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Chapter 5: Force Fields
Chapter 5 Goals:
• To introduce the concept of the force field through the
example of Newtonian Gravity
• To discuss the kinematics of Projectile Motion
• To generalize our understanding of work and energy in
higher dimensions, through usage of the scalar product
• To define the vector product through the examples of
the torque: the Moment of the force
• To introduce the concept of the Angular Momentum of
a body
• To quantitatively analyze circular orbits due to gravity
• To qualitatively analyze the more general case of
elliptical orbits  Kepler’ s 3 laws
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
What is a force field?
• to date, our forces act due to contact with a body
• they are sometimes hard to describe: magical
• ultimately, though, they are manifestations of one of
the four fundamental forces, which are…
1. Strong force: holds nucleus together (quarks, gluons)
2. Weak force: allows certain nuclear decays (neutrinos,
W and Z bosons)
3. Electromagnetic force: controls matter’s properties at
a macroscopic level, as well as EM radiation
(charges, photons)
4. Gravity force: controls large-scale behavior of matter
(mass, gravitons)
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
How to visualize a force field?
• example of a ‘electric’ field
• two charges, +q and –q
• they set up a vector electric field
E that permeates all of space
• so-called field line picture
• to ‘test’ the field, put a third
charge +Q anywhere you like
• test charge will feel an electric
force F = Q E
• field direction: the ‘field line’ direction at that place
• field magnitude: the denser the lines, the bigger the force
• technically, E is the force field, defined as E = F/Q
where Q is ‘tester’ of field; E does NOT depend on Q!!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The Gravitational Field
• it is created by the presence of mass (point/extended)
• it permeates all of space so it is ‘action at a distance’
• its presence is sensed by another ‘test’ point mass
• Henry Cavendish measured the force exerted by one
point mass on the other in a very exacting experiment
M
m
r
O
rhat
F
F (r )  G
mM
or F(r )  rˆG
• M is field maker: at
source point (origin)
• m is field feeler: at
field point (r)
• F(r) is the felt force
and is attractive
mM
2
2
r
r
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Cavendish’s Apparatus
G = 6.67 x 10–11 N-m2/kg2
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The Gravitational Field of a Point Mass
• gravitational field produced by M at r: g(r) := F(r)/m
M
• these are field lines
g(r) : rˆG 2
• M is a ‘point mass’, a useful
r
abstraction
rhat
• the gravitational field is
M
radial in direction
r
• the gravitional field is
O
rhat inverse-square in magnitude
because it drops off (fast) with
distance as 1/r2
• a test mass m will be attracted
to M
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The Gravitational Field near the Earth
• constant gravitational field near the earth: g = –jhat g
• the field lines point down and are
uniformly spaced, so field direction is
downward and constant
• the source of the field is all of the
mass in the earth
What did Isaac Newton prove?
• consider a spherically symmetric distribution of mass
as an extended source of the field
• break it up into thin spherical shells of mass dM
• 1) field outside the shell is exactly point-mass-like!!!
• 2) field inside the shell is precisely zero!!!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
What we will not do in this course
Sections 5.3 (Flux) and most of 5.4 (Gauss’s Law
for Gravitation) will be skipped!!
Extremely powerful mathematical concepts –
with wide applicability – read and be amused
 need ideas of vector calculus: surface integrals
What we will utilize from those sections
• 1) field outside a spherically symmetric ball is
exactly point-mass-like, as if all of the mass in the
ball is at the center!!!
• 2) field inside the ball only arises from matter
‘below’ the object
• implications of (1) and (2) – consider a hollow
“hull” of matter, spherically symmetric, of mass Mp
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Consider that spherically symmetric hull..
• center the hull at the origin
• inner radius a, outer radius Rp
• total hull mass Mp
• differing densities in layers
a
Rp
r  Rp : g (r )  G
Mp
or g(r ) : rˆG
Mp
r2
r2
Mp
Mp
r  Rp : g ( Rp )  G 2 or g(r ) : rˆG 2
Rp
r
r  a : g (r )  0
a  r  Rp : g (r )  g ( Rp ) as you ' burrow' in
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Consider a uniformly dense ball (Mp, Rp)
Rp
Mp
r  Rp : g (r )  G
Mp
r  Rp : g (r )  G
M pr
r2
Mp
r  Rp : g ( Rp )  G 2
Rp
Rp3
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Gravity as a force on a mass m
• again, for the uniform ball Mp
mM p
r  Rp : F(r )  rˆG 2
• force on m is F(r) = m g(r)
r
• note that expressions are
mM p r
r  Rp : F(r )  rˆG
symmetric in m and Mp
3
R
p
• under the surface, force is
linear in distance r
• by Newton’s third law, there is mutual attraction between
m and Mp of equal magnitude, opposite direction
• what happens if you tunnel through the earth? SHM!!
mM p
spring constant is k  G 3 and mass is m
Rp
SHM period is T  2 k m  84 minutes
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Gravity near the earth’s surface on m
• Force on m is W(r) = m g(r) = so near the earth’s surface
W
GmM E
RE2
 g ( RE )  G
ME
RE2

