Universal Gravitation
Chapter 12
Johannes Kepler
Johannes Kepler was a German mathematician,
astronomer and astrologer, and key figure in the 17th
century Scientific revolution. He is best known for his
laws of planetary motion,
During his career, Kepler was a mathematics teacher at a
seminary school in Graz, Austria, an assistant to
astronomer Tycho Brahe, He also did fundamental work
in the field of optics, invented an improved version of the
refracting telescope (the Keplerian Telescope), and helped
to legitimize the telescopic discoveries of his
contemporary Galileo Galilei.
Geocentric Model
A model of the solar
system which holds that
the earth is at the centre of
the universe and all other
bodies are in orbit around
it.
Heliocentric Model
Theory of the
universe that states
the sun is the
centre, and that the
earth revolves
around it.
Other Models
There where also a wide
variety of other different
models that tried to
explain the motion of
the planets. Many of
these where very
complicated and hard to
understand.
Kepler's Three
“LAWS”
Kepler's laws of planetary motion are
three mathematical laws that describe the
motion of planets in the Solar System
Kepler’s First Law
The path of the
planets about the sun
are elliptical in
shape, with the
centre of the sun
being located at one
focus.
Kepler’s Second Law
An Imaginary line drawn from the centre of the sun
to the centre of the planet will sweep out equal areas
in equal intervals of time.
Kepler’s Second Law
Kepler’s Third Law
The ratio of the squares of the periods of any two
plants revolving about the sun is equal to the ratio of
the cubes of their average distances from the sun.
Thus, if Ta and Tb are their periods and ra and rb are
their average distances from the sun, then we get a
following equation.
2
 Ta   ra 
   
 Tb   rb 
3
Astronomical Units
(AU)
An AU is a unit
of distance that
is defined as the
average
distance
between the
Sun and Earth.
Example problem:
(Using Keplers third law to find an orbital period)
Galileo discovered four moons of Jupiter. Io, which he measured
to be 4.2 units from the center of Jupiter, has a period of 1.8
days. He measured the radius of Ganymede’s orbit to be 10.7
units. Use Kepler’s third law to find the period of Ganymede.
Ta  ?
Tb  1.8days
ra  10.7units
rb  4.2units
2
 Ta   ra 
   
 Tb   rb 
3
 ra 
T  Tb  
 rb 
2
a
2
 10.7units 
Ta2  1.8days  

4.2
units


2
3
Ta  7.3days
3
Example problem
(Using Keplers third law to find an orbital radius)
The fourth moon of Jupiter, Callisto, has a period of
16.7 days. Find its distance from Jupiter using the
same units as Galileo used.
Ra=18.5 units
Example problem
Copernicus found the period of Saturn to be 29.5
earth years and it’s orbital radius to be 9.2 AU. Use
these measurements and units to predict the orbital
radius of Mars, whose period is 687 days.
rb=1.47 AU
Practice Problems (1-4)
Reviewing Concepts (1 & 3)
Applying Concepts (1)
Problems (1-5)
Newton's Law of Universal
Gravitation
The force of gravity is proportional to the product of the
two masses that are interacting and inversely proportional
to the square of the distance between their centres
m1m2
Fg  G 2
r
Where:
F
G
m1
m2
r
is the Gravitational Force
is the Gravitational Constant (6.67 × 10−-11 N m2 /kg2)
is the mass of first object
is the mass of second object
is the distance between the objects
Example problem
Determine the force of gravitational attraction between
the earth (m = 5.98 x 1024 kg) and a 70-kg physics
student if the student is standing at sea level, a distance
of 6.37 x 106 m from earth's centre.
F= 688 N
Example problem
A 65.0 kg astronaut is walking on the surface of the
moon, which has a mean radius of 1.74x103 km and a
mass of 7.35x1022 kg. What is the weight of the
astronaut?
105 N
Example problem
Now let’s use Newton’s law of universal gravitation to
calculate the force of gravity here on Earth.
m1m2
Fg  G 2
r
24
5.98

10
m2
11
Fg  6.67 10
(6.37 106 )2
Fg  9.8m2
As you can see
Newton’s law of
universal gravitation
is really another
version of his second
law of motion F=ma
Pg 580
#’s 1 - 8
Gravitational Fields
So far we have studied gravitational interaction in two
related manners.
First, we studied it in terms of energy
AKA. gravitational potential Energy
Then in terms of force.
AKA Weight
Yet there is another way to look at gravitational interactions. We
can study it in terms of what is called a gravitational field.
In the simplest form, we define a gravitational field as a region
in which gravitational force can be experienced.
For example here on earth at sea level we can experience the
force of gravity. More specifically we are said to be within a
gravitational field with a field intensity of 9.8 m/s2
What we have traditionally referred to as, the value of g
(g = 9.8 m/s2), is a specific case example of the strength
of the gravitational field intensity here on earth at sea
level.
Gravitational field intensity will change in strength
as the separation between the two mass changes
We have already
seen this in the case
where the value of g
is larger at the
bottom of a trench,
and smaller on top
of a mountain
The following is a diagram of the gravitational field
intensity of both the earth and moon system.
Can be seen that both the magnitude and direction of the
value g changes with location.
We can also see gravitational field intensity by looking at
Newton’s law of universal gravitation
m1m2
Fg  G 2
r
If we now substitute in the values for Earth at sea level we get
24
5.98

