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Transcript
Gravity
Unanswered questions
Galileo describes falling objects by rolling objects down a ramp.
But why does everything accelerate the same rate regardless of its mass?
Kepler describes planetary motions.
What force can account for the elliptical
paths of the planets?
Two seemingly unrelated observations, but…
Newton unites the two in one master stroke at the age of 24.
Inverse Square Law
2  R
r
ac 
T2
T 2  kR3
2
2  R 1
r
ac 
~ 2
3
kR
R
Circular motion result
Kepler’s observational law
2
FG  maC
1
~m 2
R
1
R2
GMm
FG 
R2
~ Mm
Centripetal acceleration behaves like inverse square
Newton’s 2nd Law for centripetal direction
Newton’s 3rd Law suggests both masses are important
Constant of proportionality determined by Cavendish
G  6.67  10
11 Nm 2
kg2
The Force Law
GMm
FG 
R2
M
m

Direction?
Along a line connecting the center of the two masses.
Action at a distance.
How does the force GET from the Earth to the moon?
“I feign no hypothesis regarding action at a distance.”
Potential Energy Revisited
Remember how we used to write the force of gravity.
FG  mg
From this, we derived an expression for gravitational potential energy:

UG  mgh
This only applies near surface of earth! (When GME/RE2=9.8 m/s2)
More generally, we have

UG  
GMm
R
As usual, we have a choice for where we set UG = 0. When using this equation, the
choice is made for us.

Two Applications
Terrestrial (free fall near the surface of a planet or star)
FG  mg
Celestial (circular orbit around a planet or star)

FG 
GMm
R2
This most general expression is always true, but sometimes the first expression is
simpler to implement
 (it has limited application, however, so be careful!).
Terrestrial Application
Example:
The radius of the Earth is 6.4 x 106 m and the value of g is 9.8 m/sec2. What is
the mass of the Earth?
GM E m
R2
FG  gm
GM E m
 gm
2
R
gR 2
ME 
G
FG 

9.86.4  10

6 2
6.67  10 11
 6.0  10 24 kg
Celestial Application
The ISS orbits the earth with a speed of approximately 7680 m/s.
What is the orbital radius of the station, and what is its altitude?
Fc  FG
m s me
v2
ms
G 2
R
R
me
me
2
v G
or, R  G 2
R
v
24
11 6 x10
R  6.67 x10
7680 2
 6.79 x10 6 m
Radius of the earth is about 6.38
x 106 m at the equator.
That gives the altitude above the
surface to be:
6.79x106 – 6.38x106 =
0.41x 106 m, or 410 km.
“g” ~ 8.68 m/s2
Torque
Dynamics
r
F1
r
F2
r
F3


Which applied force results in the largest angular acceleration of the bolt?

Dynamics
r
F2
r
F1
r
F3

Which applied force results in the largest angular acceleration of the bolt?

Dynamics
r
F1
r
F2


Which applied force results in a clockwise angular acceleration? A counter-clockwise
angular acceleration?
Dynamics quantified
Consider a force acting on a rigid body, some
distance away from a fixed pivot point.
r
FPerp
r
FP ar
Can split the force into components.


Only the perpendicular part contributes to rotation!
Where does the parallel part go?
r r
  r FPerp
r
r
r
O

Dynamics quantified
r
F1
r
F2
r
F3


Which applied force results in the largest angular acceleration of the bolt?

Dynamics quantified
r
F2
r
F1
r
F3

Which applied force results in the largest angular acceleration of the bolt?

Direction of torque
r
r
r
r
r
F1
r
F2

Direction of torque is the direction the
applied force tends to cause the object
to rotate.
F1 provides a clockwise torque.
F2 provides a counter-clockwise torque.


But torque is a vector, and vectors only point in a single direction.
Direction of torquer is given by right hand rule.
r
Draw r and F tip to tail
r
Point fingers of right hand in
the
direction
of
r
r
Curl fingers in direction of F
Thumb points in direction of torque: either into (clockwise, negative) or out of

positive) page
 (counter-clockwise,
Newton’s Laws
Newton’s 1st Law – If there is no net torque on an object, then the object rotates at a
constant angular velocity (could be zero angular velocity).
Newton’s 2nd Law – If the net torque on an object about a point is not zero, then the
net torque produces an angular acceleration about that point.
r
F
  rF
r
r
 rma

 rmr 
 mr 2
O

The quantity mr2 is the rotational equivalent of mass, and is called moment of inertia.

Newton’s 3rd Law – For every action, there is an equal and opposite reaction.
Example
r
FA
r
FB
A
B
r
FT  8000 N


20m

40m
A truck crosses a massless bridge supported by two piers. What force much each pier
exert when the truck is at the indicated position?
r
r
r
r
r
r
net, A  T, A  B, A
net, B  T, B  A, B
0  ()FT rT  ()FB rB
0  FT rT  FB rB
Fr
8000(20)
FB  T T 
 2667 N
rB
(60)
0  ()FT rT  ()FA rA
0  FT rT  FA rA
Fr
8000(40)
FA  T T 
 5333 N
rA
(60)
2667  5333  8000 N
Sample Torque Problems
X = __________
X = __________