Transcript Physics 1710—Chapter 13 Apps: Gravity

Physics 1710—Warm-up Quiz
If two (spherical) asteroids are in contact and are attracted with
a force of 1 Newton, how much more or less force would they
experience if they both had been twice as large (assuming the
same density)?
A.
B.
C.
D.
E.
F.
The same
Twice as much (2x)
Half as much (1/2x)
4x
8x
16 x
46%
24%
11%
9%
7%
2%
A
B
C
D
E
F
Physics 1710—Chapter 12 Apps: Gravity
Solution:
F = G M1 M2 /r2
M1 = M2 = 4π/3 ρR3
R=r
F = G (4π/3 ρR3)2 /R2
F∝R4
F2/F1 = (R2/R1)4 = 24 = 16
Physics 1710—Chapter 13 Apps: Gravity
1′ Lecture
•Fg = - řG Mm/d 2
•G = 6.673 x 10 –11 N m2 /kg2 ~ 2/3 x 10 –10 N m2 /kg2
•The gravitational force constant g is equal to
g = G M/(R+h) 2, M and R are the mass and radius of the
planet.
•The gravitation field is the force divided by the mass.
•The gravitation potential energy for a point mass is
proportional to the product of the masses and inversely
proportional to the distance between their centers.
Physics 1710—Chapter 13 Apps: Gravity
1′ Lecture (cont’d.)
•Kepler’s Laws
–The orbits of the planets are ellipses.
–The areal velocity of a planet is constant.
–The cube of the radius r 3 of a planet’s orbit
is proportional to the square of the period T 2.
Physics 1710—Chapter 13 Apps: Gravity
Which best corresponds to the actual shape of a
planet’s orbit?
A.
B.
52%
47%
C.
D.
None of the above.
2%
0%
A
B
C
D
Physics 1710—Chapter 12 Apps: Gravity
Solution:
Kepler’s First Law:
The orbits of the planets are ellipses.
Planet
ro
ro/(1+ε)
r = ro /[1+ε cos θ]
☿
ε = eccentricity ♀
ro/(1-ε)
ro
♂
♃
♄
♅
♆
♇
ε
0.2056
0.0068
0.0934
0.0483
0.0560
0.0461
0.0097
0.2482
Physics 1710—Chapter 12 Apps: Gravity
History:
Kepler’s First Law:
The orbits of the planets are ellipses.
What is the significance?
Repudiation of Aristotle
F is inverse square law
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.
• The areal velocity is a constant.
• T 2 ∝ r 3 implies F ∝ 1/ r 2, only.
T2
=
r3
⊙
Physics 1710—Chapter 12 Apps: Gravity
Comment:
Kepler’s Second Law:
The areal velocity is constant.
½ r2 ω= constant.
Why?
mr2 ω = L = constant.
L is conserved if no
torque,
i.e. F is “central force.”
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)
= 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
What is g♂ on Mars?
M ♂ = 0.107 M♁ , R ♂ = 0.532 R ♁
Think!
No Talking!
Confer!
Peer Instruction Time
Physics 1710—Chapter 13 Apps: Gravity
What is g♂ on Mars?
A.
B.
C.
D.
E.
25.9 m/sec2
9.80 m/sec2
3.70 m/sec2
1.97 m/sec2
None of the above.
80%
15%
5%
A
0%
B
0%
C
D
E
Physics 1710—Chapter 13 Apps: Gravity
What is g♂ on Mars?
g♂/g♁ =(M♂/M♁)(R♁/R ♂)2 = (0.107)(1/0.532)2
= 0.378; g♂= 3.70 m/sec2
M♁
Peer Instruction Time
M♂
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?
A.
B.
C.
D.
E.
9.8 N.
980. N
6.7 X 10 - 7 N
6.7 X 10 - 9 N
None of the above
65%
31%
0%
A
4%
B
0%
C
D
E
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
Gravitational Potential Energy
U = -∫∞R F•d r
U = -∫∞R G Mm/r 2d r
U = GMm/R
U = 0 as r →∞
Physics 1710—Chapter 13 Apps: Gravity
Total Energy
for Gravitationally Bound Mass
E=K+U
E = ½ m v 2 – GMm/r
E = L2/2mr 2 – GMm/r
Bound orbit if E ≤ 0 and
- dE/dr |ro= 2 (L2/2mro) - GMm = 0
E = - GMm/2ro
Physics 1710—Chapter 13 Apps: Gravity
Total Energy
for Gravitationally Bound Mass
E=K+U
E = - GMm/2ro
E - L2 /2mr 2 < – GMm/r
ro = - 2E/(GMm)
Physics 1710—Chapter 13 Apps: Gravity
Escape Velocity
If K is such that E > 0, then
K = ½ m v 2 = GMm/ r
Thus at r= R:
v = √[ 2GM/R]
Physics 1710—Chapter 13 Apps: Gravity
Summary:
• 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.
F = - G M m/ r 2
• The proportionality constant in the Universal Law of
Gravitation G is equal to 6.673 x 10 –11 N m2 /kg2 .
Physics 1710—Chapter 13 Apps: Gravity
Summary:
•The gravitational force constant g is equal to
G M/(R+h) 2, R is the radius of the planet.
• 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.
• The gravitation field is the force divided by the mass.
g = Fg / m
Physics 1710—Chapter 13 Apps: Gravity
Summary:
• The gravitation potential energy for a point mass is
proportional to the product of the masses and inversely
proportional to the distance between their centers:
U = GMm / r
• The escape velocity is the minimum speed a projectile
must have at the surface of a planet to escape the
gravitational field.
v = √[ 2GM/R]