UNIT 5 (end of mechanics  )
Universal Gravitation, Energy, Orbits, & SHM
Did not teach SHM 2012-13 to HP
As previously learned, two masses will attract one another
with equal and opposite force given by:
Gm1m2
Fg 
2
r
where ‘r’ is distance
from center to center
This describes how
the force of gravity
varies on as a particle
moves away or
towards another
G = 6.67x10-11 Nm2/kg2
this was measured 71yrs after
Newton’s death by Cavendish.
Gravity acts as an
inverse square law
Previous discussion of calculating force of gravity
due to Earth.
• Assumed gravity was constant near surface
• Only calculated Fg at a point in space
• Assumed Earth was uniform density
Recall from honors…What was true of
gravitational force inside the Earth? What
happened to your weight if you were placed at
center of the Earth?
Mathematical analysis of force of gravity inside a
planet. Assume planet of mass M and radius R and
uniform density. Particle of mass, m, located at r.
Since planet is of uniform density, ρ,
we can say that:
r
M M'

 '
V
V
M
4 R 3
3

M'
R
'
3
4 r 3 F  G mM  G m ( r M )  G mMr
g
3
R3
r2
r 2 R3
3
r
M M 3
R
'
This represents the mass
within the shell
Inside planet, it only matters
what mass is enclosed within
shell. Fg grows linearly as
you move towards surface
Gravity as a function of distance for a planet of
mass M and radius R.
You are heaviest
when you are at the
surface of a planet.
What if the density of a planet is NON-uniform?
Consider a spherical planet whose mass density is given by
ρ = br (where b = constant in kg/m4) and has radius, a.
Calculate the gravitational force on an object of mass, m, a
distance r (where r < a) from the center of the planet.
ρ = M/V where dM = ρ dV
r
M   dM    (r )4r dr
dV = surface area x thickness
dV = (4πr2) dr
2
0
r
M  4  (br )r dr  br
2
0
4
m1m2
m(br )
2
Fg  G 2  G
 Gbmr
2
r
r
4
for r < a
For the outside of the planet the gravitational force
would be
a
M  4  (br )r dr  ba
2
4
0
m1m2
mba
Gmba
Fg  G 2  G

2
2
r
r
r
4
for r > a
4
The Gravitational Field
• During the 19th century, the notion of the “field”
entered physics (via Michael Faraday).
• Objects with mass create an invisible disturbance in
the space around them that is felt by other massive
objects - this is a gravitational field. In Newton’s
time, there was much discussion about HOW
gravity worked - how does the Sun, for instance,
reach across empty space, with no actual contact at
all, to exert a force on the Earth?
• This spooky notion was called “action at a distance.”
• We learned last year about gravity and Einstein.
g-field
The region around a massive
object in which another object
having mass would feel a
gravitational force of attraction
is called a gravitational
field. The gravitational field's
strength at a distance r from
the center of an object of
mass, M, can be calculated
with the equation.
M
Fg  Fg
mM
mg  G 2
r
M
g G 2
r
Gravitational Potential Energy (Ug)
We must derive an expression that relates the potential
energy of an object to its position from Earth since ‘g’ is
NOT constant as you move away from Earth. All
previous problems dealt with objects ‘close’ to the earth
where changes in ‘g’ were negligible.
Also, we have referred to the PE of an ‘object’, but in
reality it is the PE associated with 2 objects…PE of a
system. We have taken h=0 to be surface of earth.
We pretended that the object possessed PE by itself
(mgh), rather it possessed PE due to its relationship with
the Earth. It is the change in PE that is meaningful,
not the value of PE at a point in space. We always
took PE to be zero at surface of earth.
Consider a particle of mass, m, a very far
distance, r1 , away from the center of the
earth (M).
r1
If we were to move that particle closer to the
earth to a new distance, r2, then gravity would
do + work on the particle.
r2
W   F (r )dr
r2
r1
We are moving the particle from very far away to a distance, r,
from the center of the planet. Therefore, our integral setup
and limits would be
U is always negative for a finite r. As
r
you move away, ΔU=+ and increases
mM
as it becomes less negative. Zero is
W  G 2 dr
defined infinitely far away.
r


