1 class #18 of 30 celestial engineering central forces dvd the power of equivalent 1-d problem and...

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1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential Kepler’s 3 rd law Orbits and Energy The earth-moon flywheel Energy and Eccentricity :02

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Page 1: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

1

Class #18 of 30

Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and

Pseudopotential Kepler’s 3rd law

Orbits and Energy The earth-moon flywheel

Energy and Eccentricity

:02

Page 2: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

2

The Story so far

:12

1 12 2

1

3

ˆ ˆ

ˆ 0

sphere shellsphere

spheresphere

sphere

Solid Sphere Spherical Shell

m M mMr R F G r F G r

r rm M

r R F G rR

r

1 2

2 2 2 1 2

21 22

;

1( )

2

:

mm

Mmm

r r Gr

m mr r G r

r

CM rel

rel

L L +L

L

For r>Rsphere, entire mass as if concentrated at a point at the center of the shell

Lagrangian breaks into a relative piece and a center of mass piece.

Page 3: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

3

Reduced two-body problem

:15

1r

2r

relativer

moonm

Earthm

Page 4: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

4

The power of

:20

21 2

2 2 2 2 21 2

2 21 2 1 2

1 2 2 2

1 1 1 1

2 2 2 2

i

i

Two real masses One reduced mass

I I I r

T T I I T I r

m v m m mmvr r r G G

r r r r

Page 5: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

5

The power of - part 2

:25

2 2

1 2 1 1 2 2

2 2

2 11 2

2 222 1

1 22 2

2 22 1 1 21 2 2

2 1

2

I I m r m r

m mm r m r

M M

m mm m rM M

m m mmmm r r

M m m

r

1r

2r

moonm

Earthm

21 2

2 2 2 2 21 2

2 21 2 1 2

2 2

1 1 1 1

2 2 2 2

i

i

Two real masses One reduced mass

I I I r

T I I T I r

m v m m mmvG G

r r r r

Page 6: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

6

Equivalent 1-D problem

:30

2 2 2 1 2

21 22

21 2

2 3

1( )

2

:

mmr r G

rm m

r r G rr

m mr G

r r

relL

Relative Lagrangian

Radial equation

Total Radial ForcetotalF

total pseudoF dr U

2r

2

1 222pseudo

mmU G

r r

Page 7: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

7

Pseudopotential and Energy

:35

21 2

22pseudo

mmU G

r r

1 2mmG

r

2

22 r

pseudoU

2 2 2 1 2

2 2 2 1 2

22 1 2

2

2

2

1( )

21

( )2

1

2 2

1

21

2

pseudo

mmT U r r G

rm m

T U r r Gr

m mE r G

r r

E r U

E r

rel

rel

rel

L

H

H

Page 8: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

8

Pseudopotential and Energy

:37

21 2

22pseudo

mmU G

r r

2

2

1

21

2

pseudoE r U

E r

0E bounded orbit

0E unbounded orbit

Page 9: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

9

Potential Wells

:25

Mass m

Spring constant k

mx kx

K > 0

K < 0

( ) sink

x t A tm

mx kx( )

k kt t

m mx t Ae Be

Page 10: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

10

Taylor Series Expansion

:35

Taylor series -- Generic

20 0 0 0 0 0

1

2!( ) ( ) ( )( ) ( )( ) ...f x x f x f x x x f x x x

20 0 0 0 0 0

1

2!( ) ( ) ( )( ) ( )( ) ...U x x U x U x x x U x x x

Taylor series -- Potential

Can be ignored or set to zero … “gauge invariance”

Is already zero for potential evaluated about a critical point x-02

0 0 0

20

1

2

1

2

( ) ( )( ) ...

( )effective

ThusU x x U x x x

k x x

0( )effectivek U x

Page 11: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

11

Orbital Energy

:47

1. Last class we derived values for omega, mu and r for the earth-moon system

2. Total energy consumption on Earth is 1000 Terajoules/day

3. If we could power human activites by stealing the angular kinetic energy from the earth-moon system, how much should omega change to give us 100 years of power?

