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Universal GravitationChapter 12

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.

Johannes Kepler

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 laws of planetary motion are

three mathematical laws that describe the

motion of planets in the Solar System

Kepler's Three

“LAWS”

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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 LawAn 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 LawThe 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 3

a a

b b

T r

T r

=

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.

?

1.8

10.7

4.2

a

b

a

b

T

T days

r units

r units

=

=

=

=

2 3

a a

b b

T r

T r

=

3

2 2 aa b

b

rT T

r

=

( )3

22 10.71.8

4.2a

unitsT days

units

=

7.3aT days=

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 problemCopernicus 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)

5ewton'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

1 2

2g

mmF G

r=

Where:

F is the Gravitational Force

G is the Gravitational Constant (6.67 × 10−-11 N m2 /kg2)

m1 is the mass of first object

m2 is the mass of second object

r 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

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Example problemNow let’s use Newton’s law of universal gravitation to

calculate the force of gravity here on Earth.

1 2

2g

mmF G

r=

2411 2

6 2

5.98 106.67 10

(6.37 10 )g

mF − ×

= ××

29.8

gF m=

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

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

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.

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.

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We can also see gravitational field intensity by looking at

Newton’s law of universal gravitation

1 2

2g

mmF G

r=

If we now substitute in the values for Earth at sea level we get

2411 2

6 2

5.98 106.67 10

(6.37 10 )g

mF − ×

= ××

Now simplified to get

29.8gF m=

We can now see that the gravitational field intensity (g) can

be found by the manner

1 2

2g

mmF G

r=

2411 2

6 2

5.98 106.67 10

(6.37 10 )g

mF − ×

= ××

29.8

gF m=

From this it can be seen that the universal formula for

gradational field intensity is

g Gm

r= 1

2

Or equivalently, if the gravitational force

(weight) is known and radius is not.

gF

m

g=

2

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

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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 .

2

p

c

m vF

r=

Here the gravitational force of the sun can be thought of as

that centripetal force which is causing the circular motion.

2

s p

g

m mF G

r=

2smG vr=

2

2

s p pm m m vG

r r=

So if the gravitational force is the centripetal force, we can

equate them to get

which gives us a formula for calculating orbital velocity

smv Gr

=

We also know that for circular motion

Therefor by substituting this in for the velocity we get

vr

T=

2π

22sm r

Gr T

π =

Then rearrange to get

3

2 24

sGmr

T π=

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?

sm

v Gr

=

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

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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

2

s p

g

m mF G

r=

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.

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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?

The moon is Earth's original satellite but there are many man-

made satellites, mostly closer to Earth.

An object that revolves around a planet in a circular or elliptical

path is termed a satellite.

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.

This is what we refer to as an object in orbit.

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.

The Earth would still cause a gravitational pull which

would have the effect of continuously changing the

rockets direction.

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In short, the rocket is always falling to the Earth,

but it keeps missing

5ewton’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 OrbitA 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.

Reviewing Concepts (10-13)

Problems (20-28)

THE E5D