Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Ast

rono

my

210

Sect

ion

1–M

WF

150

0-15

5013

4 A

stro

nom

y B

uild

ing

Thi

s C

lass

(L

ectu

re 8

):

The

ory

of P

lane

tary

Mot

ion

Nex

t Cla

ss:

The

Tw

o-B

ody

Prob

lem

Mus

ic: S

pace

Dog

–T

oriA

mos

Pla

neta

rium

Obs

ervi

ng

Star

dial

Obs

ervi

ng

Kal

erle

ctur

e at

Pla

neta

rium

, F

ri @

7pm

->

1/6

thH

W g

rade

(d

ue F

eb 1

1th)

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Out

line

•H

ow th

ings

mov

e in

a g

ravi

ty f

ield

, pa

rt 2

.

•B

egin

the

two

body

pro

blem

.

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Res

ult

•N

ow w

e kn

ow th

at th

e or

bitin

g pl

anet

s ar

e ju

st p

erpe

tual

ly

falli

ng b

odie

s. T

his

incl

udes

the

shut

tle, s

atel

lites

, etc

. •

“Wei

ghtle

ssne

ss”

is ju

st li

ke f

allin

g. T

here

is g

ravi

ty o

n th

e sh

uttle

, but

as

one

is in

fre

efal

l it i

s no

t not

icea

ble.

•K

eple

rha

d th

ough

t bri

efly

abo

ut th

is, b

ut h

e de

cide

d he

ne

eded

for

ces

alon

g th

e di

rect

ion

of th

e ve

loci

ty, n

ot

perp

endi

cula

r to

it.

•So

New

ton

real

ized

that

like

an

appl

e fa

lling

fro

m a

tree

or

a re

ally

big

tree

, the

moo

n m

ust h

ave

a fo

rce

tow

ard

the

Ear

th.

•N

ewto

n di

d no

t dis

cove

r gr

avity

, but

he

real

ized

that

it w

as

univ

ersa

l.

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

App

licat

ions

•W

e no

w h

ave

two

tool

s in

our

box

:–

New

ton'

s la

ws

of m

otio

n: h

ow th

ings

mov

e w

ith, w

ithou

t for

ces

appl

ied

–N

ewto

n's

grav

ity: t

he n

et f

orce

on

sola

r sy

stem

bo

dies

•N

ow, w

e ne

ed to

out

the

two

toge

ther

: how

do

bod

ies

mov

e un

der

the

inve

rse

squa

re

forc

e of

gra

vity

.

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Kin

etic

Ene

rgy

22

21

21r

mm

vK

E&r

==

http

://w

ww

.nap

.edu

/htm

l/one

univ

erse

/im

ages

/ene

rgy_

12_s

mal

l.jpg

http

://w

ww

.arm

y-te

chno

logy

.com

/pro

ject

s/th

aad/

thaa

d3.h

tml

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Pot

entia

l Ene

rgy

of G

ravi

ty

∫⋅

−=

∆f ir r

rdF

Ur

r

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Pot

entia

l Ene

rgy

of G

ravi

ty drrM

mG

Uf ir r∫

=∆

2

Def

ine

pote

ntia

l ene

rgy

at

r f=

∞to

be

0

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Pot

entia

l Ene

rgy

of G

ravi

ty

rMm

GU

−=

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Pot

entia

l Ene

rgy

of G

ravi

ty

rMm

GU

−=

Neg

ativ

e po

tent

ial e

nerg

y?

Wha

t’s

up w

ith th

at?

It r

eally

impl

ies

that

the

obje

ct is

bou

nd b

y th

e gr

avita

tiona

l fie

ld.

In o

rder

to e

scap

e th

e pu

ll of

gra

vity

, it h

as to

be

give

n th

at m

uch

ener

gy.

The

onl

y ph

ysic

ally

mea

ning

ful q

uant

ity is

rel

ativ

e ch

ange

s in

po

tent

ial e

nerg

y.

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Tot

al E

nerg

y in

Gra

vity

•T

E =

KE

+ P

E

rMm

Gm

vE

−=

2

21

Con

serv

ed!!

!!C

onse

rved

!!!!

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Orb

its

Can

cla

ssif

y or

bits

with

tota

l ene

rgy.