10
m2
11
Fg  6.67 10
(6.37 106 )2
Now simplified to get
Fg  9.8m2
We can now see that the gravitational field intensity (g) can
be found by the manner
m1m2
Fg  G 2
r
24
5.98

10
m2
11
Fg  6.67 10
(6.37 106 )2
Fg  9.8m2
From this it can be seen that the universal formula for
gradational field intensity is
m1
g G 2
r
Or equivalently, if the gravitational force
(weight) is known and radius is not.
g
Fg
m2
Example problem
A mass of 4.60 kg is placed 6.37x106 m from the
center of a planet and experiences a gravitational force
of attraction of 45.1 N. Calculate the gravitational
field intensity at this location.
9.8 m/s2
Example problem
An astronaut is sitting on the seat of a 1100.0 kg lunar rover,
on the surface of the Moon. The seat is 50.0 cm above the
centre of mass of the rover. What gravitational field intensity
does the rover exert on the astronaut?
2.93  107 N/kg [down]
Do #’s 15 – 19
pg 649
Tying universal gravitation to
circular motion
Since the planets are not flying off into space (ie in a straight
line) there must be a force causing them to stay in orbit, which
would have to be some sort of centripetal force .
Fc 
mpv
r
2
Here the gravitational force of the sun can be thought of as
that centripetal force which is causing the circular motion.
Fg  G
ms m p
r
2
So if the gravitational force is the centripetal force, we can
equate them to get
G
ms m p
r
2

mpv
2
r
ms
2
G
v
r
which gives us a formula for calculating orbital velocity
ms
v G
r
We also know that for circular motion
2r
v
T
Therefor by substituting this in for the velocity we get
ms  2 r 
G


r  T 
2
3
Then rearrange to get
Gms
r

2
T
4 2
Where ms is the mass of the planet or star which the object is
orbiting around
Example problem
How fast is the moon moving as it orbits Earth at a distance of
3.84 x 105 km from earth’s centre?
ms
v G
r
1.02x103 m/s
Example problem
A satellite in low Earth orbit is 225 km above the
surface. What is it’s orbital velocity?
7.78 km/s
Practice Problems (5-8)
Reviewing Concepts (4-9)
Applying Concepts (2-11)
Problems (6-19)
Weightlessness
Fact or Myth?
To help answer this lets examine the fallowing scenario.
If a space station has an orbit of 226 km, and an astronaut has
a mass of 65 kg use Newton's law of gravitation to find their
weight
Fg  G
ms m p
r
2
In actual fact there is no such thing as weightlessness,
NASA coined the phase “micro gravity” to describe the
condition of “apparent weightlessness”.
This is the feeling an object would experience during free
fall and is caused by simply not having a normal force to
counteract the force of gravity.
Simply put a person can be “weightless” right here on
Earth simply by removing their normal force.
AKA. If they are in free fall.
Question?
What keeps a satellite up in orbit?
What prevents it from falling out of the sky?
Answer:
Nothing!
It is falling!
It just keeps missing the earth.
What is a satellite?
An object that revolves around a planet in a circular or elliptical
path is termed a satellite.
The moon is Earth's original satellite but there are many manmade satellites, mostly closer to Earth.
The path that a satellite follows is termed an orbit.
An object, such as a javelin, that is projected horizontally will
fall to earth describing a parabolic arc.
A bullet fired by a rifle is projected at a higher velocity than the
javelin so will travel further but must still fall to earth
describing a parabolic arc.
The part that is different here is the fact that the Earth is in fact
round. For this reason the curvature of the Earth itself becomes
significant, and allows the bullet to gain extra range before landing.
If we could, however, fire a rocket with a large enough
velocity,
the rocket would cover enough distance in a short amount
of time, so that the curvature of the Earths would fall out
from under the rocket.
And the rocket would continually miss the surface of the
earth as it falls.
The Earth would still cause a gravitational pull which
would have the effect of continuously changing the
rockets direction.
If the direction is continuously changing while the speed
remains constant then we have circular motion where the
centripetal force is caused be the gravitational force.
This is what we refer to as an object in orbit.
In short, the rocket is always falling to the Earth,
but it keeps missing
Newton’s Cannon
Fire a pig out of a cannon from the top of a high mountain.
The pig falls towards the earth.
If too low of a initial speed, the pig noseplants into the earth.
However, there “is” a certain speed at which the pig falls toward
the earth at the same rate as the earth's surface curves away.
The pig then "misses" the earth and keeps "falling around it",
(i.e. pigs in space)
15468.0
Geostationary Orbit
A geostationary orbit is one in which a satellite orbits the earth at
exactly the same speed as the earth turns and at the same latitude,
specifically zero, the latitude of the equator. A satellite orbiting in
a geostationary orbit appears to be hovering in the same spot in
the sky, and is directly over the same patch of ground at all times.
THE END
Reviewing Concepts (10-13)
Problems (20-28)