Download Physic 231 Lecture 17

Physic 231 Lecture 17
•
•
•
Main points of last lecture:
Rotation kinematic equations
∆θ = ω t
ω = ω 0 + αt
1
1
ω = (ω + ω0 ) ∆θ = (ω + ω 0 )t
2
2
1
∆θ = ω 0t + αt 2 ω 2 = ω 02 + 2α∆θ
2
•
•
v2
ac =
r
•
•
ac
Newton’s law of universal
gravitation:
GMm
F=
r2
Rolling motion:
∆x = s = r∆θ
v = rω ; a = rα
Main points of today’s lecture:
Centripetal acceleration:
vt
GMm
PE = −
r
Kepler’s laws and the relation
between the orbital period and
orbital radius.
2
 4π
T = 
 GM
2
 3
 r

Forces and circular motion
vt
•
v v
v
∆v = v f − v i ≠ 0
•
This means there is an acceleration pointing
inward towards the center of the circle. If we look
at the component of the acceleration in this
direction:
v2
v∆θ
∆v
= vω =
= lim ∆t →0
ac = lim ∆t →0
r
∆t
∆t
•
Here we have used:
v
ω=
r
ac
Consider the mass undergoing horizontal circular
motion at the right. At any point, the
instantaneous velocity is tangential to the circle.
Because the direction of the tangent changes,
however, the direction of the velocity changes.
Example
•
vt
Consider the motion of a 1 kg mass constrained to move in a vertical
circle of radius r=2 m by a massless wire.
• a) What is the tension in this wire if the mass is at the bottom of the
circle and is moving with a instantaneous speed of 3 m/s?
• b) What is the tension in this wire if the mass is at the side of the circle
(wire) is horizontal and is moving with an instantaneous speed of 3
m/s?
• What is the tension in this wire if the mass is at the top of the circle
and is moving with an instantaneous speed of 3 m/s?
– a) –5.3N
– b) 9.8 N
v2
v2
a) T − mg = mac = m
c) T + mg = ma c = m
– c) 5.3N
r
r
2
2
–v 2 d) 0
(
v2
3m / s )
(
3m / s )
2
2
)
T = m ( − g ) = 1kg (
− 9 . 8 m / s ) T = m( g + ) = 1kg (9.8m / s +
r
2m
r
2m
T = 14.3N
T = − 5 .3 N
Result is impossible ,
v2
b) T = m = 4.5 N
r
tension cannot be negative ⇒ T = 0
ac
Conceptual question
•
A rider in a “barrel of fun” finds herself stuck with her back to the
wall.Which diagram correctly shows the forces acting on her?
a)
b)
c)
d)
e)
Newton’s law of universal gravitation
•
All objects (even light photons) feel a gravitational force attracting
them to other objects. This force is proportional to the two masses and
inversely proportional to the square of the distance between them.
m1m2
F12 = G 2
r12
G = 6.673 x 10-11 N m² /kg²
Example
• George weighs 200 lbs. on the Earth. What would his weight be on the
surface of another planet that has a planetary mass, which is 3 times
the mass of Earth and mass and a planetary radius, which is 2 times the
radius of Earth?
GM planet mGeorge
– a) 150 lbs.
W planet =
2
R planet
– b) 200 lbs.
GM Earth mGeorge
– c) 250 lbs.
WEarth =
2
R
Earth
– d) 300 lbs.
GM planet mGeorge
2
2
2


M planet REarth
W planet
R planet
1

 = 3 ⋅   = 3
=
=
GM Earth mGeorge
M Earth  R planet 
4
WEarth
2
2
REarth
3
W planet = 200lbs = 150lbs
4
Conceptual question
• Two satellites A and B of the same mass are going around Earth in
concentric circular orbits.The distance of satellite B from Earth’s
center is twice that of satellite A.What is the ratio of the centripetal
force acting on B to that acting on A?
– a) 1/8
– b) 1/4
GM E ms
– c) 1/2
Fc = ma c =
r2
– d) 1 2
GM E ms
– e) 1
2
2
 rA 
1
=
=
=   =   = 1 / 4
2
GM
m
Fc, A
E s
 2
rB  rB 
rA2
Fc, B
rB2
rA2
Conceptual quiz
• Suppose Earth had no atmosphere and a ball were fired from the top of
Mt. Everest in a direction tangent to the ground. If the initial speed
were high enough to cause the ball to travel in a circular trajectory
around Earth, the ball’s acceleration would
– a) be much less than g (because the ball doesn’t fall to the ground).
– b) be approximately g.
– c) depend on the ball’s speed.
Kepler’s laws
•
Johannes Kepler proposed three laws of planetary
motion:
1. All planets moved in elliptical orbits with the
Sun at one of the focal points.
2. Planetary orbits sweep out equal areas in equal
times. (This is a consequence of angular
momentum conservation.)
area =
1
1 ∆θ
1
∆t ≈ r 2ω∆t
r ⋅ r∆θ = r 2
2
2 ∆t
2
2a
∆θ
r
⇒ r 2ω = const.
3. The square of the orbital period of any planet is proportional to the cube
of the average distance from the planet to the Sun. For a circular orbit:
GM star m planet
2
mv 2
GM
v
2
star
=
=
Fgrav =
ma
⇒
=
=
ω
c
r2
r
r2 3
r2
2π
2
GM star  2π 
ωT = 2π ⇒ ω =

 3
4
π
2
⇒
=



r
⇒T =
T
r3
T
 
 GM star 
4. For an elliptical orbit:
 4π 2  3
a
⇒T =
 GM star 


2
Example
• One satellite is in an orbit about Jupiter of radius r1 and a period of 100
days. If a second satellite is placed in an orbit with 4 times the radius
(i.e. r2=4r1), what is the period for the orbit of the second satellite?
 4π 2  3
 r1
T =
 GM

Jupiter 

2
1
2
3
 4π 2  3  T2   r2 
2
 r2 ⇒   =   = (4 )3 = 64
T2 = 
T  r 

 GM
Jupiter
 1  1


T2
⇒ = 8 ⇒ T2 = 8 ⋅ 100 days = 800 days
T1
Gravitational Potential Energy
• PE = mg∆y is valid only near the earth’s
surface
• For objects high above the earth’s surface, an
alternate expression is needed
M Em
PE2 = −G
r
– Zero potential energy is defined to be the value
infinitely far from the earth