θθ

cos

1

)1(

)(

2

e

ae

r+−

=

Can

wri

te a

cir

cle

or e

llips

e in

pol

ar c

oord

inat

es a

s:

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Orb

its

Con

side

r a

circ

le w

ith e

=0 rM

Gv c

=2

that

R

ecal

l

2

21

som

vK

E=

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Orb

its

Con

side

r a

circ

le w

ith e

=0

rMm

GP

E−

=B

ut

The

n, K

E =

-PE

/2

And

TE

= K

E +

PE

= P

E/2

TE

< 0

!!

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Spoo

ky O

rbits

Wha

t?N

egat

ive

tota

l ene

rgy?

Is th

at d

ue to

spo

oky

antim

atte

r?

•N

o it

mea

ns th

at y

ou h

ave

to s

uppl

y en

ergy

to s

yste

m to

bre

ak it

apa

rt.

•W

hen

part

icle

s at

res

t, fa

r aw

ay, K

E =

PE

= 0

. To

get t

o th

is s

ituat

ion,

hav

e to

inpu

t ene

rgy

to a

n or

bitin

g sy

stem

.•

Sinc

e en

ergy

is c

onse

rved

, can

't sp

onta

neou

sly

get t

his

ener

gy, s

o sy

stem

is

⇒⇒bo

und!

boun

d!•

Not

e: K

E =

-1/

2PE

gen

eral

ly tr

ue f

or g

ravi

tatin

g sy

stem

s in

equ

ilibr

ium

: “v

iria

lthe

orem

"

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Gen

eral

Orb

its

Elli

pse:

0 <

e <

1

•It

turn

s ou

t tha

t TE

dep

ends

onl

y on

a, n

ot o

n e

•Fr

om c

onse

rvat

ion

of e

nerg

y ca

n se

e:

a

Mm

Gm

vrMm

GT

E2

212

−=

+−

=<

0!!

⇒B

ound

!F

eb 4

, 200

5A

stro

nom

y 21

0 Sp

ring

200

5

Gen

eral

Orb

its

Elli

pse:

0 <

e <

1

•It

turn

s ou

t tha

t TE

dep

ends

onl

y on

a, n

ot o

n e

•Fr

om c

onse

rvat

ion

of e

nerg

y ca

n se

e:

a

Mm

Gm

vrMm

GT

E2

212

−=

+−

=

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Gen

eral

Orb

its

Elli

pse:

0 <

e <

1

•It

turn

s ou

t tha

t TE

dep

ends

onl

y on

a, n

ot o

n e

•Fr

om c

onse

rvat

ion

of e

nerg

y ca

n se

e:

−=

ar

GM

v1

22

Thi

s is

cal

led

the

“Vis

Viv

a” e

quat

ion.

T

he m

ost h

andy

orb

ital e

quat

ion.

http

://gr

oups

.msn

.com

/And

ysU

sing

Scie

nceF

ictio

nFor

Edu

catio

nPag

e/vi

sviv

aequ

atio

n.m

snw

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Gen

eral

Orb

its

Para

bola

: e =

1

•C

an w

rite

as:

•W

here

p is

dis

tanc

e at

clo

sest

app

roac

h.

•Ju

st u

nbou

nd s

yste

m (

can

bare

ly e

scap

e to

r=∞

)

E =

0 o

r K

E =

-PE

θco

s1

+=

pr

rMG

v esca

pe2

=

For

Ear

th. V

esca

pe=

11k

m/s

= 2

5000

mph

.

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Gen

eral

Orb

its

Hyp

erbo

la: e

> 1

•C

an w

rite

as:

•C

ompl

etel

y un

boun

d sy

stem

(es

cape

s to

r=∞

)

E >

0

•T

hat i

mpl

ies

that

v >

0 a

t r=∞

θθ

cos

1

)1(

)(

2

e

ae

r+−

=

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Lif

e is

a 2

Bod

y P

robl

em

•So

far

, I h

ave

chea

ted

a lit

tle…

or

sim

plif

ied

•N

ot th

e pl

anet

s, n

or th

e Su

n is

nai

led

dow

n.

•Fr

om N

ewto

n’s

thir

d, th

e Su

n m

ust m

ove!

(b

ut m

ore

mas

s so

acc

eler

atio

n is

sm

alle

r)

How

do

bina

ry s

tars

orb

it?

We

have

to u

nder

stan

d a

2 bo

dy s

yste

m.

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Con

side

r a

Box

•A

box

in s

pace

with

mas

s aw

ay f

rom

any

gra

vity

fi

eld.

•W

hat i

f th

e bo

x ha

d m

any

piec

es in

side

of

it, h

ow

wou

ld it

mov

e?

•H

ow w

ould

it m

ove?

