Transcript Physics 1710—Chapter 13 Apps: Gravity

Physics 1710—Warm-up Quiz
I estimate my grade on exam 2 to be:
17% 17% 17% 17% 17% 17%
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Physics 1710—Chapter 13 Apps: Gravity
Center of Gravity (comparison)
CM
Ball
Fg = - mg
CM
CG
Moon
Fg = - GmM/r2
Physics 1710—Chapter 13 Apps: Gravity
What is the mass of the earth?
How do you know?
Think!
No Talking!
Confer!
Peer Instruction Time
Physics 1710—Chapter 13 Apps: Gravity
What is the mass of the earth?
How do you know? g?
Fg = - mg
Peer Instruction Time
Fg = - mg′
Physics 1710—Chapter 13 Apps: Gravity
1′ Lecture
•The moduli of elasticity (Y, E, B) characterizes the stress-
strain relation:
• stress= modulus • strain; σ = Y ε
•The force of attraction between two bodies with mass M
and m, respectively, is proportional to the product of their
masses and inversely proportional to the distance
between their centers squared.
• Fg = - řG Mm/d 2
•G = 6.673 x 10 –11 N m2 /kg2 ~ 2/3 x 10 –10 N m2 /kg2
Physics 1710—Chapter 13 Apps: Gravity
Elasticity
Definitions:
•
Stress σ : the deforming force per unit area.
•
Strain ε : the unit deformation.
Stress = modulus x strain
σ = F/A = Y ε
Physics 1710—Chapter 13 Apps: Gravity
Elasticity
Stress σ – Strain ε “Curve”
Elastic limit
Stress σ (N/m2)
•
σ=Yε
Strain ε = ΔL/L (%)
Failure
Physics 1710—Chapter 13 Apps: Gravity
Elasticity
•
•
Stress σ : the deforming force per unit area.
Strain ε : the unit deformation.
Tensile/Compressive Stress
Young’s Modulus E
Stress = modulus x strain
σ = F/A = E ε = E ΔL/L
L
ΔL
σ=Eε
Physics 1710—Chapter 13 Apps: Gravity
Elasticity
•
•
Stress σ : the deforming force per unit area.
Strain ε : the unit deformation.
Shear Modulus G
Stress = modulus x strain
σ = F/A = G ε = G Δx/h
L
h
Δx
σ
Physics 1710—Chapter 13 Apps: Gravity
Elasticity
•
•
Stress σ : the deforming force per unit area.
Strain ε : the unit deformation.
Hydraulic Stress:
Bulk Modulus B
Stress = modulus x strain
σ = F/A = p = B ε = B ΔV/V
ΔV
V
p
Physics 1710—Chapter 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.
k larger for shorter spring.
k larger for longer spring.
k the same for both springs.
None of the above.
41%
29%
28%
2%
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Physics 1710—Chapter 12 Apps: Gravity
Solution:
F = -k x?
σ = F/A = E ε = E ΔL/L
F = - A E x/ L
k = A E/L
What about a thicker (bigger A) wire?
Physics 1710—Chapter 13 Apps: Gravity
Isaac Newton’s
Universal Law of Gravitation
F = - G M m/ d 2
d moon
d apple
Physics 1710—Chapter 13 Apps: Gravity
Kepler’s Laws:
•The orbits are ellipses. (Contrary to Aristotle and
Ptolemy.)
A central force: F ∝ 1/ r
2
or r
2
• The areal velocity is a constant.
Angular momentum is conserved:
½ v r ∆t = constant implies that
rmv = L = constant.
⊙
• T 2 ∝ r 3 implies F ∝ 1/ r 2, only.
T
2
=
r3
Physics 1710—Chapter 12 Apps: Gravity
How did Newton figure out UL of G?
• Fact: a moon circling a planet has an acceleration of
a = v 2 /r
• Fact: a = F/m.
• Fact: Kepler had found that the square of the period
T was proportional to the cube of the radius of
the orbit r :
Thus:
T2=kr3.
v = 2π r / T
Physics 1710—Chapter 12 Apps: Gravity
And
Thus:
T 2= (2πr) 2 /(F r /m) = k r 3
F = (2π) 2 m/(k r 2 )
An “inverse square law,” with k = 1/ [(2π) 2G M]
F = G Mm/ r 2 ,
But what value is G?
Physics 1710—Chapter 13 Apps: Gravity
Isaac Newton’s
Universal Law of Gravitation
F = - G M m/ d 2
d moon
d apple
F=–gm
g =G M ♁ / R♁ 2
Physics 1710—Chapter 12 Apps: Gravity
Gravitation
g =G M / R♁2
G M ⊗ = gR ⊗ 2 = (9.80 N/kg)(6.37x10 6 m)2
= 3.99x10 14 N m 2/kg
Need to know G or M.
Physics 1710—Chapter 13 Apps: Gravity
Henry Cavendish
And the
Cavendish Experiment
M
m
d
Physics 1710—Chapter 13 Apps: Gravity
G = 6.673 x 10 -11 N ‧m 2 /kg 2
G ≈ 2/3 x 10 -10 N ‧m 2 /kg 2
(to an accuracy of 0.1%)
So,
M ⊗ = (3.99x10 14 N m 2/kg)/(6.673 x 10-11 N ‧m 2/kg 2)
= 5.98 x10 24 kg
Physics 1710—Chapter 13 Apps: Gravity
Gravitational Force:
F = - G M m/ d 2
What is the order of magnitude of the
attraction between two people (m~ 100 kg)
separated by a distance of ~ 1m?
M
m
d
Physics 1710—Chapter 13 Apps: Gravity
What is the order of magnitude of the attraction
between two people (m~ 100 kg) separated by
a distance of ~ 1m?
10
Answer Now !
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B.
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9.8 N.
980. N
6.7 X 10 - 7 N
6.7 X 10 - 9 N
None of the above
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Physics 1710—Chapter 13 Apps: Gravity
Gravitational Force:
F = - G M m/ d 2
F = - (6.67 x10 –11 N ‧m 2 /kg 2)(100 kg)(100kg )/(1 m) 2
F = - 6.67 x10 –7 N
Equivalent weight = F/g = 67 ng
M
m
d
Physics 1710—Chapter 13 Apps: Gravity
Summary:
•Kepler’s Laws
–The orbits of the planets are ellipses.
–The areal velocity of a planet is constant.
–The cube of the radius of a planet’s orbit
is proportional to the square of the period.
•F
= - G M m/ d 2