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.
