Transcript Gravitational Fields (AIS) - Atlanta International School Moodle

Planetary Motion &
Gravitational Fields
Historical
Background
Our ancestors envisioned the earth
at the center of the universe and
considered the stars to be fixed on
a great revolving crystal sphere.
The observation of planetary
motions depends on the frame of
reference of the viewer.
Early Theories
Geocentric Theory: Ptolemy~140 AD
• Motion of planets from the point of
view of the observer on Earth
Heliocentric Theory: Copernicus, 1543.
• Same motions but from the point of
view of the observer on the Sun
We use the heliocentric theory due to its
simpler mathematics.
Tycho Brahe (1546-1601)
The Royal Astronomer for King
Fredrick II of Denmark,
•measured accurately the positions
of the planets
•Charted planets motion for 20
years
•Accurate to 1/60 of a degree
•“Greatest Observational Genius”.
•35 years before the telescope.
Johannes Kepler
• Imperial Mathematician for Holy
Roman Emperor Rudolf II
• In 1600 became Brahe’s assistant.
• Stole Brahe’s data after his death in
1601, spent 15 years analyzing it.
• Deduced 3 laws of planetary motion
• Apply to any system composed of a
body revolving about a more massive
body
• Examples: moons, satellites, comets.
Kepler’s 1st Law
The Law of Orbits
Planets move in elliptical orbits
with the Sun at one of the focal
points.
Copyright ©2007 Pearson Prentice Hall, Inc.
Kepler’s
dA
k
dt
where k is a
constant
nd
2
Law
The Law of Areas
A line from the Sun to a
planet sweeps out equal
areas in equal lengths of
time.
Copyright ©2007 Pearson Prentice Hall, Inc.
Kepler’s
rd
3
Law
The Law of Periods
(the Harmonic Law)
The square of the orbital period of a
planet is directly proportional to the
cube of the average distance of the
planet from the sun.
2
3
T  r
• The ratio of the squares of the
periods of any two planets is equal to
the ratio of the cubes of their mean
distances from the sun.  T1  2  r1 3
    
 T2 
 r2 
Isaac Newton
• Recognized that a force must be
acting on the planets; otherwise, their
paths would be straight lines.
• Force  Sun (Kepler’s 2nd Law)
• Force decreases with the square of
the distance (Kepler’s 3rd Law)
Newton Continued…
Gravity
• Compared the fall of an apple with the fall of
the moon.
• The moon falls in the sense that it falls
away from the straight line it would
follow if there were no forces acting on it.
• Therefore, the motion of the moon and
the apple were the same motion.
• Showed, everything in the universe follows a
single set of physical laws.
• Law of Universal Gravitation.
Law of Universal Gravitation
(1666)
Every particle in the universe attracts every other
particle with a force.
This force is:
• proportional to the product of their masses
• inversely proportional to the square of the distance
between them.
• Acts on a line joining the two particles.
Gm1m 2
F
2
-11
2
2
r
G = 6.67 x 10 Nm /kg
Discovered by Henry Cavendish in 1798
Law of Universal Gravitation
What is the force of attraction between a 4.8
kg Physics book and a 2.4 kg Math book if
they are 5.0 cm apart?
Gm1m2
Fg 
2
r
G = 6.67 x 10-11 Nm2/kg2
Rule of 1’s Practice
Gm1m2
Fg 
r2
a. If m1 is doubled, how is F changed?
b. If neither of the masses were changed but r was
doubled, how would F change?
c. If r is not changed but both masses are doubled,
how would F change?
d. If r is halved and both masses doubled, how would F
change?
Kepler’s Law of Periods Proof
Assumptions:
• Must conform to equations for circular motion
• Newton’s Universal Law of Gravity
• Planet rotates in a circular (elliptical) path
Fnet
mv 2
 Fc  mac 
r
Newton’s 3rd Law  symmetry
Fgravity  Fnet  Fc
Recall,
2

T
GMm mv 2
Fg  2 
r
r
so
2r
v  r 
T
Therefore, GMm  m4 r
2
2
2
r
T
Law of Periods
T
4
rearranging 3 
R
GM
2
2
Testing the Inverse Square Law
of Gravitation
• The acceleration due to gravity at the surface of
the earth is 9.8m/s2.
• If the inverse square relationship for gravity
(Fg~1/r2) is correct , then, at a distance ~60 times
further away from the center of the earth, the
9.8
m
 2.72  10
acceleration due to gravity should be 60
s
• The centripetal acceleration of the moon is given
by a  vr where the radius of the moon’s orbit is r
= 3.84 × 108 m and the time period of the
2r
moon’s orbital motion is T = 27.3 days.
v
3
2
2
c
ac 
4 r
T2
2
ac 
4 2 (3.84  10 8 )