r
r
1
2
W  GMm  2 dr   r dr
r



U  GMm  r

1 r

GMm
U (r )  
r
Where U(∞) = 0
Don’t get hung up on the minus.
Even with mgh, you can have
negative PE if you assign 0 to be
above you on earth
This is the PE associated with
any 2 particles. When particles
are separated by a distance r, an
external agent would have to
supply an energy at least equal to
+GmM/r in order to separate them
by an infinite distance.
If we had integrated in the opposite
direction, then formula would have been
positive, however FG would have opposed
movement where minus sign would have
to be factored back in due to the work
done yielding same result.

r

F
All values of the potential energy are negative,
approaching the value of zero (becoming less
negative) as R (distance from planet) approaches
infinity. Because of this we say that the mass is
trapped in an "energy well" - that is, it will have to be
given additional energy from an external source if it
is to escape gravity's attraction.
P.E. formula from before, U = mgh
Using new formula, we should be able to get the old
expression for PE of mass, m, near the surface of earth.
Consider moving from 1st floor to 2nd floor (Δh),
1 1
U  GM E m(  )
rf ro
r f  ro
 GM E m(
)
ro rf
Algebra removes negative sign from G
where
rf  ro  h
rf ro  R
2
E
Since distance above Earth is small
compared to RE
h
 GM E m( 2 )
RE
Recall
GM E
gE 
2
RE
U  mgh
tadah!
Another way to see
if new U(g) formula makes sense…
dU
RECALL
F 
dx
dU (r )
FG  
dr
d  GMm
GMm
FG  
 2
dr
r
r
- means FG points opposite of increasing r
Previous problems involving potential energy of ‘an
object’ involved an object associated with the earth
where each was assumed a point mass, uniform
spherical objects, or having large distances between the
COM of 2 objects. It turns out that only uniform
spherical mass distributions act like point masses where
COM can be used. This is what took Isaac Newton
years to prove. (you can look up derivations of this in my calc-based
textbooks (university physics) which goes beyond scope of this class).
What if we want to find the potential energy of a
system between non-spherical situation (instead
of usual object and the earth)?
Ug between thin rod and sphere (pt mass)
A thin, uniform rod has length, L, and mass, M. A point mass,
m, is placed a distance, x, from one end of the rod, along the
axis of the rod. Calculate the gravitational potential energy of
the rod-sphere system. X is not >> L.
dM
Consider just a small slice of the rod (no rod present), so small
that we can treat it as a point mass, dM. Since we are essentially
dealing with two point masses, the potential energy is given by
dM
Gm(dM )
dU g  
r
r
Now, imagine we move more slices
(dM) from far way into position,
creating the full rod again. We could
sum up the energy dU associated
with each small slice dM by
integrating. It takes work to do this.
GmdM
 dU    r
Can’t use center of mass for rod…only
works for spheres and pt masses
The reason why COM can be used when dealing with
spheres BUT NOT a rod:
You could argue that a rod would be like a slice through the
diameter of the sphere. Therefore, why does a sphere reduce to
a point mass and rod does not?
Envision a sphere squeezed to a pancake from top to bottom,
then squeezed from left to right. The shape would be rod-like,
but it would be uneven in density. Meaning, it would be more
dense near the center and less dense at the ends assuming
diameter stays as the length. The more mass near the middle
makes up for the fact that the mass at the ends is not same
distance from the pt mass….evens it out to allow for using COM.
Rod can’t do that.
Since force varies as inverse square with same mass
throughout for rd, it doesn’t work for a rod by using
COM.
dM
 Gm
r
 Gm
x L

