Transcript Slide 1

Chapter 13 Gravitation

Newton’s law of gravitation

•

Any two (or more) massive bodies attract each other

•

Gravitational force (Newton's law of gravitation)

F

G m

1

m

2

r

2

r

ˆ •

Gravitational constant 6.67*10 –11

G

= 6.67*10 –11 N*m 2 /kg 2 m 3 /(kg*s 2 ) – universal constant =

Gravitation and the superposition principle

•

For a group of interacting particles, the net gravitational force on one of the particles is

F

1 ,

net

i n

  2 

F

1

i

•

For a particle interacting with a continuous arrangement of masses (a massive finite object) the sum is replaced with an integral

F

1 ,

body

 

body

d F

Chapter 13 Problem 5 Three uniform spheres of mass 2.00 kg, 4.00 kg and 6.00 kg are placed at the corners of a right triangle. Calculate the resultant gravitational force on the 4.00-kg object, assuming the spheres are isolated from the rest of the Universe.

Shell theorem

•

For a particle interacting with a uniform spherical shell of matter

F

1 ,

shell

 

shell d

F

•

Result of integration: a uniform spherical shell of matter attracts a particle that is outside the shell as if all the shell's mass were concentrated at its center

Gravity force near the surface of Earth

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Earth can be though of as a nest of shells, one within another and each attracting a particle outside the Earth’s surface

•

Thus Earth behaves like a particle located at the center of Earth with a mass equal to that of Earth

F

1 ,

Earth

 

G m Earth m

1 2

R Earth

ˆ

j

   

Gm Earth

2

R Earth

 

m

1

j

ˆ    

m

1

j

ˆ

g

= 9.8 m/s 2

•

This formula is derived for stationary spherical shape and uniform density Earth of ideal

Gravity force near the surface of Earth In reality

g

is not a constant because: Earth is rotating , Earth is approximately an ellipsoid with a non-uniform density

Gravitational field

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A gravitational field exists at every point in space

•

When a particle is placed at a point where there is gravitational field, the particle experiences a force

•

The field exerts a force on the particle

•

The gravitational field is defined as:

g

 

F g m

•

The gravitational field is the gravitational force experienced by a test particle placed at that point divided by the mass of the test particle

Gravitational field

•

The presence of the test particle is not necessary for the field to exist

•

The source particle creates the field

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The gravitational field vectors point in the direction of the acceleration a particle would experience if placed in that field

Gm Earth

2

R Earth

g

•

The magnitude is that of the freefall acceleration at that location

Gravitational potential energy

•

Gravitation is a conservative is path-independent ) force (work done by it

U

•

For conservative forces (Ch. 8):

 

r

f

 

r i

F

d r

  

r r i

f

  

Gm

1

m Earth r

2

dr

   

Gm

1

m Earth

    1

r i

 1

r f

   

Gravitational potential energy

U

U f

U i

Gm

1

m Earth r

1

i

 1

r f

•

To remove a particle from initial position to infinity

U

 

U i

Gm

1

m Earth

  1

r i

 1    

Gm

1

m Earth r i

•

Assuming

U ∞ =

0

U i

(

r i

)  

Gm

1

m Earth r i U

(

r

)  

Gm

1

m

2

r

Gravitational potential energy

U

(

r

)  

Gm

1

m

2

r

Escape speed

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Accounting for the shape of Earth, projectile motion (Ch. 4) has to be modified:

a c

v

2

R

g

v

gR

Escape speed

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Escape speed : speed required for a particle to escape from the planet into infinity (and stop there)

K i

U i

K f

U f m

1

v

2 2 

Gm

1

m planet R planet

 0  0

v escape

 2

Gm planet R planet

Escape speed

•

If for some astronomical object

v escape

 2

Gm object R object

 3  10 8

m

/

s

c

•

Nothing (even light) can escape from the surface of this object – a black hole

Chapter 13 Problem 30 (a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape the solar system, if it starts at the Earth’s orbit? (b) Voyager 1 achieved a maximum speed of 125 000 km/h on its way to photograph Jupiter. Beyond what distance from the Sun is this speed sufficient to escape the solar system?

Kepler’s laws

Tycho Brahe/ Tyge Ottesen Brahe de Knudstrup Johannes Kepler • (1546-1601)

Three Kepler’s laws 1.

The law of orbits : All planets move in

(1571-1630)

elliptical

•

orbits, with the Sun at one focus 2.

The law of areas : A line that connects the planet

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to the Sun sweeps out equal areas planet’s orbit in equal time intervals in the plane of the 3.

The law of periods : The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit

First Kepler’s law

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Elliptical orbits of planets are described by a semimajor axis

a

and an eccentricity

e

•

For most planets, the eccentricities are very small (Earth's

e

is 0.00167)

Second Kepler’s law

L

•

For a star-planet system, the total angular

 •

momentum is constant

rp

  (

r

)(

mv

 )  (

(no external torques)

r

)(

m

r

) 

mr

For the elementary area swept by vector

2 

r dA

 1 2 (

r

)(

rd

 )

dA dt

r

2 2

d

dt

 

const r

2  2

dA

dt L

2

m

Third Kepler’s law

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For a circular orbit and the Newton’s Second law

F

ma GMm

 (

m

)(

r

 2 )

r

2  2 

GM r

3

T

• 

From the definition of a period

2   

T

2  4   2 2

T

2  4 

GM

2

r

3 •

For elliptic orbits

T

2  4 

GM

2

a

3

Satellites

•

F

For a circular

ma

GMm r

orbit and the Newton’s Second law

2  (

m

)  

v r

2   •

Kinetic energy of a satellite

K

mv

2 2 

GMm

 2

r

U

2 •

Total mechanical energy of a satellite

E

K

U

GMm

2

r

GMm r

 

GMm

2

r

 

K

Satellites

•

For an elliptic orbit it can be shown

E

 

GMm

2

a

•

Orbits with different

e

total mechanical energy but the same

a

have the same

Chapter 13 Problem 26 At the Earth’s surface a projectile is launched straight up at a speed of 10.0 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.

Questions?

Answers to the even-numbered problems Chapter 13 Problem 2 2.67 × 10 −7 m/s 2

Answers to the even-numbered problems Chapter 13 Problem 4 3.00 kg and 2.00 kg

Answers to the even-numbered problems Chapter 13 Problem 10 (a) 7.61 cm/s 2 (b) 363 s (c) 3.08 km (d) 28.9 m/s at 72.9

° below the horizontal

Answers to the even-numbered problems Chapter 13 Problem 24 (a) −4.77 × 10 9 J (b) 569 N down (c) 569 N up