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Gravitation

Reading: pp. 124-129; 133-137

Syll. State.: 9.2.1-9.2.9—due

Friday, October 3

Newton’s Law of Universal

Gravitation

“Every material particle in the Universe

attracts every other material particle

with a force that is directly proportional

to the product of the masses of the

particles and that is inversely

proportional to the square of the

distance between them.” Principia, Sir Isaac Newton

Gravitation

The gravitational forces obey Newton’s 3rd

law of motion—they are a pair of

action/reaction forces

EVERY particle in the universe obeys this

law

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Gravitation

F12 = strength of gravitational force body 1 exerts on body 2

F21 = strength of gravitational force body 2 exerts on body 1

m1 and m2 = masses of the two bodies

r = separation distance between the center of mass for each of the bodies

G = Universal Gravitation constant

= 6.67 x 10-11 N m2 kg-2

1 212 212

Gm mF F

r

Sample Problem #1

Find the force between the sun and the

earth.

11

24

30

1.5 x 10 m

m = 5.98 x 10 kg

M = 1.99 x 10 kg

R

2

GmMF

R

11 24 30

11 2

(6.67 10 N)(5.98 x 10 kg)(1.99 x 10 kg)

(1.5 x 10 m)

xF

223.5 x 10 NF

If a spherical object is

uniform in mass, we can safely

assume that the gravitational

force due to this mass is exactly

the same as if all the mass were

concentrated at the very center.

Therefore, for something at the surface of this mass, we

can deduce that the following is true:

Thus we can conclude that the acceleration due to the

gravity of a uniform mass is related to the radius of the

large mass (in this example, Earth…)

Me Re

m

2

( )

( )

eg

e

M mG F mg

R

2

( )

( )

e

e

Mg G

R

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Sample problem #2

Determine the acceleration due to gravity

at the surface of a planet that is 10 times

more massive than the earth and has a

radius 20 times larger.

2 2

(10 ) 10

(20 ) 400

E E

e e

M Mg G G

R R

2

1 1( )

40 40

Ee

e

Mg G g

R

-219.80 0.25 m s

40g

2

( )

( )

Mg G

R

Gravitational Field Strength

Universal gravitation is VERY similar to Coulomb’s Law (review electrostatics!), except there is no such thing as negative mass…

Therefore, Gravitational forces will always be attractive forces.

The gravitational field is a “field of influence radiating outward from a particle of mass m”

Gravitational Field Strength at a given point is the force exerted per unit mass on a particle of small mass placed at that point. The strength of the gravitational field is essentially equal to the

acceleration of the particle (g)

Units for the field strength, although they are equal to m s-2, are typically listed as N kg-1

Gravitational Field (diagram)

Gravitational field lines are vectors that always

are directed radially toward the center of the

mass.

(remember…the field lines (vectors) will indicate the

direction a smaller mass will be pulled as a result of

the gravitational force caused by the field…)

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Gravitational Potential Energy

Energy that is “stored” in an object’s gravitational field.

Work must be done in order to change the position of a second mass from an infinitely far away point in space to a position relatively near the first mass.

Remember: Work-Energy Theorem

The amount of work done on an object is equal to the change in energy of that object.

In this case, the energy change is all potential.

Gravitational Potential Energy

Work is done when moving the mass from a distance infinitely far from the source of a gravitational field to some distance R from the source

This work is stored in the gravitational field between the two masses: Gravitational Potential Energy

p

GMmW E

R

Gravitational Potential (V)

A field, defined at every

point in space, but a

scalar quantity

Defined as:

The work done per unit

mass to bring a small

point mass (m) from

infinity to a point P.

r

GMV

m

r

GMm

V

m

E

m

WV

p

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Total Energy of an Object in Orbit

Total energy, as usual, is the sum of the

orbiting object’s gravitational potential and

kinetic energies:

However, we can simplify this…

212

GMmE mv

r

Total Energy…

simplified! Solve for v2 using

Newton’s laws:

Substitute into

energy equation:

2

2

2

2

2

Mm mvG

r r

GMmrv

r m

GMv

r

21

12

22

GM GMm

GMmE mv

E mr

r

r

Graphing Mechanical Energy:

Graph of gravitational potential energy as a

function of distance from a planet’s surface

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Orbital Motion

What kind of force causes objects to move

along a circular path?

Centripetal Force!

To determine the orbital speed of a mass

that is orbiting around a larger central

mass, we must first recognize that the

centripetal force keeping it in orbit is

actually the gravitational force, so…

Orbital Motion

And…

So it follows that:

The orbital speed of a satellite, therefore,

can be determined by using:

2

MmF G

r

2

c

mvF

r

2

2

Mm mvG

r r

2MG v

r

Orbital Motion

The previous example shows us how the orbital speed of a satellite depends on its orbital radius.

The farther out a satellite is from Earth’s surface, the slower it can go and still remain in orbit

How far out (what orbital radius) must a satellite have to remain in geosynchronous orbit?

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Kepler’s laws of Planetary Motion

1st Law: Planets move in ellipses around the sun, the sun being at one of the foci of each ellipse.

2nd Law: The line from the sun to he planet sweeps out equal areas in equal times.

3rd Law: The time taken to complete one full revolution is proportional to the 3/2 power of the mean distance from the sun.

Or: the orbital period squared is directly proportional to the cube of the mean distance from the sun.

Orbital Period, related to Kepler’s

third law of planetary motion…

Assume a circular orbit…the average orbital speed is equal to the distance traveled by the planet or satellite in one orbital period:

Which Means…

Using the relationship we previously deduced:

We can relate Newton’s law of gravitation to Kepler’s 3rd law of planetary motion (see problem #7 from your homework assignment)

2MG v

r

dv

t

2 rv

T

Weightlessness

The forces that act on an astronaut (or anything else) in orbit in a spacecraft, some distance r from the center of the earth are:

His weight (gravitational force from Earth)

The normal force (N) from the “floor” of the spacecraft.

What is the net force acting on the astronaut?

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Fnet = (GMm) – N = mv2

r2 r

(GMm) – mv2 = N

r2 r

N = (m/r){(GM/r) – (v2)}

When we looked at Orbital motion, we showed that

(GM/r) = v2 …so our equation above mathematically

shows that the Normal force that the astronaut feels is

equal to 0 N...

Therefore: “Weightlessness” implies not that there is no

gravity, but rather that there is no reaction force exerting

the upward force that we sense as our weight.