x
where dM = λdr since λ is linear
mass density
dr
r
GmM x  L
 xL
U  Gm ln 
ln

L
x
 x 
B) Using this expression, use calculus to find the
expression for the force of gravity between the rod
and the sphere using F = - dU / dx
dU
F 
dx
GMm
Fg  
( x  L) x
GMm
F 2
r
You can’t use center of mass of rod and just plug
in…not the same answer. Formula on left reduces to
formula on right when ‘x’ is very large where the
length of rod is insignificant.
Calc derivation in notes
Kepler's 1st Law: The Law of Elliptical Orbits
All planets trace out elliptical
orbits. When the planet is
located at point P, it is at the
perihelion position. When the
planet is located at point A it is
at the aphelion position. The
distance PA = RP + RA is called
the major axis.
Semi-major axis is half of the major axis. Eccentricity is
a measure of how "oval" an ellipse is.
Kepler's 2nd Law: Law of Areas
A line from the planet to the
sun sweeps out equal areas
of space in equal intervals of
time where dA/dt = constant
At the perihelion, the
planet’s speed is a
maximum.
At the aphelion, the planet’s
speed is a minimum.
Angular momentum is conserved since net torque is
zero. Gravity lies along the moment arm or position
vector, r. vARA = vPRP
Kepler's 3rd Law: Law of Periods
The square of the period of any planet is proportional to the
cube of the avg distance from the sun.
Fg  mac
GMm mv

2
r
r
2
After some trivial
substitution…
Since the orbits of the planets in our
solar system are EXTREMELY close to
being circular in shape (the Earth's
eccentricity is 0.0167), we can set the
centripetal force equal to the force of
universal gravitation and,
4 3
T 
r
GM
2
3
T  kr
2
2
Energy Considerations for circular Satellite Motion
E  K U
1 2
Mm
E  mv  G
2
r
K will never be negative,
however, both U and E can be
negative. No deep significance
to this, just a consequence of
choosing infinity to be zero PE.
Furthermore, if we consider the system to be isolated, then
energy is conserved. We can show that the total energy of a
satellite system is negative.
Fg  mac
GMm mv

2
r
r
2
1 GMm 1 2
 mv
2 r
2
Multiplying both sides by ½ r
Subbing above into K
term in the E eqn
Note that K is
positive and
equal to half the
magnitude of U.
GMm
E
2r
Changing the orbit of a satellite
Calculate the work required to move an earth satellite of mass m
from a circular orbit of radius 2RE to one of radius 3RE.
Using our expression from before for total energy:
We get the energies at both orbits:
GM E m
Eo  
4 RE
GMm
E
2r
GM E m
Ef  
6 RE Makes sense work
Solving for the work done:
GM E m
W  E  E f  Eo 
12 RE
by external source
is positive since
FEXT and d are both
away from planet
How was the energy distributed in changing the orbit?
The initial and final
potential energies were:
GMm
Uo  
2R
GMm
Uf 
3R
Therefore,
GMm
U  
6R
The kinetic
energies were:
Meaning, we only
looked at TOTAL
E on last slide
1 2 GMm
mv 
K
2
2r
GMm
Ko 
4R
GMm
Kf 
6R
GMm
K  
12 R
Part of the
work goes into
increasing U
and the other
part goes into
decreasing K.
Wkshts energy and
orbits
Escape Velocity
If an object of mass m is launched vertically upwards with speed
vo we can use energy to determine minimum speed to escape
planet’s gravitational field. Our parameters are when the object
can just reach infinity with zero speed.
M pm
M pm
1 2
mvo  G
 G
2
Rp
rmax
1
1
v  2GM p ( 
)
R p rmax
2
o
0
Rp is the distance between
object and earth at launch
Where the rmax = infinity
2GM
vescape 
Rp
*If object is given initial speed of escape speed, its total energy is
equal to zero since both K and U will be zero. If vo > vesc than
object will have residual K at r = ∞.
Example: Two satellites, both of mass m, orbit a planet
of mass M. One satellite is elliptical and one is circular.
Elliptical satellite has energy E = -GMm/6r.
a) Find energy, E, of circular
satellite
1 2 GMm
R
E  mv 
M
2
r
GMm
mv
Fg  2  Fc 
r
r
2
Solve for v2 and sub into K term to get
1 GM GMm
GMm
E  m(
)