2

N
m
 6.67x10 -11
 5.98x10 24 kg
2 

kg

g
6.38x10 m
6
2
N
m
 9.79
 9.79 2
kg
s
• well THAT’s reassuring!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

What kind of motion can occur near the Earth?
• object is projected with an initial velocity v0
• g and v0 define a plane in space: the xy plane
• vertical component v0y horizontal component v0x
•let projection angle be q0 so v0 = {v0 ,q0}
• v0y = v0 sin q0 and v0x = v0 cos q0 so v0 = <v0x , v0y>
• assume projectile starts at the origin: <x0, y0> = <0,0>
•now allow x and y motion to unfold by components
v y t   v0 y  gt
1
2
y(t )  v0 y t  gt 2
v x t   v0 x
(constant!!)
xt   v0 x t (steady!!)
• We still don’t know the ‘shape’ of the path, but we
see that the motion remains in the xy plane forever
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Some questions about Projectile motion on level
ground (ending y is same as y0 = 0)
• what is the time to maximum height? (call it T)
v y T   0  v0 y
v0 sin q 0
 gT  T 

g
g
voy
• clearly, the time to return to the ground is 2T
• what is the maximum height ? (call it ym)
 v0 sin q 0  1  v0 sin q 0 
1
2
  g 

ym  y (T )  v0 yT  2 gT  v0 sin q 0 
2
g
g




v02 sin 2 q 0 v02 sin 2 q 0 v02 sin 2 q 0



g
2g
2g
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2
More questions about Projectile motion on level
ground (ending y is same as y0 = 0)
• what is the range – the maximum distance along the
ground? (call it R) . . . This is the x value at t = 2T
 2v0 sin q 0  2vo2 cosq 0 sin q 0
 
R : x(2T )  v0 x 2T  v0 cosq 0 
g


this can be rewritten using double - angle identity :
sin 2   2 sin  cos
v02 sin 2q 0 
R
g
• note that R = 0 for q0 = 0° and 90°
• note that the range for q0 is the same as the
range for 90 ° q0
• the maximum range is at q0 = 45°
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
g
What is the actual Trajectory in Projectile
Motion? What is y(x)?
• solve the x(t) equation for t and insert into y(t) equation
x
1
2
t
into y (t )  v0 sin q 0t  2 gt
v0 cosq 0

 1 

x
x
  g 

 y ( x)  v0 sin q 0 
2
v
cos
q
0
 0
 v0 cosq 0 
 x tanq 0 
gx
2
2
2v02 cos 2 q 0
• this is a parabolic path in space!!
• Gravity’s Rainbow… {show Active figure 04_07}
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Side view of the parabolic trajectory, as seen in the
plane of g and v0
v0 r
q0
rm y
m
r
• note that the position vector is pretty complicated
• confusingly, we use q for the angle of the velocity
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The velocity vector is less complicated
q
vx0
vx0
q
qq0
vx0
vx0
q0
0
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example of a problem involving projectile motion
You have locked yourself out of the dorm and your roommate
throws your keys to you from a window that is 12 m above the
ground, with a launch speed of 4 m/s, at a launch angle of 50°.
a) how much higher do the keys rise?
b) where should you stand to catch the keys?
2
2
2
 2
vo sin q 0 4 m/s  sin 50
a) ym 
2g



