i estimate my grade on exam 2 to be:

Physics 1710Physics 1710—Warm-up Quiz—Warm-up Quiz

I estimate my grade on exam 2 to be:I estimate my grade on exam 2 to be:

A B C D F

17% 17% 17%17%17%17%

A.A. A A B.B. BBC.C. C C D.D. DDE.E. N/AN/AF.F. FF

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Center of Gravity (comparison)Center of Gravity (comparison)

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

BallBallFFg g = - mg= - mg

MoonMoonFFg g = - GmM/r= - GmM/r22

CM

CMCG

No Talking!No Talking!

Think!Think!

Confer!Confer!

Peer Instruction Peer Instruction TimeTime

What is the mass of the earth? What is the mass of the earth? How do you know?How do you know?

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

Peer Instruction Peer Instruction TimeTime

What is the mass of the earth? What is the mass of the earth? How do you know? g?How do you know? g?

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

Fg = - mg Fg = - mg′

11′′ Lecture Lecture

•The moduli of elasticity (Y, E, B) characterizes the The moduli of elasticity (Y, E, B) characterizes the stress-strain relation:stress-strain relation:

• stress= modulus • strainstress= modulus • strain; ; σ = Y εσ = Y ε

•The force of attraction between two bodies with The force of attraction between two bodies with mass M and m, respectively, is proportional to the mass M and m, respectively, is proportional to the productproduct of their of their massesmasses and and inversely proportionalinversely proportional to the distance between their centers to the distance between their centers squaredsquared..

• FFgg = - = - řřG Mm/d G Mm/d 22

•G = 6.673 x 10 G = 6.673 x 10 –11–11 N m N m22 /kg /kg22 ~ 2/3 x 10 ~ 2/3 x 10 –10–10 N m N m22 /kg/kg22

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ElasticityElasticityDefinitions:Definitions:

• Stress Stress σσ : the deforming force per unit : the deforming force per unit area.area.

• Strain Strain εε : the unit deformation. : the unit deformation.

Stress = modulus x strainStress = modulus x strainσ = σ = F/A = Y F/A = Y εε

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ElasticityElasticity

• Stress Stress σσ – Strain – Strain εε “Curve” “Curve”

Strain Strain ε = ΔL/L (%)ε = ΔL/L (%)

Str

ess

S

tress

σ

σ

(N

/m(N

/m22))

σ = σ = Y Y εε

Elastic limit

Failure

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

ElasticityElasticity

• Stress Stress σ σ : the deforming force per unit : the deforming force per unit area.area.

• Strain Strain ε ε : the unit deformation.: the unit deformation.

Tensile/Compressive Stress Tensile/Compressive Stress Young’s ModulusYoung’s Modulus EE

Stress = modulus x strainStress = modulus x strainσ = σ = F/A = E F/A = E ε = E ε = E ΔL/LΔL/L

σ σ = E = E ε ε ΔLΔLLL

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

ElasticityElasticity

• Stress Stress σ σ : the deforming force per unit : the deforming force per unit area.area.

• Strain Strain ε ε : the unit deformation.: the unit deformation.

Shear Modulus GShear Modulus GStress = modulus x strainStress = modulus x strain

σ = σ = F/A = G F/A = G ε = G ε = G Δx/hΔx/h

σσΔxΔxLL

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

h

ElasticityElasticity

• Stress Stress σ σ : the deforming force per unit : the deforming force per unit area.area.

• Strain Strain ε ε : the unit deformation.: the unit deformation.

Hydraulic Stress: Hydraulic Stress: Bulk Modulus BBulk Modulus B

Stress = modulus x strainStress = modulus x strainσ = σ = F/A = p = B F/A = p = B ε = B ε = B ΔV/VΔV/V

ppVV

ΔVΔV

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

Which spring should have the larger spring constant, a short spring (0.1 m long) or a longer spring (0.3 m long)? F = -k x

A B C D

41%

2%

28%29%

A.A. k larger for shorter spring. k larger for shorter spring. B.B. k larger for longer spring. k larger for longer spring.

C.C. k the same for both springs.k the same for both springs.

D.D. None of the above.None of the above.

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Solution:Solution:

F = -k x?

σ = σ = F/A = E F/A = E ε = E ε = E ΔL/LΔL/LF = - A E x/ L

k = A E/L

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What about a thicker (bigger A) wire?

Isaac Newton’s Isaac Newton’s Universal Universal Law of Law of GravitationGravitation

F = - G M m/ d F = - G M m/ d 22

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

d moon

d d appleapple

Kepler’s Laws:Kepler’s Laws:

•The orbits are ellipses. (Contrary to Aristotle and The orbits are ellipses. (Contrary to Aristotle and Ptolemy.)Ptolemy.)