2
r
r
2r
r
b) Find speed of elliptical satellite at closest approach
(perihelion)
1 2 GMm
Eelliptical  mv 
2
r
GMm 1 2 GMm

 mvr 
6r
2
r
R
r
Given in problem
5GMm 1 2
 mvr
6r
2
5 GM
vr 
3 r
c) Derive an equation that would allow you to solve for R
but do not solve it.
Can’t use Fg = Fc since orbit isn’t circular
5 GM
vr 
3 r
GMm 1 2 GMm

 mvR 
6r
2
R
Lr  LR
mvr r  mvR R
r
vr  v R
R
Subbing vr answer from
part b
5GM r
vR 
3r R
2
GMm 1 5GM r
GMm

 m

2
6r
2
3r R
R
1 5 r
1



2
6r 6 R
R
Orbit wkshts
Example: A satellite of mass ms is in an elliptical orbit
around a planet of mass mp which is located at one
focus of the ellipse. The satellite has a velocity va at the
distance ra when it is furthest from the planet.
Note: rp and vp not given
Derive expression for the magnitude of the velocity vp of
the satellite when it is closest to the planet? Express
your answer in terms of ms, mp, G, va, and ra as needed.
This will incorpate your algebra skills!
LL
mv p rp  mva ra
va ra
rp 
vp
EE
K U  K U
1 2 GMm 1 2 GMm
mv p 
 mva 
2
rp
2
ra
1 2 1 2
GM GM
v p  va  

2
2
ra
rp
2GM 2GM
v v  

ra
rp
2
p
2
a
2GM 2GM
v v  

va ra
ra
vp
2
p
2
a
2GMva 2GMv p
(v p  va )(v p  va )  

va ra
va ra
(v p  va )(v p  va ) 
2GM (v p  va )
va ra
2GM
v p  va 
va ra
2GM
vp 
 va
va ra
Black hole + into wormhole dvd on
hawking dvd
• Phet lab
Periodic Motion & S.H.M.
Repeating oscillation or
vibration is called periodic
Since oscillating mass repeats itself after a
period, where t = T, it must be at the same position
and moving in the same direction as it was at t = 0.
Since cosine or sine repeats itself every 2πrad,
we can equate that with
ω = 2πf
ωt = θ
ωT = 2π rads
where ω is the angular
frequency
We can relate periodic motion with an object
undergoing uniform circular motion. Consider a
mass sitting on top of a rotating platform where
lights cast a shadow of mass on wall.
Shadow of rotating mass
will move back and forth on
wall exhibiting periodic
motion.
As the mass rotates with a
constant angular speed, ω,
through an angle θ in a
time t, we can describe the
linear movement across
the screen as
x = Acosθ
Since θ = ωt we can relate
the position of the object with
respect to time and angular
movement as
x(t) = Acos(ωt)
x(t )  A cos(t   )
(full version)
where Ф is called the phase angle which
represents the initial angular position at t = 0
(ωt + Ф ) = the phase of motion
A = amplitude from equilibrium
NOTE: ALL calculations with above
equation must be done in RADIAN
MODE.
Trig Calculus
d
(sin x)  cos x
dx
d
(cos x)   sin x
dx
d
d
(cos ax)   a sin ax
(sin ax)  a cos ax
dx
dx
 sin xdx   cos x  C  cos xdx  sin x  C
1
 sin( ax)dx   a cos ax  C
1
 cos(ax)dx  a sin ax  C
Expressions for velocity and
acceleration:
x(t )  A cos(t   )
dx
 v(t )  A sin( t   )
dt
dv
2
 a(t )   A cos(t   )
dt
NOTE: ALL calculations with above equation
must be done in RADIAN MODE.
SIMPLE HARMONIC MOTION
A special case of periodic
When periodic motion possesses a restoring force
that is directly proportional to the negative of
displacement, this is said to be SHM.
x=0
Consider a mass oscillating on
a frictionless surface
Fs
v
 F  ma
2
d x
 kx  m 2
dt
Recall minus means
acc opposes disp
This is called a 2nd
order differential
equation.
Will the equation of position, x(t), satisfy
the above condition? If so, then those
equations of motion are simple harmonic.
2
d x
 kx  m 2
dt
Inserting our expressions for both position and
acceleration into the differential equation above…
x(t )  A cos(t   )
2
a(t )   A cos(t   )
 k ( A cos(t   ))   m A cos(t   )
2
Yields…
k
 