2


2 10 m/s
b) trajectory equation : y ( x)  x tanq 0 
 .47 m
gx2
2v02 cos 2 q 0
We know y   12.0 m and so putting all # s in we get
 12.0  1.19 x  .76 x 2 or .76 x 2  1.19 x  12.0  0
x

1.19 
1.19 2  4.76 (12.0)
2.76 

1.19  1.42  36.48
1.52
1.19  6.16
 4.84 m out from the base of the wall
1.52
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Another example
A projectile is launched at a certain angle with a certain speed.
a) suppose q0 corresponds to the maximum range (call it
Rm). What is ym in terms of Rm?
b) suppose the angle corresponds to the situation where
the range is the same as the maximum height ym. What is q0?


v02 sin 2  45 v02
a) The launch angle is q 0  45 so Rm 

g
g
1
v02 sin 2 45 v02 2 Rm
 ym 


2g
g 2
4
v02 sin 2q 0  v02 sin 2 q 0
b) equate expressions for R and ym :

g
2g

 
2v02 sin q 0 cosq 0 v02 sin 2 q 0
use sin/cos form for R :

g
2g
 tan q 0  4  q 0  Arctan4   76 
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The (re)definition of Work
• consider a vector force that acts on a body, F(r)
• the force may vary with a body’s position r
• we drop the argument r at times from now on…
• if the body moves a small vector displacement ds, so
small that F doesn’t change much, the small bit of work
done by the force is dW := F ∙ds = |F|| ds| cos q
• for a non-small displacement (from rA to rB), the work
done by the force is the integral of dW
rB

W AB : F (r )  ds
rA
• how can this be thought of as
an area under a graph?
• it is, in 3d, the sum of 3 areas!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The (re)definition of Work
• write the force, and the arbitrary small
diplacement, using component/unit vectors
F  F ˆi  F ˆj  F kˆ and ds  dx ˆi  dy ˆj  dz kˆ
x
y
z
• using the fact that the unit vectors are perpendicular,
only three of the nine possible terms are non-zero:
rB

rB

rB

rB

rB

WAB : dW  F(r )  ds  F dx  F dy  F dz
rA
rA
rA
rA
rA
• to execute these integrals one must choose a path in space
• the vector version is a line (path) integral
• the result may well be path-dependent… but in special
cases it is not: conservative forces
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Here the forces are constant. Which ones do
positive work? Which negative work? Which
none?
W  F  r  ' force dotted into displacement'
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The (~same ) definition of Kinetic Energy,
and the Work-Energy Theorem survives!!
r
B
• If the force is the net force we get WAB,net :  Fnet (r)  ds
r
• Since Fnet = ma = m dv/dt we have
A
W AB,net
rB
rB
rB
rA
rA
rA
dx
dv
 m v  dv
 m  ds  m dv 
dt
dt




rB


 m v x dvx  2 more terms, for y and z 

rA
m 2
v xB  v 2yB  v z2B  v x2A  v 2yA  v z2A

2
m 2
m
m
 v B  v B   v A  v A   K where K  v
2
2
2


 
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
Some new mathematics, that generalizes the 1d
derivative with respect to x
• consider a scalar function of position f(r)
• for example, f is the temperature at any point in a room
• we ask ‘how does f change as you vary x, or y, or z?’
• three answers: hold y and z constant, and vary x only
• you can guess the the other two answers!
• conceptually it is the partial derivative; Notationally
f ( x, y, z )
is rate of change of f with x (hold y and z constant)
x
f ( x, y, z )
f ( x, y, z )
So,
and
are rates of change along y and z
y
z
• in which direction does f change most steeply?
• that direction is in the direction of the gradient of f,
also known as the vector derivative of a scalar function f
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The gradient – the vector derivative
f ˆ f ˆ f ˆ
 the gradient of f is f :
i  j k
x
y
z
• it is a vector that points in the direction of greatest
change of f (this is a bit tricky to prove but we’ll try)
• at some point r we move a small displacement ds
ˆ
 write this ' line element' as ds  dx ˆi  dy ˆj  dz k
• how much does f change because of ds?
• In 1d calculus, df = f’(x) dx (fundamental theorem)
f
f
f
 in 3d , we need 3 terms : df 
dx  dy  dz
x
y
z
 using
the scalar product, this is df  f  ds  f ds cosq
• we are finding how “parallel” the gradient is to the
displacement, and that is a maximum if q = 0!! QED
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The Potential Energy of an r-dependent force,
and its relation to its associated Force
• we generalize our ideas in straightforward fashion:
xB

