1

Class #20 of 30

Celestial engineering Clarification on scattering

angles

Rotating reference frames Vector angular velocity Newton’s laws on rotating

frame Two “fictitious forces”

:02

2

Class #19 Windup

Epsilon is function of L and E and GMm

:60

2

2

1 2

2

1 2

21

( )(1 cos )

EGmm

rGmm

KEPLER1st Law – Planets move in ellipses 2nd law dA/dt=const3rd law Period goes as semi-minor radius to 3/2 power

3

Class #20 Windup

No class or office hours on 11/5 and 11/6Office hours Thurs 11/7. HW 11 due 11/12 in class. HW 12 due 11/14 as usual.

:60

rv

2 ( )external mr m rmr F

0S Sr r r

4

Planetary Scattering Angle – New … improved!!

:37

Epsilon -1/eps 2*arccos Omega1 -1.00 360.0 0.0

1.001 -1.00 354.9 5.11.02 -0.98 337.3 22.71.05 -0.95 324.5 35.51.1 -0.91 310.8 49.21.3 -0.77 280.6 79.42 -0.50 240.0 120.05 -0.20 203.1 156.920 -0.05 185.7 174.350 -0.02 182.3 177.7100 -0.01 181.2 178.8

max max

2

1 2

1 cos 0

( )(1 cos )

( )

rGmm

r

( )r

( )r

max2

1360 2arccos

impactrSketch for epsilon=2

5

Vector angular velocity

:60

rv

v

r

Theta=90-Latitude = “CoLatitude”

Earth obeys right-hand rule

Omega points North Earth spins clockwise Sun rises in east

6

Vector velocity – 1 minute problem

:60

rv v

r

Latitudes –

Socorro – 34 N

Nome, Alaska – 64 N

Hilo, Hawaii – 19 N

Hobart, Tazmania – 43 S

What is circumferential velocity of each of these four cities?

57.3 10 /earth rad s

7

Vector velocity – 1 minute problem

:60

rv v

r

Latitudes –

Socorro – (.38 km/s=880 mph)

Nome (.20 km/s=473 mph)

Hilo – (.44 km/s=1021 mph)

Hobart (.34 km/s=790 mph)

57.3 10 /earth rad s

8

Vectors expressed in

rotating frames

:60

r

Imagine same axes (x,y,z) expressed in two frames S0 stationary) and S (fixed to earth).

0

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

0,

ˆˆ

ˆ

ˆ

ˆ

i iSi

iiS

i

iiS

ii

i

i ii

i i i ii

i

i

ii

ii

i

Si

iiS

i

r re

drr e

dt

drr e

dt

Because e is any arbitrary FIXED

vector in frame S it t

derdt

r e

r

ransforms

dein S as e

dtdr

r edt

drr e

dt

e re r

0S

rr r

9 :60

0

0

0

ˆ

ˆ

.

( )

( )

(2

) (

)

iiS

i

iiS

i

S S

S S

S S

We know

drr e

dt

drdr e

dt dt

r r r

r r r

r

We differentiate again with

same procedure

r

r r

dr

dt

r

rr

r

Derive Newton in rotating frame

10

Newton in a rotating frame

:60

2 ( )external mr m rmr F

Coriolis Term Centrifugal TermNewtonian Term                                 

11 :60

r

r

Alternate form of Left-Hand Rule

2 ( )external mr m rmr F

12

Class #20 Windup

No class or office hours on 11/5 and 11/6Office hours Thurs 11/7. HW 11 due 11/12 in class. HW 12 due 11/14 as usual.

:60

rv

2 ( )external mr m rmr F

0S Sr r r