Download Lecture 15 Final (with examples)

Orbiting Earth
“The secret to flying is to throw yourself at the earth and miss.”
Hitch Hiker’s Guide to the Galaxy
The centripetal force on the satellite
is provided by the gravitational force
from the earth.


GME m
v2 ˆ
Fc  m ac  m r  
rˆ
2
r
So:
GME
v
r
r
The smaller the radius, the greater the speed.
Synchronous orbit: period = 24 hours
– satellite stays above same part of earth (above the equator)
– used by communications satellites
– what is the radius of the orbit?
Synchronous orbit – what is the radius?
“The period of an orbit is:
From previous
So: T
circumfere nce 2r
T 

speed
v
GME
page: v 
r
r
1
3/2
 2r
 2r
GME
GME
2
 T  Kepler’s 3rd law of planetary motion: T 2 r
r  GME  
 2 
3
3
With T = 24  3600 s, r = 4.23  10 7 m = 42,300 km from centre
of earth, The speed of the satellite is:
v
2r
T
 3070 m/s  11,000km/h
Problem 5.32
Earth orbits the sun once per year at a distance of 1.51011 m.
Venus orbits the sun at a distance of 1.08  1011 m.
What is the length of the year on Venus?
Problem 5.32
A satellite has a mass of 5850 kg and is in circular orbit 4.1105 m
above the surface of a planet. The period of the orbit is 2 hours.
The radius of the planet is 4.15106 m. What is the true weight of
the satellite when it is at rest on the planet’s surface?
Free Fall – Weightlessness
An orbiting satellite is in free fall – there’s nothing hold it up.
Only its forward speed lets it “miss the earth” and keep orbiting.
Everything in the satellite is accelerated toward the centre of the
earth at the same rate.
An object exerts no force on the bathroom scales as the scales are
also being accelerated toward the centre of the earth.
Artificial Gravity
A space station is rotating about
its axis to provide an artificial
gravity.
At what speed must the surface
of the space station move
so that the astronaut experiences
a push on his feet equal to
his weight on earth?
The radius is 1700 m.
v2
Fc  m  mg
r
Solve for the velocity
v  rg 
1700 m 9.80 m s 2   129 m s
Artificial Gravity
Problem 5.28:
Problems of motion sickness start to appear in a rotating
environment when the rotation rate is greater than 2
revolutions / minute.
Find the minimum radius of the station to allow
an artificial gravity of one g (ac = 9.8 m/s2)
while avoiding motion sickness.
v2
v  rg  rac
Fc  m  mg
r
2r 2  2r meter 2r
v


 rac
t
60 sec
30
900ac
r
 223 m
2
4
Vertical Circular Motion
1)
2)
3)
4)
v12
FN 1  mg  m
r
v 22
FN 2  m
r
v 32
FN 3  mg  m
r
v 42
FN 4  m
r
force toward center
weight larger than mg
force toward center
force toward center
rider falls off if FN 3
0
force toward center
v 32
mg  m
r
The rider falls off if
v3  rg
Problem 5.40:
A motorcycle is traveling up one side of a hill and down the
other side. The crest is a circular arc with a radius of 45 m.
Determine the maximum speed that the motorcycle can have
while moving over the crest without losing contact with the
road.
The net downward force on the bike at
the crest of the hill allows the motorbike
to remain in contact with the ground.
Then FN  0.
That is:
net downward force = centripetal force, mv2/r.
Chapter 6
Work and Energy
• Work done by a constant force
• Work-energy theorem, kinetic energy
• Gravitational potential energy
• Conservation of mechanical energy
• Conservative and non-conservative forces
• Work-energy theorem and
non-conservative forces
• Power
• Work done by a variable force
IMPORTANT:
YOU NEED TO KNOW HOW TO APPLY THESE
1)
v  vo  at
2)
1 2
x  xo  vot  at
2
3)
1
x  xo  v vo t
2
4)
2a x  xo   v 2 vo2
F 
2ax  v vo  2 x
m 
1
1
2
 Fx  mv  mv o2
2
2
2
2
Work-Energy theorem
1
Kinetic Energy : EK  mv 2
2
Work done = change in kinetic energy
1
1
2
Fx  mv  mv o2
2
2
Fx = “work” done by the force = force times displacement
The work changes the speed of the object thereby increasing its
kinetic energy.
Initial kinetic energy:
1
KE o  mv o2
2
Final kinetic energy:
1
KE  mv 2
2
Work done:
W  KE  KEo  KE
The unit of work and energy: Joule (J)
1 J  1 N m