rB

U ( xB )  U ( xA )  F ( x)dx  U (rB )  U (rA )  F(r )  ds
xA
rA
• we have a potential energy function U(r) = (x,y,z)
•it should be clear that there are THREE integrals in there
• if the force is not a specified function of r, can’t do it!
• so, in the same way that the derivative is the ‘undoing’
of the integral in 1d, the gradient is the ‘undoing’ of the
line integral in higher dimensions: get the force F(r) via
U ( x, y, z )
F(r)  U (r)  Fx (r)  
, etc. [notation shift]
x
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Mechanical Energy and Energy conservation
• if a force is expressible as a gradient of a potential
energy, it is said to be conservative
• find U := U(rB) – U(rA) along two different paths
• on path I: UI ; on path II: UII
I
B
• now consider the closed path formed by
A
I, followed by a reversed II
• the reversed path II has –UII
II
• U for the closed path must be zero  0 = UI – UI I
• thus UI = UI I : the integral is termed path-independent
• conservative means that if the body moves in a closed
path its potential energy doesn’t change…
• Mechanical Energy is Conserved: E = 0
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Potential energy example: Gravity
• revisit: gravitational potential energy U = mg(y-y0)
force W( x, y, z )  U ( x, y, z )  mg ( y  y0 )
W  mg ˆi (0)  ˆj(1)  kˆ (0)  mgˆj familiar!!
 gravity


• consider: Newtonian gravity F(r)  G
mM p
r
2
rˆ
• what is the associated potential energy?
• we want to line-integrate this from infinity ‘down’ to r
• because the force is conservative, any path will work
• our ‘line element’ ds will have one radial component dr
and two other perpendicular ‘angular components’ , but
since the force is purely radial, F∙ds = Fr dr
• only the radial part of the path contributes!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Potential Energy of Newtonian Gravity
• setting up ther line integral, wer have


U (r )  U ()  F(r ' )  ds '  0  Fr (r ' ) dr'


r
mM

p 
    G 2  dr'  GmM p r ' 2 dr'
r' 


r


r
GmM p
GmM p 
0

 
r
r' 
• the potential energy is indeed 0 at r = ∞, and becomes
increasingly negative as the body approaches the ‘planet’
mM p
mM p
U
U(r)  G
and so Fr (r )  
 G 2
r
r
r
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Potential Energy of Newtonian Gravity graphed
mM E
U(r)  G
r
• at the surface of the Earth
U is a large negative value
• to escape the earth one
must ‘climb the potential’
to reach r = ∞
• it turns out that U
continues to drop below
the surface quadratically,
reaching 3/2 of its surface
value at the center of Earth
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Example to show how one can use either
Cartesian or radial/tangential coordinates
Express the gravitational potential in cartesian coordinatees, take
minus its gradient explicitly, and confirm the result for the force
mM p
mM p
U
U(r)  G
and so Fr (r )  
 G 2
r
r
r
GmM p
U (r )  U ( x, y, z )  

x2  y2  z 2

so Fx  

U
etc. for y and z
x




 2
1 2
2
2 1 / 2
2
2 3 / 2 
but
x y z
 2 x y z
x2  y2  z 2
x
x
x
x
 2
2
2


since
x

y

z
 2x
3
/
2
3
x
r
x2  y2  z 2
 2
y
2
2 1 / 2
and
x y z
 3 and similar for z
y
r
 GmM p
 GmM p
 GmM p rˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
 F  Fx i  Fy j  Fz k 
xi  yj  zk 
r
3
3
r
r
r2







Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example to calculate the ‘escape velocity’
Find the value of the potential energy
per unit mass at the surface of the
earth, and find the escape speed.

mM E
U(r)  G
r

2

N
m
 6.67x10 -11
 5.98x10 24 kg
kg 2 
U ( RE )  GM E 
MJ


 62.5
6
m
RE
kg
6.38x10 m


• this is exactly the energy in ½ gallon of gasoline!
• escape speed: conserve energy in a flight from RE to ∞
 GmM E 1 2
U ( RE )  K ( RE )  U ()  K () 
 mv esc  0  0
RE
2
GmM E 1 2
2GM E

 mvesc  vesc 
 11,400 m/s
R
E
RE
2
 we never said U
 K . We said U  K !
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New concept: the moment arm of a vector
• an origin-dependent quantity: given a vector, we
have the line of the vector in space
• the distance from the origin to the line of the vector
quantity is called the moment arm of the vector
• we will see soon enough why this is a new
separately important concept, by applying it to the
force, and then to the momentum
F  dp dt
• goal: to derive an analog to N2 for angular motion
• one says that the torque on a body is the time rate
of change of its angular momentum
• for point objects, nothing really new is gained
• for finite-sized objects (extended bodies) a lot is
gained by this new approach!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
A definition: the cross product of two vectors
• the vector product of a vector with a vector: the
cross(vector(outer(wedge))) product A x B
• the cross product of two vectors is itself a vector!!
magnitude of A  B is A  B  A B sin q
direction of A  B is normal to plane of A and B
• which normal? Use Right Hand Rule: swing A into B
through the smaller tail-to-tail angle, according to how
your right hand operates; its thumb points along A x B
•so we have a completely new ‘thing’, in a completely
new direction, derived from A and B, which couldn’t be
captured by the dot product
• A x B = sort of ‘signed’ area of parallelogram spanned
by A and B; a measure of their ‘perpendicularness’
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
More about the cross product of two vectors
A
|A|sin q
q
B
• magnitude of cross product is
magnitude of B times
‘moment’ of A [ |A| sin q ]
from perpendicular projection
of A along line of B
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Now, let the first vector be the vector
from the origin to the point in space (r)
where the second vector (A) ‘lives’
r
|r| sin q
q
tail-to-tail, A
O
in standard position
A
|r| sin q
r
q
tip-to-tail,
where A lives
• the moment arm is b = |r| sin q, with units of length
• example: if F is a force acting at some point, the
moment of the force is M: = r x F with units of N-m
• also called the torque ; written as T, or M, or N, or t
• the moment of the force is perpendicular to the force
• the magnitude is |M| = (force)(moment arm) = Fb
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example: the moment of the momentum
• if p is the momentum of a point body, the moment of
the momentum is L: = r x p with units of kg-m2/s
• also called the angular momentum
• angular momentum is perpendicular to the momentum
The angular version of N2
• N2 reads F = dp/dt… Consider dL/dt :
dL d
dr
dp
 r  p    p  r 
 mv v rF
dt dt
dt
dt
dL

 M angular version of N2
dt
• if the line of the force goes through O, then b = 0, so
M = 0: angular momentum is constant about that origin
• subtlety here is that the origin is an important choice
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The decision about where to put the origin
• a trivial example: a body m moves at constant velocity v
• obviously, the momentum is constant since p = mv
• let L and L’ be referred to origins O and O’
r’
p
O’
r
b
O
• referred to origin O, |L| = |r x p| = mvb = constant
• referred to origin O’, |L’| = |r’ x p| = 0 = constant
• in both cases, L is conserved as seen from either origin
• using L is overkill for this situation!
• it can happen that L is constant as seen from one
origin by not constant as seen from another
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Planetary orbits I: central forces conserve L
• put a very massive body M centered at O, so massive that
it does not move; for example, the sun
• M gravitationally attracts a much less massive body m, at
a distance r; for example, a planet
mM
Fr (r )  G 2
M
r
r m hat
r
O
F
• line of the force passes through O (a central force) so
the moment of the force (M) about that origin is zero
•  L is conserved as seen from O
• the orbit may be closed (circular, elliptical) or open
(parabolic, hyperbolic) depending on total energy
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Planetary orbits II: the circular orbit
• a circular orbit has constant r = R and constant v =V
• gravity force is purely centripetal in this case, so
mV 2
mM
 Fcent (r )  Fr (r )  
 G 2  V  GM
R
R
R
• example: if M is the earth and R is its radius (‘treetop’
orbit) we get Vtt ≈ 7800 m/s and note that vesc = √2 Vtt
GmM
GmM
GmM
 energetics K

