Physics 203 – College Physics I
Department of Physics – The Citadel
Physics 203
College Physics I
Fall 2012
S. A. Yost
Chapter 5 Part 2
Circular Motion,
Universal Gravitation
Physics 203 – College Physics I
Department of Physics – The Citadel
Announcements
Chapter 5’s homework, HW05, is due next Tuesday.
You should have finished reading the chapter,
except for “Nonuniform Circular Motion”, which we
will skip (sec. 5-4). Sec. 5-5 is optional, but an
interesting application.
Today: circular motion, banked turns, and universal
gravitation (orbits).
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz: Question 1
The net force vector on an object in uniform circular
motion points
(A) away from the center of the circular path.
(B) in the direction opposite to the motion of the
object.
(C) toward the center of the circular path.
(D) in the same direction as the motion of the
object.
(E) nowhere. The net force is zero because the
object is moving with a constant speed.
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz: Question 2
Two objects attract each other gravitationally. If the
distance between them decreases by a factor of 2,
what happens to the gravitational force?
(A) It increases by a factor of 2.
(B) It decreases by a factor of 2.
(C) It increases by a factor of 4.
(D) It decreases by a factor of 4.
(E) It doesn’t change.
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz: Question 3
The orbits of planets are
(A) circular with the sun at the center.
(B) circular with the sun off-center.
(C) elliptical with the sun at the center.
(D) elliptical with the sun off-center.
(E) something else I haven’t thought of.
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz: Question 4
Why does a satellite in a circular orbit travel at a
constant speed?
(A) The net force acting on the satellite is zero.
(B) There is a force acting opposite to the direction
of the motion of the satellite.
(C) The gravitational force acting on the satellite is
balanced by the centrifugal force acting on the
satellite.
(D) There is no component of force acting along the
direction of motion of the satellite.
Physics 203 – College Physics I
Department of Physics – The Citadel
Vertical Circular Motion Question
A mass (red) is
rotated at a
constant speed in
a vertical circle on
the end of the rod.
A
B
Which vector (green)
could possibly
show the force of
the rod on the
mass?
C
D
Physics 203 – College Physics I
Department of Physics – The Citadel
Vertical Circular Motion Question
The forces could look
like this…
Fr
Fc
q
Fg
Physics 203 – College Physics I
Department of Physics – The Citadel
The Rotor
People ride in a Rotor ride at a fair. When it spins
fast enough, the floor drops out, and the people
feel like they are pressed against the wall.
Physics 203 – College Physics I
Department of Physics – The Citadel
The Rotor
Is there really an outward force pushing the people
against the wall? A Yes
B
No
Physics 203 – College Physics I
Department of Physics – The Citadel
The Rotor
Which picture shows the correct forces on a
passenger in the Rotor?
A
B
C
D
E
Ff
FN
mg
What is the name of each of the three forces?
Physics 203 – College Physics I
The Rotor
If the barrel’s radius is 3.0
m and the coefficient of
friction is 1.2, how
many RPM must the
barrel make to keep the
woman from sliding
down?
Department of Physics – The Citadel
Physics 203 – College Physics I
Department of Physics – The Citadel
Banked Curve
Suppose a car is going
around a track with radius
200 m banked at an angle
of 25o.
What is the minimum
coefficient of friction for
the tires if the car can go
around the track at
50 m/s without skidding?
v
R
200 m
Physics 203 – College Physics I
Department of Physics – The Citadel
Universal Gravitation
Newton’s Law of Universal Gravitation:
Any two masses exert an attractive force on
each other in proportion to their masses and
inversely proportional to the distance
between them:
F12 = G m1 m2 /R2
Newton’s Gravitational Constant:
G = 6.67 x 10-11 N m2/kg2
Physics 203 – College Physics I
Department of Physics – The Citadel
Orbital Velocity Example
We can use the inverse
square law to find the
orbital velocity of a
spaceship at a height
2Re above the Earth’s
surface.
(Re = 6380 km)
2Re
Re
Physics 203 – College Physics I
Department of Physics – The Citadel
Orbital Velocity Example
The spaceship is 3 times
as far from the center
of the Earth as the
ground is, so at this
height,
a = g/9 = 1.1 m/s2
I don’t need to use G or Me!
2Re
Re
Physics 203 – College Physics I
Department of Physics – The Citadel
Orbital Velocity Example
Setting the gravitational
acceleration equal to
the centripetal
acceleration gives
a = g/9 = v2/R
= v2 / (3Re).
2Re
Re
Physics 203 – College Physics I
Department of Physics – The Citadel
Orbital Velocity Example
Setting these equal,
g/9 =
v2
/ (3Re)
which implies
v = √ gRe/3
= √ (9.8 m/s2)(6380 km)/3
= 144 m/s
2Re
Re
Physics 203 – College Physics I
Department of Physics – The Citadel
Newton’s Law and Orbits
Determine the mass of the sun using the properties
of Earth’s orbit.
Newton’s Gravitational law:
Newton’s 2nd Law:
F = G Ms Me / R2.
F = Mea = Mev2/R
v = 2pR / T
G Ms Me / R2 = Me (2p/T)2 R
Ms = (2p/T)2 R3 / G
Physics 203 – College Physics I
Department of Physics – The Citadel
Newton’s Law and Orbits
Ms = 4p2 R3/(GT2)
R = 1.50 x 1011 m
T = 1 year = 3.16 x 107 s
G = 6.67 x 10–11 Nm2/kg2
The numbers give Ms = 2.00 x 1030 kg
Physics 203 – College Physics I
Department of Physics – The Citadel
Kepler’s Laws
Law 1: All planets move in
elliptical orbits.
P = perihelion
A = aphelion
Physics 203 – College Physics I
Department of Physics – The Citadel
Kepler’s Laws
Law 2: A line joining any planet to the sun sweeps
out equal area in equal times.
Physics 203 – College Physics I
Department of Physics – The Citadel
Kepler’s Laws
Kepler’s Third Law:
The square of a planet’s year, divided by the
cube of its mean distance from the sun, is
the same for all planets:
T2/R3 = TE2/RE3
for Earth
Physics 203 – College Physics I
Department of Physics – The Citadel
Kepler’s Third Law
R3 / T2 = constant
If R is measured in astronomical
units (1 AU = the distance from
the Earth to the sun) and T in
Earth years, then R = 1 and T
= 1 for the Earth.
2R
Physics 203 – College Physics I
Department of Physics – The Citadel
Kepler’s Third Law
Therefore,
R3 / T2 = 1
if R is in AU and T in Earth
years.
1 AU = 1.5 x 1011 m
1 year = 3.16 x 107 s
2R