A central force: A central force: F ∝ 1/ rF ∝ 1/ r 2 2 or r or r 2 2

• The areal velocity is a constant.The areal velocity is a constant.Angular momentum is conserved:Angular momentum is conserved:

½ v r ½ v r ∆t = constant implies that∆t = constant implies thatrmv = rmv = L = constant.L = constant.

• TT 2 2 ∝ r∝ r 3 3 implies implies F ∝ 1/ rF ∝ 1/ r 2 2, only., only.

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⊙

T T 22 = r r33

How did Newton figure out UL of G?How did Newton figure out UL of G?

• Fact:Fact: a moon circling a planet has an a moon circling a planet has an acceleration of acceleration of a = v a = v 22 /r /r

• Fact:Fact: a = F/m. a = F/m.

• Fact:Fact: Kepler had found that the square of the Kepler had found that the square of the period period T was proportional to the cube of T was proportional to the cube of the radius of the radius of the orbit r :the orbit r :

T T 22 = k r = k r 33 . .Thus:Thus:

v = 2v = 2ππ r / T r / T

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AndAnd

T T 22= (2= (2ππr) r) 22 /(F r /m) = k r /(F r /m) = k r 33

Thus:Thus:

F = (2F = (2ππ)) 2 2 m/(k m/(k r r 22 ) )

An “inverse square law,” with k = 1/ [(2An “inverse square law,” with k = 1/ [(2ππ)) 2 2G M]G M]

F = G Mm/ r F = G Mm/ r 22 , ,

But what value is G? But what value is G?

Physics 1710Physics 1710—C—Chapter 12 Apps: Gravityhapter 12 Apps: Gravity

Isaac Newton’s Isaac Newton’s Universal Universal Law of Law of GravitationGravitation

F = - G M m/ d F = - G M m/ d 22

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

d moon

d d appleapple

F = – g m

g =G M G M ♁♁ / R / R♁♁ 22

GravitationGravitation

g =G M / RG M / R♁♁22

G M G M ⊗⊗ = gR R ⊗ ⊗ 2 2 = (9.80 N/kg)(6.37x10 = (9.80 N/kg)(6.37x10 66 m) m)22

= 3.99x10 = 3.99x10 1414 N m N m 22/kg/kg

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Need to know G or M.

Henry Cavendish Henry Cavendish And the And the

Cavendish ExperimentCavendish Experiment

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

mM

d

G = 6.673 x 10 G = 6.673 x 10 -11-11 N N ‧‧mm 2 2 /kg /kg 22

G G ≈ 2/3 ≈ 2/3 x 10 x 10 -10-10 N N ‧‧mm 2 2 /kg /kg 22 (to an accuracy of 0.1%)(to an accuracy of 0.1%)

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

So,

M ⊗⊗ = (3.99x10 3.99x10 1414 N m N m 22/kg)/(6.673 x 10/kg)/(6.673 x 10-11-11 N N ‧‧mm

22/kg /kg 22))

= 5.98 x10 = 5.98 x10 2424 kg kg

Gravitational Force: Gravitational Force:

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

mM

d

F = - G M m/ d F = - G M m/ d 22

What is the order of magnitude of the attraction between two people (m~ 100 kg)

separated by a distance of ~ 1m?

What is the order of magnitude of the attraction between two people (m~ 100 kg) separated by a distance of ~ 1m?

1010

A B C D E

20% 20% 20%20%20%A.A. 9.8 N. 9.8 N.

B.B. 980. N980. N

C.C. 6.7 X 10 6.7 X 10 - 7- 7 N N

D.D. 6.7 X 10 6.7 X 10 - 9- 9 N N

E.E. None of the aboveNone of the above

Answer Answer Now !Now !

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21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

0% 0 of 1

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

Gravitational Force: Gravitational Force:

Physics 1710Physics 1710—C—Chapter 13 Apps: Gravityhapter 13 Apps: Gravity

mM

d

F = - G M m/ d F = - G M m/ d 22

F = - (6.67 x10 F = - (6.67 x10 –11–11 N N ‧‧mm 2 2 /kg /kg 22))(100 kg)(100kg )/(1 (100 kg)(100kg )/(1 m)m) 22

F = - 6.67 x10 F = - 6.67 x10 –7–7 N N

Equivalent weight = F/g = 67 ngEquivalent weight = F/g = 67 ng

Summary:Summary:•Kepler’s LawsKepler’s Laws

–The orbits of the planets are ellipses.The orbits of the planets are ellipses.–The areal velocity of a planet is The areal velocity of a planet is constant.constant.–The cube of the radius of a planet’s The cube of the radius of a planet’s orbit orbit is proportional to the square of the is proportional to the square of the period.period.

• F = - G M m/ d F = - G M m/ d 22

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