m
2
which is a solution to the
differential equation.
Note that frequency and period do
not depend on the amplitude. The
frequency at which particle
oscillates is called the natural
frequency or resonant frequency.
Since
2
d x
 kx  m 2
dt
And
  k/m
2
k
m
 
then for harmonic
oscillators,
a   x
2
•At maximum displacement,
Fs = max, v = 0 & a = max
•At equilibrium Fs = 0,
v = max & a = 0
•Repeat of part ‘a’
Energy considerations for SHM
A mass displaced a
Total energy could also be
distance A (amplitude) will
written as
oscillate about that distance
assuming no friction.
2
E = ½ mv
E = ½ kA2
max
E = K + U where
½ kA2 = ½mv2 + ½kx2
v at some arbitrary position, x
Example: A 0.500-kg mass is vibrating in a system in
which the restoring constant is 100 N/m; the amplitude
of vibration is 0.20 m. Find:
a. the energy of the system
b. the maximum kinetic energy and maximum velocity
c. the PE and KE when x = 0.100 m
d. the maximum acceleration
e. the equation of motion if x = A at t =0
a) E = ½
kA2
= 2.0J
b) Kmax = E = 2.0J
b) vmax = 2.83m/s
c) U = ½ kx2 = 0.5J
K = E - U = 1.5J
d) amax =| -ω2 A |= 40m/s2
e) x(t) = 0.20 cos (14.1t)
Does a vertical spring obey SHM due to
gravitational force?
If a mass is at rest, suspended
from a spring, the mass is in
equilibrium. The only forces
acting on the mass are a
gravitational force down, mg,
and the spring force up, kxo,
where xo is the stretch of the
spring at equilibrium.
If the mass is now
displaced from that
equilibrium position by
some displacement x, the
net force acting on the
mass would be the extra
force due to this further
displacement x. Newton's
second law would yield
Fnet  kx  ma
a  kx / m
This fits the form for SHM. Gravitational force
doesn’t affect the SHM of the mass
Mathematically, kd = mg when mass is in equilibrium
where d = stretch. If mass is displaced downward a distance x
beyond d, the restoring force within spring would be
Fs = k (d + x)
Fs-mg = ma
2nd Law would follow as
and substitution yields
k (d + x) – mg = ma
where kd + kx –mg = ma
Since kd = mg from above
mg + kx – mg = ma and kx =ma where it is obvious that
–(k/m)x = a where force opposes displacement
A body of mass m is suspended from a spring with spring constant
k in configuration (i) and the spring is stretched 0.1m . If two
identical bodies of mass m/ 2 are suspended from a spring with
the same spring constant k in configuration (ii), how much will the
spring stretch? Explain your answer.
Spring on left
Spring on right stretches
stretches x=mg/k
x/2 =mg/2k. Why?
Recall scale between 2 masses hanging from pulleys.
Mass on right holds it in place. Mass on left stretches
spring
5 worksheets
A block is moving undergoing SHM having amplitude
A0. At instant block passes thru equilibrium, putty of
mass M is dropped vertically onto block and it sticks.
Find new amplitude and period
k
To find new amplitude
Get
expression for
speed at x=0
2
Eo = ½ mvo = ½ kA0
2,
k
so vo = A0
m
It’s a collision where x-component of momentum
is conserved where new speed is…
m
v
mvo + 0 = (m + M)v
mM
vo
K is not conserved due to inelastic collision
Energy in system
is now
2
1
(
M