U 
E
2R
R
2R
 angular momentum L : r  p so L  RmV (r is perp to p)
circumference
2R
2R 3 / 2
 Period 


1/ 2
speed
GM R GM 
1
2
 mV
2
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Planetary orbits III: Kepler’s Laws 1 and 3
2 3
4 R
2
 Kepler's Third Law : T 
GM
• Kepler’s First Law: orbits are conic sections: circles,
ellipses (if closed) and parabolas, hyperbolas (if open)
with the sun at one focus
• closed: replace R by a (‘semimajor axis’ of ellipse)
 squaring
•
f
r
p
c
a
M, f, O
• v goes down as r goes up
• rmin: perihelion/perigee
• rmax: aphelion/apogee
• circular orbit with a given a has the most negative Ecirc
• ellipses with the same a have E with Ecirc < E < 0
• a parabola has E = 0, and hyperbolas have E > 0
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Planetary orbits IV: Kepler’s Law 2
• Kepler’s Second Law: Angular Momentum conserved
• in short time dt, r  r + dr
rhat
dr
•
dr
is
along
v
and
p
ds dq
r
• angle change: dq
r+dr
• arc length: ds = vtan dt ,
that
where by ‘tan’ we now mean
perpendicular to r (‘rad’)
• area = base ∙ height/2 ; height = ds = vtan dt = wr dt
and base = r  d(area)= wr2 dt/2  d(area)/dt = wr2/2
•we can safely ignore the change in length during dt of r
• L = mv x r = so L = m vtan r = mw r2 = 2 m d(area)/dt
• Thus, the constancy of L  Kepler’s Second Law: equal
area swept out by r in equal times
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Problem 5.9
A particle of mass 0.35 kg is fastened to one end of a light string
of length 2.5 m which is fixed at the other end. The particle is
caused to move in a vertical circle of that radius (2.5 m). If the
speed at the top is 12 m/s, find the speed and the tension when
the string is horizontal.
Speed: conserve energy so Etop = Ehoriz
1
1
2
2
m
v

m
gy

m
v
top
top 2
horiz  mgyhoriz
2
2
 vhoriz  vtop
 gyhoriz 
take ytop  0 m
12 2  10  2.5  13.0 m/s
Tension: tension is purely centripetal, but gravity is perp
T
2
mv horiz
R
2



.
35
13
.
0

2.5
 23.6 N [Weight W = mg = 3.5 N]
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Problem 5.12
The ‘Lennard-Jones’ interatomic
potential is often used in atomic
physics, to capture the longdistance attraction and shortdistance repulsion of a pair of
atoms. Its form is (a is a distance,
and U0 is a constant with units of
energy)
  a 12  a  6 
U (r )  U 0     2  
 r 
 r  

Note that for small r, the first term dominates and U is repulsive,
whereas for larger r the second term dominates and U is attractive
Determine the equilibrium spacing rmin, and the force between the
atoms when the spacing is a/2.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Problem 5.12 cont’d

dU (r )
13
7
 0 for r  rmin  0  U 0  12 a12rmin
 12 a 6 rmin
dr

6
 a 6 rmin
 1  rmin  a
dU (r )
Force is 
 0 for r  a / 2  F  U 0   12a12
dr

 F  12U 0

8192 128
 a
a
  97768U
0

/a
Book gives numbers: a = 0.30 nm and U0 = 3 x 10-21 J
We get F = 9.7 x 10-7 N. Is this a lot? YES! Areal
density of atoms is 1/(.3 nm)2 = 1.1x1019 atoms/m2 so
total pressure is 1.1x1013 Pascals. Atmospheric
pressure is only 1.01x105 Pa.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

a 13
6 a 7 
 12a

2
2
