Transcript Thomson Template - University of North Texas

Physics 1710—Warm-up Quiz
In the movie “Armaggedon” a comet “the size of Texas’” (800
miles or 1250 km in diameter) was heading for Earth. If the
density were 800 kg/m3, the mass would be about 8.x10 20 kg.
How far would a ball on the surface fall in 1.0 second?
A.
B.
C.
D.
About 5 m
About 7 cm
About 2 cm
None of the above
78%
16%
4%
A
2%
B
C
D
Physics 1710—Chapter 12 Apps: Gravity
Gravitation
g =G M/ R2
g=(6.67x10-11 N m2/kg2)(8x1020 kg)/(625x103 m)2
g = 0.14 N/kg
d = ½ g t 2 = 0.5 (0.14 m/s2)(1 sec)2
= 0.0 7 m = 7 cm
Physics 1710—Chapter 13 Apps: Gravity
1′ Lecture
•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.
•The escape velocity is the initial velocity at the surface
required to leave a body.
vescape = √[ 2GM/R]
Physics 1710—Chapter 13 Apps: Gravity
Does the position of the moon affects us as
astrology claims?
Think!
No Talking!
Confer!
Peer Instruction Time
Physics 1710—Chapter 13 Apps: Gravity
Does the position of the moon affects us as
astrology claims?
A.
B.
C.
D.
Yes, I strongly agree
Yes, I Agree
No, I disagree
No, I strongly disagree
58%
23%
17%
2%
A
B
C
D
Physics 1710—Chapter 13 Apps: Gravity
Astrology: Does the moon affect us?
F = - G M m/ d 2
M = 7.36x1023 kg; d = 3.84x108 m
|F/m|= (7x10-11)(7x1023)/(4x108)2
= 3x10-4 N/kg ~ 30 μ g
d moon=3.84x108 m
M =7.36x1023 kg
Physics 1710—Chapter 13 Apps: Gravity
Gravitational Potential Energy
U = -∫∞R F•d r
U = -∫∞R G Mm/r 2d r
U = GMm/R = mgR
U = 0 as r →∞
F
dr
Physics 1710—Chapter 13 Apps: Gravity
Escape Velocity
If K is such that E = 0 as r →∞, then
and
KR = ½ m v 2 = UR = GMm/ R
vescape = √[ 2GM/R] = √[2gR]
UR = GMm/R
K = ½ m v2
U∞ = 0
K∞ = 0
Physics 1710—Chapter 13 Apps: Gravity
Were the astronauts in danger of jumping off the
“Armaggedon” comet? (Hint: What is the escape
velocity from our “Armaggedon” comet? M = 8x1020 kg,
R =6.25x10 5 m, g = 0.14 N/kg)
A.
B.
C.
D.
Yes. Definitely.
Maybe.
No. Definitely not.
Not enough information.
48%
35%
13%
4%
A
B
C
D
Physics 1710—Chapter 13 Apps: Gravity
Escape Velocity
vescape = √[ 2GM/R] = √[2gR]
= √[2(0.14 N/kg)(625 km)] ~ 400 m/s
~ 900 mph. Safe!
vescape = √[ 2GM/R] = √[2gR]??
Physics 1710—Chapter 13 Apps: Gravity
Escape Velocity of Earth?
vescape = √[ 2GM/R] = √[2gR]
= √[2(9.8 N/kg)(6.3 x10 6 m)] ~ 11,000 m/s
25,000 mph. Stuck!
vescape = √[ 2GM/R] = √[2gR]??
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
r
E<0
ro = - 2E/(GMm)
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.
vescape = √[ 2GM/R]
• Total Energy E is conserved for two body geavitational
problem; bodies are bound for E ≤ 0
E = L2/2mr 2 – GMm/r