m
)
m
m
1 2
2
2
vo 
( mvo )
E = ½ (M+m) v =
2
(m  M ) 2
2 (m  M )
m
E
Eo
(m  M )
1 2
m
1 2
1 2
kA  (
)( kAo )
E  kA
2
mM 2
2
1 2
m
1 2
kA  (
)( kAo )
2
mM 2
A  Ao
m
(m  M )
Amplitude is less
ω = 2πf = 2π/T
1/ω = T/2π where  
T is greater. More
inertia.
m
M m
T  2
 2
k
k
k
m
Is a simple pendulum a S.H.O.?
It is if the restoring force is proportional to x or θ
ΣF = ma
Fgx = mgsinθ
In the above equation, Fgx is
proportional to sinθ, not θ,
therefore not SHM.
However, when θ is small,
sinθ is very nearly equal to
θ (in radians) where arc
length s=Lθ is nearly the
same as the chord length
(x=Lsin θ)….therefore
x
F = -mgsinθ ≈ -mgθ where θ = x / L
Where – sign indicates force tends to decrease θ
F = - mgx / L
F ≈ -x we can say it is SHM for small angles
We let (mg/L) represent ‘k’ in Hooke’s law
equation where
mg
if k 
L
then

k

m
2
L
since  
then T p  2
T
g
mg

Lm
g
L
Physical pendulum
Any rigid body that is suspended from a fixed
point that does not pass through the COM and
oscillates about that point.
 net  I
Fr  I
r  d sin 
Pivot
dsinθ
d
d
 mgd sin   I 2
dt
2
dsinθ
Where – sign indicates torque tends to decrease θ
Again, if we assume small angles
d
 mgd sin   I 2
dt
2
mgd
d

 2
I
dt
2
2
then
Since
d x
2
  x
2
dt
mgd
2
(
)   
I
and
Finally
mgd

T
2

I
 2
I
mgd
d = Distance
from pivot
Example: A uniform rod of mass M and length L is
pivoted about one end and oscillates in a vertical plane.
Find the period of oscillation if the amplitude is small.
d
T  2
I
mgd
1
2
I  ML
3
Mg
1
2
ML
2L
3
T  2
 2
L
3
g
Mg ( )
2
Torsional Pendulum
A mass suspended to a fixed
support by a thin wire can be
made to twist about its axis. This
is known as a torsional pendulum.
The mass attached to the wire
rotates in the horizontal plane
where θ is the angle of rotation. It
is the twisting of the wire that
creates a restoring torque due
to the resistance of the wire to
the deformation.
For small angles of θ the magnitude of the
restoring torque is proportional to the angle θ
  
Where κ is the torsion
constant of support wire
d
    I  I 2
dt
2
d
   I 2
dt
2
Again we have a format that matches SHM

d
2
   2   
I
dt
2
T  2 I

Journey through the Earth!
What happens if you could jump into a
tunnel in the earth? Assume the Earth
to be a solid sphere of uniform density,
ρ, mass, M, and radius R. A hole is
drilled through from pole to pole. A
person of mass m is dropped in the
hole with zero initial velocity. The
force acting on you as derived
earlier…
Mm
F G 3 r
R
Does this force act as simple harmonic?
Mm
 F G R 3 r  ma
2
M
d x
 (G 3 )r  2
R
dt
2
M
d x
 (G 3 )r  2
R
dt
2
M
d x
2
 G 3 x  2   x
R
dt
M
2
G 3 
R
It turns out that you would behave like a harmonic
oscillator. Determine the period of oscillation from
pole to pole in minutes.

M
G 3   2  1.52 x10 6
R
T /2

 42.3 min
Turns out that it takes the same time to fall through
tunnel whether its pole to pole or LA to NY…doesn’t
matter…always 42.3 minutes!
Now determine the period
of low orbit satellite
skimming just above
surface of earth, ignoring
friction.
Using Kepler’s 3rd
Law we get…
4 3
T 
r
GM
2
2
In fact we end up showing circular motion mimics
the motion of mass inside earth
Set up this demo.
2 springs tied
together. Another
string tied to
crossbar is tied to
bottom spring.
String tied to top
spring is tied to
mass.
http://video.mit.edu/watch/spring-paradox6471/