24

22

22

6

8

6.0 10

7.3 10

7.2 10

2.6 10 /

3.9 10

Earth

moon

m kg

m kg

kg

rad s

r m

Page 12: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

12

Converting .

:30

21 2

2 3

2

1 22

2

1 2

1;

( ) ( )

( )(1 cos )

mmr G

r r

d d d u dr

u dt dt d d

Gm mu u

rGm m

( ) ( )r t to r

Page 13: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

13

Properties of ellipses

:30

2

1 2

min max

2

2

1 2 22

2

1 2

;1 1

1

12

( )(1 cos )

cGm m

c cr r

b

a

Gm mE

rGmm

Page 14: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

14

Energy and Eccentricity

:30

2

1 2 22

2

1 2

12

( )(1 cos )

GmmE

rGmm

1

1

0

0 1 Eccentricity

Energy Orbit

E<0 Circle

E<0 Ellipse

E=0 Parabola

E>0 hyperbola

0

1 1

0 1

Page 15: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

15

Class #18 Windup

HW Assignment #10 posted – Do supplement first.

:60

2

2 2

21 2

2

1

2

r

T r

m mvG

r r

21 2

22pseudo

mmU G

r r

21

2 pseudoE r U

Page 16: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

16

Exam 2-3

:72

In Figure (a), a spring with spring constant “k” has been added to an Atwood machine that is glued to the top of the pulley. [If the pulley turned a large fraction of a revolution, the spring would no longer be horizontal, which makes it pretty messy. Ignore this complication.] Assume the wheel only turns a few degrees from the position in which it is shown so that we may assume that the string compresses or stretches but remains horizontal. a) Define an appropriate generalized coordinate. Write down constraints. Write down T and U and finally the Lagrangian for this case. The pulley is massless and has a radius “R” (if you feel you need to know). (10)b) Write down Lagrange’s equation(s) to find the differential equation for the acceleration of m1 (8)c) Initially, at time t=0, we are told that the spring is unstretched (it has length L), and that the masses are at the same height (just as drawn). What is the acceleration of mass m1 at t=0? (5)d) The system has an equilibrium point at which the spring length is not L. How far are the masses displaced from their initial positions shown in the figure at this equilibrium point? (You don’t need to solve the differential equation to get this answer!) (5)

Page 17: 1 Class #18 of 30 Celestial engineering Central Forces DVD The power of Equivalent 1-D problem and Pseudopotential  Kepler’s 3 rd law Orbits and Energy

17

Exam 2-2, 2-4[A hoop and a sphere both start from rest at height h, and race down two identical ramps as shown below. They have identical radius “R” and identical mass “m” (and the mass is distributed uniformly as you should expect in a physics problem!)]a) Define the generalized coordinate you will use. Define which direction is positive. Write the equation of constraint (6). b) Write down the Lagrangian for the hoop AND the Lagrangian for the sphere in terms of the generalized coordinate. You may use "I-hoop" and “I-sphere” as the moments of inertia (8 pts)c) Write down Lagrange's equations, still using I as the moment of inertia. (6)d) Use Lagrange's equation to get the acceleration of the hoop and the sphere along the ramp using. . [You are not required to solve for position as a function of time. Just get acceleration.] 5)e) What is the outcome of the race? (3)f) Why? (One sentence) (2)

2 22

5sphere hoopI MR and I MR

A bead of mass “m” is constrained to slide on a rod. The rod is rotating about the z-axis such that its angle “phi” at any time is omega-t. The rod is not horizontal or vertical, it is inclinded at an angle “theta” (which doesn’t change).  a) The correct generalized coordinates for this problem are where is the angle the rod makes with respect to the z-axis (the axis of rotation), as shown and s is the position “s” of the bead defined as its distance on the rod from the origin. Write x, y and z in terms of . (hint …this is very similar to using spherical coordinates). (5)b) Write down the Lagrangian for the system. (Gravity matters) (5)c) Write down Lagrange’s equation for “s” (only).(5) Extra Credit –There is an equilibrium position for the bead at any . What is it? (5)

, and s

t

, and s

t