24h (3600s) 
(27.3days )(1day ) 1h 


T
2
a c  2.72  10 3
m
s2
2
Acceleration due to Gravity
When only gravity is present,
Fg = Fnet where Fnet = ma (Newton’s 2nd law)
Fnet = ma =
So,
ma =
Note: a = g
GMm
2
r
Gm1m2
Fg 
2
r
GM
g 2
r
Changes in
Gravity
Acceleration due to Gravity on
different Planets
Consider an object thrown upward
v f  vo  2 ad
2
2
at the maximum height vf=0
v o   2 gd
2
If the same object was thrown upward with the same
initial speed gd  constant
So,
g 1d 1  g 2 d 2
g1 d 2

g 2 d1
Acceleration due to Gravity on
different Planets
• Example
A person can jump 1.5m on the earth.
How high could the person jump on a
planet having the twice the mass of the
earth and twice the radius of the earth?
Gravitational Field
Strength (g.f.s)
The force acting on a 1 kg mass at a
specific point in a gravitational field.
F
g
m
Units: N/kg
http://www.chrisbence.com/projects/powerplant/img/gravity_zoomed_mass.gif
From Newton’s 2nd law F/m = a
g.f.s is the acceleration due to gravity (g)
GM
g 2
r
Variation of g
With distance from a point mass
• g.f.s at a point p is
GM
g 2
r
• Variation of g with distance from the
center of a uniform spherical mass of
radius, R
Variation of g
GM
g 2
r
From
• Close to earth the acceleration due to gravity
changes slowly.
• As the distance from earth increases, the
change in the acceleration due to gravity
increases.
www.physicsclassroom.com/Class/energy/U5L1b.cfm
Variation of g
On a line joining the centers of two
point masses
• If m1 > m2 then
Note: when g = 0
the gravitational
fields cancel each
other out
Gravitational Potential Energy
(GPE) of two point sources
G P E   W   Fgd
• Gravitational force is not constant but
decreases as the separation of masses
increase.
• Requires calculus to compute the total
R
G Mm
work done.
W 
G Mm
W 
r
R



G Mm

R
r
2
dr
G Mm
GPE  
R
Gravitational Potential (V)
Background:
The potential at a point in a gravitational field is
equal to the work done bringing a 1kg mass
from infinity to that point
W G PE
V 

m
m
G Mm
V 
mr
Recall,
G Mm
G PE  
r
GM
V 
r
Units: J/kg
Gravitational Potential (V)
Potential (V) is the GPE per unit mass in a
gravitational field.
GM
V 
Why the negative sign?
r
• A body at infinity, has zero gravitational
potential.
• A body normally falls to its lowest state
of potential (energy)  as r decreases,
the potential decreases.
• We can do this by including a negative
sign in the above equation.
• As r decreases, V decreases
(becomes a greater negative quantity).
Source: http://www.saburchill.com/physics/chapters/0007.html
Equipotential Surfaces
Equipotential Surface – points that all
have the same potential (V) where
GM
V 
r
•
Formula shows that the
points will all be the same
distance from the mass or
lie on a sphere whose
center is that of a
spherical mass.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Equipotential Surfaces
Equipotential Surface – points that all
have the same potential (V)
• This can be shown two
dimensionally where the
blue lines represent the
equipotential lines
Similar to topographical maps
http://www.iop.org/activity/education/Projects/Teaching%20Advanced%20Physics/Fields/Images%20400/img_tb_4719.gif
Equipotential surfaces
due to two equal masses.
• Note that the field becomes
more spherical further away
from the masses
• At a far enough distance the field appears
to come form one mass with the combined
mass of the individual masses.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Equipotential Surfaces
Equipotential Surfaces
• Equipotential surfaces due to two
unequal masses
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Field Lines
Field Lines – Indicate the direction of
gravitational force that would act on a
point mass at that location.
1. For a gravitational field,
the lines will always
point inward as gravity
is always an attractive
force.
2. Note that the field lines
are perpendicular to
the equipotential lines.
http://www.intuitor.com/student/Planet%27s_g-field.jpg
http://www.chrisbence.com/projects/powerplant/img/gravity_zoomed_mass.gif
www.chrisbence.com/projects/powerplant/index.htm
Field Lines
Consider moving a test mass from point a to b as
shown where the distance between a and b is
r and the difference in potential is ΔV.
The work W required is
W  mV
W  Fg r  m gr
Setting these expressions
equal to each other
V
g
r
A more precise result based on using Calculus results in
dV
g 
dr
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
The work done could also
be calculate as
Field Lines
V
g
r
This expression implies that on a graph of
the variation of gravitational potential with
distance (V vs r), the slope would be equal
to the gravitational field strength g
This also implies that the
equipotential surfaces are
always perpendicular to the
gravitational field lines.
• As shown for a single mass
• True for any shape field lines
and equipotential surfaces
http://www.intuitor.com/student/Planet%27s_g-field.jpg
Escape Velocity
• The velocity needed to overcome
(escape) the gravitational pull of a
planet.
• As the body is moving away from the
planet, it is losing kinetic energy and
gaining potential energy.
• To completely escape from the
gravitational attraction of the planet, the
body must be given enough kinetic
energy to take it to a position where its
potential energy is zero.
Escape Velocity
• The potential energy possessed by a body
of mass m, in a gravitational field is given
by
GPE = Vm
• If the field is due to a planet of mass M and
radius R, then the escape velocity can be
calculated as follows:
ΔKE = ΔGPE
• So,
2 GM
vesc 
R
• as g = GM/R²
v es c 
2 gR
1 m v 2 esc  G Mm
2
rE
Satellites: Orbits & Energy
G Mm using Newton’s second law (F=ma)
U 
r
G Mm
v2
F 
m
2
r
r
Solving for v2
where v2/r is the
centripetal acceleration.
GM
v 
r
2
GMm
KE  mv 
So,
2r
1
2
2
U
KE  
2
Total Mechanical Energy of a satellite is:
GMm GMm
ME  KE  U 

