.9 rev/min

 11) 2.5 m/s 2

 13) 126 rad/s, 145 m/s, 1.82 × 10

4

10 3 g m/s

2 , 1.85 ×

 16) 31.4 rad/s, 118 m/s 2 , 384 m/s 2 , comments

 18) 0.524 km/s, 29.7 km/s

 19) 0.313 rad/s

6.3 Centripetal Force

6.4 Fictitious Forces: The Coriolis Effect

 Newton’s 2 nd Law

 Whenever there is acceleration there is a force to cause it

 F = ma

 F c

= ma c

𝐹

𝐶

= 𝑚𝑣

2

𝑟

= 𝑚𝑟𝜔

2



 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



 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 F c

?

 F c

= 1000 N

 Tension in the string



 What affects F c 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.



 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

𝐹

𝑁 sin 𝜃 = 𝑚𝑣 2 𝑟

 y component (up) cancels the weight of the car

𝐹

𝑁 cos 𝜃 = 𝑚𝑔



 Divide the x by the y

𝐹

𝑁

𝐹

𝑁 𝑚𝑣 2 sin 𝜃 = 𝑟 cos 𝜃 = 𝑚𝑔

 Gives 𝑣 2 tan 𝜃 = 𝑟𝑔

 Notice mass is not involved



 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



 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



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.



 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



 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 × 10 3 N

 25) 4.14°

 26) 18.9 m/s

 27) 24.6 m, 36.6 m/s 2 , 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/s 2 ?

 It’s related to Newton’s Law of Universal Gravitation

 Watch Eureka 6


 Every particle in the universe exerts a force on every other particle 𝑚𝑀

𝐹 𝑔

= 𝐺 𝑟 2

 G = 6.673 x 10 -11 N m 2 /kg 2

 m and M are the masses of the particles

 r = distance between the particles


 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


 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?

 F g

= 2.5 x 10 -7 N

 = 2.6 x 10 -8 lbs of force


 Weight is Gravitational Force the earth exerts on an object

 Unit: Newton (N)

 Remember!!!

 Weight is a Force

 Watch Eureka 7


 Weight

𝑊 = 𝐺 𝑚𝑀 𝑟 2

𝑊 = 𝑚𝑔

𝑀 𝑔 = 𝐺 𝑟 2

 r is usually R

E

 So g = 9.80 m/s 2


 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


 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 × 10 24 kg

 34) 3.33 × 10 −5 m/s 2 , 5.93 × 10 −3 m/s 2 , 178

 35) 1.62 m/s 2 , 3.75 m/s 2

 36) 274 m/s 2 , 28.0 times

 37) 3.42 × 10 −5 m/s 2 , 3.34 × 10 −5 m/s 2

 41) 1.66 × 10 −10 m/s 2 , 2.17 × 10 5 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


1.

The orbit of each planet about the Sun is an ellipse with the sun at one focus.


2.

Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times.


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.

𝑇

1

2

𝑇

2

2

= 𝑟 𝑟

1

3

2

3

 These laws work for all satellites

 For circular orbits

𝑇 2 =

4𝜋

2 𝑟 3

𝐺𝑀



𝑇

2 𝑟 3

=

4𝜋

2

𝐺𝑀

 Table 6.2 gives data about the planets and moons


 Use the data of Mars in Table 6.2 to find the mass of sun.

Mars, 𝑟 = 2.279 × 10 8 km, 𝑇 = 1.881

y

Day 23 Homework

 Draw an ellipse around your answers to these problems

 6P43-47, 48a-b

 Answers

 43) 4.23 × 10 4 km

 44) 1.98 × 10 30 kg

 45) 1.89 × 10 27 kg

 46) 316

 47) 3 × 10 8 y, 2 × 10 13 solar masses

 48) 7401 m/s, 1.05 × 10 4 m/s

02-Forces and Uniform Circular Motion

Forces and Uniform

Circular Motion

𝑓

≤ 𝜇

𝐹

6.1 Rotation Angle and Angular Velocity

6.2 Centripetal Acceleration

𝐹

= 𝑚𝑣

𝑟

= 𝑚𝑟𝜔

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02-Forces and Uniform Circular Motion

Forces and Uniform

Circular Motion

𝑓

≤ 𝜇

𝐹

6.1 Rotation Angle and Angular Velocity

6.2 Centripetal Acceleration

𝐹

= 𝑚𝑣

𝑟

= 𝑚𝑟𝜔

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