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Con

side

r a

Box

•L

et o

ur b

ox h

ave

2 ob

ject

s w

ith m

asse

s m

1an

d m

2

•H

ow w

ould

it m

ove?

No

mat

ter

wha

t th

e pi

eces

do,

th

e bo

x st

ill a

cts

as a

fre

e bo

dy.

⇒C

onst

ant

velo

city

.

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Con

side

r a

Box

•If

you

are

mov

ing

with

the

box,

at t

he s

ame

velo

city

:

?

Can

we

say

anyt

hing

abo

ut w

here

the

seco

nd m

ass

has

to b

e?

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Red

uce

Pro

blem

m1

m2

M =

m1

+ m

2

If w

e tr

eat a

s a

sing

le o

bjec

t of

mas

s M

with

no

exte

rnal

for

ces

(onl

y gr

avity

bet

wee

n th

em),

then

how

doe

s th

e bo

x m

ove?

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Cen

ter

of M

ass

(CO

M)

21

22

11

mm

rm

rm

R++

=r

rr

1rr

2rr

m1

m2

Rrrr

M

12

rr

rr

rr

−=

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Cen

ter

of M

ass

(CO

M)

m1

m2

Rr

M

0R

cons

tant

0=

=⇒

=&r

&r&&r

RR

With

no

exte

rnal

for

ces,

can

sho

w th

at

Can

pic

k in

ertia

fra

me

whe

re

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Cen

ter

of M

ass

(CO

M)

1rrm

1m

2

M

0R

cons

tant

0=

=⇒

=r

&r&&r

RR

With

no

exte

rnal

for

ces,

can

sho

w th

at

Can

pic

k re

fere

nce

fram

e w

here

2rr

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Cen

ter

of M

ass

(CO

M)

m1

m2

Rr

M

Wha

t we

are

real

ly d

oing

is t

radi

ng o

ne p

air

of v

ecto

rs (

posi

tions

of

each

mas

s) f

or a

dif

fere

nt p

air

(CO

M a

nd r

elat

ive

sepa

ratio

n) w

hich

ar

e m

ore

conv

enie

nt a

nd n

atur

al...

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

CO

M•

So c

an th

ink

of tw

o ob

ject

s or

bitin

g th

e ce

nter

of

mas

s M

= m

1+

m2

•E

ach

star

has

its

own

orbi

t with

the

sam

e ec

cent

rici

ty a

nd a

dif

fere

nt,

but r

elat

ed s

emim

ajor

axis

.

•T

he s

tars

are

alw

ays

oppo

site

eac

h ot

her

abou

t the

cen

ter-

of-m

ass.

B

oth

star

s w

ill r

each

per

istr

on(c

lose

st a

ppro

ach)

at s

ame

time.

•If

m1>

> m

2, th

en w

e’re

bac

k to

12

1

12

1

22

11

mm

mM

rm

m

rm

rm

R

≈+

=

≈++

=r

rr

r

m1

m2

RrM

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

For

ce in

the

Box

rrm

mG

Fˆ

32

121

−=

rr

rmm

Gr

32

1=

m1

m2

2112

FF

rr

−=

rr

rrm

mG

Fˆ

22

112

=r

New

ton’

s 3rd

Law

!

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Red

uced

Mas

s

1rr

2rr

m1

m2

212

2

121

1

Fr

m

Fr

mr

&&r

r&&r

==

rr

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Red

uced

Mas

s

1rr

2rr

m1

m2

rr

+=

21

21

21m

m

mm

Fr

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Red

uced

Mas

s

ssre

duce

d m

am

m

mm

defi

ne

2

1

21 +

=µ 1rr

2rr

m1

m2

rr

21

Fr

then

r&&r=

µ

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Red

uced

Mas

s

rm

mmm

mth

enm

mif

note

&&r&&r

221

21

21

12

For

:

==

≈<<

µ

1rr

2rr

m1

m2

rr

21Fr

r&&r=

µId

entic

al to

sim

plif

ied

case

, but

now

mas

s is

mod

ifie

d.

Feb

4, 2

005

Ast

rono

my

210

Spri

ng 2

005

Red

uced

Mas

s Mµ

rr

21Fr

r&&r=

µ•

Syst

em c

an b

e th

ough

t of

a te

st p

arti

cle

of r

educ

ed

mas

s th

at is

orb

iting

a m

ass

M =

m1

+ m

2

•D

iffe

rent

ref

eren

ce f

ram

es.