2r
r
ME  
G Mm
2r
Satellites: Orbits & Energy
ME = -KE
This is for a circular orbit.
• For a satellite in an elliptical orbit with a
semi-major axis a
G Mm
E 
2a
where a was substituted for r
Energy Graphs
• As shown the energies for a satellite are,
G Mm
KE 
2r
• Graphing these,
http://www.opencourse.info/astronomy/introduction/06.motion_gravity_laws/energy_elliptical.gif
G Mm
U 
r
G Mm
ME  
2r
Energy Wells
• If we rotate the potential energy
graph, we can generate a 3D picture
known as an energy well.
http://www.opencourse.info/astronomy/introduction/06.motion_gravity_laws/
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Energy Changes in a Gravitational
Field
• A mass placed in a gravitational field
experiences a force.
– If no other force acts, the total energy will
remain constant but energy might be
converted from g.p.e. to kinetic energy.
• If the mass of the planet is M and the
radius of the orbit of the satellite is r, then
it can easily be shown that the speed of
the satellite, v, is given by
– if r decreases, v must increase.
Energy Changes in a Gravitational
Field
• If the satellite’s mass is m, then the kinetic
energy, K, of the satellite is K  1 m GM 
2  r 
• the potential energy of the satellite, U, is
These equations show:
 GM 
U  m

 r 
– that if r decreases, K increases but U decreases
(becomes a bigger negative number)
– the decrease in U is greater than the increase in K.
• Therefore, to fall from one orbit to a lower orbit,
the total energy must decrease.
– work must be done to decrease the energy of the
satellite if it is to fall to a lower orbit.
Energy Changes in a Gravitational
Field
• The work done, W, is equal to the change in the
total energy of the satellite, W = ΔK + ΔP.
• This work results in a conversion of energy
from gravitational potential energy to
internal energy of the satellite (it makes it
hot!).
• Air resistance reduces the speed of the satellite
along its orbit. This allows the satellite to fall
towards the planet. As it falls, it gains speed.
• So, if a viscous drag (air resistance) acts on a
satellite, it will
– decrease the radius of the orbit
– increase the speed of the satellite in it’s new orbit.
Energy Changes in a Gravitational
Field
• In principle, the satellite could settle in a
lower, faster orbit.
• In practice it will usually be falling to a
region where the drag is greater. It will
therefore continue to move towards the
planet in a spiral path.
Energy Changes in a
Gravitational Field
• Looking at this with a fbd of the satellite
spiraling downward due to air resistance.
– Note that the velocity is no longer perpendicular to
the weight.
– Due to this there is a component of gravity which is
now acting in the direction of the velocity providing
a net force which is accelerating the satellite.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Weightlessness
Astronauts on the orbiting space station are
weightless because...
a. there is no gravity in space and they do
not weigh anything.
b. space is a vacuum and there is no gravity
in a vacuum.
c. space is a vacuum and there is no air
resistance in a vacuum.
d. the astronauts are far from Earth's surface
at a location where gravitation has a
minimal affect.
e. None of the above
Source: http://www.physicsclassroom.com/Class/circles/u6l4d.cfm
Weightlessness
Source: Kirk, Tim, Physics for the IB Diploma, Oxford University Press 2007
Astronauts on the orbiting space station are weightless
because...
The astronaut and the space station are both free-falling
together - Apparent Weightlessness
• The gravitational force on the astronaut provided the
needed centripetal acceleration for the astronaut to
stay in orbit.
• The space station remains in orbit because of the
gravitational force on it.
• However, since there is no contact force between the
satellite and the astronaut here is an apparent
weightlessness
Weightlessness
Source: Kirk, Tim, Physics for the IB Diploma, Oxford University Press 2007
Weightlessness
Astronauts on the orbiting space station are
weightless because...with calculations
• Summing the forces on the astronaut
F  F
g
• For circular motion
(Will learn in the next unit)
•
•
•
•
 FN  Fnet
Fnet
v2
m
r
Mm
v2 m  M

N G
 m  G
 v2 
r2
r
r r

Mm
v2
 F G r 2  FN  m r
So,
or
Recall for a satellite, v  GM
r
Substituting, N  m  G M  G M   0
r r
r 
With no reaction force (normal) the astronaut
feels weightless
2
Source: Physics for the IB Diploma 5th Edition (Tsokos) 2008