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Gravity & Planets
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 next 15 years
analyzing it .
• Deduced three laws of planetary motion
• Apply to any system composed of a body revolving about a more
massive body
•Examples: moons, satellites, comets.
Kepler’s
st
1
Law
Kepler’s first law
( the Law of Orbits)
Planets move in elliptical orbits with the Sun at one of the
focal points.
Kepler’s
nd
2
Law
Kepler’s second law ( the Law of Areas)
A line from the Sun to a planet sweeps out equal areas in
equal lengths of time.
dA = k
dt
Kepler’s
rd
3
Law
The Law of Periods or 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.
T2  r3
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/ T2) 2 = ( r1/ r2) 3
Finding Kepler’s Third Law
with a Graphing Calculator
James
Physics
Teacher,
April
Kepler struggled for 15
years toMetz,
learn theThe
relationship
between
the period
(T) 2000
Kepler’s
and the radius (r) of the orbit of the planets, but
you canValues
easily determine this
relationship with a graphing calculator.
Planet
Radius (r) of orbit of
Period (T) in days
planet in A.U.
Using a TI-83 The calculator should return y = 364.2740081x1.5
Mercury
0.389
87.77
Press
the STAT key
Select
EDIT Since the period is y and the 0.724
Venus
radius is x, T = 364.2740081x3/2
Enter the radius of orbit of each planet in the table below in L1
Earth
1.000
The
corresponding
in L2of the a planet
Thus, period
the period
is directly proportional to the
3/2 power of its orbital radius,
Kepler’s Third Law of Planetary
Mars
1.524
After enteringMotion.
the data
Press
STAT
Jupiter
5.200
Select CALC
Saturnthe power regression equation 9.510
Request
224.70
365.25
686.98
4332.62
10 759.20
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.
Universal Law of Gravitation
Proof
Assumptions:
• Must conform to equations for circular motion
• Used Kepler’s laws as evidence
Fnet = ma
Newton’s 3rd Law  symmetry
Force of gravity ~ to both the masses
Fnet = Fgravity = G m1m2 = m1 a
ac = v2
r2
r
Universal Law of Gravitation Proof
Gravitational force
Gm1m2
2
r
Where
centripetal force
m1v
r
=
2r
v
T
So,
Gm1m2 m1 4 r

2
2
r
T
2
T
4

Rearranging,

3
R
Gm
2
2
2
Law of Periods
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.
G = 6.67 x 10-11 Nm2/kg2
Gm1m2

Discovered by Henry CavendishFin 1798
2
r
Principle of Superposition
For a group of particles, the net effect is the
sum of the individual effects.
Find the individual gravitational force
Then find the net force vectorially.
F=ΣF
Gravitation Near Earth’s surface
F =
F
ma =
Note: a = g
GMm
2
r
GM
g 2
r
Acceleration due to Gravity on
different Planets
Consider an object thrown upward
v f  vo  2ad
2
2
at the maximum height vf=0
vo  2 gd
2
If the same object was thrown upward with the same
initial speed gd  constant
So,
g1d1  g2d2
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 Potential Energy (GPE) of two point
sources
For particles not on the Earth’s surface, the
GPE decreases when the separation decreases
U = - GMm
rE
Gravitational Field Strength
(g.f.s)
The force acting on a 1kg mass at a
specific point in a gravitational field.
Units: N/kg
From Newton’s 2nd law F/m = a
So, another name for g.f.s is the
acceleration due to gravity (g)
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 on a line joining
the centers of two point masses
• If m1 > m2 then
Gravitational Potential, V
The potential at a point in a gravitational field is
equal to the work done bringing a 1kg mass from
infinity to that point.
• Units: J/kg
• W = Fd (the force is not of constant magnitude).
• The force varies with distance from the body (Newton’s law
of universal gravitation) and it can be shown*
• where, M is the mass of the body (the earth, in this case).
*a very useful phrase if you a) don't know how to do something or b) can't be bothered to do it !
Gravitational Potential, V
• 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
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
G.P.E. = Vm
• If the field is due to a planet of mass M and
radius R, then the escape velocity can be
calculated as follows:
ΔK.E. = ΔG.P.E.
• So,
2GM
vesc 
R
• as g = GM/R²
vesc  2 gR
1 mv2 esc  GMm
2
r
E
Satellites
Satellites: Orbits & Energy
GMm using Newton’s second law (F=ma)
U 
r
2
GMm
v
F  2 m
r
r
Solving for v2
where v2/r is the
centripetal acceleration.
GM
2
v 
r
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
GMm
ME  
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
GMm
E
2a
where a was substituted for r
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 k.e.
• 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.
• If the satellite’s mass is m, then the kinetic energy, K,
possessed by the satellite is given by
Energy Changes in a Gravitational Field
• the potential energy, P, possessed by the satellite is
given by
• These equations show:
• that if r decreases, K increases but P decreases
(becomes a bigger negative number)
• the decrease in P is greater than the increase in K.
• Therefore, to fall from one orbit to a lower orbit, the
total energy must decrease. In other words, some
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 can thus reduce 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 but 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.